Std12-Stat-EM - STATISTICS HIGHER SECONDARY - SECOND YEAR...

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Unformatted text preview: STATISTICS HIGHER SECONDARY - SECOND YEAR Untouchability is a sin Untouchability is a crime Untouchability is inhuman TAMILNADU TEXTBOOK CORPORATION College Road, Chennai- 600 006 i ©Government of Tamilnadu First Edition – 2005 Reprint – 2006 Chairperson Dr. J. Jothikumar Reader in Statistics Presidency College Chennai - 600 005. Reviewers and Authors Thiru K.Nagabushanam Dr. R.Ravanan S.G.Lecturer in Statistics Presidency College Chennai - 600 005. Reader in Statistics Presidency College Chennai - 600 005. Authors Thiru G.Gnana Sundaram Tmt.N.Suseela P.G.Teacher S.S.V. Hr. Sec. School Parktown, Chennai - 600 003. P.G.Teacher Anna Adarsh Matric HSS Annanagar, Chennai -600 040 Tmt.S.Ezhilarasi P.G.Teacher P.K.G.G. Hr. Sec. School Ambattur, Chennai -600 053. Thiru A. S. Sekar P.G.Teacher O.R.G.N. Govt Boys HSS Redhills, Chennai - 600 052 Price: Rs. This book has been prepared by the Directorate of School Education on behalf of the Government of Tamilnadu. This book has been printed on 60 G.S.M paper Printed by offset at: ii PREFACE We take great pleasure in presenting the book on Statistics to the students of the Second year Higher Secondary classes. This book has been written in conformity with the revised syllabus. The book is designed to be self-contained and comprises of ten chapters and includes two new chapters Association of attributes and Decision Theory. Besides the additional (new) topics covered in this book, all the chapters have been completely rewritten and simplified in many ways. The book covers the theoretical, practical and applied aspects of statistics as far as possible in a clear and exhaustive manner. Every chapter in this book explains the principles through appropriate examples in a graded manner. A set of exercise concludes each chapter to provide an opportunity for the students to reinforce what they learn, to test their progress and increase their confidence. The book is very helpful to the students who take their higher studies and the professional courses like Charted Accountants and ICWA At the end of this textbook, necessary statistical tables are included for the convenience of the students. We welcome suggestions from students, teachers and academicians so that this book may further be improved upon. We thank everyone who has a helping hand in the lent preparation of this book. Dr. J. Jothikumar Chairperson Writing team iii CONTENTS Probability 1. Page 1 1.0 Introduction 1 1.1 Definition and basic concepts 1 1.2 Definitions of Probability 3 1.3 Addition theorem on probability 5 1.4 Conditional Probability 8 1.5 Multiplication theorem on Probability 9 1.6 Bayes’ theorem 10 1.7 11 2.0 Basic principles of permutation and combination Random Variable and Mathematical Expectation Introduction 2.1 Random Variable 37 2.2 Probability mass function 39 2.3 Properties of distribution function 41 2.4 45 2.5 An introduction to elementary calculus Mathematical Expectation 2.6 Moment generating function 61 2.7 Characteristic function 62 3.1 Some important Theoretical Distributions Binomial Distribution 3.2 Poisson Distribution 2. 3. iv 37 37 53 68 68 78 3.3 4. Normal Distribution 87 Test of significance (Basic Concepts) 110 4.0 Introduction 110 4.1 Parameter and statistic 110 4.2 Sampling Distribution 110 4.3 Standard Error 111 4.4 4.5 Null hypothesis and Alternative hypothesis Level of significance and critical value 4.6 One Tailed and Two Tailed Tests 115 4.7 Type I and Type II errors 117 4.8 Test procedure 118 Test of Significance (Large Samples) 122 5.0 Introduction 122 5.1 Large Samples (n >30) 122 5.2 Test of significance for proportion 123 5.3 Test of significance for difference between two proportions 126 5.4 Test of significance for Mean 131 5.5 Test of significance for difference between two means Test of significance (Small Samples) 135 6.0 Introduction 143 6.1 t-statistic definition 143 6.2 Test of significance for Mean 146 5. 6. v 112 113 143 6.3 Test of significance for difference between two means 150 6.4 Chi-square distribution 159 6.5 161 6.6 Testing the Goodness of fit (Binomial and Poisson distribution) Test of Independence 6.7 Test for Population variance 172 6.8 F-statistic definition 174 Analysis of Variance 186 7.0 Introduction 186 7.1 Definition 186 7.2 Assumptions 187 7.3 One-way classification 187 7.4 Test Procedure 188 7.5 Two-way Classification 194 7.6 Test Procedure for two-way classification 195 Time Series 207 8.0 Introduction 207 8.1 Definition 207 8.2 Components of Time Series 208 8.3 Method of Least Square 217 8.4 Seasonal Variation 222 8.5 Forecasting 227 7. 8. vi 165 9. 235 9.0 Introduction 235 9.1 Notations 235 9.2 Classes and class frequencies 235 9.3 Consistency of the data 237 9.4 Independence of Attributes 238 9.5 10. Theory of Attributes Yules’ co-efficient of Association 241 Decision Theory 249 10.0 Introduction 249 10.1 Pay-off 252 10.2 Decision making under certainty (without probability) 10.3 Decision making under risk (with probability) 10.4 Decision Tree Analysis 256 vii 262 266 1. PROBABILITY 1.0 Introduction: The theory of probability has its origin in the games of chance related to gambling such as tossing of a coin, throwing of a die, drawing cards from a pack of cards etc. Jerame Cardon, an Italian mathematician wrote ‘ A book on games of chance’ which was published on 1663. Starting with games of chance, probability has become one of the basic tools of statistics. The knowledge of probability theory makes it possible to interpret statistical results, since many statistical procedures involve conclusions based on samples. Probability theory is being applied in the solution of social, economic, business problems. Today the concept of probability has assumed greater importance and the mathematical theory of probability has become the basis for statistical applications in both social and decision-making research. Probability theory, in fact, is the foundation of statistical inferences. 1.1 Definitions and basic concepts: The following definitions and terms are used in studying the theory of probability. Random experiment: Random experiment is one whose results depend on chance, that is the result cannot be predicted. Tossing of coins, throwing of dice are some examples of random experiments. Trial: Performing a random experiment is called a trial. Outcomes: The results of a random experiment are called its outcomes. When two coins are tossed the possible outcomes are HH, HT, TH, TT. 1 Event: An outcome or a combination of outcomes of a random experiment is called an event. For example tossing of a coin is a random experiment and getting a head or tail is an event. Sample space: Each conceivable outcome of an experiment is called a sample point. The totality of all sample points is called a sample space and is denoted by S. For example, when a coin is tossed, the sample space is S = { H, T } . H and T are the sample points of the sample space S. Equally likely events: Two or more events are said to be equally likely if each one of them has an equal chance of occurring. For example in tossing of a coin, the event of getting a head and the event of getting a tail are equally likely events. Mutually exclusive events: Two or more events are said to be mutually exclusive, when the occurrence of any one event excludes the occurrence of the other event. Mutually exclusive events cannot occur simultaneously. For example when a coin is tossed, either the head or the tail will come up. Therefore the occurrence of the head completely excludes the occurrence of the tail. Thus getting head or tail in tossing of a coin is a mutually exclusive event. Exhaustive events: Events are said to be exhaustive when their totality includes all the possible outcomes of a random experiment. For example, while throwing a die, the possible outcomes are {1, 2, 3, 4, 5, 6} and hence the number of cases is 6. Complementary events: The event ‘ A occurs’ and the event ‘ A does not occur’ are called complementary events to each other. The event ‘ A does not occur’ is denoted by A′ o rA or Ac. The event and its complements are mutually exclusive. For example in throwing a die, the event of getting odd numbers is { 1, 3, 5 } and getting even numbers is 2 { 2, 4, 6} .These two events are mutually exclusive and complement to each other. Independent events: Events are said to be independent if the occurrence of one does not affect the others. In the experiment of tossing a fair coin, the occurrence of the event ‘ head’ in the first toss is independent of the occurrence of the event ‘ head’ in the second toss, third toss and subsequent tosses. 1.2 Definitions of Probability: There are two types of probability. They are Mathematical probability and Statistical probability. 1.2.1 Mathematical Probability (or a priori probability): If the probability of an event can be calculated even before the actual happening of the event, that is, even before conducting the experiment, it is called Mathematical probability. If the random experiments results in exhaustive, mutually exclusive and equally likely cases, out of which are favourable to the occurrence of an event A, then the ratio m/n is called the probability of occurrence of event A, denoted by P(A), is given by m Number of cases favourable to the event A P(A) = = n Total number of exhaustive cases Mathematical probability is often called classical probability or a priori probability because if we keep using the examples of tossing of fair coin, dice etc., we can state the answer in advance (prior), without tossing of coins or without rolling the dice etc., The above definition of probability is widely used, but it cannot be applied under the following situations: (1) If it is not possible to enumerate all the possible outcomes for an experiment. (2) If the sample points(outcomes) are not mutually independent. (3) If the total number of outcomes is infinite. (4) If each and every outcome is not equally likely. 3 Some of the drawbacks of classical probability are removed in another definition given below: 1.2.2 Statistical Probability (or a posteriori probability): If the probability of an event can be determined only after the actual happening of the event, it is called Statistical probability. If an event occurs m times out of n, its relative frequency is m/n. In the limiting case, when n becomes sufficiently large it corresponds to a number which is called the probability of that event. In symbol, P(A) = Limit (m/n) n→∞ The above definition of probability involves a concept which has a long term consequence. This approach was initiated by the mathematician Von Mises . If a coin is tossed 10 times we may get 6 heads and 4 tails or 4 heads and 6 tails or any other result. In these cases the probability of getting a head is not 0.5 as we consider in Mathematical probability. However, if the experiment is carried out a large number of times we should expect approximately equal number of heads and tails and we can see that the probability of getting head approaches 0.5. The Statistical probability calculated by conducting an actual experiment is also called a posteriori probability or empirical probability. 1.2.3 Axiomatic approach to probability: The modern approach to probability is purely axiomatic and it is based on the set theory. The axiomatic approach to probability was introduced by the Russian mathematician A.N. Kolmogorov in the year 1933. Axioms of probability: Let S be a sample space and A be an event in S and P(A) is the probability satisfying the following axioms: 4 (1) (2) (3) The probability of any event ranges from zero to one. i.e 0 ≤ P(A) ≤ 1 The probability of the entire space is 1. i.e P(S) = 1 If A1, A2,…is a sequence of mutually exclusive events in S, then P (A1 ∪ A2 ∪ … = P(A1) + P(A2) +... ) Interpretation of statistical statements in terms of set theory: S ⇒ Sample space A ⇒ A does not occur A ∪ A = S A ∩ B = φ ⇒ A and B are mutually exclusive. A ∪ B ⇒ Event A occurs or B occurs or both A and B occur. (at least one of the events A or B occurs) A ∩ B ⇒ Both the events A and B occur. A ∩ B ⇒ Neither A nor B occurs A ∩ B ⇒ Event A occurs and B does not occur A ∩ B ⇒ Event A does not occur and B occur. 1.3 Addition theorem on probabilities: We shall discuss the addition theorem on probabilities for mutually exclusive events and not mutually exclusive events. 1.3.1 Addition theorem on probabilities for mutually exclusive events: If two events A and B are mutually exclusive, the probability of the occurrence of either A or B is the sum of individual probabilities of A and B. ie P(AUB) = P(A) + P(B) This is clearly stated in axioms of probability. A B 5 1.3.2 Addition theorem on probabilities for not-mutually exclusive events: If two events A and B are not-mutually exclusive, the probability of the event that either A or B or both occur is given as P(AUB) = P(A) + P(B) – P(A I B) Proof: Let us take a random experiment with a sample space S of N sample points. Then by the definition of probability , P(AUB) = n (AUB) n (AUB) = n (S) N S B A A I B AI B AI B From the diagram, using the axiom for the mutually exclusive events, we write n( A) + n( A I B) P(AUB) = N Adding and subtracting n( A I B ) in the numerator, n( A) + n( A I B) + n( A I B) − n( A I B) = N n( A) + n( B) − n( A I B) = N n( A) n( B) n( A I B) = + − N N N P(AUB) = P(A) + P(B) – P(A I B) 6 Note: In the case of three events A,B,C, P(AUBUC) = P(A) + P(B) + P(C) – P( A I B ) – P( A I B ) – P( B I C ) + P ( A I B I C ) Compound events: The joint occurrence of two or more events is called compound events. Thus compound events imply the simultaneous occurrence of two or more simple events. For example, in tossing of two fair coins simultaneously, the event of getting ‘ atleast one head’ is a compound event as it consists of joint occurrence of two simple events. Namely, Event A = one head appears ie A = { HT, TH} and Event B = two heads appears ie B = {HH} Similarly, if a bag contains 6 white and 6 red balls and we make a draw of 2 balls at random, then the events that ‘ both are white’ or one is white and one is red’ are compound events. The compound events may be further classified as (1) Independent event (2) Dependent event Independent events: If two or more events occur in such a way that the occurrence of one does not affect the occurrence of another, they are said to be independent events. For example, if a coin is tossed twice, the results of the second throw would in no way be affected by the results of the first throw. Similarly, if a bag contains 5 white and 7 red balls and then two balls are drawn one by one in such a way that the first ball is replaced before the second one is drawn. In this situation, the two events, ‘ the first ball is white’ and ‘ second ball is red’ , will be independent, since the composition of the balls in the bag remains unchanged before a second draw is made. Dependent events: If the occurrence of one event influences the occurrence of the other, then the second event is said to be dependent on the first. 7 In the above example, if we do not replace the first ball drawn, this will change the composition of balls in the bag while making the second draw and therefore the event of ‘ drawing a red ball’ in the second will depend on event (first ball is red or white) occurring in first draw. Similarly, if a person draw a card from a full pack and does not replace it, the result of the draw made afterwards will be dependent on the first draw. 1.4 Conditional probability: Let A be any event with p(A) >0. The probability that an event B occurs subject to the condition that A has already occurred is known as the conditional probability of occurrence of the event B on the assumption that the event A has already occurred and is denoted by the symbol P(B/A) or P(B|A) and is read as the probability of B given A. The same definition can be given as follows also: Two events A and B are said to be dependent when A can occur only when B is known to have occurred (or vice versa). The probability attached to such an event is called the conditional probability and is denoted by P(B/A) or, in other words, probability of B given that A has occurred. If two events A and B are dependent, then the conditional probability of B given A is P(B/A) = P(A I B) P ( A) Similarly the conditional probability of A given B is given as P(A/B) = P(A I B) P(B) Note: If the events A and B are independent, that is the probability of occurrence of any one of them P(A/B) = P(A) and P(B/A) = P(B) 8 1.5 Multiplication theorem on probabilities: We shall discuss multiplication theorem on probabilities for both independent and dependent events. 1.5.1 Multiplication theorem on probabilities for independent events: If two events A and B are independent, the probability that both of them occur is equal to the product of their individual probabilities. i.e P(A I B) = P(A) .P(B) Proof: Out of n1 possible cases let m1 cases be favourable for the occurrence of the event A. m1 ∴P(A) = n1 Out of n2 possible cases, let m2 cases be favourable for the occurrence of the event B m ∴ P(B) = 2 n2 Each of n1 possible cases can be associated with each of the n2 possible cases. Therefore the total number of possible cases for the occurrence of the event ‘ A’ and ‘ B’ is n1 × n2 . Similarly each of the m1 favourable cases can be associated with each of the m2 favourable cases. So the total number of favourable cases for the event ‘ A’ and ‘ B’ is m1 × m2 mm ∴ P(A I B) = 1 2 n1 n 2 m m2 = 1. n1 n2 = P(A).P(B) Note: The theorem can be extended to three or more independent events. If A,B,C…. … be independent events, then P(A I B I C….) = P(A).P(B).P(C)…. … … 9 Note: If A and B are independent then the complements of A and B are also independent. i.e P( A I B ) = P( A ) . P( B ) 1.5.2 Multiplication theorem for dependent events: If A and B be two dependent events, i.e the occurrence of one event is affected by the occurrence of the other event, then the probability that both A and B will occur is P(A I B) = P(A) P(B/A) Proof: Suppose an experiment results in n exhaustive, mutually exclusive and equally likely outcomes, m of them being favourable to the occurrence of the event A. Out of these n outcomes let m1 be favourable to the occurrence of another event B. Then the outcomes favourable to the happening of the events ‘ A and B’ are m1. m ∴ P(A I B) = 1 n m1 m m1 m = = × n m nm m1 m = × n m ∴ P(A I B) = P(A) . P(B/A) Note: In the case of three events A, B, C, P(A I B I C) = P(A). P(B/A). P(C/A I B). ie., the probability of occurrence of A, B and C is equal to the probability of A times the probability of B given that A has occurred, times the probability of C given that both A and B have occurred. 1.6 BAYES’ Theorem: The concept of conditional probability discussed earlier takes into account information about the occurrence of one event to 10 predict the probability of another event. This concept can be extended to revise probabilities based on new information and to determine the probability that a particular effect was due to specific cause. The procedure for revising these probabilities is known as Bayes theorem. The Principle was given by Thomas Bayes in 1763. By this principle, assuming certain prior probabilities, the posteriori probabilities are obtained. That is why Bayes’ probabilities are also called posteriori probabilities. Bayes’ Theorem or Rule (Statement only): Let A1, A2, A3, ….Ai, …An be a set of n mutually … … exclusive and collectively exhaustive events and P(A1), P(A2)… , P(An) are their corresponding probabilities. If B is another event such that P(B) is not zero and the priori probabilities P(B|Ai) i =1,2…n are also known. Then , P(B | A i ) P(A i ) P(Ai | B) = k ∑ P(B | A i ) P(A i ) i =1 1.7 Basic principles of Permutation and Combination: Factorial: The consecutive product of first n natural numbers is known as factorial n and is denoted as n! o r ∠n That is n! = 1 × 2 × 3 × 4 × 5 ×...× n 3! = 3 × 2 × 1 4! = 4 × 3 × 2 × 1 5! = 5 × 4 × 3 × 2 × 1 Also 5! = 5 × ( 4 × 3 × 2 × 1 ) = 5 × ( 4! ) Therefore this can be algebraically written as n! = n × (n – 1)! Note that 1! = 1 and 0! = 1. Permutations: Permutation means arrangement of things in different ways. Out of three things A, B, C taking two at a time, we can arrange them in the following manner. AB BA 11 AC CA BC CB Here we find 6 arrangements. In these arrangements order of arrangement is considered. The arrangement AB and the other arrangement BA are different. The number of arrangements of the above is given as the number of permutations of 3 things taken 2 at a time which gives the value 6. This is written symbolically, 3P2 = 6 Thus the number of arrangements that can be made out of n things taken r at a time is known as the number of permutation of n things taken r at a time and is denoted as nPr. The expansion of nPr is given below: nPr = n(n-1)(n-2)……… – ( r – 1)] ……[n The same can be written in factorial notation as follows: n! nPr = (n − r )! For example, to find 10P3 we write this as follows: 10P3 = 10(10-1)(10-2) = 10 × 9 × 8 = 720 [To find 10P3, Start with 10, write the product of 3 consecutive natural numbers in the descending order] Simplifying 10P3 using factorial notation: 10P3 = 10! 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = (10 − 3)! 7 × 6 × 5 × 4 × 3 × 2 ×1 = 10 × 9 × 8 = 720 Note that nPo = 1, nP1 = n, nPn = n! Combinations: A combination is a selection of objects without considering the order of arrangements. 12 For example, out of three things A,B,C we have to select two things at a time. This can be selected in three different ways as follows: AB AC BC Here the selection of the object A B and B A are one and the same. Hence the order of arrangement is not considered in combination. Here the number of combinations from 3 different things taken 2 at a time is 3. This is written symbolically 3C2 = 3 Thus the number of combination of n different things, taken n Pr r at a time is given by nCr = r! n! Or nCr = (n − r )!r! Note that nC0 =1, Find 10C3. Find 8C 4. nC1 = n, nCn = 1 10 × 9 × 8 10 P3 = = 120 10C3 = 3! 1× 2 × 3 8C 4 = 8×7×6×5 = 70 1× 2 × 3 × 4 [ To find 8 C 4 : In the numerator, first write the product of 4 natural numbers starting with 8 in descending order and in the denominator write the factorial 4 and then simplify.] Compare 10C 8 and 10 C 2 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 10 × 9 = 45 = 10 C 8 = 1× 2 × 3 × 4 × 5 × 6 × 7 × 8 1× 2 10 × 9 = 45 10 C 2 = 1× 2 From the above, we find 10C 8 = 13 10 C2 This can be got by the following method also: 10C 8 = 10 C (10 – 8) = 10 C 2 This method is very useful, when the difference between n and r is very high in nCr. This property of the combination is written as nCr = nC (n-r). To find 200C198 we can use the above formula as follows: 200 × 199 = 19900. 200C198 = 200C(200 – 198) = 200C2 = 1× 2 Example: Out of 13 players, 11 players are to be selected for a cricket team. In how many ways can this be done? Out of 13 players, 11 players are selected in 13 C 11 ways 13 x 12 i.e. 13 C 11 = 13 C 2 = -------- = 78. 1x2 Example 1: Three coins are tossed simultaneously Find the probability that (i) no head (ii) one head (iii) two heads (iv) atleast two heads. (v) atmost two heads appear. Solution: The sample space for the 3 coins is S = { HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} ; n(S) = 8 (i) No head appear A = {TTT}; n(A) = 1 1 ∴P(A) = 8 (ii) One head appear B = {HTT, THT, TTH}; n (B) = 3 3 ∴ P(B) = 8 (iii) Two heads appear C = {HHT, HTH, THH}; n(c)=3 3 ∴ P(C) = 8 (iv) Atleast two heads appear D = { HHT, HTH, THH, HHH}; n(D) = 4 14 4 = 1/2 8 (v) Atmost two heads appear E = { TTT, HTT, THT, TTH,HHT, HTH,THH} n(E)= 7 7 ∴P(E) = 8 Example 2: When two dice are thrown, find the probability of getting doublets (Same number on both dice) ∴ P(D) = Solution: When two dice are thrown, the number of points in the sample space is n(S) = 36 Getting doublets: A = {(1,1) , (2,2) , (3,3) , (4,4) , (5,5) , (6,6)} 61 ∴P(A) = = 36 6 Example 3: A card is drawn at random from a well shuffled pack of 52 cards. What is the probability that it is (i) an ace (ii) a diamond card Solution: We know that the Pack contains 52 cards ∴n(S)= 52 (i) There are 4 aces in a pack. n(A) = 4 4 1 ∴P(A) = = 52 13 (ii) There are 13 diamonds in a pack ∴n(B) = 13 13 1 ∴ P(B) = = 52 4 Example 4: A ball is drawn at random from a box containing 5 green, 6 red, and 4 yellow balls. Determine the probability that the ball drawn is (i) green (ii) Red (iii) yellow (iv) Green or Red (v) not yellow. 15 Solution: Total number of balls in the box = 5+6+4 = 15 balls 51 (i) Probability of drawing a green ball = = 15 3 62 (ii) Probability of drawing a red ball = = 15 5 4 (iii) Probability of drawing a yellow ball = 15 (iv) Probability of drawing a Green or a Red ball 5 6 11 = + = 15 15 15 (v) Probability of getting not yellow = 1 − P (yellow) 4 =1 − 15 11 = 15 Example 5: Two dice are thrown, what is the probability of getting the sum being 8 or the sum being 10? Solution: Number of sample points in throwing two dice at a time is n(S)=36 Let A= {the sum being 8} 5 ∴ A= {(6,2), (5,3) , (4,4), (3,5) , (2,6)}; P(A) = 36 B = { the sum being 10} 3 P(B) = ∴ B = {(6,4), (5,5) (4,6)} ; 36 A I B = { 0 } ; n(A I B) = 0 ∴ The two events are mutually exclusive ∴P(AUB) = P(A) + P(B) 5 3 + = 36 36 16 = 8 2 = 36 9 Example 6 : Two dice are thrown simultaneously. Find the probability that the sum being 6 or same number on both dice. Solution: n(S) = 36 The total is 6: ∴ A = {(5,1) , (4,2), (3,3) , (2,4) , (1,5)}; P(A) = 5 36 Same number on both dice: ∴B = {(1,1) (2,2), (3,3), (4,4), (5,5), (6,6)}; A I B = {(3,3)} ; 6 36 1 P(A B) = 36 P(B) = Here the events are not mutually exclusive. ∴ P(AUB) = P(A) + P(B) – P(A I B) 5 6 1 + = − 36 36 36 5 + 6 −1 = 36 11 − 1 = 36 10 = 36 5 = 18 Example 7: Two persons A and B appeared for an interview for a job. The probability of selection of A is 1/3 and that of B is 1/2. Find the probability that (i) both of them will be selected (ii) only one of them will be selected (iii) none of them will be selected 17 Solution: 1 1 P(A)= , P(B) = 3 2 2 1 P( A ) = and P( B ) = 3 2 Selection or non-selection of any one of the candidate is not affecting the selection of the other. Therefore A and B are independent events. (i) Probability of selecting both A and B P(A I B) = P(A).P(B) 1 1 =× 3 2 1 = 6 (ii) Probability of selecting any one of them = P (selecting A and not selecting B) + P(not selecting A and selecting B) _ _ _ _ i.e P (A I B) + P (A I B) = P(A). P(B) + P(A). P(B) 1 1 2 1 = + × × 3 2 3 2 1 2 + = 6 6 1 3 = = 6 2 (iii) Probability of not selecting both A and B i.e P( A I B ) = P( A ). P( B ) 21 . = 32 1 = 3 18 Example 8: There are three T.V programmes A , B and C which can be received in a city of 2000 families. The following information is available on the basis of survey. 1200 families listen to Programme A 1100 families listen to Programme B 800 families listen to Programme C 765 families listen to Programme A and B 450 families listen to Programme A and C 400 families listen to Programme B and C 100 families listen to Programme A, B and C Find the probability that a family selected at random listens atleast one or more T.V Programmes. Solution: Total number of families n(S)= 2000 Let n(A) = 1200 n(B) = 1100 n(C) = 800 n(A I B) = 765 n(A I C) = 450 n(B I C) = 400 n(A I B I C) = 100 Let us first find n(AUBUC). n(AUBUC) = n(A) + n(B)+ n(C) – n(A I B)–n(A I C)− n (B I C) + n(A I B I C) = 1200 + 1100 + 800 − 765 − 450 − 400 + 100 n(AUBUC) = 1585 n (AUBUC) now P(AUBUC) = n (S) 1585 = = 0.792 2000 Therefore about 79% chance that a family selected at random listens to one or more T.V. Programmes. Example 9: A stockist has 20 items in a lot. Out of which 12 are nondefective and 8 are defective. A customer selects 3 items from the 19 lot. What is the probability that out of these three items (i) three items are non-defective (ii) two are non defective and one is defective Solution: (i) Let the event, that all the three items are non-defective, be denoted by E1. There are 12 non-defective items and out of them 3 can be selected in 12C3 ways ie n(E1)=12C3 Total number of ways in which 3 items can be selected are 20C3 i.e n(S) = 20C3 12C 3 n (E 1 ) = ∴ P(E1) = n (S) 20C 3 12 × 11 × 10 = 20 × 19 × 18 = 0.193 ii) Let the event, that two items are non-defective and one is defective be denoted by E2. Two non-defective items out of 12 can be selected in12C2 ways. One item out of 8 defective can be selected in 8C1 ways. Thus n(E2) =12C2 . 8C1 n (E 2 ) 12C 2 . 8C1 = Then the probability P(E2) = n (S) 20C 3 12 × 11 × 8 × 3 = 20 × 19 × 18 = 0.463 Example 10: A test paper containing 10 problems is given to three students A,B,C. It is considered that student A can solve 60% problems, student B can solve 40% problems and student C can solve 30% problems. Find the probability that the problem chosen from the test paper will be solved by all the three students. Solution: Probability of solving the problem by A = 60% Probability of solving the problem by B = 40% Probability of solving the problem by C = 30% 20 Solving the problem by a student is independent of solving the problem by the other students. Hence, P(A I B I C) = P(A). P(B). P(C) 60 40 30 = × × 100 100 100 = 0.6 × 0.4 × 0 .3 = 0.072 Example 11: From a pack of 52 cards, 2cards are drawn at random. Find the probability that one is king and the other is queen. Solution: From a pack of 52 cards 2 cards are drawn n(S)=52C2 Selection of one king is in 4C1 ways Selection of one queen is in 4C1 ways Selection of one king and one queen is in 4C1.4C1 ways ie n(E) = 4C1.4C1 n (E) 4C1 .4C1 ∴ P(E) = = n (S) 52C 2 52 × 51 = 4×4÷ 1× 2 4× 4× 2 = 52 × 51 8 = 663 Example 12: An urn contains 4 black balls and 6 white balls. If 3 balls are drawn at random, find the probability that (i) all are black (ii) all are white Solution: Total number of balls = 10 Total number ways of selecting 3 balls = 10C3 (i) Number of ways of drawing 3 black balls = 4C3 4C 3 Probability of drawing 3 black balls = 10C 3 21 4 × 3 × 2 10 × 9 × 8 ÷ 1× 2 × 3 1 × 2 × 3 4× 3× 2 = 10 × 9 × 8 1 = 30 (ii) Number of ways of drawing 3 white balls = 6C3 6C 3 Probability of drawing 3 white balls = 10C 3 6× 5× 4 = 10 × 9 × 8 1 = 6 Example 13: A box containing 5 green balls and 3 red colour balls. Find the probability of selecting 3 green colour balls one by one (i) without replacement (ii) with replacement = Solution: (i) Selection without replacement Selecting 3 balls out of 8 balls = 8C3 ways i.e n(S) = 8C3 Selecting 3 green balls in 5C3 ways 5C 3 5× 4× 3 5 = = ∴ P(3 green balls) = 8×7×6 28 8C 3 (ii) Selection with replacement When a ball is drawn and replaced before the next draw, the number of balls in the box remains the same. Also the 3 events of drawing a green ball in each case is independent. ∴Probability 5 of drawing a green ball in each case is 8 The event of selecting a green ball in the first, second and third event are same, 22 ∴ Probability of drawing 5 5 5 125 3 green balls = × ×= 8 8 8 512 Example 14: A box contains 5 red and 4 white marbles. Two marbles are drawn successively from the box without replacement and it is noted that the second one is white. What is the probability that the first is also white? Solution: If w1, w2 are the events ‘ white on the first draw’ , ‘ white on the second draw’ respectively. Now we are looking for P(w1/w2) P(w 1 I w 2 ) P(w 1 ).P( w 2 ) P(w1/w2) = = P( w 2 ) P( w 2 ) (4 / 9)(3 / 8) = (3 / 8) 4 = 9 Example 15: A bag contains 6 red and 8 black balls. Another bag contains 7 red and 10 black balls. A bag is selected and a ball is drawn. Find the probability that it is a red ball. Solution: There are two bags 1 2 Let A denote the first bag and B denote the second bag. 1 Then P(A) = P(B)= 2 Bag ‘ A’ contains 6 red and 8 black balls. 6 ∴ Probability of drawing a red ball is 14 ∴ probability of selecting a bag = 23 Probability of selecting bag A and drawing a red ball from that bag 1 6 3 is P(A). P(R/A) = = × 2 14 14 Similarly probability of selecting bag B and drawing a red ball 1 7 7 from that bag is P(B). P(R/B) = = × 2 17 34 All these are mutually exclusive events ∴ Probability of drawing a red ball either from the bag A or B is P(R) = P(A) P(R/A) + P(B) P(R/B) 3 7 + = 14 34 17 × 3 + 7 × 7 = 238 51 + 49 = 238 100 50 = = 238 119 Example 16: If P(A I B) = 0.3, P(A) = 0.6, P(B) = 0.7 Find the value of P(B/A) and P(A/B) Solution: P(A I B) P(B/A) = P (A ) 0.3 = 0.6 1 = 2 P(A I B) P(A/B) = P(B) 0.3 = 0.7 3 = 7 24 Example 17: In a certain town, males and females form 50 percent of the population. It is known that 20 percent of the males and 5 percent of the females are unemployed. A research student studying the employment situation selects unemployed persons at random. What is the probability that the person selected is (i) a male (ii) a female? Solution: Out of 50% of the population 20% of the males are 50 20 10 unemployed. i.e × = = 0.10 100 100 100 Out of 50% the population 5% of the females are unemployed. 50 5 25 = 0.025 i.e × = 100 100 1000 Based on the above data we can form the table as follows: Employed Unemployed Total Males 0.400 0.100 0.50 Females 0.475 0.025 0.50 Total 0.875 0.125 1.00 Let a male chosen be denoted by M and a female chosen be denoted by F Let U denotes the number of unemployed persons then P(M ∩ U) 0.10 (i) P(M/U) = = = 0.80 P(U) 0.125 P(F ∩ U) 0.025 (ii) P(F/U) = = = 0.20 P(U) 0.125 Example 18: Two sets of candidates are competing for the positions on the Board of directors of a company. The probabilities that the first and second sets will win are 0.6 and 0.4 respectively. If the first set wins, the probability of introducing a new product is 0.8, and the corresponding probability if the second set wins is 0.3. What is the probability that the new product will be introduced? 25 Solution: Let the probabilities of the possible events be: P(A1) = Probability that the first set wins = 0.6 P(A2) = Probability that the second set wins = 0.4 P(B) = Probability that a new product is introduced P(B/A1) = Probability that a new product is introduced given that the first set wins = 0.8 P(B/A2) = Probability that a new product is introduced given that the second set wins = 0.3 Then the rule of addition gives: P(new product) = P(first set and new product) + P(second set and new product) i.e P(B) = P(A1 B) + P(A2 B) = P(A1) P(B/A1) + P(A2).P(B/A2) = 0.6 × 0.8 + 0.4 × 0.3 = 0.60 Example 19: Three persons A, B and C are being considered for the appointment as the chairman for a company whose chance of being selected for the post are in the proportion 4:2:3 respectively. The probability that A, if selected will introduce democratization in the company structure is 0.3 the corresponding probabilities for B and C doing the same are respectively 0.5 and 0.8. What is the probability that democratization would be introduced in the company? Solution: Let A1 and A2 and A3 denote the events that the persons A, B and C respectively are selected as chairman and let E be the event of introducing democratization in the company structure. Then we are given 4 2 3 P(A1) = P(A2) = P(A3) = 9 9 9 P(E/A1) = 0.3 P(E/A2) = 0.5 26 P(E/A3) = 0.8 The event E can materialize in the following mutually exclusive ways: (i) Person A is selected and democratization is introduced ie A1 I E happens (ii) Person B is selected and democratization is introduced ieA2 I E happens (iii) Person C is selected and democratization is introduced ie A3 I E happens Thus E = (A1 I E) U (A2 I E) U(A3 I E) , where these sets are disjoint Hence by addition rule of probability we have P(E) = P(A1 I E) + P(A2 I E) + P(A3 I E) = P(A1) P(E/A1) + P(A2) P(E/A2) + P(A3) P(E/A3) 4 2 3 = × 0.3 + × 0.5 + × 0.8 9 9 9 46 23 = = 90 45 Example 20: In a bolt factory machines A1, A2, A3 manufacture respectively 25%, 35% and 40% of the total output. Of these 5, 4, and 2 percent are defective bolts. A bolt is drawn at random from the product and is found to be defective. What is the probability that it was manufactured by machine A2 ? Solution: P(A1 ) = P( that the machine A1 manufacture the bolts) = 25% = 0.25 Similarly P(A2) = 35% = 0.35 and P(A3) = 40% = 0.40 Let B be the event that the drawn bolt is defective. P(B/ A1 ) = P (that the defective bolt from the machine A1 ) = 5 % = 0.05 Similarly, P(B/ A2 ) = 4% = 0.04 And P(B/ A3) = 2% = 0.02 27 We have to find P(A2/ B). Hence by Bayes’ theorem, we get P ( A 2 ) P( B / A 2 ) P(A2/ B).= P ( A 1 ) P ( B / A 1 ) + P ( A 2 ) P ( B / A 2 ) + P ( A 3 ) P( B / A 3 ) (0.35)(0.04) = (0.25)(0.05) + (0.35)(0.04) + (0.4)(0.02) 28 = 69 = 0.4058 Example 21: A company has two plants to manufacture motorbikes. Plant I manufactures 80 percent of motor bikes, and plant II manufactures 20 percent. At Plant I 85 out of 100 motorbikes are rated standard quality or better. At plant II only 65 out of 100 motorbikes are rated standard quality or better. (i) What is the probability that the motorbike, selected at random came from plant I. if it is known that the motorbike is of standard quality? (ii) What is the probability that the motorbike came from plant II if it is known that the motor bike is of standard quality? Solution: Let A1 be the event of drawing a motorbike produced by plant I. A2 be the event of drawing a motorbike produced by plant II. B be the event of drawing a standard quality motorbike produced by plant I or plant II. Then from the first information, P(A1) = 0.80, P(A2) = 0.20 From the additional information P(B/A1) = 0.85 P(B/A2) = 0.65 The required values are computed in the following table. The final answer is shown in last column of the table. 28 A1 0.80 0.85 0.68 Posterior (revised) probability P(Ai/B) = P(A ∩ B) P(B) 0.68 68 = 0.81 81 A2 0.20 0.65 0.13 0.13 13 = 0.81 81 Total 1.00 Event Prior probability P(Ai) Conditional probability of event B given Ai Joint probability P(Ai I B) = P(Ai)P(B/Ai) P(B/Ai) P(B) = 0.81 1 Without the additional information, we may be inclined to say that the standard motor bike is drawn from plant I output, since P(A1) = 80% is larger than P(A2) =20% Remark: The above answer may be verified by actual number of motorbikes as follows: Suppose 10,000 motorbikes were produced by the two plants in a given period, the number of motorbikes produced by plant I is 10,000 × 80% = 8000 and number of motorbikes produced by plant II is 10000 × 20% = 2000 =2000 The number of standard quality motorbikes produced by plant I is 85 8000 × = 6800 100 And by plant II is 2000 × 65 = 1300 100 29 The probability that a standard quality motor bike was produced by plant I is 6800 6800 68 = = 6800 + 1300 8100 81 And that by plant II is 1300 1300 13 = = 6800 + 1300 8100 81 The process of revising a set of prior probabilities may be repeated if more information can be obtained. Thus Bayes’ theorem provides a powerful method in improving quality of probability for aiding the management in decision making under uncertainty. Exercise - 1 I. Choose the best answer : 1. Probability is expressed as (a) ratio (b) percentage (c) Proportion (d) all the above 2. Probability can take values from (a) - ∞ to +∞ (b) - ∞ to 1 (c) 0 to +1 (d) –1 to +1 3. Two events are said to be independent if (a) each out come has equal chance of occurrence (b) there is the common point in between them (c) one does not affect the occurrence of the other. (d) both events have only one point 4. Classical probability is also known as (a) Statistical probability (b) A priori probability (c) Empirical probability (d) None of the above 5. When a coin and a die are thrown, the number of all possible cases is (a) 7 (b) 8 (c)12 (d) 0 30 6. Probability of drawing a spade queen from a well shuffled pack of cards is 1 1 4 (a) (b) (c) (d) 1 13 52 13 7. Three dice are thrown simultaneously the probability that sum being 3 is (a) 0 (b) 1 / 216 (c) 2 / 216 (d) 3 / 216 8. An integer is chosen from 1 to 20. The probability that the number is divisible by 4 is 1 1 1 1 (a) (b) (c) (d) 4 3 2 10 9. The conditional probability of B given A is P(A ∩ B) P(A ∩ B) (b) (a) P(B) P(B) P(A) P(AUB) P(AUB) (c) (d) P(B) P ( A) 10. P(X) = 0.15, P(Y) = 0.25, P(X I Y) = 0.10 then P(XUY) is (a) 0.10 (b) 0.20 (c) 0.30 (d) 0.40 11. If P(A) = 0.5, P(B) = 0.3 and the events A and B are independent then P(A I B) is (a) 0.8 (b) 0.15 (c) 0.08 (d) 0.015 12. If P(A) = 0.4 P(B) = 0.5 and P(A I B) = 0.2 then P(B/A) is 1 1 4 2 (b) (c) (d) (a) 2 3 5 5 13. A coin is tossed 6 times. Find the number of points in the sample space. (a) 12 (b)16 (c) 32 (d) 64 14. When a single die is thrown the event of getting odd number or even number are (a) Mutually exclusive events (b) Not-mutually exclusive events (c) Independent event (d) Dependent event 31 15. The probability of not getting 2, when a die is thrown is 1 2 1 5 (a) (b) (c) (d) 3 3 6 6 II. Fill in the blanks: 16. The probability of a sure event is __________ 17. The probability of an impossible event is _________ 18. Mathematical probability is also called a ______________ probability. 19. The joint occurrence of two or more events is called ___________ 20. If A and B are mutually exclusive events, then P(AUB) = ____ 21. If A and B are independent events then P(A I B) = _________ 22. If A and B are dependent events then P(A/B) = ___________ 23. If A and B are mutually exclusive events P(A I B) = _________ 24. When three coins are tossed the probability of getting 3 heads is _________ 25. When three dice are thrown the probability of sum being 17 is __________ 26. The probability getting the total is 11 when two dice are throws ____________ III Answer the following: 27. Define the following terms: Event, equally likely events, mutually exclusive events, exhaustive events, sample space. 28. Define dependent and independent events. 29. Define mathematical probability. 30. Define statistical probability. 31. State the axioms of probability. 32. Explain addition theorem on probability for any two events. 33. State the multiplication theorem on probability. 34. Define conditional probability. 35. State Bayes’ Rule. 32 36. There are 5 items defective in a sample of 30 items. Find the probability that an item chosen at random from the sample is (i) defective, (ii) non-defective 37. Four coins are tossed simultaneously what is the probability of getting (i) 2 heads (ii) 3 heads (iii) atleast 3 heads 38. Two dice are thrown. What is probability of getting (i) the sum is 10 (ii) atleast 10 39. Three dice are rolled once. What is the chance that the sum of the face numbers on the dice is (i) exactly 18 (ii) exactly 17 (iii) atmost 17. 40. An integer is chosen from 20 to 30. Find the probability that it is a prime number. 41. An integer is chosen from 1 to 50. Find the probability that it is multiple of 5 or multiply of 7 42. From a pack of cards find the probability of drawing a spade card or a diamond card. 43. Find the probability that a leap year selected at random will contain 53 Sundays. 44. Find the probability that a non-leap year selected at random will contain either 53 Sundays or 53 Mondays. 45. If two events A and B are not mutually exclusive and are not connected with one random experiment P(A)= 1/4, P(B) =2/5 and P(AUB) = 1/2 then find the value of P(B/A) 46. For two independent events A and B for which P(A) = 1/2 and P(B) = 1/3. Find the probability that one of them occur. 47. For two events A and B, P(A) =1/3 = P( B ), P(B/A) =1/4 find P(A/B) 48. A box contains 4 red pens and 5 black pens. Find the probability of drawing 3 black pens one by one (i) with replacement (ii) without replacement 49. An urn contains 5 red and 7 green balls. Another urn contains 6 red and 9 green balls. If a ball is drawn from any one of the two urns, find the probability that the ball drawn is green. 50. Two cards are drawn at random from a pack of 52 cards. Find the probability that the cards drawn are (i) a diamond and a spade (ii) a king and a queen (iii) 2 aces 33 51. A problem in statistics is given to two students A and B. The probability that A solves the problem is 1/2 and that of B’ s to solve it is 2/3. Find the probability that the problem is solved. 52. A bag contains 6 white, 4 green and 10 yellow balls. Two balls are drawn at random. Find the probability that both will be yellow. 53. In a certain class there are 21 students in subject A, 17 in subject B and 10 in subject C. Of these 12 attend subjects A and B, 5 attend subjects B and C, 6 attend subjects A and C. These include 2 students who attend all the three subjects. Find the probability that a student studies one subject alone. 54. If P(A) = 0.3, P(B) = 0.2 and P(C) = 0.1 and A,B,C are independent events, find the probability of occurrence of at least one of the three events A , B and C 55. The odds that A speaks the truth are 3:2 and the odds that B speaks the truth 5:3. In what percentage of cases are they likely to contradict each other on an identical point? 56. The chances of X, Y and Z becoming managers of a certain company are 4:2:3. The probabilities that bonus scheme will be introduced if X, Y and Z become managers are 0.3, 0.5 and 0.4 respectively. If the bonus scheme has been introduced what is the probability that Z is appointed as the manager? 57. A manufacturing firm produces steel pipes in three plants with daily production volumes of 500, 1000 and 2000 units respectively. According to past experience, it is known that the fractions of defective outputs produced by the three plants are respectively, 0.005, 0.008 and 0.010. If a pipe is selected from days total production and found to be defective, what is the probability that it came from the (i) first plant (ii) the second plant (iii) the third plant? 34 Answers: I. 1. (d) 6. (b) 11. (b) 2. (c) 7. (b) 12.(a) II. 16. 1 17. zero 20. P(A) +P(B) P(A ∩ B) 22. P(B) 3 25. 216 3. (c) 8. (a) 13.(d) 4. (b) 9. (b) 14.(a) 5. (c) 10. (c) 15. (d) 18. priori 19. Compound events 21. P(A). P(B) 1 23. 0 24. 8 1 26. 18 III 15 , 66 2 40. 11 36. 31 5 ;; 8 4 16 8 41. 25 37. 11 ; 12 6 1 42. 2 38. 1 3 215 , , 216 216 216 2 2 43. 44. 7 7 39. 45. P(A I B) = 3/20 ; P(B/A) = 3/5 1 2 125 5 , 48. 729 42 5 51. 6 46. 54. 0.496 56. 6/17 47. P(A I B) = 1/12 71 120 9 52. 38 3 3 2 5 19 55. × + × = 5 8 5 8 40 124 57. (a) ; ; 777 49. 35 P(A/B)=1/8 13 8 1 ; ; 102 663 221 8 53. 27 50. = 47.5% (b) 5 16 40 ;; 61 61 61 Activity: We know that, when a coin is tossed, the probability of getting a head is 0.5 mathematically. Now do the following experiment. 1. Toss a fair coin 10 times. Record the event that the number of heads occur in this experiment 2. Toss a fair coin 100 time with the help of your friends group and record the same event that the number of heads appearing. 3. Now compare all the three above mentioned and write your inference. 36 2. RANDOM VARIABLE AND MATHEMATICAL EXPECTATION 2.0 Introduction: It has been a general notion that if an experiment is conducted under identical conditions, values so obtained would be similar. Observations are always taken about a factor or character under study, which can take different values and the factor or character is termed as variable. These observations vary even though the experiment is conducted under identical conditions. Hence, we have a set of outcomes (sample points) of a random experiment. A rule that assigns a real number to each outcome (sample point) is called random variable. From the above discussion, it is clear that there is a value for each outcome, which it takes with certain probability. Hence a list of values of a random variable together with their corresponding probabilities of occurrence, is termed as Probability distribution. As a tradition, probability distribution is used to denote the probability mass or probability density, of either a discrete or a continuous variable. The formal definition of random variable and certain operations on random variable are given in this chapter prior to the details of probability distributions. 2.1 Random variable: A variable whose value is a number determined by the outcome of a random experiment is called a random variable. We can also say that a random variable is a function defined over the sample space of an experiment and generally assumes different values with a definite probability associated with each value. Generally, a random variable is denoted by capital letters like X, Y, Z… where as the values of the random variable are .., denoted by the corresponding small letters like x, y, z …. … 37 Suppose that two coins are tossed so that the sample space is S = {HH, HT, TH, TT} Suppose X represent the number of heads which can come up, with each sample point we can associate a number for X as shown in the table below: Sample point X HH HT TH TT 2 1 1 0 Thus the random variable X takes the values 0, 1,2 for this random experiment. The above example takes only a finite number of values and for each random value we can associate a probability as shown in the table. Usually, for each random variable xi, the probability of respective random variable is denoted by p(xi) or simply pi . X x1 = 0 x2 = 1 x3 = 2 p(xi) p(xi) = 1 4 p(xi) = 2 4 p(xi) = 1 4 Observe that the sum of the probabilities of all the random variable 121 is equal to one. ie p(x1) + p(x2) + p(x3) = + + = 1 444 Thus the probability distribution for a random variable provides a probability for each possible value and that these probabilities must sum to 1. Similarly if 3 coins are tossed, the random variable for getting head will be X=0, X=1, X=2, X=3 and sum of their respective probabilities i.e Σp(xi) =1 If two dice are rolled then the sample space S consists of 36 sample points. Let X denote the sum of the numbers on the two dice. Then X is a function defined on S by the rule X(i,j) = i+j . Then X is a random variable which can takes the values 2,3,4…12. That is the range of X is {2,3,4…12} … … 38 2.1.1 Discrete random variable: If a random variable takes only a finite or a countable number of values, it is called a discrete random variable. For example, when 3 coins are tossed, the number of heads obtained is the random variable X assumes the values 0,1,2,3 which form a countable set. Such a variable is a discrete random variable. 2.1.2 Continuous random variable: A random variable X which can take any value between certain interval is called a continuous random variable. Note that the probability of any single value at x, value of X is zero. i.e P(X = x) = 0 Thus continuous random variable takes value only between two given limits. For example the height of students in a particular class lies between 4 feet to 6 feet. We write this as X = {x|4 ≤ x ≤ 6} The maximum life of electric bulbs is 2000 hours. For this the continuous random variable will be X = {x | 0 ≤ x ≤ 2000} 2.2 Probability mass function: Let X be a discrete random variable which assumes the values x1, x2, ...xn with each of these values, we associate a number called the probability Pi= P(X=xi), i = 1,2,3… This is called n probability of xi satisfying the following conditions. (i) Pi ≥ 0 for all i, ie Pi’ s are all non-negative (ii) p Σpi = p1 + p2 + …n =1 ie the total probability is one. This function pi or p(xi) is called the probability mass function of the discrete random variable X. The set of all possible ordered pairs (x, p(x)) is called the probability distribution of the random variable X. Note: The concept of probability distribution is similar to that of frequency distribution. Just as frequency distribution tells us how the total frequency is distributed among different values (or classes) of the variable, a probability distribution tells us how total 39 probability 1 is distributed among the various values which the random variable can take. It is usually represented in a tabular form given below: X x1 x2 x3 … . xn P(X = x) P(x1) P(x2) P(x3) … . P(xn) 2.2.1 Discrete probability distribution: If a random variable is discrete in general, its distribution will also be discrete. For a discrete random variable X, the distribution function or cumulative distribution is given by F(x) and is written as F(x) = P(X ≤ x) ; - ∞ < x < ∞ Thus in a discrete distribution function, there are a countable number of points x1, x2,… and their probabilities pi such .. that F(xi) = ∑ pi , i = 1, 2, ….n … xi < x Note: For a discrete distribution function, F(xj) – F(xj-1) = p(xj) 2.2.2 Probability density function (pdf): A function f is said to be the probability density function of a continuous random variable X if it satisfies the following properties. (i) f(x) ≥ 0 −∞ < x < ∞ ∞ (ii) ∫ f ( x) dx = 1 −∞ Remark: In case of a discrete random variable, the probability at a point ie P(x = a) is not zero for some fixed ‘ a’ However in case of continuous random variables the probability at a point is always zero a ie P(X = a) = ∫ f ( x) dx =0 a 40 Hence P( a ≤ X ≤ b) = P(a < X < b) = P(a ≤ X < b) = P(a < X ≤ b) The probability that x lies in the interval (a,b) is given by b P( a < X < b) = ∫ f ( x) dx a Distribution function for continuous random variable. If X is a continuous random variable with p.d.f f(x), then the distribution function is given by x ∫ (i) F(x) = f ( x ) dx = P(X ≤ x) ; -∞ < x < ∞ −∞ b (ii) F(b) – F(a) = ∫ f ( x) dx = P(a ≤ X ≤ b) a 2.3 Properties of distribution function: Suppose that X be a discrete or continuous random variable, then (i) F(x) is a non - decreasing function of x (ii) 0 ≤ F(x) ≤ 1 , −∞ < x < ∞ (iii) F(- ∞) = limit F(x) =0 xà−∞ (iv) F(∞) = limit F(x) =1 xà∞ (v) If F(x) is the cumulative distribution function of a continuous random variable X with p.d.f f(x) then F′(x) = f(x) Example 1: A random variable has the following probability distribution Values of X P(x) 0 1 2 3 4 5 6 7 8 a 3a 5a 7a 9a 11 a 13 a 15 a 17 a (1) Determine the value of a (2) Find (i) P( x < 3) (ii) P(x ≤ 3) (iii) (iv)P( 2 ≤ x ≤ 5), (v) P(2 < x <5) (3) Find the cumulative distribution function of x. 41 P(x >7) Solution: (1) Since pi is the probability mass function of discrete random variable X, We have Σpi = 1 ∴ a + 3 a + 5a + 7a + 9a +11a + 13a + 15a + 17a = 1 81a = 1 a = 1/81 (2) (i) P(x <3) = P(x=0) + P(x=1) + P(x=2) = a + 3 a + 5a = 9a 1 =9( ) 81 1 = 9 (ii) P(x ≤ 3) = P (x=0) + P(x=1) + P(x=2) +P(x=3) =a+3a+5a+7a = 16 a 16 = 81 iii) P(x >7) = P(x = 8) = 17 a 17 = 81 iv) P ( 2 ≤ x ≤ 5) = P(x=2) +P(x=3) + P( x = 4) +P(x=5) = 5 a + 7a +9a +11a = 32a 32 = 81 v) P(2 < x < 5 ) = P(x = 3) + P(x = 4) = 7a + 9a = 16a 16 = 81 42 3) The distribution function is as follows: X=x 0 F(x)= a P (X≤ x) (or) 1 F(x) 81 1 4a 2 9a 3 16a 4 25a 5 36a 6 49a 7 64a 8 81a 4 81 9 81 16 81 25 81 36 81 49 81 64 81 81 =1 81 Example 2: Find the probability distribution of the number of sixes in throwing two dice once. Solution: When two dice are thrown the total number of sample points are 36. Let X denote the number of sixes obtained in throwing two dice once. Then X is the random variable, which can take the values 0,1,2. Let A denote the success of getting a six in throwing a die and A denote not getting a six. Then probability getting a six P(A) = 1 6 Probability not getting a six P( A ) = 5 6 No sixes: ∴ P(x = 0) = P( A and A ) = P( A ) . P( A ) 5 5 = . 6 6 25 = 36 43 P(x = 1) = P(A and A ) or P( A and A) = P(A) . P( A ) + P( A ) .P(A) 15 51 =. + 66 66 5 5 = + 36 36 10 = 36 5 = 18 P(x = 2) = P( A and A) = P(A) .P(A) 11 =. 66 1 = 36 Hence the probability distribution of X is given by X= x P(X = x) 0 1 2 25 36 10 36 1 36 Example 3: An urn contains 6 red and 4 white balls. Three balls are drawn at random. Obtain the probability distribution of the number of white balls drawn. Solution: The total number of balls in the urn is 10 Let X denote the number of white balls drawn If three balls are drawn, the random variable takes the value X= 0, 1, 2, 3 Probability of getting white balls from the urn containing 10 balls (red and white) with the following combination are 44 4C 0 6C 3 1 × 120 5 = = 720 30 10C 3 4C1 .6C 2 15 = P (1 white, 2 red) = 30 10C 3 4C 2 6C1 9 = P (2 white, 1 red) = 30 10C 3 4C 3 6C 0 1 P (3 white, no red) = = 30 10C 3 Hence the probability distribution of X is given by C 0 1 2 3 P(X=x) 5 15 9 1 30 30 30 30 2.4 An introduction to elementary calculus: Before going to see the problems on continuous random variables, we need to know some fundamental ideas about differentiation and integration, which are part of calculus in higherlevel mathematics. Hence we introduce some simple techniques and formulae to calculate the problems in statistics, which involve calculus methods. 2.4.1 Differentiation: 1. Functional value is an exact value. For some function f(x), when x = a, we obtain the functional value as f(a) = k. 2. Limiting value is an approximate value. This value approaches the nearest to the exact value k. Suppose the exact value is 4. Then the limiting value will be 4.000000001 or 3.999999994. Here the functional value and limiting value are more or less same. Hence in many occasions we use the limiting values for critical problems. The limiting value of f(x) when x approaches a number 2 is given as Limit f(x) = f(2) = l xà 2 45 P (no white, 3 red balls) = f (x + h ) − f ( x ) is h →0 h called the derivative of the function f with respect to x and is denoted by f ′(x). If y is the function x then we say the differential coefficient of y with respect to x and is denoted dy as dx 4. Some rules on differentiation: (i) Derivative of a constant function is zero. f ′(c)=0 where c is some constant. (ii) If u is a function of x and k is some constant and dash denotes the differentiation, [ku]′ = k[u]′ (iii) (u ± v)′ = u′ ± v′ (iv) (uv)′ = u′v +uv′ 3. The special type of existing limit, limit ′ u ' v − uv' u (v) = v2 v 5. Important formulae: (i) (xn)′ = nxn-1 (ii) (ex)′ = ex 1 (iii) (logx)′ = x Example 4: Evaluate the following limits: x 2 + 5x x 2 −1 (i) Limit (ii) Limit x →1 x+2 x −1 x→2 Solution: (i) Limit x→2 x 2 + 5x x+2 = (2) 2 + 5(2) 4 + 10 14 7 = = = 2+2 4 4 2 x 2 −1 12 − 1 0 = = . This is an indeterminate form. x →1 x −1 1−1 0 Therefore first factorise and simplify and then apply the same limit to get the limiting value 46 (ii) Limit x2 −1 ( x − 1)( x + 1) = = x +1 x −1 ( x − 1) ∴ ∴ Limit x →1 x 2 −1 = Limit (x+1) = 1+1 = 2 x →1 x −1 Example 5: Find the derivative of the following with respect to x. x2 +1 12 4 2 3 x (i) x + 7 (ii) (x + 4x –5) (iii) (x ) (e ) (iv) x −5 Solution: (i) Let y = x12 + 7 dy = 12x12-1 + 0 = 12x11 ∴ dx (ii) Let y = x3 + 4x2 –5 y′ = 4x3 +4(2x) – 0 = 4x3 +8 (iii) Let y = x3 ex (uv)′= u′v +uv′ = [x3]′ (ex) + (x3) [ex]′ = 3x2 ex +x3 ex = ex (3x2 + x3) ′ (iv) y = u ' v − uv' x2 +1 u . This is of the type = x −5 v2 v ′ ′ dy x 2 + 1 (x − 5) − (x 2 + 1)[x − 5 ] ∴ = dx ( x − 5) 2 [ [2x (]x − 5) − ( x 2 + 1)[1] = ( x − 5) 2 2x 2 − 10 x − x 2 − 1 = (x − 5) 2 x 2 − 10 x − 1 = ( x − 5) 2 47 2.4.2 Integration: Integration is known as the reverse process of differentiation. Suppose the derivative of x3 is 3x2. Then the integration of 3x2 with respect to x is x3 . We write this in symbol as follows: d4 3 ( x ) = 4x3 ⇒ 4 ∫ x dx = x4 dx Similarly d8 ( x ) = 8x7 dx dx (e ) = ex dx ⇒8 ⇒ ∫x ∫e x 7 dx = x8 dx = ex Note: While differentiating the constant term we get zero. But in the reverse process, that is on integration, unless you know the value of the constant we cannot include. That is why we include an arbitrary constant C to each integral value. Therefore the above examples, we usually write x 7 x 8 ∫ e dx = e +c and ∫ 8x dx = x + c These integrals are also called improper integrals or indefinite integrals Rules and formulae on integration: (i) ∫ k dx = kx + xn 1 (ii) ∫ x dx = n +1 x (iii) ∫ e dx = ex n 1 ∫ x dx = log x (v) ∫ (u ± v)dx = ∫ u dx ± ∫ v dx (iv) 48 Example 6: Integrate the following with respect to x: + x6 1 x7 6 = (i) ∫ x dx = +c 6 +1 7 −+ − 11 1 x51 x4 -5 =− +c = =(ii) ∫ x dx = 4 4x − 5 +1 −4 4x 4 1 (iii) ∫ dx = log x +c x + 2 3/ 2 x1/ 2 1 x3/ 2 1/ 2 (iv) ∫ x dx = ∫ x dx = x +c = = 1 3 3 +1 2 2 4 2 (v) ∫ (x +2x + 4x + 8) dx x5 x3 x2 +2 +4 + 8x + c 5 3 2 (ex + x4 + 1/x3 +10) dx = (vi) = ex + x5 /5 − 1/2x2 + 10x + c The above discussed integrals are known as improper integrals or indefinite integrals. For the proper or definite integrals we have the limiting point at both sides. ie on the lower limit and the upper limit. This integral ∫ f ( x ) dx is an indefinite integral Integrating the same function within the given limits a and b is known as the definite integral. b ie ∫ f (x )dx = k (a constant value) is a definite integral where a is a known as lower limit and b is known as the upper limit of the definite integral. To find the value of definite integral we use the formulae: b Suppose ∫ f ( x ) dx = F(x) then ∫ f ( x ) dx = F(b) – F(a) a 49 An important note to the Teachers and students As per as statistics problems concerned, the differentiation and integration methods restricted to simple algebraic functions only. Example 7: Evaluate the following definite integrals. 4 3 (i) ∫ 3x 2 dx (ii) 0 3 ∫ x dx 1 Solution: 4 3x 3 34 (i) ∫ 3x dx = = [x ]0 3 0 0 3 = 4 –03 = 64 4 2 3 3 (ii) ∫ x dx 3 1 x4 = 4 1 1 43 [ x ]1 4 1 = [3 4 − 14 ] 4 1 = [81 − 1] 4 1 = [80] 4 = 20 = 5 x2 (iii) ∫ xdx = 2 2 2 1 = [5 2 − 2 2 ] 2 1 21 = [25 − 4] = 2 2 5 50 5 (iii) ∫ xdx 2 Example 8: Examine whether f(x) = 5x4 , 0 < x < 1 can be a p.d.f of a continuous random variable x. Solution: ∞ For a probability density function, to show that ∫ f ( x ) dx =1 −∞ 1 That is to show that ∫ 5( x) 4 dx = 1 0 1 x5 5( x) dx =5 ∫ 0 5 0 1 5 = x5 0 5 = [15 –0] =1 ∴ f(x) is a p.d.f 1 4 Example 9: A continuous random variable x follows the rule f(x) = Ax2, 0 < x < 1. Determine A Solution: Since f(x) is a p.d.f, ∫ ∞ −∞ f ( x ) dx =1 1 Therefore ∫ Ax2 dx = 1 0 1 x3 A 3 0 A 31 x0 3 A [1] 3 A =1 [ ]= 1 =1 =3 51 Example 10: Let f(x) = c(1-x) x2 , 0 < x < 1 be a probability density function of a random variable x. Find the constant c Solution: f(x) = c(1-x)x2 , 0 < x < 1 since f(x) is a p.d.f ∫ ∞ −∞ f ( x ) dx = 1 1 ∴ ∫ c(x2 –x3)dx =1 0 1 x3 x4 =1 c − 3 4 0 13 14 c − − (0 − 0) = 1 3 4 1 1 c − =1 3 4 4 −1 c =1 12 1 c =1 12 c = 12 Example 11: A random variable x has the density function 1 -2< x<2 , f(x) = 4 0, else where obtain (i) P (−1 < x < 2) (ii) P (x >1) Solution: 2 (i) P(-1 < x < 2) = ∫ f(x) dx −1 52 2 +2 1 1 dx = [ x]−1 ∫4 4 −1 1 = [2 – (-1)] 4 1 = [3] 4 3 = 4 (ii) Here the upper limit of the p.d.f is 2∴the probability for the given random variable. 2 1 P( x > 1) =∫ dx 4 1 1 2 = [x ] 1 4 1 [2 –1] = 4 1 = [1] 4 1 = 4 2.5 Mathematical Expectation: Expectation is a very basic concept and is employed widely in decision theory, management science, system analysis, theory of games and many other fields. Some of these applications will be discussed in the chapter on Decision Theory. The expected value or mathematical expectation of a random variable X is the weighted average of the values that X can assume with probabilities of its various values as weights. Thus the expected value of a random variable is obtained by considering the various values that the variable can take multiplying these by their corresponding probabilities and summing these products. Expectation of X is denoted by E(X) 53 2.5.1 Expectation of a discrete random variable: Let X be a discrete random variable which can assume any of the values of x1, x2, x3…..xn with respective probabilities p1, … p2, p3…pn. Then the mathematical expectation of X is given by … E(x) = x1p1 + x2p2 + x3p3 +……npn …x n n = ∑ i =1 xipi , where ∑ pi = 1 i =1 Note: Mathematical expectation of a random variable is also known as its arithmetic mean. We shall give some useful theorems on expectation without proof. 2.5.2 Theorems on Expectation: 1. For two random variable X and Y if E(X) and E(Y) exist, E(X + Y) = E(X) + E(Y) . This is known as addition theorem on expectation. 2. For two independent random variable X and Y, E(XY) = E(X).E(Y) provided all expectation exist. This is known as multiplication theorem on expectation. 3. The expectation of a constant is the constant it self. ie E(C) = C 4. E(cX) = cE(X) 5. E(aX+b) = aE(X) +b 6. Variance of constant is zero. ie Var(c) = 0 7. Var(X+c) = Var X Note: This theorem gives that variance is independent of change of origin. 8. Var (aX) = a2 var(X) Note: This theorem gives that change of scale affects the variance. 9. Var (aX+b) = a2Var(X) 10. Var (b-ax) = a2 Var(x) Definition: Let f(x) be a function of random variable X. Then expectation of f(x) is given by E(f(x)) = Σ f(x) P(X=x) , where P(X=x) is the probability function of x. 54 Particular cases: 1. If we take f(x) = Xr, then E(Xr) = Σxrp(x) is defined as the rth moment about origin or rth raw moment of the probability distribution. It is denoted by µ′r Thus µ′r = E(Xr) µ′1 = E(X) µ′2 = E(X2) Hence mean = X = µ′1 = E(X) 2 Σ X 2 ΣX - N N 2 = E ( X ) – [E (X)]2 = µ′2 – (µ′1)2 Variance is denoted by µ2 2. If we take f(x) = (X – X )r then E(X – X )r = Σ(X – X )r p(x) which is µr, the rth moment about mean or rth central moment. In particular if r = 2, we get µ2 = E (X – X )2 = Σ (X – X )2 p(X) = E [X – E(X)]2 These two formulae give the variance of probability distribution in terms of expectations. Variance = Example 12: Find the expected value of x, where x represents the outcome when a die is thrown. Solution: Here each of the outcome (ie., number) 1, 2, 3, 4, 5 and 6 1 occurs with probability . Thus the probability distribution of 6 X will be x 1 2 3 4 5 6 P(x) 1 6 1 6 1 6 55 1 6 1 6 1 6 Thus the expected value of X is E(X) = Σxipi = x1p1 + x2p2 + x3p3 + x4p4+ x5p5 + x6p6 1 1 1 1 E(X) = 1× + 2 × + 3 × + 4 × 6 6 6 6 1 1 + 5 × + 6 × 6 6 7 = 2 E(X) = 3.5 Remark: In the games of chance, the expected value of the game is defined as the value of the game to the player. The game is said to be favourable to the player if the expected value of the game is positive, and unfavourable, if value of the game is negative. The game is called a fair game if the expected value of the game is zero. Example 13: A player throws a fair die. If a prime number occurs he wins that number of rupees but if a non-prime number occurs he loses that number of rupees. Find the expected gain of the player and conclude. Solution: Here each of the six outcomes in throwing a die have been assigned certain amount of loss or gain. So to find the expected gain of the player, these assigned gains (loss is considered as negative gain) will be denoted as X. These can be written as follows: Outcome on a die 1 2 3 4 5 6 Associated gain to -1 2 3 -4 5 -6 the outcome (xi) 1 1 1 1 1 1 P(xi) 6 6 6 6 6 6 56 Note that 2,3 and 5 are prime numbers now the expected gain is 66 E (x)∑ EΣ1 xi pi == i =1 1 1 1 1 1 1 = (–1) + (2) + (3) + (– 4) + (5) + (– 6) 6 6 6 6 6 6 1 =– 6 Since the expected value of the game is negative, the game is unfavourable to the player. Example 14: An urn contains 7 white and 3 red balls. Two balls are drawn together at random from the urn. Find the expected number of white balls drawn. Solution: From the urn containing 7 white and 3 red balls, two balls can be drawn in 10C2 ways. Let X denote the number of white balls drawn, X can take the values 0, 1 and 2. The probability distribution of X is obtained as follows: P(0) = Probability that neither of two balls is white. = Probability that both balls drawn are red. 3× 2 1 3C 2 = = = 10 × 9 15 10C 2 P(1) = Probability of getting 1 white and 1 red ball. 7 × 3× 2 7 = 7C1 × 3C1 = = 10 × 9 15 10C 2 P(2) = Probability of getting two white balls 7× 6 7 7C 2 = = = 10C 2 10 × 9 15 Hence expected number of white balls drawn is 1 7 7 E(x) = Σ xi p(xi) = 0 × + 1 × + 2 × 15 15 15 7 = = 1.4 5 57 Example 15: A dealer in television sets estimates from his past experience the probabilities of his selling television sets in a day is given below. Find the expected number of sales in a day. Number of TV Sold in a day Probability 0 1 2 3 4 5 6 0.02 0.10 0.21 0.32 0.20 0.09 0.06 Solution: We observe that the number of television sets sold in a day is a random variable which can assume the values 0, 1,2, 3,4,5,6 with the respective probabilities given in the table. Now the expectation of x = E(X) = Σxipi = x1p1 + x2p2 + x3p3 + x4p4+ x5p5 + x6p6 = (0) (0.02) + (1) (0.010) + 2(0.21) + (3) (0.32) + 4(0.20) +(5) (0.09) + (6) (0.06) E(X) = 3.09 The expected number of sales per day is 3 Example 16: Let x be a discrete random variable with the following probability distribution X -3 6 1/2 P(X= x) 1/6 Find the mean and variance Solution: E (x) 9 1/3 = Σ xi pi 1 1 1 = (-3) + (6) + (9) 6 2 3 11 = 2 58 = Σ xi2 pi 1 1 1 93 = (-3)2 + (6)2 + (9)2 = 6 2 3 2 2 2 Var (X) = E (X ) - [E(X)] E(x2) 2 93 11 = - 2 2 93 121 = - 2 4 186 − 121 = 4 65 = 4 2.5.3 Expectation of a continuous random variable: Let X be a continuous random variable with probability density function f(x), then the mathematical expectation of x is defined as ∞ E(x) = ∫ xf ( x)dx , provided the integral exists. −∞ Remark: If g(x) is function of a random variable and E[g(x)] exists, ∞ then E[(g(x)] = ∫ g(x) f(x)dx −∞ Example 17: Let X be a continuous random variable with p.d.f given by f(x) = 4x3, 0 < x < 1. Find the expected value of X. Solution: ∞ We know that E(X) = ∫ xf ( x)dx −∞ 1 In this problem E(X) = ∫ x(4 x ) dx 3 0 59 1 = 4 ∫ x ( x 3 )dx 0 1 x5 =4 5 0 1 4 x5 = 0 5 4 = [ 15 - 05] 5 4 = [1] 5 4 = 5 Example 18: Let x be a continuous random variable with pdf. given by f(x) = 3x2 , 0 < x < 1 Find mean and variance Solution: ∞ E(x) ∫ xf (x ) dx = −∞ 1 E(x) = ∫ x[3x 2 dx 0 1 =3 ∫ x dx 3 0 1 = = = = x4 3 4 0 3 41 x0 4 34 1 −0 4 3 4 [ 60 ∞ ∫x 2 E(x ) = 2 f ( x ) dx −∞ 1 = ∫x 2 [3x 2 ] dx 0 1 =3 ∫x 4 dx 0 1 x5 =3 5 0 1 3 = x5 0 5 3 = 15 − 0 5 3 = 5 Variance = E(x2) – [E(x)]2 [ 2 3 3 − 5 4 3 9 = − 5 16 48 − 45 3 = = 80 80 2.6 Moment generating function (M.G.F) (concepts only): To find out the moments, the moment generating function is a good device. The moment generating function is a special form of mathematical expectation and is very useful in deriving the moments of a probability distribution. Var (x)= Definition: If X is a random variable, then the expected value of etx is known as the moment generating functions, provided the expected value exists for every value of t in an interval, - h < t < h , where h is some positive real value. 61 The moment generating function is denoted as Mx(t) For discrete random variable Mx(t) = E(etx) = Σ etx p(x) (tx ) 2 (tx ) 3 = Σ 1 + tx + + + ....... px(x) 2! 3! 2 3 ∞ tr t t 1 + tµ1 '+ µ 2 '+ µ 3 '+....... = ∑ µ r ' Mx(t) = = r! 2! 3! r0 In the above expression, the rth raw moment is the tr coefficient of in the above expanded sum. To find out the r! moments differentiate the moment generating function with respect to t once, twice, thrice… and put t = 0 in the first, second, third, … … derivatives to obtain the first, second, third,….. moments. .. … From the resulting expression, we get the raw moments about the origin. The central moments are obtained by using the relationship between raw moments and central moments. 2.7 Characteristic function: The moment generating function does not exist for every distribution. Hence another function, which always exists for all the distributions is known as characteristic function. It is the expected value of eitx, where i = − 1 and t has a real value and the characteristic function of a random variable X is denoted by φx(t) For a discrete variable X having the probability function p(x), the characteristic function is φx(t) = Σ eitx p(x) For a continuous variable X having density function f(x), b such that a < x < b , the characteristic function φx(t) = ∫ a 62 eitx f(x)dx Exercise - 2 I. Choose the best answer: n 1. ∑ p( x ) is equal to i =1 i (a) 0 (b) 1 (c) –1 (d) ∞ 2. If F(x) is distribution function, then F(-∞) is (a) –1 (b) 0 (c) 1 (d) -∞ 3. From the given random variable table, the value of a is (a) 1 X=x pi 1 (b) 2 0 a 1 2a (c) 4 2 a (d) 1 4 4. E(2x+3) is (a) E(2x) (b) 2E(x) +3 (c) E(3) (d) 2x+3 5. Var(x+8) is (a) var (8) (b) var(x) (c) 8 var(x) (d) 0 6. Var(5x+2) is (a) 25 var (x) (b) 5 var (x) (c) 2 var (x) (d) 25 7. Variance of the random variable X is (a) E(x2) - [E(x)]2 (b) [E(x)]2 - E(x2) 2 (c) E(x ) (d) [E(x)]2 1 8. Variance of the random variable x is ; its standard 16 deviation is 1 1 (a) (b) 256 32 1 1 (c) (d) 64 4 9. A random variable X has E(x) = 2 and E(x2) = 8 its variance is (a) 4 (b) 6 (c) 8 (d) 2 63 10. If f(x) is the p.d.f of the continuous random variable x, then E(x2) is ∞ (a) ∞ ∫ f ( x) dx (b) ∫x (d) −∞ ∞ (c) 2 ∫ xf (x ) dx −∞ ∞ f ( x ) dx ∫ f (x 2 ) dx −∞ −∞ II. Fill in the blanks: 11. If f(x) is a distribution function, then F(+∞) is equal to ________ 12. If F(x) is a cumulative distribution function of a continuous random variable x with p.d.f f(x) then F′(x) = __________ 13. f(x) is the probability density function of a continuous ∞ random variable X. Then ∫ f ( x) dx is equal to ________ −∞ 14. Mathematical expectation of a random variable X is also known as _____________ 15. Variance of a constant is _____________ 16. Var (12x) is _____________ 17. Var (4x+7) is _________ 18. If x is a discrete random variable with the probabilities pi , then the expected value of x2 is ________ 19. If f(x) is the p.d.f of the continuous random variable X, then the expectation of X is given by __________ 20. The moment generating function for the discrete random variable is given by ____________ III. Answer the following: 21. Define random variable. 22. Define discrete random variable 23. Define continuous random variable 24. What is probability mass function? 25. What is discrete probability distribution? 26. Define probability density function. 64 27. Write the properties of distribution function. 28. Define mathematical expectation for discrete random variable. 29. Define the expectation of a continuous random variable. 30. State the moment generating function. 31. State the characteristic function for a discrete random variable. 32. State the characteristic function for the continuous random variable. 33. Write short note on moment generating function. 34. Write a short note on characteristic function. 35. Find the probability distribution of X when 3 coins are tossed, where x is defined as getting head. 36. Two dice are thrown simultaneously and getting three is termed as success. Obtain the probability distribution of the number of threes. 37. Three cards are drawn at random successively, with replacement, from a well shuffled pack of 52 cards. Getting a card of diamond is termed as success. Obtain the probability distribution of the number of success. 38. A random variable X has the following probability distribution Value of x 0 1 2 3 4 P(X=x) 3a 4a 6a 7a 8a (a) determine the value of a (b) Find p( 1 < x < 4 ) (c) P(1 ≤ x ≤ 4) (d) Find P(x >2) (e) Find the distribution function of x 39. A random variable X has the following probability function. Values 0 1 2 3 4 5 6 7 of X, x P(x) 0k (i) Find k (iii) Find p(x ≤ 6) 2k 2k 3k k2 (ii) Find p(0 < x < 5) 65 2k2 7k2+k 40. Verify whether the following are probability density function (i) f(x) = 6x5 , 0<x<1 2x (ii) f(x) = , 0<x<3 9 41. A continuous random variable x follows the probability law. f(x) = Ax3, 0 < x < 1 determine A 42. A random variable X has the density function f(x) = 3x2 , 0 < x < 1 Find the probability between 0.2 and 0.5 43. A random variable X has the following probability distribution X=x 5 2 1 1 1 1 4 2 4 Find the expected value of x 44. A random variable X has the following distribution x -1 0 1 2 P(x) 1 1 3 6 2 Find E(x) , E(x ) and Var (x) P(x) 45. A random variable X has E(x) = 1 6 1 3 1 1 and E(x2) = find its 2 2 variance and standard deviation. 46. In a continuous distribution, whose probability density 3 function is given by f(x) = x(2-x) , 0 < x < 2. Find the 4 expected value of x. 47. The probability density function of a continuous random x variable X is given by f(x) = for 0 < x < 2. Find its mean 2 and variance 66 Answers I. 1. (b) 6. (a) II. 11. 1 15. zero 18. Σxi2 pi 2. (b) 7. (a) 3. (d) 8. (d) 12. f(x) 16. 144 var(x) 4. (b) 9. (a) 13. 1 ∞ 19. ∫ x f (x ) dx −∞ 5. (b) 10. (c) 14. Mean 17. 16 var(x) ∞ tr 20. ∑ µ!r r = 0 r! III. 35. X=x 0 1 2 3 P(xi) 1/8 3/8 3/8 1/8 36. X=x 0 37. X=x P(xi) 0 2 25 36 P(x=x) 1 10 36 1 36 1 2 3 27 27 9 1 64 64 64 64 38. (i) a = 1/28 (ii) 13/28 (iii) 25/28 (iv) 15/28 (v) x 0 1 2 3 4 F(x) 28 3 7 13 20 =1 28 28 28 28 28 39. (i) k = 1/10 (ii) 4/5 (iii) 83/100 40. (i) p.d.f (ii) p.d.f 41. A = 4 42. P(0.2 < x , 0.5) = 0.117 43. 2.5 44. E(x) = 1/2 , var (x) = 19/12 45. 1/4 , 1/2 46. E(x) = 1 47. E(x) = 4/3 , var (x) = 2/9 67 3. SOME IMPORTANT THEORETICAL DISTRIBUTIONS 3.1 BINOMIAL DISTRIBUTION 3.1.0 Introduction: In this chapter we will discuss the theoretical discrete distributions in which variables are distributed according to some definite probability law, which can be expressed mathematically. The Binomial distribution is a discrete distribution expressing the probability of a set of dichotomous alternative i.e., success or failure. This distribution has been used to describe a wide variety of process in business and social sciences as well as other areas. 3.1.1 Bernoulli Distribution: A random variable X which takes two values 0 and 1 with probabilities q and p i.e., P(x=1) = p and P(x=0) = q, q = 1−p, is called a Bernoulli variate and is said to be a Bernoulli Distribution, where p and q takes the probabilities for success and failure respectively. It is discovered by Swiss Mathematician James Bernoulli (1654-1705). Examples of Bernoulli’ s Trails are: 1) Toss of a coin (head or tail) 2) Throw of a die (even or odd number) 3) Performance of a student in an examination (pass or fail) 3.1.2 Binomial Distribution: A random variable X is said to follow binomial distribution, if its probability mass function is given by nCx px qn-x ; x = 0, 1,2, … ,n 0 ; otherwise Here, the two independent constants n and p are known as the ‘ parameters’ of the distribution. The distribution is completely determined if n and p are known. x refers the number of successes. 68 P (X = x) = P(x) = If we consider N sets of n independent trials, then the number of times we get x success is N(nCx px qn-x). It follows that the terms in the expansion of N (q + p)n gives the frequencies of the occurrences of 0,1,2,...,x,...,n success in the N sets of independent trials. 3.1.3 Condition for Binomial Distribution: We get the Binomial distribution under the following experimental conditions. 1) The number of trials ‘ n’ is finite. 2) The trials are independent of each other. 3) The probability of success ‘ p’ is constant for each trial. 4) Each trial must result in a success or a failure. The problems relating to tossing of coins or throwing of dice or drawing cards from a pack of cards with replacement lead to binomial probability distribution. 3.1.4 Characteristics of Binomial Distribution: 1. Binomial distribution is a discrete distribution in which the random variable X (the number of success) assumes the values 0,1, 2, … where n is finite. .n, 2. Mean = np, variance = npq and standard deviation σ = npq , q− p Coefficient of skewness = , npq 1 - 6pq , clearly each of the npq probabilities is non-negative and sum of all probabilities is 1 ( p < 1 , q < 1 and p + q =1, q = 1− p ). 3. The mode of the binomial distribution is that value of the variable which occurs with the largest probability. It may have either one or two modes. 4. If two independent random variables X and Y follow binomial distribution with parameter (n1, p) and (n2, p) respectively, then their sum (X+Y) also follows Binomial distribution with parameter (n1 + n2, p) 69 Coefficient of kurtosis = 5. If n independent trials are repeated N times, N sets of n trials are obtained and the expected frequency of x success is N(nCx px qn-x). The expected frequencies of 0,1,2… n success are the successive terms of the binomial distribution of N(q + p)n. Example 1: Comment on the following: “ The mean of a binomial distribution is 5 and its variance is 9” Solution: The parameters of the binomial distribution are n and p We have mean ⇒ np = 5 Variance ⇒ npq = 9 npq 9 = ∴q= np 5 9 q = >1 5 Which is not admissible since q cannot exceed unity. Hence the given statement is wrong. Example 2: Eight coins are tossed simultaneously. Find the probability of getting atleast six heads. Solution: Here number of trials, n = 8, p denotes the probability of getting a head. 1 1 and q = ∴p= 2 2 If the random variable X denotes the number of heads, then the probability of a success in n trials is given by P(X = x) = ncx px qn-x , x = 0 , 1, 2, ..., n x 1 1 = 8Cx 2 2 1 = 8 8Cx 2 8− x 1 = 8Cx 2 70 8 Probability of getting atleast six heads is given by P(x ≥ 6) = P(x = 6) + P(x = 7) + P(x = 8) 1 1 1 = 8 8C6 + 8 8C7 + 8 8C8 2 2 2 1 = 8 [ 8C6 + 8C7 + 8C8] 2 1 37 = 8 [ 28 +8 +1] = 256 2 Example 3: Ten coins are tossed simultaneously. Find the probability of getting (i) atleast seven heads (ii) exactly seven heads (iii) atmost seven heads Solution: 1 2 1 q = Probability of not getting a head = 2 The probability of getting x heads throwing 10 coins simultaneously is given by P(X = x) = nCx px qn-x. , x = 0 , 1, 2 , ..., n p = Probability of getting a head x 10 − x 1 1 = 10Cx 2 2 = = 1 10Cx 210 i) Probability of getting atleast seven heads P(x ≥ 7) = P (x = 7) + P(x = 8) + P (x = 9) + P (x =10) 1 = 10 [ 10C7 + 10C8 + 10C9+ 10C10] 2 1 176 = [ 120 + 45 + 10 +1] = 1024 1024 ii) Probability of getting exactly 7 heads 1 1 P ( x = 7) = 10 10C7 = 10 (120) 2 2 120 = 1024 71 iii) Probability of getting atmost 7 heads P( x ≤ 7) = 1 – P(x > 7) = 1 − { P(x = 8) + P (x = 9) + P(x = 10)} 1 = 1− 10 {10C8 + 10C9 + 10C10} 2 1 = 1 − 10 [45 +10+1] 2 56 =1− 1024 968 = 1024 Example 4: 20 wrist watches in a box of 100 are defective. If 10 watches are selected at random, find the probability that (i) 10 are defective (ii) 10 are good (iii) at least one watch is defective (iv) at most 3 are defective. Solution: 20 out of 100 wrist watches are defective 20 1 Probability of defective wrist watch , p = = 100 5 ∴ q = 1− p = Since 10 watches are selected at random, n =10 P(X = x) = nCx px qn-x , x = 0 , 1, 2, ...,10 10 − x x 1 4 = 10Cx 5 5 i) Probability of selecting 10 defective watches 10 P( x =10) = 10C10 = 1. 1 5 1 .1 510 0 4 5 1 = 10 5 72 4 5 ii) Probability of selecting 10 good watches (i.e. no defective) 0 10 1 4 P(x = 0) = 10C0 5 5 10 10 4 4 = 1.1. = 5 5 iii) Probability of selecting at least one defective watch P(x ≥ 1) = 1 – P(x < 1) = 1 – P(x = 0) 0 10 1 4 = 1 − 10C0 5 5 10 4 =1− 5 iv) Probability of selecting at most 3 defective watches P (x ≤ 3) = P (x = 0) + P(x =1) + P(x = 2) + P(x = 3) 0 10 1 9 2 1 4 1 4 1 4 = 10C0 +10C1 + 10C2 5 5 5 5 5 5 3 1 4 +10C3 5 5 10 7 1 9 2 10.9 1 4 4 1 4 = 1.1. + 10 + 1.2 5 5 5 5 5 3 8 8 7 10.9.8 1 4 + 1.2.3 5 5 = 1. (0.107) + 10 (0.026) + 45 (0.0062) + 120 (0.0016) = 0.859 (approx) Example 5: With the usual notation find p for binomial random variable X if n = 6 and 9P(X = 4) = P(X = 2) Solution: The probability mass function of binomial random variable X is given by P(X = x) = nCx px qn-x. , x = 0 , 1, 2, ...,n 73 ∴ P(X = x) = 6Cx px q6-x P (x = 4) = 6C4 p4 q2 P (x = 2) = 6C2 p2 q4 Given that, 9. P(x = 4) = P(x = 2) 9. 6C4 p4q2 = 6C2 p2q4 ⇒ 9 × 15p2 = 15q2 9p2 = q2 Taking positive square root on both sides we get, 3p = q = 1− p 4p = 1 1 = 0.25 ∴p= 4 Here n = 6 3.1.5 Fitting of Binomial Distribution: When a binomial distribution is to be fitted to an observed data, the following procedure is adopted. Σfx 1. Find Mean = x = = np Σf x where n is number of trials ⇒p= n 2. Determine the value, q = 1− p. 3. The probability function is P(x) = nCx px qn-x put x = 0, we set P(0) = qn and f(0) = N × P(0) 4. The other expected frequencies are obtained by using the recurrence formula is given by n-x p f(x+1) = f(x) x+1 q Example 6: A set of three similar coins are tossed 100 times with the following results Number of heads : 0 1 2 3 Frequency : 36 40 22 74 2 Solution: X 0 1 2 3 f fx 36 0 40 40 22 44 2 6 Σf =100 Σfx = 90 90 Σfx = Mean = x = = 0.9 100 Σf x p= n 0.9 = 0.3 = 3 q = 1 –0.3 = 0.7 The probability function is P(x) = nCx px qn-x Here n = 3, p = 0.3 q = 0.7 ∴P(x) = 3Cx (0.3)x (0.7)3-x P(0) = 3C0 (0.3)0 (0.7)3 = (0.7)3 = 0.343 ∴ f(0) = N × P(0) = 0.343 × 100 = 34.3 The other frequencies are obtained by using the recurrence formula n-x p f(x+1) = f(x). By putting x = 0, 1, 2 the expected x+1 q frequencies are calculated as follows. 3-0 p f(1) = × 34.3 0 +1 q = 3 × (0.43) × 34.3 = 44.247 3 -1 p f(2) = f(1) 1+1 q 2 = (0.43) × 44.247 2 = 19.03 75 3− 2 p f(2) 2 +1 q 1 = (0.43) × 19.03 3 = 2.727 The observed and theoretical (expected) frequencies are tabulated below: Total Observed 36 40 22 2 100 frequencies Expected 34 44 19 3 100 frequencies f(3) = Example 7: 4 coins are tossed and number of heads noted. The experiment is repeated 200 times and the following distribution is obtained . x: Number of heads 0 1 2 3 4 f: frequencies 62 85 40 11 2 Solution: X f fx 0 62 0 1 85 85 2 40 80 3 11 33 206 Σfx = = 1.03 Σf 200 x 1.03 p= = = 0.2575 n 4 ∴ q = 1− 0.2575 = 0.7425 Here n = 4 , p = 0.2575 ; q = 0.7425 The probability function of binomial distribution is P(x) = nCx px qn-x Mean = x = 76 4 2 8 Total 200 206 The binomial probability function is P(x) = 4Cx (0.2575)x (0.7425)4-x P(0) = (0.7425)4 = 0.3039 ∴ f(0) = NP(0) = 200 × 0.3039 = 60.78 The other frequencies are calculated using the recurrence formula n-x p f(x+1) = f(x). By putting x = 0,1, 2, 3 then the expected x+1 q frequencies are calculated as follows: Put x = 0, we get 4-0 f(1) = (0.3468) (60.78) 0 +1 = 84.3140 4 -1 f(2) = (0.3468) (84.3140) 1+1 = 43.8601 4-2 f(3) = (0.3468) (43.8601) 2 +1 = 10.1394 4-3 f(4) = (0.3468) (10.1394) 3 +1 = 0.8791 The theoretical and expected frequencies are tabulated below: Total Observed frequencies Expected frequencies 62 85 40 11 2 200 61 84 44 10 1 200 77 3.2 POISSON DISTRIBUTION: 3.2.0 Introduction: Poisson distribution was discovered by a French Mathematician-cum-Physicist Simeon Denis Poisson in 1837. Poisson distribution is also a discrete distribution. He derived it as a limiting case of Binomial distribution. For n-trials the binomial distribution is (q + p)n ; the probability of x successes is given by P(X=x) = nCx px qn-x . If the number of trials n is very large and the probability of success ‘ p’ is very small so that the product np = m is non – negative and finite. The probability of x success is given by e−m m x P( X = x ) = for x = 0,1,2, … x! 0 ; otherwise Here m is known as parameter of the distribution so that m >0 Since number of trials is very large and the probability of success p is very small, it is clear that the event is a rare event. Therefore Poisson distribution relates to rare events. Note: 1) e is given by e = 1 + 1 11 + + +… = 2.71828 .. 1! 2! 3! e− m m 0 , 0! = 1 and 1! = 1 0! e− m m1 3) P(X=1) = 1! Some examples of Poisson variates are : 1. The number of blinds born in a town in a particular year. 2. Number of mistakes committed in a typed page. 3. The number of students scoring very high marks in all subjects 4. The number of plane accidents in a particular week. 5. The number of defective screws in a box of 100, manufactured by a reputed company. 6. Number of suicides reported in a particular day. 2) P(X=0) = 78 3.2.1 Conditions: Poisson distribution is the limiting case of binomial distribution under the following conditions: 1. The number of trials n is indefinitely large i.e., n à ∞ 2. The probability of success ‘ p’ for each trial is very small; i.e., p à 0 3. np = m (say) is finite , m > 0 3.2.2 Characteristics of Poisson Distribution: The following are the characteristics of Poisson distribution 1. Discrete distribution: Poisson distribution is a discrete distribution like Binomial distribution, where the random variable assume as a countably infinite number of values 0,1,2 … . 2. The values of p and q: It is applied in situation where the probability of success p of an event is very small and that of failure q is very high almost equal to 1 and n is very large. 3. The parameter: The parameter of the Poisson distribution is m. If the value of m is known, all the probabilities of the Poisson distribution can be ascertained. 4. Values of Constant: Mean = m = variance; so that standard deviation = m Poisson distribution may have either one or two modes. 5. Additive Property: If X and Y are two independent Poisson distribution with parameter m1 and m2 respectively. Then (X+Y) also follows the Poisson distribution with parameter (m1 + m2) 6. As an approximation to binomial distribution: Poisson distribution can be taken as a limiting form of Binomial distribution when n is large and p is very small in such a way that product np = m remains constant. 7. Assumptions: The Poisson distribution is based on the following assumptions. i) The occurrence or non- occurrence of an event does not influence the occurrence or non-occurrence of any other event. 79 ii) iii) The probability of success for a short time interval or a small region of space is proportional to the length of the time interval or space as the case may be. The probability of the happening of more than one event is a very small interval is negligible. Example 8: Suppose on an average 1 house in 1000 in a certain district has a fire during a year. If there are 2000 houses in that district, what is the probability that exactly 5 houses will have a fire during the year? [given that e-2 = 0.13534] 1 Mean, x = np , n = 2000 and p = 1000 1 = 2000 × 1000 m=2 The Poisson distribution is e−m m x P(X=x) = x! −2 5 e2 ∴P(X =5) = 5! (0.13534) × 32 = 120 = 0.036 (Note: The values of e-m are given in Appendix ) Example 9: In a Poisson distribution 3P(X=2) = P(X=4) parameter ‘ m’ . Solution: e−m m x Poisson distribution is given by P(X=x) = x! Given that 3P(x=2) = P(x= 4) 80 Find the e−m m2 e−m m4 = 2! 4! 3× 4! m2 = 2! ∴ m= ±6 Since mean is always positive ∴ m = 6 3. Example 10: If 2% of electric bulbs manufactured by a certain company are defective. Find the probability that in a sample of 200 bulbs i) less than 2 bulbs ii) more than 3 bulbs are defective.[e-4 = 0.0183] Solution: 2 = 0.02 100 Given that n = 200 since p is small and n is large We use the Poisson distribution mean, m = np = 200 × 0.02 = 4 e−m m x Now, Poisson Probability function, P(X = x) = x! i) Probability of less than 2 bulbs are defective = P(X<2) = P(x = 0) + P(x = 1) e−4 40 e−4 41 = + 0! 1! -4 -4 = e + e (4) = e- 4 (1 + 4) = 0.0183 × 5 = 0.0915 ii) Probability of getting more than 3 defective bulbs P(x > 3) = 1− P(x ≤ 3) = 1− {P(x = 0) + P(x =1) + P(x=2) + P(x=3)} 4 2 43 -4 = 1− e {1+ 4+ +} 2! 3! = 1− {0.0183 × (1 + 4 + 8 + 10.67)} = 0.567 81 The probability of a defective bulb = p = 3.2.3 Fitting of Poisson Distribution: The process of fitting of Poisson distribution for the probabilities of x = 0, 1,2,... success are given below : ∑ fx i) First we have to calculate the mean = x = =m ∑f ii) The value of e-m is obtained from the table (see Appendix ) e − m .m x iii) By using the formula P(X=x) = x! Substituting x = 0, P(0) = e-m Then f(0) = N×P(0) The other expected frequencies will be obtained by using the recurrence formula m f(x+1) = f(x) ; x = 0,1,2, … x +1 Example 11: The following mistakes per page were observed in a book. Number of mistakes ( per page) 0 1 2 3 4 Number of pages 211 90 19 5 0 Fit a Poisson distribution to the above data. Solution: xi 0 1 2 3 4 fi 211 90 19 5 0 N = 325 fx N 143 = = 0 .44 = m 325 Then e-m ⇒ e- 0.44 = 0.6440 82 Mean = x = fixi 0 90 38 15 0 fx = 143 Probability mass function of Poisson distribution is mx P(x) = e-m x! 0 -0.44 44 Put x = 0, P(0) = e 0! = e-0.44 = 0.6440 ∴ f(0) = N P(0) = 325 × 0.6440 = 209.43 The other expected frequencies will be obtained by using the recurrence formula m f(x+1) = f(x). By putting x = 0,1,2,3 we get the x +1 expected frequencies and are calculated as follows. f(1) = 0.44 × 209.43 = 92.15 0.44 f(2) = × 92.15 = 20.27 2 0.44 f(3) = × 20.27 = 2.97 3 0.44 = 0.33 f(4) = × 2.97 4 Total Observed 211 90 19 5 0 325 frequencies Expected 210 92 20 3 0 325 frequencies Example 12: Find mean and variance to the following data which gives the frequency of the number of deaths due to horse kick in 10 corps per army per annum over twenty years. X 0 1 2 3 4 Total F 109 65 22 3 1 200 83 Solution: Let us calculate the mean and variance of the given data xi fi fixi fixi2 0 109 0 0 1 65 65 65 2 22 44 88 3 3 9 27 4 1 4 16 Total N = 200 fx = 122 fx2 = 196 i Mean = x = fix N 122 = 200 = 0.61 Variance = σ2 = Hence, i f i2x () −x 2 N 196 = − (0.61)2 200 = 0.61 mean = variance = 0.61 Example 13: 100 car radios are inspected as they come off the production line and number of defects per set is recorded below No. of defects No. of sets 0 1 2 3 4 79 18 2 1 0 Fit a Poisson distribution and find expected frequencies 84 Solution: x 0 1 2 3 4 f fx 79 0 18 18 2 4 1 3 0 0 N = 100 f x = 25 fx Mean = x = N 25 = 100 ∴m = 0.25 Then e-m = e- 0.25 = 0.7788 = 0.779 Poisson probability function is given by e− m m x P(x) = x! −0.25 e (0.25)0 P(0) = = (0.779) 0! ∴ f(0) = N.P(0) = 100 × (0.779) = 77.9 Other frequencies are calculated using the recurrence formula m f(x+1) = f(x). x +1 By putting x = 0,1,2,3, we get the expected frequencies and are calculated as follows. m f(1) = f(0+1) = f(0) 0+1 0.25 (77.9) f(1) = 1 = 19.46 0.25 f(2) = (19.46) 2 = 2.43 85 0.25 (2.43) 3 = 0.203 0.25 f(4) = (0.203) 4 = 0.013 f(3) = Observed frequencies Expected frequencies 79 18 2 1 0 100 78 20 2 0 0 100 Example 14: Assuming that one in 80 births in a case of twins, calculate the probability of 2 or more sets of twins on a day when 30 births occurs. Compare the results obtained by using (i) the binomial and (ii) Poisson distribution. Solution: (i) Using Binomial distribution 1 = 0.0125 80 ∴ q = 1− p = 1 – 0.0125 = 0.9875 n = 30 Binomial distribution is given by P(x) = nCx px qn-x P(x ≥ 2) = 1 – P(x < 2) = 1 – {P(x = 0) + P(x =1)} = 1 – {30C0(0.0125)0 (0.9875)30 + 30C1 (0.0125)1(0.9875)29} = 1– {1.1(0.9875)30 + 3 (0.125) (0.9875)29} = 1 – { 0.6839 + 0.2597} = 1 – 0.9436 P( x ≥ 2) = 0.0564 Probability of twins birth = p = 86 (ii) By using Poisson distribution: The probability mass function of Poisson distribution is given by e− m m x P(x) = x! Mean = m = np = 30 (0.0125) = 0.375 P(x ≥2) = 1− P(x <2) = 1 − { P( x = 0) + P( x = 1)} e −0.375 (0.375)0 e −0.375 (0.375)1 =1–{ + } 0! 1! = 1 − e- 0.375 ( 1 + 0.375) = 1 – (0.6873) (1.375) = 1 – 0.945 = 0.055 3.3 NORMAL DISTRIBUTION: 3.3.0 Introduction: In the preceding sections we have discussed the discrete distributions, the Binomial and Poisson distribution. In this section we deal with the most important continuous distribution, known as normal probability distribution or simply normal distribution. It is important for the reason that it plays a vital role in the theoretical and applied statistics. The normal distribution was first discovered by DeMoivre (English Mathematician) in 1733 as limiting case of binomial distribution. Later it was applied in natural and social science by Laplace (French Mathematician) in 1777. The normal distribution is also known as Gaussian distribution in honour of Karl Friedrich Gauss(1809). 3.3.1 Definition: A continuous random variable X is said to follow normal distribution with mean µ and standard deviation σ, if its probability density function 1 x −µ 2 − 1 2 σ ;−∞ < x < ∞ , − ∞ < µ < ∞, σ > 0. f(x) = e σ 2π 87 Note: The mean µ and standard deviation σ are called the parameters of Normal distribution. The normal distribution is expressed by X ∼ N(µ, σ2) 3.3.2 Condition of Normal Distribution: i) Normal distribution is a limiting form of the binomial distribution under the following conditions. a) n, the number of trials is indefinitely large ie., nà ∞ and b) Neither p nor q is very small. ii) Normal distribution can also be obtained as a limiting form of Poisson distribution with parameter m à ∞ iii) Constants of normal distribution are mean = µ, variation =σ2, Standard deviation = σ. 3.3.3 Normal probability curve: The curve representing the normal distribution is called the normal probability curve. The curve is symmetrical about the mean (µ), bell-shaped and the two tails on the right and left sides of the mean extends to the infinity. The shape of the curve is shown in the following figure. -∞ x=µ 88 ∞ 3.3.4 Properties of normal distribution: 1. The normal curve is bell shaped and is symmetric at x = µ. 2. Mean, median, and mode of the distribution are coincide i.e., Mean = Median = Mode = µ 3. It has only one mode at x = µ (i.e., unimodal) 4. Since the curve is symmetrical, Skewness = β1 = 0 and Kurtosis = β2 = 3. 5. The points of inflection are at x = µ ± σ 6. The maximum ordinate occurs at x = µ and 1 its value is = σ 2π 7. The x axis is an asymptote to the curve (i.e. the curve continues to approach but never touches the x axis) 8. The first and third quartiles are equidistant from median. 9. The mean deviation about mean is 0.8 σ 10. Quartile deviation = 0.6745 σ 11. If X and Y are independent normal variates with mean µ1 and µ2, and variance σ12 and σ22 respectively then their sum (X + Y) is also a normal variate with mean (µ1 + µ2) and variance (σ12 + σ22) 12. Area Property P(µ - σ < × < µ + σ) = 0.6826 P(µ - 2σ < × < µ + 2σ) = 0.9544 P(µ - 3σ < × < µ + 3σ) = 0.9973 3.3.5 Standard Normal distribution: Let X be random variable which follows normal distribution with mean µ and variance σ2 .The standard normal variate is X−µ defined as Z = which follows standard normal distribution σ with mean 0 and standard deviation 1 i.e., Z ∼ N(0,1). The standard 1 −1 2 Z 2 ; - ∞ < z< ∞ e 2π The advantage of the above function is that it doesn’ t contain any parameter. This enable us to compute the area under the normal probability curve. 89 normal distribution is given by φ(z) = 3.3.6 Area properties of Normal curve: The total area under the normal probability curve is 1. The curve is also called standard probability curve. The area under the curve between the ordinates at x = a and x = b where a < b, represents the probabilities that x lies between x = a and x = b i.e., P(a ≤ x ≤ b) -∞ x = µ x=a x=b +∞ To find any probability value of x, we first standardize it by X−µ using Z = , and use the area probability normal table. (given σ in the Appendix). For Example: The probability that the normal random variable x to lie in the interval (µ−σ , µ+σ) is given by -∞ x=µ−σ z = -1 x=µ x=µ+σ z=0 z=+1 90 +∞ P( µ − σ < x < µ+σ) = P(−1 ≤ z ≤ 1 ) = 2P(0 < z < 1) = 2 (0.3413) (from the area table) = 0.6826 P( µ - 2σ < x < µ+2σ) = P(-2 < z < 2 ) = 2P(0 < z < 2) = 2 (0.4772) = 0.9544 -∞ x=µ−2σ z = -2 x=µ z=0 x=µ+2σ z = +2 +∞ P(µ − 3σ < x < µ + 3σ) = P(−3 < z < 3 ) = 2P(0 < z < 3) = 2 (0.49865) = 0.9973 -∞ x=µ−3σ z = -3 x=µ z =0 91 x=µ+3σ z=+3 +∞ The probability that a normal variate x lies outside the range µ ± 3σ is given by P(|x −µ | > 3σ) = P(|z| >3) = 1 – P(−3 ≤ z ≤ 3) = 1 − 0.9773 = 0.0027 Thus we expect that the values in a normal probability curve will lie between the range µ ± 3σ, though theoretically it range from − ∞ to ∞. Example 15: Find the probability that the standard normal variate lies between 0 and 1.56 Solution: 0.4406 -∞ z =0 z = 1.56 +∞ P(0<z<1.56) = Area between z = 0 and z = 1.56 = 0.4406 (from table) Example 16: Find the area of the standard normal variate from –1.96 to 0. Solution: 0.4750 -∞ z = -1.96 z=0 92 +∞ Area between z = 0 & z =1.96 is same as the area z = −1.96 to z = 0 P(-1.96 < z < 0) = P(0 < z < 1.96) (by symmetry) = 0.4750 (from the table) Example 17: Find the area to the right of z = 0.25 Solution: 0.4013 -∞ +∞ z = 0 z = 0.25 P(z >0.25) = P(0<z < ∞) – P(0<z<0.25) = 0.5000 - 0.0987 (from the table) = 0.4013 Example 18: Find the area to the left of z = 1.5 Solution: 0.9332 -∞ z=0 z = 1.5 +∞ P(z < 1.5) = P( − ∞ < z < 0 ) + P( 0 < z < 1.5 ) = 0.5 + 0.4332 (from the table) = 0.9332 93 Example 19: Find the area of the standard normal variate between –1.96 and 1.5 Solution: 0.9082 -∞ z= -1.96 z=0 z=1.5 +∞ P(-1.96 < z < 1.5) = P(-1.96 < z < 0) + P(0 < z < 1.5) = P(0 < z < 1.96) + P(0 < z < 1.5) = 0.4750 + 0.4332 (from the table) = 0.9082 Example 20: Given a normal distribution with µ = 50 and σ = 8, find the probability that x assumes a value between 42 and 64 Solution: 0.8012 -∞ z= -1 z=0 Given that µ = 50 and σ = 8 The standard normal variate z = 94 z=1.75 x−µ σ +∞ 42 − 50 − 8 = −1 = 8 8 64 − 50 14 If X = 64, Z2 = = = 1.75 8 8 ∴ P(42 < x < 64) = P(−1 < z <1.75) = P(−1< z < 0) + P(0 < z <1.95) = P(0<z<1) + P (0 < z <1.75) (by symmetry) = 0.3413 +0 .4599 (from the table) = 0 .8012 Example 21: Students of a class were given an aptitude test. Their marks were found to be normally distributed with mean 60 and standard deviation 5. What percentage of students scored. i) More than 60 marks (ii) Less than 56 marks (iii) Between 45 and 65 marks If X = 42 , Z1 = Solution: Given that mean = µ = 60 and standard deviation = σ = 5 x−µ i) The standard normal varaiate Z = σ 0.5 -∞ z=0 z>0 +∞ x−µ 60 − 60 = =0 5 σ ∴P(x > 60) = P(z > 0) = P(0 < z < ∞ ) = 0.5000 Hence the percentage of students scored more than 60 marks is 0.5000(100) = 50 % If X = 60, Z = 95 ii) If X = 56, Z = 56 − 60 − 4 = = − 0.8 5 5 0.2119 -∞ z= -0.8 +∞ z=0 P(x < 56) = P(z < −0.8) = P(- ∞ < z < 0) – P(−0.8 < z < 0) (by symmetry) = P(0 < 2 < ∞) – P(0 < z < 0.8) = 0.5 − 0.2881 (from the table) = 0.2119 Hence the percentage of students score less than 56 marks is 0.2119(100) = 21.19 % 45 − 60 − 15 iii) If X = 45, then z = = = −3 5 5 0.83995 -∞ z= -3 z=0 z=1 65 − 60 5 = =1 5 5 P(45 < x < 65) = P(−3 < z < 1) = P(−3 < z < 0 ) + P ( 0 < z < 1) X = 65 then z = 96 +∞ = P(0 < z < 3) + P(0 < z < 1) ( by symmetry) = 0.4986 + 0.3413 (from the table) = 0.8399 Hence the percentage of students scored between 45 and 65 marks is 0.8399(100) = 83.99 % Example 22: X is normal distribution with mean 2 and standard deviation 3. Find the value of the variable x such that the probability of the interval from mean to that value is 0.4115 Solution: Given µ = 2, σ = 3 Suppose z1 is required standard value, Thus P (0 < z < z1) = 0.4115 From the table the value corresponding to the area 0.4115 is 1.35 that is z1 = 1.35 x−µ Here z1 = σ x−2 1.35 = 3 x = 3(1.35) + 2 = 4.05 + 2 = 6.05 Example 23: In a normal distribution 31 % of the items are under 45 and 8 % are over 64. Find the mean and variance of the distribution. Solution: Let x denotes the items are given and it follows the normal distribution with mean µ and standard deviation σ The points x = 45 and x = 64 are located as shown in the figure. i) Since 31 % of items are under x = 45, position of x into the left of the ordinate x = µ ii) Since 8 % of items are above x = 64 , position of this x is to the right of ordinate x = µ 97 -∞ z = -z1 z=0 x = 45 x = µ z = z2 x = 64 +∞ x−µ 45 − µ = = − z1 (say) σ σ Since x is left of x = µ , z1 is taken as negative 64 − µ = z2 (say) When x = 64, z = σ From the diagram P(x < 45) = 0.31 P(z < - z1) = 0.31 P(- z1 < z < 0) = P(- ∞ < z < 0) – p(- ∞ < z < z1) s = 0.5 - 0.31 = 0.19 P(0 < z < z1) = 0.19 (by symmetry) z1 = 0.50 (from the table) Also from the diagram p(x > 64) = 0.08 P(0 < z < z2) = P(0 < z < ∞) – P(z2 < z < ∞) = 0.5 - 0.08 = 0.42 z2 = 1.40 (from the table) Substituting the values of z1 and z2 we get 45 − µ 64 − µ = − 0.50 and = 1.40 σ σ Solving µ - 0.50 σ = 45 ----- (1) µ + 1.40 σ = 64 ----- (2) (2) – (1) ⇒ 1.90 σ = 19 ⇒ σ = 10 Substituting σ = 10 in (1) µ = 45 + 0.50 (10) = 45 + 5.0 = 50.0 Hence mean = 50 and variance = σ2 = 100 When x = 45, z = 98 Exercise – 3 I. Choose the best answer: 1. Binomial distribution applies to (a) rare events (b) repeated alternatives (c) three events (d) impossible events 2. For Bernoulli distribution with probability p of a success and q of a failure, the relation between mean and variance that hold is (a) mean < variance (b) mean > variance (c) mean = variance (d) mean < variance 3. The variance of a binomial distribution is (a) npq (b) np (c) npq (d) 0 x 4. 5. 6. 7. 15 − x 2 1 The mean of the binomial distribution 15Cx 3 3 2 in which p = is 3 (a) 5 (b) 10 (c) 15 (d) 3 The mean and variance of a binomial distribution are 8 and 4 respectively. Then P(x = 1) is equal to 1 1 1 1 (a) 12 (b) 4 (c) 6 (d) 8 2 2 2 2 If for a binomial distribution , n = 4 and also P(x = 2) = 3P(x=3) then the value of p is 9 1 (a) (b) 1 (c) (d) None of the above 11 3 The mean of a binomial distribution is 10 and the number of trials is 30 then probability of failure of an event is (a) 0.25 (b) 0.333 (c) 0.666 (d) 0.9 99 8. The variance of a binomial distribution is 2. Its standard deviation is (a) 2 (b) 4 (c) 1/2 (d) 2 9. In a binomial distribution if the numbers of independent trials is n, then the probability of n success is (a) nCxpxqn-x (b) 1 (c) pn (d)qn 10. The binomial distribution is completely determined if it is known (a) p only (b) q only (c) p and q (d) p and n 11. The trials in a binomial distribution are (a) mutually exclusive (b) non-mutually exclusive (c) independent (d) non-independent 12. If two independent variables x and y follow binomial distribution with parameters,(n1, p) and (n2, p) respectively, their sum(x+y) follows binomial distribution with parameters (a) (n1 + n2, 2p) (b) (n, p) (c) (n1 + n2, p) (d) (n1 + n2, p + q) 13. For a Poisson distribution (a) mean > variance (b) mean = variance (c) mean < variance (d) mean < variance 14. Poisson distribution correspondents to (a) rare events (b) certain event (c) impossible event (d) almost sure event 15. If the Poisson variables X and Y have parameters m1 and m2 then X+Y is a Poisson variable with parameter. (a) m1m2 (b) m1+m2 (c) m1−m2 (d)m1/m2 16. Poisson distribution is a (a) Continuous distribution (b) discrete distribution (c) either continuous or discrete (d) neither continue nor discrete 100 17. Poisson distribution is a limiting case of Binomial distribution when (a) n à ∞ ; pà 0 and np = m (b) n à 0 ; pà ∞ and p=1/m (c) n à ∞ ; pà ∞ and np=m (d) n à ∞ ; pà 0 ,np=m 18. If the expectation of a Poisson variable (mean) is 1 then P(x < 1) is (a) e-1 (b) 1-2e-1 (c) 1- 5/2e-1 (d) none of these 19. The normal distribution is a limiting form of Binomial distribution if (a) nà ∞ pà0 (b) nà0 , pàq (c) nà∞ , pà n (d) nà ∞ and neither p nor q is small. 20. In normal distribution, skewness is (a) one (b) zero (c) greater than one (d) less than one 21. Mode of the normal distribution is 1 (a) σ (b) (c) µ (d) 0 2π 22. The standard normal distribution is represented by (a) N(0,0) (b) N(1,1) (c) N(1,0) (d) N(0,1) 23. Total area under the normal probability curve is (a) less than one (b) unity (c) greater than one (d) zero 24. The probability that a random variable x lies in the interval (µ - 2σ , µ + 2σ) is (a) 0.9544 (b) 0.6826 (c) 0.9973 (d) 0.0027 25. The area P(- ∞ < z < 0) is equal to (a) 1 (b) 0.1 (c) 0.5 (d) 0 26. The standard normal distribution has (a) µ =1, σ = 0 (b) µ = 0, σ = 1 (c) µ = 0 ,σ = 0 (d) µ =1, σ = 1 101 27. The random variable x follows the normal distribution − 1 ( x −100 ) 2 2 25 then the value of C is 1 1 (a) 5 2π (c) (d) 5 (b) 5 2π 2π 28. Normal distribution has (a) no mode (b) only one mode (c) two modes (d) many mode 29. For the normal distribution (a) mean = median =mode (b) mean < median < mode (c) mean > median > mode (d) mean > median < mode 30. Probability density function of normal variable f(x) = C. e P(X = x) = 1 5 2π − e 1 ( x −30 ) 2 25 2 ; -α < x < α then mean and variance are (a) mean = 30 variance = 5 (b) mean = 0, variance = 25 (c) mean = 30 variance = 25 (d) mean = 30, variance = 10 31. The mean of a Normal distribution is 60, its mode will be (a) 60 (b) 40 (c) 50 (d) 30 2 32. If x is a normal variable with µ =100 and σ = 25 then P(90 < x < 120) is same as (a) P(-1 < z < 1) (b) P(-2 < z < 4) (c) P(4 < z < 4.1) (d) P(-2 < z < 3) 33. If x is N(6, 1.2) and P(0 ≤ z ≤1) = 0.3413 then P(4.8 ≤ x ≤ 7.2) is (a) 0.3413 (b) 0.6587 (c) 0.6826 (d) 0.3174 II. Fill in the blanks: 34. The probability of getting a head in successive throws of a coin is _________ 35. If the mean of a binomial distribution is 4 and the variance is 2 then the parameter is __________ 102 9 2 1 36. + refers the binomial distribution and its standard 3 3 deviation is _________ 37. In a binomial distribution if number of trials to be large and probability of success be zero, then the distribution becomes ________. 38. The mean and variance are _______ in Poisson distribution 39. The mean of Poisson distribution is 0.49 and its standard deviation is ________ 40. In Poisson distribution, the recurrence formula to calculate expected frequencies is ______. 2 2 ∑ fx 41. The formula − x is used to find ________ N 42. In a normal distribution, mean takes the values from __________to ________ 43. When µ = 0 and σ = 1 the normal distribution is called ________ 44. P( − ∞ < z < 0) covers the area ______ 45. If µ = 1200 and σ = 400 then the standard normal variate z for x = 800 is _________ 46. At x = µ ± σ are called as __________ in a normal distribution. 47. P(−3 < z < 3) takes the value __________ 48. X axis be the ________to the normal curve. () III. Answer the following 49. Comment the following “ For a binomial distribution mean = 7 and variance = 16 50. Find the binomial distribution whose mean is 3 and variance 2 51. In a binomial distribution the mean and standard deviation are 12 and 2 respectively. Find n and p 52. A pair of dice is thrown 4 times. If getting a doublet is considered a success, find the probability of 2 success. 53. Explain a binomial distribution. 103 54. State the characteristics of a binomial distribution. 55. State the conditions for a binomial variate. 56. Explain the fitting of a binomial distribution. 57. For the binomial distribution (0.68+0.32)10 find the probability of 2 success. 58. Find the mean of binomial distribution of the probability of occurrence of an event is 1/5 and the total number of trials is 100 59. If on an average 8 ships out of 10 arrive safely at a port, find the mean and standard deviation of the number of ships arriving safely out of total of 1600 ships. 60. The probability of the evening college student will be a graduate is 0.4. Determine the probability that out of 5 students (i) none (ii) one (iii) atleast one will be a graduate 61. Four coins are tossed simultaneously. What is the probability of getting i) 2 heads and 2 tails ii) atleast 2 heads iii) atleast one head. 62. 10% of the screws manufactured by an automatic machine are found to be defective. 20 screws are selected at random. Find the probability that i) exactly 2 are defective ii) atmost 3 are defective iii) atleast 2 are defective. 63. 5 dice are thrown together 96 times. The numbers of getting 4, 5 or 6 in the experiment is given below. Calculate the expected frequencies and compare the standard deviation of the expected frequencies and observed frequencies. Getting 4 ,5 or 6 : 0 1 2 3 4 5 Frequency : 1 10 24 35 18 8 64. Fit a binomial distribution for the following data and find the expected frequencies. X: 0 1 2 3 4 f 18 35 30 13 4 65. Eight coins are tossed together 256 times. Number of heads observed at each toss is recorded and the results are given 104 below. Find the expected frequencies. What are the theoretical value of mean and standard deviation? Calculate also mean and standard deviation of the observed frequencies. Number of heads: 0 1 2 3 4 5 6 7 8 32 10 1 Frequencies : 2 6 39 52 67 56 66. Explain Poisson distribution. 67. Give any two examples of Poisson distribution. 68. State the characteristics of Poisson distribution. 69. Explain the fitting of a Poisson distribution 70. A variable x follows a Poisson distribution with mean 6 calculate i) P(x = 0) ii) P(x = 2) 71. The variance of a Poisson Distribution is 0.5. Find P(x = 3). [e- 0.5 = 0.6065] 72. If a random variable X follows Poisson distribution such that P(x =1) = P(x = 2) find (a) the mean of the distribution and P(x = 0). [e-2 = 0.1353] 73. If 3% of bulbs manufactured by a company are defective then find the probability in a sample of 100 bulbs exactly five bulbs are defective. 74. It is known from the past experience that in a certain plant there are on the average 4 industrial accidents per month. Find the probability that in a given year there will be less than 3 accidents. Assume Poisson distribution.[e-4 = 0.0183] 75. A manufacturer of television sets known that of an average 5% of this product is defective. He sells television sets in consignment of 100 and guarantees that not more than 4 sets will be defective. What is the probability that a television set will fail to meet the guaranteed quality? [e-5 = 0.0067] 76. One fifth percent of the blades produced by a blade manufacturing factory turns out to be a defective. The blades are supplied in pockets of 10. Use Poisson distribution to calculate the approximate number of pockets containing i) no defective (ii) all defective (iii) two defective blades respectively in a consignment of 1,00,000 pockets. 105 77. A factory employing a huge number of workers find that over a period of time, average absentee rate is three workers per shift. Calculate the probability that in a given shift i) exactly 2 workers (ii) more than 4 workers will be absent. 78. A manufacturer who produces medicine bottles finds that 0.1 % of the bottles are defective. They are packed in boxes containing 500 bottles. A drag manufactures buy 100 boxes from the producer of bottles. Using Poisson distribution find how many boxes will contain (i) no defective ii) exactly 2 (iii) atleast 2 defective. 79. The distribution of typing mistakes committed by a typist is given below: Mistakes per page: 0 1 2 3 4 5 No of pages : 142 156 69 57 5 1 Fit a Poisson distribution. 80. Fit a Poisson distribution to the following data: 01 2 3 45 6 7 8 Total x: 229 325 257 119 50 17 2 1 0 1000 f: 81. The following tables given that number of days in a 50, days period during which automatically accidents occurred in city. Fit a Poisson distribution to the data No of accidents : 0 1 2 3 4 No of days : 21 18 7 3 1 82. Find the probability that standard normal variate lies between 0.78 and 2.75 83. Find the area under the normal curve between z = 0 and z = 1.75 84. Find the area under the normal curve between z = -1.5 and z = 2.6 85. Find the area to the left side of z = 1.96 86. Find the area under the normal curve which lies to the right of z = 2.70 87. A normal distribution has mean = 50 and standard deviation is 8. Find the probability that x assumes a value between 34 and 62 88. A normal distribution has mean = 20 and S.D = 10. Find area between x =15 and x = 40 106 89. Given a normal curve with mean 30 and standard deviation 5. Find the area under the curve between 26 and 40 90. The customer accounts of a certain departmental store have an average balance of Rs.1200 and a standard deviation of Rs.400. Assuming that the account balances are normally distributed. (i) what percentage of the accounts is over Rs.1500? (ii) What percentage of the accounts is between Rs.1000 and Rs.1500? iii) What percentage of the accounts is below Rs.1500? 91. The weekly remuneration paid to 100 lecturers coaching for professional entrance examinations are normally distributed with mean Rs.700 and standard deviation Rs.50. Estimate the number of lecturers whose remuneration will be i) between Rs.700 and Rs.720 ii) more than Rs.750 iii) less than Rs.630 92. x is normally distributed with mean 12 and standard deviation 4. Find the probability of the following i) x ≥ 20 ii) x ≤ 20 iii) 0 < x < 12 93. A sample of 100 dry cells tested to find the length of life produced the following results µ =12 hrs, σ = 3 hrs. Assuming the data, to be normally distributed. What percentage of battery cells are expressed to have a life. i) more than 15 hrs ii) between 10 and 14 hrs as iii) less than 6 hrs?. 94. Find the mean and standard deviation of marks in an examination where 44 % of the candidates obtained marks below 55 and 6 % got above 80 marks. 95. In a normal distribution 7 % of the items are under 35 and 89 % of the items are under 63. Find its mean as standard deviation. Note: For fitting a binomial distribution in the problem itself, if it is given that the coin is unbiased, male and female births are equally probable, then we consider p = q = ½ . All other cases we have to find the value of p from the mean value of the given data. 107 Answers I. 1. b 2. b 3. a 4. b 5. a 6. c 7. c 8. d 9. c 10. d 11.c 12.c 13. b 14. a 15. b 16. b 17. d 18. a 19. d 20. b 21. c 22. d 23. b 24. a 25. c 26. b 27. b 28. b 31. a 32. b 33. c 29. a 30. c 1 1 34. 35. (8, ) 2 2 38. equal 39. 0.7 36. 37. Poisson distribution m 40. f(x+1) = f(x) 41. variance 42. - ∞, + ∞ x +1 43. Standard normal distribution 44. 0.5 45. –1 46. Point of inflections 47. 0.9973 48. Asymptote 16 49. This is not admissible . Since q = >1 7 9 2 1 2 1 50. + , p = , q = and n = 9 3 3 3 3 2 25 51. n = 18, p = . 52. 3 216 57. 10 C2 (0.32)2+(0.68)8 58. 20 59. 1280 3 11 15 60. i) 0.08 ii) 0.259 iii) 0.92 61. i) ii) iii) 8 16 16 18 9 1 62. (i) 190 × 20 (ii) 20 [920 + 20 × 919 +190× 918 +1140 × 917] 10 10 1 (iii) 1- 20 [920 + 20 × 919 +190× 918] 10 2 108 63. Observed S.D. = 1.13 and expected S.D.= 1.12 65. Observed mean = 4.0625 and S.D. = 1.462 70. i).0.00279 ii) 0.938 71. 0.0126 72. a) Mean = 2 b) P(x=0) = 0.1353 73. P(x = 5) = 0.1008 76. i) 98,020 ii)1960 78. i) 61 ii) 76 80. P(x) = e 74. 0.2379 iii) 20 77. i) 0.2241 ii) 0.1846 − e 1 1x 79. P(x) = x! − e 0.9 (0.9) x 81. P(x) = x! 83. 0.4599 iii) 9 −1. 5 (1.5) x x! 82. 0.2147 84. 0.9285 85. 0.9750 86. 0.0035 87. 0.9104 88. 0.6687 89. 0.7653 90. i) 22.66 % ii) 46.49 % iii) 77.34 % 91. i) 16 iii) 8 ii)16 75. 0.9598 92. i) 0.0228 ii) 0.9772 iii) 0.4987 93. i) 15.87 % ii) 49.72 % iii) 2.28 % 94. Mean = 57.21 and SD = 14.71 95. Mean = 50.27 and SD = 10.35 109 4. TEST OF SIGNIFICANCE (Basic Concepts) 4.0 Introduction: It is not easy to collect all the information about population and also it is not possible to study the characteristics of the entire population (finite or infinite) due to time factor, cost factor and other constraints. Thus we need sample. Sample is a finite subset of statistical individuals in a population and the number of individuals in a sample is called the sample size. Sampling is quite often used in our day-to-day practical life. For example in a shop we assess the quality of rice, wheat or any other commodity by taking a handful of it from the bag and then to decide to purchase it or not. 4.1 Parameter and Statistic: The statistical constants of the population such as mean, (µ), variance (σ2), correlation coefficient (ρ) and proportion (P) are called ‘ Parameters’ . Statistical constants computed from the samples corresponding to the parameters namely mean ( x ), variance (S2), sample correlation coefficient (r) and proportion (p) etc, are called statistic. Parameters are functions of the population values while statistic are functions of the sample observations. In general, population parameters are unknown and sample statistics are used as their estimates. 4.2 Sampling Distribution: The distribution of all possible values which can be assumed by some statistic measured from samples of same size ‘ n’ randomly drawn from the same population of size N, is called as sampling distribution of the statistic (DANIEL and FERREL). Consider a population with N values .Let us take a random sample of size n from this population, then there are 110 N! = k (say), possible samples. From each of n!(N - n)! these k samples if we compute a statistic (e.g mean, variance, correlation coefficient, skewness etc) and then we form a frequency distribution for these k values of a statistic. Such a distribution is called sampling distribution of that statistic. For example, we can compute some statistic t = t(x1, x2,… n) for each of these k samples. Then t1, t2 … tk ..x .., determine the sampling distribution of the statistic t. In other words statistic t may be regarded as a random variable which can take the values t1, t2 … tk and we can compute various statistical constants .., like mean, variance, skewness, kurtosis etc., for this sampling distribution. The mean of the sampling distribution t is 1k 1 t = [t1 + t2 + ..... + tk ] = ∑ ti K i =1 K 1 (t1 - t ) 2 + (t2 - t) 2 +......... + (tk - t )2 and var (t) = K 1 = ( ti - t ) 2 K 4.3 Standard Error: The standard deviation of the sampling distribution of a statistic is known as its standard error. It is abbreviated as S.E. For example, the standard deviation of the sampling distribution of the mean x known as the standard error of the mean, x + x + ...........xn Where v( x ) = v 1 2 n v( x1 ) v( x2 ) v( x ) = + 2 + ....... + 2n 2 n n n 2 2 2 nσ 2 σ σ σ = 2 + 2 + ...... + 2 = 2 n n n n σ ∴ The S.E. of the mean is n NCn = 111 The standard errors of the some of the well known statistic for large samples are given below, where n is the sample size, σ2 is the population variance and P is the population proportion and Q = 1−P. n1 and n2 represent the sizes of two independent random samples respectively. Sl.No Statistic Standard Error 1. σ Sample mean x n 2. Observed sample proportion p PQ n 3. Difference between of two samples means ( x 1 − x 2) σ112 σ2 2 + n1 n2 4. Difference of two proportions p1 – p2 P1Q1 P 2Q 2 + n1 n2 sample Uses of standard error i) Standard error plays a very important role in the large sample theory and forms the basis of the testing of hypothesis. ii) The magnitude of the S.E gives an index of the precision of the estimate of the parameter. iii) The reciprocal of the S.E is taken as the measure of reliability or precision of the sample. iv) S.E enables us to determine the probable limits within which the population parameter may be expected to lie. Remark: S.E of a statistic may be reduced by increasing the sample size but this results in corresponding increase in cost, labour and time etc., 4.4 Null Hypothesis and Alternative Hypothesis Hypothesis testing begins with an assumption called a Hypothesis, that we make about a population parameter. A hypothesis is a supposition made as a basis for reasoning. The conventional approach to hypothesis testing is not to construct a 112 single hypothesis about the population parameter but rather to set up two different hypothesis. So that of one hypothesis is accepted, the other is rejected and vice versa. Null Hypothesis: A hypothesis of no difference is called null hypothesis and is usually denoted by H0 “ Null hypothesis is the hypothesis” which is tested for possible rejection under the assumption that it is true “ by Prof. R.A. Fisher. It is very useful tool in test of significance. For example: If we want to find out whether the special classes (for Hr. Sec. Students) after school hours has benefited the students or not. We shall set up a null hypothesis that “H0: special classes after school hours has not benefited the students”. Alternative Hypothesis: Any hypothesis, which is complementary to the null hypothesis, is called an alternative hypothesis, usually denoted by H1, For example, if we want to test the null hypothesis that the population has a specified mean µ0 (say), i.e., : Step 1: null hypothesis H0: µ = µ0 then 2. Alternative hypothesis may be i) H1 : µ ≠ µ0 (ie µ > µ0 o r µ < µ0) ii) H1 : µ > µ0 iii) H1 : µ < µ0 the alternative hypothesis in (i) is known as a two – tailed alternative and the alternative in (ii) is known as right-tailed (iii) is known as left –tailed alternative respectively. The settings of alternative hypothesis is very important since it enables us to decide whether we have to use a single – tailed (right or left) or two tailed test. 4.5 Level of significance and Critical value: Level of significance: In testing a given hypothesis, the maximum probability with which we would be willing to take risk is called level of significance of the test. This probability often denoted by “ α” is generally specified before samples are drawn. 113 The level of significance usually employed in testing of significance are 0.05( or 5 %) and 0.01 (or 1 %). If for example a 0.05 or 5 % level of significance is chosen in deriving a test of hypothesis, then there are about 5 chances in 100 that we would reject the hypothesis when it should be accepted. (i.e.,) we are about 95 % confident that we made the right decision. In such a case we say that the hypothesis has been rejected at 5 % level of significance which means that we could be wrong with probability 0.05. The following diagram illustrates the region in which we could accept or reject the null hypothesis when it is being tested at 5 % level of significance and a two-tailed test is employed. Accept the null hypothesis if the sample statistics falls in this region Reject the null hypothesis if the sample Statistics falls in these two region Note: Critical Region: A region in the sample space S which amounts to rejection of H0 is termed as critical region or region of rejection. Critical Value: The value of test statistic which separates the critical (or rejection) region and the acceptance region is called the critical value or significant value. It depends upon i) the level of 114 significance used and ii) the alternative hypothesis, whether it is two-tailed or single-tailed For large samples the standard normal variate corresponding to the t - E (t) statistic t, Z = ~ N (0,1) S.E. (t) asymptotically as n à ∞ The value of z under the null hypothesis is known as test statistic. The critical value of the test statistic at the level of significance α for a two - tailed test is given by Zα/2 and for a one tailed test by Zα. where Zα is determined by equation P(|Z| >Zα)= α Zα is the value so that the total area of the critical region α α on both tails is α . ∴ P(Z > Zα) = . Area of each tail is . 2 2 Zα is the value such that area to the right of Zα and to the α left of − Zα is as shown in the following diagram. 2 /2 /2 4.6 One tailed and Two Tailed tests: In any test, the critical region is represented by a portion of the area under the probability curve of the sampling distribution of the test statistic. One tailed test: A test of any statistical hypothesis where the alternative hypothesis is one tailed (right tailed or left tailed) is called a one tailed test. 115 For example, for testing the mean of a population H0: µ = µ0, against the alternative hypothesis H1: µ > µ0 (right – tailed) or H1 : µ < µ0 (left –tailed)is a single tailed test. In the right – tailed test H1: µ > µ0 the critical region lies entirely in right tail of the sampling distribution of x , while for the left tailed test H1: µ < µ0 the critical region is entirely in the left of the distribution of x . Right tailed test: Left tailed test: Two tailed test: A test of statistical hypothesis where the alternative hypothesis is two tailed such as, H0 : µ = µ0 against the alternative hypothesis H1: µ ≠µ0 (µ > µ0 and µ < µ0) is known as two tailed test and in such a case the critical region is given by the portion of the area lying in both the tails of the probability curve of test of statistic. 116 For example, suppose that there are two population brands of washing machines, are manufactured by standard process(with mean warranty period µ1) and the other manufactured by some new technique (with mean warranty period µ2): If we want to test if the washing machines differ significantly then our null hypothesis is H0 : µ1 = µ2 and alternative will be H1: µ1 ≠ µ2 thus giving us a two tailed test. However if we want to test whether the average warranty period produced by some new technique is more than those produced by standard process, then we have H0 : µ1 = µ2 and H1 : µ1 < µ2 thus giving us a left-tailed test. Similarly, for testing if the product of new process is inferior to that of standard process then we have, H0 : µ1 = µ2 and H1 : µ1>µ2 thus giving us a right-tailed test. Thus the decision about applying a two – tailed test or a single –tailed (right or left) test will depend on the problem under study. Critical values (Zα) of Z 0.05 or 5% 0.01 or 1% Level of significance Left Right Left Right α Critical 2.33 values of 1.645 −2.33 −1.645 Zα for one tailed Tests Critical 2.58 values of 1.96 −2.58 −1.96 Zα/2 for two tailed tests 4.7 Type I and Type II Errors: When a statistical hypothesis is tested there are four possibilities. 1. The hypothesis is true but our test rejects it ( Type I error) 2. The hypothesis is false but our test accepts it (Type II error) 3. The hypothesis is true and our test accepts it (correct decision) 4. The hypothesis is false and our test rejects it (correct decision) 117 Obviously, the first two possibilities lead to errors. In a statistical hypothesis testing experiment, a Type I error is committed by rejecting the null hypothesis when it is true. On the other hand, a Type II error is committed by not rejecting (accepting) the null hypothesis when it is false. If we write , α = P (Type I error) = P (rejecting H0 | H0 is true) β = P (Type II error) = P (Not rejecting H0 | H0 is false) In practice, type I error amounts to rejecting a lot when it is good and type II error may be regarded as accepting the lot when it is bad. Thus we find ourselves in the situation which is described in the following table. Accept H0 Reject H0 H0 is true Correct decision Type I Error H0 is false Type II error Correct decision 4.8 Test Procedure : Steps for testing hypothesis is given below. (for both large sample and small sample tests) 1. Null hypothesis : set up null hypothesis H0. 2. Alternative Hypothesis: Set up alternative hypothesis H1, which is complementry to H0 which will indicate whether one tailed (right or left tailed) or two tailed test is to be applied. 3. Level of significance : Choose an appropriate level of significance (α), α is fixed in advance. 4. Test statistic (or test of criterian): t - E (t) Calculate the value of the test statistic, Z = under S.E. (t) the null hypothesis, where t is the sample statistic 5. Inference: We compare the computed value of Z (in absolute value) with the significant value (critical value) Zα/2 (or Zα). If |Z| > Zα, we reject the null hypothesis H0 at α % level of significance and if |Z| ≤ Zα, we accept H0 at α % level of significance. 118 Note: 1. Large Sample: A sample is large when it consists of more than 30 items. 2. Small Sample: A sample is small when it consists of 30 or less than 30 items. Exercise -4 I. Choose the best answers: 1. A measure characterizing a sample such as x or s is called (a). Population (b). Statistic (c).Universe (d).Mean 2. The standard error of the mean is n σ σ (a). σ2 (b). (d). (c). n σ n 3. The standard error of observed sample proportion “P” is P(1 − Q) PQ (1 − P)Q PQ (a). (b). (d). (c). n n n n 4. Alternative hypothesis is (a). Always Left Tailed (b). Always Right tailed (c). Always One Tailed (d). One Tailed or Two Tailed 5. Critical region is (a). Rejection Area (b). Acceptance Area (c). Probability (d). Test Statistic Value 6. The critical value of the test statistic at level of significance α for a two tailed test is denoted by (b).Zα (c). Z2α (a). Zα/2 (d). Zα/4 7. In the right tailed test, the critical region is (a). 0 (b). 1 (c). Lies entirely in right tail (d). Lies in the left tail 8. Critical value of |Zα| at 5% level of significance for two tailed test is (a). 1.645 (b). 2.33 (c). 2.58 (d). 1.96 119 9. Under null hypothesis the value of the test statistic Z is t - S.E. (t) t + E(t) t - E (t) PQ (a). (b). (c). (d). E (t) S.E. (t) S.E. (t) n 10. The alternative hypothesis H1: µ ≠ µ0 (µ >µ0 or µ < µ0) takes the critical region as (a). Right tail only (b). Both right and left tail (c). Left tail only (d). Acceptance region 11. A hypothesis may be classified as (a). Simple (b). Composite (c). Null (d). All the above 12. Whether a test is one sided or two sided depends on (a). Alternative hypothesis (b). Composite hypothesis (c). Null hypothesis (d). Simple hypothesis 13. A wrong decision about H0 leads to: (a). One kind of error (b). Two kinds of error (c). Three kinds of error (d). Four kinds of error 14. Area of the critical region depends on (a). Size of type I error (b). Size of type II error (c). Value of the statistics (d). Number of observations 15. Test of hypothesis H0 : µ = 70 vs H1 = µ > 70 leads to (a). One sided left tailed test (b). One sided right tailed test (c). Two tailed test (d). None of the above 16. Testing H0 : µ = 1500 against µ < 1500 leads to (a). One sided left tailed test (b). One sided right tailed test (c). Two tailed test (d). All the above 17. Testing H0 : µ = 100 vs H1: µ ≠ 100 lead to (a). One sided right tailed test (b). One sided left tailed test (c). Two tailed test (d). None of the above II. Fill in the Blanks 18. n1 and n2 represent the _________ of the two independent random samples respectively. 19. Standard error of the observed sample proportion p is _______ 20. When the hypothesis is true and the test rejects it, this is called _______ 120 21. When the hypothesis is false and the test accepts it this is called _______ 22. Formula to calculate the value of the statistic is __________ III. Answer the following 23. Define sampling distribution. 24. Define Parameter and Statistic. 25. Define standard error. 26. Give the standard error of the difference of two sample proportions. 27. Define Null hypothesis and alternative hypothesis. 28. Explain: Critical Value. 29. What do you mean by level of significance? 30. Explain clearly type I and type II errors. 31. What are the procedure generally followed in testing of a hypothesis ? 32. What do you mean by testing of hypothesis? 33. Write a detailed note on one- tailed and two-tailed tests. Answers: I. 1. (b) 2. (c) 3. (b) 4.(d) 5. (a) 6. (a) 7. (c) 8. (d) 9. (c) 10 (b) 11.(d) 12.(a) 13.(b) 14.(a) 15.(b) 16.(a) II. 17.(c) 18. size 19. PQ n 20. Type I error 21. Type II error t - E (t) 22. Z = S.E. (t) 121 5. TEST OF SIGNIFICANCE (Large Sample) 5.0 Introduction: In practical problems, statisticians are supposed to make tentative calculations based on sample observations. For example (i) The average weight of school student is 35kg (ii) The coin is unbiased Now to reach such decisions it is essential to make certain assumptions (or guesses) about a population parameter. Such an assumption is known as statistical hypothesis, the validity of which is to be tested by analysing the sample. The procedure, which decides a certain hypothesis is true or false, is called the test of hypothesis (or test of significance). Let us assume a certain value for a population mean. To test the validity of our assumption, we collect sample data and determine the difference between the hypothesized value and the actual value of the sample mean. Then, we judge whether the difference is significant or not. The smaller the difference, the greater the likelihood that our hypothesized value for the mean is correct. The larger the difference the smaller the likelihood, which our hypothesized value for the mean, is not correct. 5.1 Large samples (n > 30): The tests of significance used for problems of large samples are different from those used in case of small samples as the assumptions used in both cases are different. The following assumptions are made for problems dealing with large samples: (i) Almost all the sampling distributions follow normal asymptotically. (ii) The sample values are approximately close to the population values. The following tests are discussed in large sample tests. (i) Test of significance for proportion (ii) Test of significance for difference between two proportions 122 (iii) Test of significance for mean (iv) Test of significance for difference between two means. 5.2 Test of Significance for Proportion: Test Procedure Set up the null and alternative hypotheses H0 : P =P0 H1 = P ≠ P0 (P>P0 or P <P0) Level of significance: Let α = 0 .05 or 0.01 Calculation of statistic: Under H0 the test statistic is Z0 = p−P PQ n Expected value: p−P ∼ N (0, 1) PQ n = 1.96 for α = 0.05 (1.645) = 2.58 for α = 0.01 (2.33) Ze = Inference: (i) If the computed value of Z0 ≤ Ze we accept the null hypothesis and conclude that the sample is drawn from the population with proportion of success P0 (ii) If Z0 > Ze we reject the null hypothesis and conclude that the sample has not been taken from the population whose population proportion of success is P0. Example 1: In a random sample of 400 persons from a large population 120 are females.Can it be said that males and females are in the ratio 5:3 in the population? Use 1% level of significance 123 Solution: We are given n = 400 and x = No. of female in the sample = 120 p = observed proportion of females in the sample = 120 = 0.30 400 Null hypothesis: The males and females in the population are in the ratio 5:3 3 i.e., H0: P = Proportion of females in the population = = 0.375 8 Alternative Hypothesis: H1 : P ≠ 0.375 (two-tailed) Level of significance: α = 1 % o r 0.01 Calculation of statistic: Under H0, the test statistic is Z0 = = = p−P PQ n 0.300 − 0.375 0.375 × 0.625 400 0.075 0.075 = 3.125 = 0.024 0.000586 Expected value: Ze = p−P PQ n 124 ∼ N(0,1) = 2.58 Inference : Since the calculated Z0 > Ze we reject our null hypothesis at 1% level of significance and conclude that the males and females in the population are not in the ratio 5:3 Example 2: In a sample of 400 parts manufactured by a factory, the number of defective parts was found to be 30. The company, however, claimed that only 5% of their product is defective. Is the claim tenable? Solution: We are given n = 400 x = No. of defectives in the sample = 30 p= proportion of defectives in the sample x 30 = = 0.075 = n 400 Null hypothesis: The claim of the company is tenable H0: P= 0.05 Alternative Hypothesis: H1 : P > 0.05 (Right tailed Alternative) Level of significance: 5% Calculation of statistic: Under H0, the test statistic is Z0 = = = p−P PQ n 0.075 − 0.050 0.05 × 0.95 400 0.025 0.0001187 125 = 2.27 Expected value: p−P ∼ N(0, 1) PQ n = 1.645 (Single tailed) Ze = Inference : Since the calculated Z0 > Ze we reject our null hypothesis at 5% level of significance and we conclude that the company’ s claim is not tenable. 5.3 Test of significance for difference between two proportion: Test Procedure Set up the null and alternative hypotheses: H0 : P1 =P2 = P (say) H1 : P1 ≠ P2 (P1>P2 or P1 <P2) Level of significance: Let α = 0.05 or 0.01 Calculation of statistic: Under H0, the test statistic is Z0 = p1 − p 2 P1Q1 P2 Q 2 + n1 n2 = p1 − p 2 (P1 and P2 are known) (P1 and P2 are not known) 1 1 PQ + n n 2 1 n p + n 2 p 2 x1 + x 2 = where P = 1 1 n1 + n 2 n1 + n 2 Q =1− P 126 Expected value: Ze = p1 − p 2 S.E(p1 − p 2 ) ∼ N(0,1) Inference: (i) If Z0 ≤ Ze we accept the null hypothesis and conclude that the difference between proportions are due to sampling fluctuations. (ii) If Z0 > Ze we reject the null hypothesis and conclude that the difference between proportions cannot be due to sampling fluctuations Example 3: In a referendum submitted to the ‘ student body’ at a university, 850 men and 550 women voted. 530 of the men and 310 of the women voted ‘ yes’ . Does this indicate a significant difference of the opinion on the matter between men and women students? Solution: We are given n1= 850 n2 = 550 530 310 p1 = = 0.62 p2 = = 0.56 850 550 x + x2 530 + 310 = P= 1 = 0.60 n1 + n 2 1400 x1= 530 x2=310 Q = 0.40 Null hypothesis: H0: P1= P2 ie the data does not indicate a significant difference of the opinion on the matter between men and women students. Alternative Hypothesis: H1 : P1≠ P2 (Two tailed Alternative) Level of significance: Let α = 0.05 127 Calculation of statistic: Under H0, the test statistic is Z0 = = p1 − p 2 1 1 PQ + n n2 1 0.62 − 0.56 1 1 0.6 × 0.4 + 850 550 0.06 = = 2.22 0.027 Expected value: Ze = p1 − p 2 1 1 PQ + n 1 n2 ∼ N(0,1) = 1.96 Inference : Since Z0 > Ze we reject our null hypothesis at 5% level of significance and say that the data indicate a significant difference of the opinion on the matter between men and women students. Example 4: In a certain city 125 men in a sample of 500 are found to be self employed. In another city, the number of self employed are 375 in a random sample of 1000. Does this indicate that there is a greater population of self employed in the second city than in the first? Solution: We are given n1= 500 n2 = 1000 x1 = 125 x2 = 375 128 125 375 = 0.25 p2 = = 0.375 500 1000 x + x2 125 + 375 P= 1 = n 1 + n 2 500 + 1000 500 1 = = 1500 3 12 Q = 1− = 33 p1 = Null hypothesis: H0: P1= P2 There is no significant difference between the two population proportions. Alternative Hypothesis: H1 : P1< P2 (left tailed Alternative) Level of significance: Let α = 0.05 Calculation of statistic: Under H0, the test statistic is Z0 = = p1 − p 2 1 1 PQ + n 1 n2 0.25 − 0.375 1 2 1 1 × + 3 3 500 1000 = 0.125 = 4.8 0.026 Expected value: Ze = p1 − p 2 1 1 PQ + n n2 1 129 ∼ N(0,1) = 1.645 Inference : Since Z0 > Ze we reject the null hypothesis at 5% level of significance and say that there is a significant difference between the two population proportions. Example 5: A civil service examination was given to 200 people. On the basis of their total scores, they were divided into the upper 30% and the remaining 70%. On a certain question 40 of the upper group and 80 of the lower group answered correctly. On the basis of this question, is this question likely to be useful for discriminating the ability of the type being tested? Solution: We are given 30 × 200 70 × 200 n1 = = 60 n2 = = 140 100 100 x1 = 40 x2 = 80 40 2 80 4 p1= p2 = = = 60 3 140 7 x + x2 40 + 80 P= 1 = n 1 + n 2 60 + 140 120 6 = = 200 10 14 Q = 1− P = 1− = 6 10 Null hypothesis: H0: P1= P2 (say) The particular question does not discriminate the abilities of two groups. Alternative Hypothesis: H1 : P1≠ P2 (two tailed Alternative) Level of significance: Let α = 0.05 Calculation of statistics Under H0, the test statistic is 130 Z0 = = = p1 − p 2 1 1 PQ + n n2 1 24 − 37 6 41 1 ×+ 10 10 60 140 10 21 3 = 1.3 Expected value: Ze = p1 − p 2 1 1 PQ + n n2 1 = 1.96 for α=0.05 ∼ N(0,1) Inference : Since Z0 < Ze we accept our null hypothesis at 5% level of significance and say that the particular question does not discriminate the abilities of two groups. 5.4 Test of significance for mean: Let xi (i = 1,2… be a random sample of size n from a ..n) population with variance σ2, then the sample mean x is given by 1 (x1 + x2 +… n) ..x x= n E( x ) = µ 1 V( x ) = V [ (x1 + x2 +… n)] ..x n 131 1 [(V(x1) + V(x2) +… ..V(xn)] n2 2 1 σ = 2 nσ2 = n n σ ∴ S.E ( x ) = n = Test Procedure: Null and Alternative Hypotheses: H0:µ = µ0. H1:µ ≠ µ0 (µ > µ0 o r µ < µ0) Level of significance: Let α = 0.05 or 0.01 Calculation of statistic: Under H0, the test statistic is x −µ x − E( x ) Z0 = = S.E( x ) σ/ n Expected value: x −µ Ze = ∼ N(0,1) σ/ n = 1.96 for α = 0.05 (1.645) or = 2.58 for α = 0.01 (2.33) Inference : If Z0 < Ze, we accept our null hypothesis and conclude that the sample is drawn from a population with mean µ = µ0 If Z0 > Ze we reject our H0 and conclude that the sample is not drawn from a population with mean µ = µ0 Example 6: The mean lifetime of 100 fluorescent light bulbs produced by a company is computed to be 1570 hours with a standard deviation of 120 hours. If µ is the mean lifetime of all the bulbs produced by the company, test the hypothesis µ=1600 hours against 132 the alternative hypothesis µ ≠ 1600 hours using a 5% level of significance. Solution: We are given x = 1570 hrs µ = 1600hrs s =120 hrs n=100 Null hypothesis: H0: µ= 1600.ie There is no significant difference between the sample mean and population mean. Alternative Hypothesis: H1: µ ≠ 1600 (two tailed Alternative) Level of significance: Let α = 0.05 Calculation of statistics Under H0, the test statistic is x −µ Z0 = s/ n 1570 − 1600 120 100 30 × 10 = 120 = 2.5 = Expected value: x −µ ∼ N(0,1) s/ n = 1.96 for α = 0.05 Z0 = Inference : Since Z0 > Ze we reject our null hypothesis at 5% level of significance and say that there is significant difference between the sample mean and the population mean. 133 Example 7: A car company decided to introduce a new car whose mean petrol consumption is claimed to be lower than that of the existing car. A sample of 50 new cars were taken and tested for petrol consumption. It was found that mean petrol consumption for the 50 cars was 30 km per litre with a standard deviation of 3.5 km per litre. Test at 5% level of significance whether the company’ s claim that the new car petrol consumption is 28 km per litre on the average is acceptable. Solution: We are given x = 30 ; µ =28 ; n=50 ; s=3.5 Null hypothesis: H0: µ = 28. i.e The company’ s claim that the petrol consumption of new car is 28km per litre on the average is acceptable. Alternative Hypothesis: H1: µ < 28 (Left tailed Alternative) Level of significance: Let α = 0.05 Calculation of statistic: Under H0 the test statistics is x −µ Z0 = s/ n = 30 − 28 3.5 50 2 × 50 3.5 = 4.04 = Expected value: x −µ ∼ N(0,1) at α = 0.05 s/ n = 1.645 Ze = 134 Inference : Since the calculated Z0 > Ze we reject the null hypothesis at 5% level of significance and conclude that the company’ s claim is not acceptable. 5.5 Test of significance for difference between two means: Test procedure Set up the null and alternative hypothesis H0: µ1 = µ2 ; H1: µ1 ≠ µ2 (µ1 > µ2 or µ1 < µ2) Level of significance: Let α% Calculation of statistic: Under H0 the test statistic is Z0 = x1 − x 2 σ1 σ +2 n1 n2 2 2 If σ12 = σ22 = σ2 (ie) If the samples have been drawn from the population with common S.D σ then under H0 : µ1 = µ2 Z0 = x1 − x 2 1 1 σ + n1 n 2 Expected value: Ze = x1 − x 2 ∼N(0,1) S.E( x 1 − x 2 ) Inference: (i) If Z0 ≤ Ze we accept the H0 (ii) If Z0 > Ze we reject the H0 Example 8: A test of the breaking strengths of two different types of cables was conducted using samples of n1 = n2 = 100 pieces of each type of cable. 135 Cable I Cable II x 1 =1925 x 2 = 1905 σ1= 40 σ2 = 30 Do the data provide sufficient evidence to indicate a difference between the mean breaking strengths of the two cables? Use 0.01 level of significance. Solution: We are given x1 =1925 x 2 = 1905 σ1= 40 σ2 = 30 Null hypothesis H0:µ1 = µ2 .ie There is no significant difference between the mean breaking strengths of the two cables. H1 : µ1 ≠ µ2 (Two tailed alternative) Level of significance: Let α = 0.01 or 1% Calculation of statistic: Under H0 the test statistic is Z0 = = x1 − x 2 σ1 σ +2 n1 n2 2 2 1925 − 1905 40 2 30 2 + 100 100 = 20 =4 5 Expected value: Ze = x1 − x 2 σ1 σ +2 n1 n2 2 2 136 ∼ N (0,1) = 2.58 Inference: Since Z0 > Ze, we reject the H0. Hence the formulated null hypothesis is wrong ie there is a significant difference in the breaking strengths of two cables. Example 9: The means of two large samples of 1000 and 2000 items are 67.5 cms and 68.0cms respectively. Can the samples be regarded as drawn from the population with standard deviation 2.5 cms. Test at 5% level of significance. Solution: We are given n1 = 1000 ; n2 = 2000 x 1 = 67.5 cms ; x 2 = 68.0 cms σ = 2.5 cms Null hypothesis H0: µ1 = µ2 (i.e.,) the sample have been drawn from the same population. Alternative Hypothesis: H1: µ1 ≠ µ2 (Two tailed alternative) Level of significance: α = 5% Calculation of statistic: Under Ho the test statistic is Z0 = = x1 − x 2 1 1 σ + n1 n 2 67.5 − 68.0 1 1 + 1000 2000 0.5 × 20 2.5 = 2.5 3 / 5 = 5.1 137 Expected value: Ze = x1 − x 2 1 1 σ + n1 n 2 ∼ N(0,1) = 1.96 Inference : Since Z0 > Ze we reject the H0 at 5% level of significance and conclude that the samples have not come from the same population. Exercise – 5 I. Choose the best answer: 1. Standard error of number of success is given by pq np (a) (c) npq (d) (b) npq n q 2. Large sample theory is applicable when (a) n > 30 (b) n < 30 (c) n < 100 (d) n < 1000 3. Test statistic for difference between two means is x −µ p−P (a) (b) PQ σ/ n n x1 − x 2 p1 − p 2 (d) (c) 2 2 1 1 σ1 σ2 PQ + + n n2 n1 n2 1 4. Standard error of the difference of proportions (p1-p2) in two classes under the hypothesis H0:p1 = p2 with usual notation is 1 1 1 1 (b) p( + ) (a) pq ( + ) n1 n 2 n1 n 2 (c) pq ( 1 1 +) n1 n 2 (d) p 1q 1 p 2 q 2 + n1 n2 138 5. Statistic z = x−y 1 1 σ + n1 n 2 is used to test the null hypothesis (a) H0: µ1 + µ 2 = 0 (c) H0: µ = µ 0 ( a constant) (b) H0: µ1 − µ 2 = 0 (c) none of the above. II. Fill in the blanks: 2 6. If P = , then Q = _________ 3 7. If z0 < ze then the null hypothesis is ____________ 8. When the difference is __________, the null hypothesis is rejected. 9. Test statistic for difference between two proportions is ________ 10. The variance of sample mean is ________ III. Answer the following 11. In a test if z0 ≤ ze, what is your conclusion about the null hypothesis? 12. Give the test statistic for (a) Proportion (b) Mean (c) Difference between two means (d) Difference between two proportions 13. Write the variance of difference between two proportions 14. Write the standard error of proportion. 15. Write the test procedure for testing the test of significance for (a) Proportion (b) mean (c) difference between two proportions (d) difference between two mean 16. A coin was tossed 400 times and the head turned up 216 times. Test the hypothesis that the coin is unbiased. 17. A person throws 10 dice 500 times and obtains 2560 times 4, 5 or 6. Can this be attributed to fluctuations of sampling? 139 18. In a hospital 480 female and 520 male babies were born in a week. Do these figure confirm the hypothesis that males and females are born in equal number? 19. In a big city 325 men out of 600 men were found to be selfemployed. Does this information support the conclusion that the majority of men in this city are self-employed? 20. A machine puts out 16 imperfect articles in a sample of 500. After machine is overhauled, it puts out 3 imperfect articles in a batch of 100. Has the machine improved? 21. In a random sample of 1000 persons from town A , 400 are found to be consumers of wheat. In a sample of 800 from town B, 400 are found to be consumers of wheat. Do these data reveal a significant difference between town A and town B, so far as the proportion of wheat consumers is concerned? 22. 1000 articles from a factory A are examined and found to have 3% defectives. 1500 similar articles from a second factory B are found to have only 2% defectives. Can it be reasonably concluded that the product of the first factory is inferior to the second? 23. In a sample of 600 students of a certain college, 400 are found to use blue ink. In another college from a sample of 900 students 450 are found to use blue ink. Test whether the two colleges are significantly different with respect to the habit of using blue ink. 24. It is claimed that a random sample of 100 tyres with a mean life of 15269kms is drawn from a population of tyres which has a mean life of 15200 kms and a standard deviation of 1248 kms. Test the validity of the claim. 25. A sample of size 400 was drawn and the sample mean B was found to be 99. Test whether this sample could have come from a normal population with mean 100 and variance 64 at 5% level of significance. 26. The arithmetic mean of a sample 100 items drawn from a large population is 52. If the standard deviation of the population is 7, test the hypothesis that the mean of the population is 55 against the alternative that the mean is not 55. Use 5% level of significance. 140 27. A company producing light bulbs finds that mean life span of the population of bulbs is 1200 hrs with a standard deviation of 125hrs. A sample of 100 bulbs produced in a lot is found to have a mean life span of 1150hrs. Test whether the difference between the population and sample means is statistically significant at 5% level of significance. 28. Test the significance of the difference between the means of the samples from the following data Size of sample Mean 100 150 50 51 Sample A Sample B Standard deviation 4 5 29. An examination was given to two classes consisting of 40 and 50 students respectively. In the first class the mean mark was 74 with a standard deviation of 8, while in the second class the mean mark was 78 with a standard deviation of 7. Is there a significant difference between the performance of the two classes at a level of significance of 0.05? 30. If 60 M.A. Economics students are found to have a mean height of 63.60 inches and 50 M.Com students a mean height of 69.51 inches. Would you conclude that the commerce students are taller than Economics students? Assume the standard deviation of height of post-graduate students to be 2.48 inches. Answers: I. 1. (b) 2.(a) 3.(c) 7.accepted 8.significant II. 6. 1 3 141 4.(a) 5.(b) 9. p1 − p 2 1 1 PQ + n n2 1 σ n 2 10. III. 16. z = 1.6 Accept H0 17. z = 1.7 Accept H0 18. z = 1.265Accept H0 19. z = 2.04 Accept H0 20. z = 0.106 Accept H0 21. z = 4.247 Reject H0 22. z = 1.63 Accept H0 23. z = 6.38 Reject H0 24. z = 0.5529 Accept H0 25. z = 2.5 Reject H0 26. z = 4.29 Reject H0 27. z = 4 Reject H0 28. z = 1.75 Accept H0 29. z = 2.49 Reject H0 30. z = 12.49 Reject H0 142 6. TESTS OF SIGNIFICANCE (Small Samples) 6.0 Introduction: In the previous chapter we have discussed problems relating to large samples. The large sampling theory is based upon two important assumptions such as (a) The random sampling distribution of a statistic is approximately normal and (b) The values given by the sample data are sufficiently close to the population values and can be used in their place for the calculation of the standard error of the estimate. The above assumptions do not hold good in the theory of small samples. Thus, a new technique is needed to deal with the theory of small samples. A sample is small when it consists of less than 30 items. ( n< 30) Since in many of the problems it becomes necessary to take a small size sample, considerable attention has been paid in developing suitable tests for dealing with problems of small samples. The greatest contribution to the theory of small samples is that of Sir William Gosset and Prof. R.A. Fisher. Sir William Gosset published his discovery in 1905 under the pen name ‘ Student’ and later on developed and extended by Prof. R.A.Fisher. He gave a test popularly known as ‘ t-test’ . 6.1 t - statistic definition: If x1, x2, …xn is a random sample of size n from a normal … population with mean µ and variance σ2, then Student’ s t-statistic is x−µ defined as t = S n where x = ∑x is the sample mean n 143 2 1 ∑( x − x ) n −1 is an unbiased estimate of the population variance σ2 It follows student’ s t-distribution with ν = n −1 d.f and S2 = 6.1.1 Assumptions for students t-test: 1. The parent population from which the sample drawn is normal. 2. The sample observations are random and independent. 3. The population standard deviation σ is not known. 6.1.2 Properties of t- distribution: 1. t-distribution ranges from −∞ to ∞ just as does a normal distribution. 2. Like the normal distribution, t-distribution also symmetrical and has a mean zero. 3. t-distribution has a greater dispersion than the standard normal distribution. 4. As the sample size approaches 30, the t-distribution, approaches the Normal distribution. Comparison between Normal curve and corresponding t curve: 144 6.1.3 Degrees of freedom (d.f): Suppose it is asked to write any four number then one will have all the numbers of his choice. If a restriction is applied or imposed to the choice that the sum of these number should be 50. Here, we have a choice to select any three numbers, say 10, 15, 20 and the fourth number is 5: [50 − (10 +15+20)]. Thus our choice of freedom is reduced by one, on the condition that the total be 50. therefore the restriction placed on the freedom is one and degree of freedom is three. As the restrictions increase, the freedom is reduced. The number of independent variates which make up the statistic is known as the degrees of freedom and is usually denoted by ν (Nu) The number of degrees of freedom for n observations is n − k where k is the number of independent linear constraint imposed upon them. For the student’ s t-distribution. The number of degrees of freedom is the sample size minus one. It is denoted by ν = n −1 The degrees of freedom plays a very important role in χ2 test of a hypothesis. When we fit a distribution the number of degrees of freedom is (n– k−1) where n is number of observations and k is number of parameters estimated from the data. For e.g., when we fit a Poisson distribution the degrees of freedom is ν = n – 1 −1 In a contingency table the degrees of freedom is (r−1) (c −1) where r refers to number rows and c refers to number of columns. Thus in a 3 × 4 table the d.f are (3−1) (4−1) = 6 d.f In a 2 × 2 contingency table the d.f are (2−1) (2−1) = 1 In case of data that are given in the form of series of variables in a row or column the d.f will be the number of observations in a series less one ie ν = n−1 Critical value of t: The column figures in the main body of the table come under the headings t0.100, t0.50, t0.025, t0.010 and t0.005. The subscripts 145 give the proportion of the distribution in ‘ tail’ area. Thus for twotailed test at 5% level of significance there will be two rejection areas each containing 2.5% of the total area and the required column is headed t0.025 For example, tν (.05) for single tailed test = tν (0.025) for two tailed test tν (.01) for single tailed test = tν (0.005) for two tailed test Thus for one tailed test at 5% level the rejection area lies in one end of the tail of the distribution and the required column is headed t0.05. Critical value of t – distribution -tα t=0 tα +∞ ∞ 6.1.4 -Applications of t-distribution: The t-distribution has a number of applications in statistics, of which we shall discuss the following in the coming sections: (i) t-test for significance of single mean, population variance being unknown. (ii) t-test for significance of the difference between two sample means, the population variances being equal but unknown. (a) Independent samples (b) Related samples: paired t-test 6.2 Test of significance for Mean: We set up the corresponding null and alternative hypotheses as follows: 146 H0: µ = µ0; There is no significant difference between the sample mean and population Mean. H1: µ ≠ µ0 ( µ < µ0 (or) µ > µ0) Level of significance: 5% or 1% Calculation of statistic: Under H0 the test statistic is x −µ x−µ or S s / n −1 n ∑x where x = is the sample mean n 2 2 1 1 and S2 = ∑( x − x ) (or) s2 = ∑( x − x ) n −1 n Expected value : t0 = te = x−µ ∼ student’ s t-distribution with (n-1) d.f S n Inference : If t0 ≤ te it falls in the acceptance region and the null hypothesis is accepted and if to > te the null hypothesis H0 may be rejected at the given level of significance. Example 1: Certain pesticide is packed into bags by a machine. A random sample of 10 bags is drawn and their contents are found to weigh (in kg) as follows: 50 49 52 44 45 48 46 45 49 45 Test if the average packing can be taken to be 50 kg. Solution: Null hypothesis: H0 : µ = 50 kgs in the average packing is 50 kgs. 147 Alternative Hypothesis: H1 : µ ≠ 50kgs (Two -tailed ) Level of Significance: Let α = 0.05 Calculation of sample mean and S.D X d = x –48 50 2 49 1 52 4 44 –4 45 –3 48 0 46 –2 45 –3 49 +1 45 –3 Total –7 ∑d x =A+ n −7 = 48 + 10 = 48–0.7 =47.3 1 (∑ d ) 2 [∑d 2 − S2 = n −1 n 2 1 (7 ) = [69 − 9 10 64.1 = = 7.12 9 Calculation of Statistic: Under H0 the test statistic is : t0 = x −µ S2 / n 148 d2 4 1 16 16 9 0 4 9 1 9 69 = = 47.3 − 50.0 7.12 / 10 2.7 0.712 = 3.2 Expected value: te = x −µ follows t distribution with (10–1) d.f S2 / n = 2.262 Inference: Since t0 > te , H0 is rejected at 5% level of significance and we conclude that the average packing cannot be taken to be 50 kgs. Example 2: A soap manufacturing company was distributing a particular brand of soap through a large number of retail shops. Before a heavy advertisement campaign, the mean sales per week per shop was 140 dozens. After the campaign, a sample of 26 shops was taken and the mean sales was found to be 147 dozens with standard deviation 16. Can you consider the advertisement effective? Solution: We are given n = 26; x = 147dozens; s = 16 Null hypothesis: H0: µ = 140 dozens i.e. Advertisement is not effective. Alternative Hypothesis: H1: µ > 140kgs (Right -tailed) Calculation of statistic: Under the null hypothesis H0, the test statistic is x−µ t0 = s / n −1 = 147 − 140 16 / 25 149 = 7×5 16 = 2.19 Expected value: x−µ te = follows t-distribution with (26-1) = 25d.f s / n −1 = 1.708 Inference: Since t0 > te, H0 is rejected at 5% level of significance. Hence we conclude that advertisement is certainly effective in increasing the sales. 6.3 Test of significance for difference between two means: 6.3.1 Independent samples: Suppose we want to test if two independent samples have been drawn from two normal populations having the same means, the population variances being equal. Let x1, x2,…x n1 and y1, y2, . … y n 2 be two independent random samples from the given … normal populations. Null hypothesis: H0 : µ1 = µ2 i.e. the samples have been drawn from the normal populations with same means. Alternative Hypothesis: H1 : µ1 ≠ µ2 (µ1 < µ2 o r µ1 > µ2) Test statistic: Under the H0, the test statistic is t0 = where x = x−y 1 1 S + n1 n 2 ∑x ∑y ; y= n1 n2 2 n s + n 2s 2 1 [ ∑ ( x − x ) + ∑( y − y ) 2 ] = 1 1 n1 + n 2 − 2 n1 + n 2 − 2 2 and S2 = 150 2 Expected value: te = x−y 1 1 S + n1 n 2 follows t-distribution with n1+ n2 –2 d.f Inference: If the t0 < te we accept the null hypothesis. If t0 > te we reject the null hypothesis. Example 3: A group of 5 patients treated with medicine ‘ A’ weigh 42, 39, 48, 60 and 41 kgs: Second group of 7 patients from the same hospital treated with medicine ‘ B’ weigh 38, 42 , 56, 64, 68, 69 and 62 kgs. Do you agree with the claim that medicine ‘ B’ increases the weight significantly? Solution: Let the weights (in kgs) of the patients treated with medicines A and B be denoted by variables X and Y respectively. Null hypothesis: H0 : µ1 = µ2 i.e. There is no significant difference between the medicines A and B as regards their effect on increase in weight. Alternative Hypothesis: H1 : µ1 < µ2 (left-tail) i.e. medicine B increases the weight significantly. Level of significance : Let α = 0.05 Computation of sample means and S.Ds Medicine A X x – x ( x = 46) (x – x )2 42 –4 16 39 –7 49 48 2 4 60 14 196 41 –5 25 230 0 290 151 x= y= ∑ x 230 = = 46 n1 5 Medicine B Y y – y ( y = 57) 38 –19 42 –15 56 –1 64 7 68 11 69 12 62 5 399 0 ∑y n2 = (y – y )2 361 225 1 49 121 144 25 926 399 = 57 7 2 1 [ ∑ ( x − x ) + ∑( y − y ) 2 ] n1 + n 2 − 2 1 = [ 290 + 926] = 121.6 10 Calculation of statistic: Under H0 the test statistic is S2 = t0 = = x−y 1 1 S2 + n 1 n2 46 − 57 1 1 121.6 + 5 7 152 = 11 121.6 × = 12 35 11 = 1.7 6.57 Expected value: te = x−y 1 1 S2 + n 1 n2 follows t-distribution with (5+7–2) =10 d.f = 1.812 Inference: Since t0 < te it is not significant. Hence H0 is accepted and we conclude that the medicines A and B do not differ significantly as regards their effect on increase in weight. Example 4: Two types of batteries are tested for their length of life and the following data are obtained: No of samples Mean life Variance (in hrs) Type A 9 600 121 Type B 8 640 Is there a significant difference in the two means? 144 Solution: We are given s22=144 n1=9; x 1=600hrs; s12 =121; n2 =8; x 2 =640hrs; Null hypothesis: H0 : µ1 = µ2 i.e. Two types of batteries A and B are identical i.e. there is no significant difference between two types of batteries. 153 Alternative Hypothesis: H1 : µ1 ≠ µ2 (Two- tailed) Level of Significance: Let α = 5% Calculation of statistics: Under H0, the test statistic is t0 = x−y 1 1 S2 + n 1 n2 n s + n 2s 2 where S = 1 1 n1 + n 2 − 2 9 × 121 + 8 × 144 = 9+8− 2 2241 = = 149.4 15 2 2 2 ∴ t0 = = 600 − 640 1 1 149.4 + 9 8 40 40 = 6.735 = 5.9391 17 149.4 72 Expected value: x−y te = 1 1 S + n 1 n2 = 2.131 follows t-distribution with 9+8–2 =15 d.f 2 154 Inference: Since t0 > te it is highly significant. Hence H0 is rejected and we conclude that the two types of batteries differ significantly as regards their length of life. 6.3.2 Related samples –Paired t-test: In the t-test for difference of means, the two samples were independent of each other. Let us now take a particular situations where (i) The sample sizes are equal; i.e., n1 = n2 = n(say), and (ii) The sample observations (x1, x2, …..xn) and (y1, y2, … ….yn) are not completely independent but they are … dependent in pairs. That is we are making two observations one before treatment and another after the treatment on the same individual. For example a business concern wants to find if a particular media of promoting sales of a product, say door to door canvassing or advertisement in papers or through T.V. is really effective. Similarly a pharmaceutical company wants to test the efficiency of a particular drug, say for inducing sleep after the drug is given. For testing of such claims gives rise to situations in (i) and (ii) above, we apply paired t-test. Paired – t –test: Let di = Xi – Yi (i = 1, 2, …n) denote the difference in … the observations for the ith unit. Null hypothesis: H0 : µ1 = µ2 ie the increments are just by chance Alternative Hypothesis: H1 : µ1 ≠ µ2 ( µ1 > µ2 (or) µ1 < µ2) Calculation of test statistic: d t0 = S/ n 1 (∑ d ) 1 ∑d 2 where d = [∑ d 2 − and S2 = ∑ (d − d ) = n −1 n −1 n n 2 155 Expected value: d follows t-distribution with n –1 d.f te = S/ n Inference: By comparing t0 and te at the desired level of significance, usually 5% or 1%, we reject or accept the null hypothesis. Example 5: To test the desirability of a certain modification in typists desks, 9 typists were given two tests of as nearly as possible the same nature, one on the desk in use and the other on the new type. The following difference in the number of words typed per minute were recorded: Typists A B C D E F G H I Increase in 5 number of words 2 4 0 3 −1 4 −3 2 Do the data indicate the modification in desk promotes speed in typing? Solution: Null hypothesis: H0 : µ1 = µ2 i.e. the modification in desk does not promote speed in typing. Alternative Hypothesis: H1 : µ1 < µ2 (Left tailed test) Level of significance: Let α = 0.05 Typist A B C D E F G H I d 2 4 0 3 −1 4 −3 2 5 Σd = 16 156 d2 4 16 0 9 1 16 9 4 25 Σd2 = 84 d= S= = 16 ∑d = = 1.778 n 9 1 ( ∑ d) 2 [∑ d 2 − n −1 n 1 (16) 2 [84 −= 8 9 6.9 = 2.635 Calculation of statistic: Under H0 the test statistic is t0 = d. n 1.778 × 3 = = 2.024 S 2.635 Expected value: te = d. n S follows t- distribution with 9 –1 = 8 d.f = 1.860 Inference: When t0 < te t he null hypothesis is accepted. The data does not indicate that the modification in desk promotes speed in typing. Example 6: An IQ test was administered to 5 persons before and after they were trained. The results are given below: Candidates I II III IV V IQ before 110 120 123 132 125 training IQ after 120 118 125 136 121 training Test whether there is any change in IQ after the training programme (test at 1% level of significance) Solution: Null hypothesis: H0 : µ1 = µ2 i.e. there is no significant change in IQ after the training programme. 157 Alternative Hypothesis: H1 : µ1 ≠ µ2 (two tailed test) Level of significance : α = 0.01 x 110 120 y 120 118 2 d = x−y −10 2 d 100 4 123 125 −2 4 132 136 −4 16 125 121 4 16 Total −10 140 ∑d − 10 = = −2 n 5 1 (∑ d ) 2 2 2 S= [∑ d − n −1 n 1 100 = [140 − = 30 4 5 Calculation of Statistic: Under H0 the test statistic is d t0 = S/ n d= = −2 30 / 5 2 = 2.45 = 0.816 Expected value: te = d follows t-distribution with 5 –1 = 4 d.f S2 / n = 4.604 Inference: Since t0 < te at 1% level of significance we accept the null hypothesis. We therefore, conclude that there is no change in IQ after the training programme. 158 6.4 Chi square statistic: Various tests of significance described previously have mostly applicable to only quantitative data and usually to the data which are approximately normally distributed. It may also happens that we may have data which are not normally distributed. Therefore there arises a need for other methods which are more appropriate for studying the differences between the expected and observed frequencies. The other method is called Non-parametric or distribution free test. A non- parametric test may be defined as a statistical test in which no hypothesis is made about specific values of parameters. Such non-parametric test has assumed great importance in statistical analysis because it is easy to compute. 6.4.1 Definition: The Chi- square (χ2) test (Chi-pronounced as ki) is one of the simplest and most widely used non-parametric tests in statistical work. The χ2 test was first used by Karl Pearson in the year 1900. The quantity χ2 describes the magnitude of the discrepancy between theory and observation. It is defined as n (Oi − Ei) 2 χ2 = ∑ Ei i =1 Where O refers to the observed frequencies and E refers to the expected frequencies. Note: If χ2 is zero, it means that the observed and expected frequencies coincide with each other. The greater the discrepancy between the observed and expected frequencies the greater is the value of χ2. Chi square - Distribution: The square of a standard normal variate is a Chi-square variate with 1 degree of freedom i.e., If X is normally distributed x −µ with mean µ and standard deviation σ, then is a Chiσ square variate (χ2) with 1 d.f. The distribution of Chi-square depends on the degrees of freedom. There is a different distribution for each number of degrees of freedom. 159 2 6.4.2 properties of Chi-square distribution: 1. The Mean of χ2 distribution is equal to the number of degrees of freedom (n) 2. The variance of χ2 distribution is equal to 2n 3. The median of χ2 distribution divides, the area of the curve into two equal parts, each part being 0.5. 4. The mode of χ2 distribution is equal to (n−2) 5. Since Chi-square values always positive, the Chi square curve is always positively skewed. 6. Since Chi-square values increase with the increase in the degrees of freedom, there is a new Chi-square distribution with every increase in the number of degrees of freedom. 7. The lowest value of Chi-square is zero and the highest value is infinity ie χ2 ≥ 0. 8. When Two Chi- squares χ12 and χ22 are independent χ2 distribution with n1 and n2 degrees of freedom and their sum χ12 + χ22 will follow χ2 distribution with (n1 + n2) degrees of freedom. 9. When n (d.f) > 30, the distribution of 2χ 2 approximately follows normal distribution. The mean of the distribution 2χ 2 is 2n - 1 and the standard deviation is equal to 1. 160 6.4.3 Conditions for applying χ 2 test: The following conditions should be satisfied before applying χ2 test. 1. N, the total frequency should be reasonably large, say greater than 50. 2. No theoretical cell-frequency should be less than 5. If it is less than 5, the frequencies should be pooled together in order to make it 5 or more than 5. 3. Each of the observations which makes up the sample for this test must be independent of each other. 4. χ2 test is wholly dependent on degrees of freedom. 6.5 Testing the Goodness of fit (Binomial and Poisson Distribution): Karl Pearson in 1900, developed a test for testing the significance of the discrepancy between experimental values and the theoretical values obtained under some theory or hypothesis. This test is known as χ2-test of goodness of fit and is used to test if the deviation between observation (experiment) and theory may be attributed to chance or if it is really due to the inadequacy of the theory to fit the observed data. Under the null hypothesis that there is no significant difference between the observed and the theoretical values. Karl Pearson proved that the statistic n (Oi − Ei) 2 χ2 = ∑ Ei i =1 = (O 1 − E 1 ) 2 (O 2 − E 2 ) 2 (O − E n ) 2 + + ........ n E1 E2 En Follows χ2-distribution with ν = n – k – 1 d.f. where 01, 02, ...0n are the observed frequencies, E1 , E2… n, corresponding to the E expected frequencies and k is the number of parameters to be estimated from the given data. A test is done by comparing the computed value with the table value of χ2 for the desired degrees of freedom. 161 Example 7: Four coins are tossed simultaneously and the number of heads occurring at each throw was noted. This was repeated 240 times with the following results. No. of heads 0 1 2 3 4 No. of throws 13 64 85 58 20 Fit a Binomial distribution assuming under the hypothesis that the coins are unbiased. Solution: Null Hypothesis: H0: The given data fits the Binomial distribution. i.e the coins are unbiased. p = q = 1/2 n=4 N = 240 Computation of expected frequencies: No. of P(X= x ) = 4 Cx px q n-x heads 0 1 2 3 4 1 1 1 4Co 0 4 = 2 2 16 1 1 4 4C1 1 3 = 2 2 16 1 1 6 4C2 2 2 = 2 2 16 1 1 4 4C3 3 1 = 2 2 16 1 1 1 4C4 4 0 = 2 2 16 162 Expected Frequency N. P(X = x) 1 x 240 = 15 16 4 x 240 = 60 16 6 x 240 = 90 16 4 x 240 = 60 16 1 x 240 = 15 16 240 Computation of chi square values ( O – E)2 Expected Observed Frequency Frequency E O 13 15 4 64 60 16 85 90 25 58 60 4 20 15 25 ( O - E) 2 χ0 = Σ E 2 ( O - E) 2 E 0.27 0.27 0.28 0.07 1.67 2.56 =2.56 Expected Value: ( O - E) 2 2 follows χ2-distribution with ( n−k−1) d.f. χe = Σ E (Here k = 0, since no parameter is estimated from the data) = 9.488 for ν = 5−1= 4 d.f. Inference: Since χ02 < χe2 we accept our null hypothesis at 5% level of significance and say that the given data fits Binomial distribution. Example 8: The following table shows the distribution of goals in a foot ball match. No. of goals 0 1 2 3 4 5 6 7 No. of matches 95 158 108 63 40 9 5 2 Fit a Poisson distribution and test the goodness of fit. Solution: Null Hypothesis : The given data fits the Poisson distribution. Level of significance : Let α = 0.05 163 Computation of expected frequencies: 812 m= = 1.7 480 0 – 1.7 (1 .7 ) P(0) = e = 0.183. 0! f (o) = N.P(o) = 480 x 0.183 = 87.84 The other expected frequencies will be obtained by using the recurrence formula m f (x+1) = x f(x) x +1 Putting x = 0, 1, 2, ... we obtain the following frequencies. f (1) = 1.7 x 87.84 = 149.328 1.7 f (2) = x 149.328 2 = 126.93 1.7 f (3) = x 126.93 3 = 71.927 1.7 f (4) = x 71.927 4 = 30.569 1.7 f (5) = x 30.569 5 = 10.393 1.7 f (6) = x 10.393 6 = 2.94 1.7 f (7) = x 2.94 7 = 0.719 No. of goals 0 1 2 3 4 5 6 7 Total Expected 1 480 88 149 127 72 30 10 3 frequency 164 Computation of statistic: Observed Frequency O 95 158 108 63 40 9 5 16 2 ( O – E)2 Expected Frequency E 88 150 126 72 30 10 3 14 1 49 64 324 81 100 4 ( O - E) 2 E 0.56 0.43 2.57 1.13 3.33 0.29 8.31 ( O - E) χo2 = Σ E 2 = 8.31 Expected Value: ( O - E) 2 χe2 = Σ E 2 χ -distribution with ( n – k – 1) d.f for ν = 6 −1−1 = 4 d.f. = 9.488 Inference: Since χ02 < χe2 , we accept our null hypothesis at 5% level of significance and say that the given data fits the Poisson distribution. 6.6 Test of independence Let us suppose that the given population consisting of N items is divided into r mutually disjoint (exclusive) and exhaustive classes A1, A2, … Ar with respect to the attribute A so that , randomly selected item belongs to one and only one of the attributes A1, A2, … Ar .Similarly let us suppose that the same , population is divided into c mutually disjoint and exhaustive classes B1, B2, … Bc w.r.t another attribute B so that an item , selected at random possess one and only one of the attributes B1, B2, … Bc. The frequency distribution of the items belonging to , 165 the classes A1, A2, … Ar and B1, B2, … Bc can be represented in , , the following r × c manifold contingency table. r × c manifold contingency table B B1 B2 A1 A2 . . . Ai . . . Ar Total (A1B1) (A2B1) . . . (AiB1) . . . (ArB1) (B1) (A1B2) (A2B2) . . . (AiB2) . . . (ArB2) (B2) … Bj … Bc Total … … . . . … . . . … … (A1Bc) (A2Bc) . . . (AiBc) . . . (ArBc) (Bc) (A1) (A2) . . . (Ai) . . . (Ar) A … (A1Bj) … (A2Bj) . . . . . . … (AiBj) . . . . . . … (ArBj) … (Bj) ΣAi = ΣBj = N (Ai) is the number of persons possessing the attribute Ai, (i=1,2,… (Bj) is the number of persons possing the attribute r), Bj,(j=1,2,3,… and (Ai Bj) is the number of persons possessing .c) both the attributes Ai and Bj (i=1,2,… ,j=1,2,… r c). Also ΣAi = ΣBj = N Under the null hypothesis that the two attributes A and B are independent, the expected frequency for (AiBj) is given by (Ai)( Bj) = N Calculation of statistic: Thus the under null hypothesis of the independence of attributes,the expected frequencies for each of the cell frequencies of the above table can be obtained on using the formula ( O - Ei )2 χ02 = Σ i Ei 166 Expected value: ( O - Ei )2 follows χ2-distribution with (r−1) (c−1) d.f χe2 = Σ i Ei Inference: Now comparing χ02 with χe2 at certain level of significance ,we reject or accept the null hypothesis accordingly at that level of significance. 6.6.1 2× 2 contingency table : Under the null hypothesis of independence of attributes, the value of χ2 for the 2×2 contingency table Total a c a+c b d b+d Total a+b c+d N is given by χo2 = N (ad − bc) 2 (a + c)(b + d )(a + b)(c + d ) 6.6.2 Yate’ s correction In a 2×2 contingency table, the number of d.f. is (2−1)(2−1) =1. If any one of the theoretical cell frequency is less than 5,the use of pooling method will result in d.f = 0 which is meaningless. In this case we apply a correction given by F.Yate (1934) which is usually known as “Yates correction for continuity”. This consisting adding 0.5 to cell frequency which is less than 5 and then adjusting for the remaining cell frequencies accordingly. Thus corrected values of χ2 is given as 1 1 1 1 N (a m )(d m ) − (b ± )(c ± ) 2 2 2 2 χ2 = (a + c)(b + d )(a + b)(c + d) 167 2 Example 9: 1000 students at college level were graded according to their I.Q. and the economic conditions of their homes. Use χ2 test to find out whether there is any association between economic condition at home and I.Q. Economic Conditions Rich Poor Total IQ High 460 240 700 Total Low 140 160 300 600 400 1000 Solution: Null Hypothesis: There is no association between economic condition at home and I.Q. i.e. they are independent. (A )(B) 600 × 700 E11 = = = 420 N 1000 The table of expected frequencies shall be as follows. Total 420 180 600 280 120 400 Total 700 300 1000 Observed Frequency O 460 240 140 160 ( O – E)2 Expected Frequency E 420 280 180 120 χo2 1600 1600 1600 1600 ( O - E) 2 =Σ E 168 ( O - E) 2 E 3.81 5.714 8.889 13.333 31.746 = 31.746 Expected Value: ( O - E) 2 follow χ2 distribution with (2−1) (2−1) = 1 d.f χe2 = Σ E = 3.84 Inference : χo2 > χe2, hence the hypothesis is rejected at 5 % level of significance. ∴ t here is association between economic condition at home and I.Q. Example 10: Out of a sample of 120 persons in a village, 76 persons were administered a new drug for preventing influenza and out of them, 24 persons were attacked by influenza. Out of those who were not administered the new drug ,12 persons were not affected by influenza.. Prepare (i) 2x2 table showing actual frequencies. (ii) Use chi-square test for finding out whether the new drug is effective or not. Solution: The above data can be arranged in the following 2 x 2 contingency table. Table of observed frequencies Effect of Influenza New drug Total Attacked Not attacked Administered 24 76 – 24 = 52 76 Not 44 –12 = 32 12 120 – 76 = 44 administered Total 120 – 64 = 56 52 + 12 = 64 120 24 + 32 = 56 Null hypothesis: ‘ Attack of influenza’ and the administration of the new drug are independent. 169 Computation of statistic: χo2 = N (ad − bc) 2 (a + c)(b + d )(a + b)(c + d ) 120(24 × 12 − 52 × 32) 2 56 × 64 × 76 × 44 120(−1376) 2 120(1376) 2 = = 56 × 64 × 76 × 44 56 × 64 × 76 × 44 = Anti log [log 120 + 2log1376 –(log56 +log64+log76+log44)] = Antilog (1.2777) = 18.95 = Expected value: ( O - E) 2 2 χe = Σ E = 3.84 follows χ2distribution with ( 2−1) × (2−1) d.f Inference: Since χo2 > χe2, H0 is rejected at 5 % level of significance. Hence we conclude that the new drug is definitely effective in controlling (preventing) the disease (influenza). Example 11: Two researchers adopted different sampling techniques while investigating the same group of students to find the number of students falling in different intelligence levels. The results are as follows Researchers X Y Total No. of Students Below Average average 86 60 40 33 126 93 Total Above Genius average 44 10 25 2 69 12 200 100 300 Would you say that the sampling techniques adopted by the two researchers are independent? 170 Solution: Null Hypothesis: The sampling techniques adopted by the two researchers are independent. 126 × 200 E(86) = = 84 300 93 × 200 E(60) = = 62 300 69 × 200 E(44) = = 46 300 The table of expected frequencies is given below. Below average 84 Average 62 Above average 46 Y 126 – 84 = 42 Total 126 X Genius Total 200−192 =8 200 93 – 62 = 31 69 – 46 12 – 8 = 4 = 23 100 93 69 300 12 Computation of chi-square statistic: Observed Frequency O 86 60 44 10 40 33 25 2 27 300 Expected Frequency E 84 62 46 8 42 31 23 4 27 300 ( O − E) ( O – E)2 2 −2 −2 2 −2 2 4 4 4 4 4 4 0 0 0 171 ( O - E) 2 E 0.048 0.064 0.087 0.500 0.095 0.129 0 0.923 χo2 ( O - E) 2 = Σ E = 0.923 Expected value: ( O - E) 2 χe2 =Σ E follows χ2distribution with (4−1)( 2−1) = 3 –1 = 2 df = 5.991 Inference: Since χo2 < χe2, we accept the null hypothesis at 5 % level of significance. Hence we conclude that the sampling techniques by the two investigators, do not differ significantly. 6.7 Test for population variance: Suppose we want to test if the given normal population has a specified variance σ2 = σo2 Null Hypothesis: Ho : σ2 = σo2 if x1, x2 … n x Level of significance: Let α = 5% or 1% Calculation of statistic: χo2 = n ∑ (x i i =1 Where s2 = − x) 2 σ02 = ns 2 σ0 2 1n 2 ∑ (x i − x) i =1 n Expected Value: χe2 = n ∑ (x i i =1 − x) 2 σ0 2 follows χ2 distribution with ( n –1) degrees of freedom. Inference: If χo2 ≤ χe2 we accept the null hypothesis otherwise if χo2 > χe2 we reject the null hypothesis. 172 Example 12: A random sample of size 20 from a population gives the sample standard deviation of 6. Test the hypothesis that the population Standard deviation is 9. Solution: We are given n = 20 and s = 6 Null hypothesis: H0 The population standard deviation is σ= 9. Level of significance: Let α = 5 % Calculation of statistic: Under null hypothesis H 0 : ns 2 20 × 36 2 = 8.89 χo = 2 = 9×9 σ Expected value: ns 2 2 χe = 2 follows χ2 distribution 20 –1 = 19 d.f. σ = 30.144 Inference: Since χo2 < χe2 , we accept the null hypothesis at 5 % level of significance and conclude that the population standard deviation is 9. Example 13: Weights in kgs of 10 students are given below: 38, 40, 45, 53, 47, 43, 55, 48, 52 and 49. Can we say that the variance of distribution of weights of all the students from which the above sample of 10 students was drawn is equal to 20 sq kg? Solution: Null hypothesis : H0 : σ2 = 20 173 Computation of sample variance Weight in Kg x 38 40 45 53 47 43 55 48 52 49 x − x = x −47 −9 −7 −2 6 0 −4 8 1 5 2 Sample mean is (x − x )2 81 49 4 36 0 16 64 1 25 4 280 470 ∑x = = 47 n 10 Calculation of statistic: Test statistic is 2 ns 2 280 ∑( x − x ) 2 = 14 = χo = 2 = 2 20 σ σ Expected value: ns 2 2 χe = 2 follows χ2 distribution with 10 – 1 = 9 d.f. σ = 16.919 Inference: Sinceχ02 < χe2 we accept Ho and we conclude that the variance of the distribution of weights of all the students in the population is 20 sq. kgs. 6.8 F – Statistic Definition: If X is a χ2 variate with n1d.f. and Y is an independent χ2X / n1 variate with n2 d.f., then F - statistic is defined as F = Y / n2 x= 174 i.e. F - statistic is the ratio of two independent chi-square variates divided by their respective degrees of freedom. This statistic follows G.W. Snedocor’ s F-distribution with ( n1, n2) d.f. 6.8.1 Testing the ratio of variances: Suppose we are interested to test whether the two normal population have same variance or not. Let x1, x2, x3 … x n 1 , be a .. random sample of size n1, from the first population with variance σ12 and y1, y2, y3 … y n 2 , be random sample of size n2 form the second population with a variance σ22. Obviously the two samples are independent. Null hypothesis: H0 = σ12 = σ22 = σ2 i.e. population variances are same. In other words H0 is that the two independent estimates of the common population variance do not differ significantly. Calculation of statistics: Under H0, the test statistic is 2 S F0= 1 2 S2 2 ns 1 2 Where S1 = ∑( x − x ) = 1 1 n1 − 1 n1 − 1 2 2 ns 1 2 ∑ ( y − y) = 2 2 n 2 −1 n 2 −1 It should be noted that numerator is always greater than the denominator in F-ratio L arg er Variance F= Samller Variance ν1 = d.f for sample having larger variance ν2 = d.f for sample having smaller variance Expected value : 2 S1 Fe = 2 follows F- distribution with ν1 = n1 – 1 , ν2 = n2−1 d.f S2 S2 2 = 175 The calculated value of F is compared with the table value for ν1 and ν2 at 5% or 1% level of significance If F0 > Fe then we reject H0. On the other hand if F0 < Fe we accept the null hypothesis and it is a inferred that both the samples have come from the population having same variance. Since F- test is based on the ratio of variances it is also known as the variance Ratio test. The ratio of two variances follows a distribution called the F distribution named after the famous statisticians R.A. Fisher. Example 14: Two random samples drawn from two normal populations are : Sample I: 20 16 26 27 22 23 18 24 19 25 Sample II: 27 33 42 35 32 34 38 28 41 43 30 37 Obtain the estimates of the variance of the population and test 5% level of significance whether the two populations have the same variance. Solution: Null Hypothesis: H0: σ12 = σ22 i.e. The two samples are drawn from two populations having the same variance. Alternative Hypothesis: H1: σ12 ≠ σ22 (two tailed test) ∑ x1 x1 = n1 220 = 10 = 22 ∑ x2 x2 = n2 420 = 12 = 35 176 x1 x1 − x 1 −2 −6 4 5 0 1 −4 2 −3 3 0 20 16 26 27 22 23 18 24 19 25 220 (x1 − x 1 )2 4 36 16 25 0 1 16 4 9 9 120 x2 27 33 42 35 32 34 38 28 41 43 30 37 420 x2 − x 2 −8 −2 7 0 −3 −1 3 −7 6 8 −5 2 0 (x2− x 2 )2 64 4 49 0 9 1 9 49 36 64 25 4 314 Level of significance : 0.05 The statistic F is defined by the ratio 2 S1 F0 = 2 S2 Where S12 = ∑(x 1 − x 1 ) n1 − 1 2 = 120 = 13.33 9 ∑( x 2 − x 2 ) 314 = = 28.54 S2 = 11 n2 −1 Since S22 > S12 larger variance must be put in the numerator and smaller in the denominator 28.54 = 2.14 ∴F0 = 13.33 Expected value: 2 S2 follows F- distribution with Fe = 2 S1 ν1 = 12−1 = 11 ; ν2 = 10−1= 9 d.f = 3.10 2 2 177 Inference : Since F0 < Fe we accept null hypothesis at 5% level of significance and conclude that the two samples may be regarded as drawn from the populations having same variance. Example 15: The following data refer to yield of wheat in quintals on plots of equal area in two agricultural blocks A and B Block A was a controlled block treated in the same way as Block B expect the amount of fertilizers used. No of plots Mean yield Variance Block A 8 60 50 Block B 6 51 40 Use F test to determine whether variance of the two blocks differ significantly? Solution: We are given that n1 = 8 n2 =6 x 1 = 60 x 2 = 51 s12 =50 s22 = 40 Null hypothesis: H0: σ12 = σ22 ie there is no difference in the variances of yield of wheat. Alternative Hypothesis: H1: σ12 ≠ σ22 (two tailed test) Level of significance: Let α = 0.05 Calculation of statistic: 2 ns 8 × 50 S1 2 = 1 1 = 7 n1 − 1 = 57.14 2 n 2 s2 6 × 40 2 = S2 = 5 n 2 −1 = 48 178 Since S12 > S22 2 S 57.14 F0 = 1 2 = = 1.19 48 S2 Expected value: 2 S1 Fe = 2 follows F- distribution with ν1 = 8−1 =7 ν2 = 6−1 = 5 d.f S2 = 4.88 Inference: Since F0 < Fe, we accept the null hypothesis and hence infer that there is no difference in the variances of yield of wheat. Exercise 6 I. Choose the best answer: 1. Student’ s ‘ t’ distribution was pioneered by (a) Karl Pearson (b) Laplace (c)R.A. Fisher (d) William S.Gosset 2. t - distribution ranges from (a) − ∞ to 0 (b) 0 to ∞ (c) − ∞ to ∞ (d) 0 to 1 3. The difference of two means in case of a small samples is tested by the formula x1 − x 2 x1 − x 2 n1 + n2 (a) t = (b) s s n1 + n2 x1 − x 2 n1n2 n1n2 (d) t = s n1 + n2 n1 + n2 4. While testing the significance of the difference between two sample means in case of small samples, the degree of freedom is (a) n1+n2 (b) n1+n2 –1 (c) n1+n2 –2 (d) n1+n2 +2 5. Paired t-test is applicable when the observations in the two samples are (a) Paired (b) Correlated (c) equal in number (d) all the above 179 (c) t = 6. The mean difference between a paired observations is 15.0 and the standard deviation of differences is 5.0 if n = 9, the value of statistic t is (a) 27 (b) 9 (c) 3 (d) zero 7. When observed and expected frequencies completely coincide χ2 will be (a) –1 (b) +1 (c) greater than 1 (d) 0 2 8. For ν =2, χ 0.05 equals (a) 5.9 (b) 5.99 (c) 5.55 (d) 5.95 2 9. The calculated value of χ is (a) always positive (b) always negative (c ) can be either positive or negative (d) none of these 10. The Yate’ s corrections are generally made when the cell frequency is (a) 5 (b) < 5 (c) 1 (d) 4 2 11. The χ test was derived by (a) Fisher (b) Gauss (c) Karl Pearson (d) Laplace 12. Degrees of freedom for Chi-square in case of contingency table of order (4 ×3) are (a) 12 (b) 9 (c) 8 (d) 6 13. Customarily the larger variance in the variance ratio for Fstatistic is taken (a) in the denominator (b) in the numerator (c) either way (d) none of the above 2 S 14. The test statistic F = 1 2 is used for testing S2 (a) H0: µ1 = µ2 (b) H0: σ12 = σ22 (c) H0: σ1 = σ2 (d) H0: σ2 = σ02 15. Standard error of the sample mean in testing the difference between population mean and sample mean under t- statistic s σ2 (b) (a) n n σ s (d) (c) n n 180 II. Fill in the blanks: 16. The assumption in t- test is that the population standard deviation is ____________ 17. t- values lies in between ____________ 18. Paired t- test is applicable only when the observations are _____________ 19. Student t- test is applicable in case of ___________ samples 20. The value of χ2 statistic depends on the difference between __________ and _________ frequencies 21. The value of χ2 varies from ___________to ___________ 22. Equality of two population variances can be tested by ___________ 23. The χ2 test is one of the simplest and most widely used _________test. 24. The greater the discrepancy between the observed and expected frequency _________the value of χ2 25. In a contingency table ν _________ 26. The distribution of the χ2 depends on the __________ 27. The variance of the χ2 distribution is equal to _______ the d.f 28. One condition for application of χ2 t est is that no cell frequency should be _________ 29. In a 3 × 2 contingency table, there are ________ cells 30. F- test is also known as __________ ratio test. III. Answer the following 31. Define students ‘ t’ – statistic 32. State the assumption of students ‘ t’ test 33. State the properties of t- distribution 34. What are the applications of t- distribution 35. Explain the test procedure to test the significance of mean in case of small samples. 36. What do you understand by paired ‘ t’ test > What are its assumption. 37. Explain the test procedure of paired – t- test 38. Define Chi square test 39. Define Chi square distribution 181 40. What is χ2 test of goodness of fit. 41. What are the precautions are necessary while applying χ2 test? 42. Write short note on Yate’ s correction. 43. Explain the term ‘ Degrees of freedom’ 44. Define non-parametric test 45. Define χ2 test for population variance 46. Ten flower stems are chosen at random from a population and their heights are found to be (in cms) 63 , 63, 66, 67, 68, 69, 70, 70, 71 and 71. Discuss whether the mean height of the population is 66 cms. 47. A machine is designed to produce insulating washers for electrical devices of average thickness of 0.025cm. A random sample of 10 washers was found to have an average thickness of 0.024cm with a standard deviation of 0.002cm. Test the significance of the deviation. 48. Two types of drugs were used on 5 and 7 patients for reducing their weight. Drug A was imported and drug B indigenous. The decrease in the weight after using the drugs for six months was as follows: Drug A : 10 12 13 11 14 Drug B : 8 9 12 14 15 10 9 Is there a significant difference in the efficiency of the two drugs? If not, which drug should you buy? 49. The average number of articles produced by two machines per day are 200 and 250 with standard deviations 20 and 25 respectively on the basis of records 25 days production. Can you conclude that both the machines are equally efficient at 1% level of significance. 50. A drug is given to 10 patients, and the increments in their blood pressure were recorded to be 3, 6, -2 , +4, −3, 4, 6, 0, 0, 2. Is it reasonable to believe that the drug has no effect on change of blood pressure? 51. The sales data of an item in six shops before and after a special promotional campaign are as under: 182 Shops: A B C D E F Before Campaign: 53 28 31 48 50 42 After Campaign: 58 29 30 55 56 45 Can the campaign be judges to be a success? Test at 5% level of significance. 52. A survey of 320 families with 5 children each revealed the following distribution. No of 5 4 3 2 1 0 boys No of 0 1 2 3 4 5 Girls No of 14 56 110 88 40 12 Families Is the result consistent with the hypothesis that the male and female births are equally probable? 53. The following mistakes per page were observed in a book. No of mistakes per page No of pages 0 1 2 3 4 Total 211 90 19 5 0 325 Fit a Poisson distribution and test the goodness of fit. 54. Out of 800 persons, 25% were literates and 300 had travelled beyond the limits of their district 40% of the literates were among those who had not travelled. Test of 5% level whether there is any relation between travelling and literacy 183 55. Fathers You are given the following Intelligent Boys 24 32 Not intelligent Total boys 12 36 32 64 Skilled father Unskilled Father Total 56 44 100 Do these figures support the hypothesis that skilled father have intelligent boys? 56. A random sample of size 10 from a normal population gave the following values 65 , 72, 68, 74, 77, 61,63, 69 , 73, 71 Test the hypothesis that population variance is 32. 57. A sample of size 15 values shows the s.d to be 6.4. Does this agree with hypothesis that the population s.d is 5, the population being normal. 58. In a sample of 8 observations, the sum of squared deviations of items from the mean was 94.5. In another sample of 10 observations, the value was found to be 101.7 test whether the difference in the variances is significant at 5% level. 59. The standard deviations calculated from two samples of sizes 9 and 13 are 2.1 and 1.8 respectively. May the samples should be regarded as drawn from normal populations with the same standard deviation? 60. Two random samples were drawn from two normal populations and their values are A 66 67 75 76 82 84 88 90 92 − B 64 66 74 78 82 85 87 92 93 95 97 Test whether the two populations have the same variance at 5% level of significance. 61. An automobile manufacturing firm is bringing out a new model. In order to map out its advertising campaign, it wants to determine whether the model will appeal most to a particular 184 age – group or equal to all age groups. The firm takes a random sample from persons attending a pre-view of the new model and obtained the results summarized below: Person who Liked the car Disliked the car Total Age groups 40-50 60 and over 48 28 Under 20 146 20-39 54 52 32 62 200 200 130 80 90 500 78 Total 300 What conclusions would you draw from the above data? Answers: I. 1. (d) 2.(c) 3. (c) 4. (c) 5. (d) 6. (b) 7. (d) 8. (b) 9. (a) 10. (c) 11. (c) 12. (d) 13. (b) 14. (b) 15. (b) II. 16. not known 17. − ∞ t o ∞ 18. paired 19 small 20.observed, expected 21. 0, ∞ 22. F- test 23.non parametric 24. greater 25. ( r−1 ) ((−1)) 26. degrees of freedom 27. d.f. twice 28. less than 5 29. 6 30. variance III. 46. t = 1.891 H0 is accepted 47. t = 1.5 H0 is accepted 48. t = 0.735 H0 is accepted 49. t = 7.65 H0 is rejected 50. t = 2, H0 is accepted 51. t= 2.58 H0 is rejected 52. χ2 = 7.16 H0 is accepted 53. χ2 = 0.068 H0 is accepted 2 54. χ = 0.0016 H0 is accepted 55. χ2 = 2.6 H0 is accepted 56. χ2 = 7.3156 H0 is accepted 57. χ2 = 24.58 H0 is rejected 58. χ2 = 24.576 H0 is rejected 59. F = 1.41 H0 is accepted 60. F = 1.415 H0 is accepted 61. χ2 = 7.82 , H0 is rejected 185 7. ANALYSIS OF VARIANCE 7.0 Introduction: The analysis of variance is a powerful statistical tool for tests of significance. The term Analysis of Variance was introduced by Prof. R.A. Fisher to deal with problems in agricultural research. The test of significance based on t-distribution is an adequate procedure only for testing the significance of the difference between two sample means. In a situation where we have three or more samples to consider at a time, an alternative procedure is needed for testing the hypothesis that all the samples are drawn from the same population, i.e., they have the same mean. For example, five fertilizers are applied to four plots each of wheat and yield of wheat on each of the plot is given. We may be interested in finding out whether the effect of these fertilizers on the yields is significantly different or in other words whether the samples have come from the same normal population. The answer to this problem is provided by the technique of analysis of variance. Thus basic purpose of the analysis of variance is to test the homogeneity of several means. Variation is inherent in nature. The total variation in any set of numerical data is due to a number of causes which may be classified as: (i) Assignable causes and (ii) Chance causes The variation due to assignable causes can be detected and measured whereas the variation due to chance causes is beyond the control of human hand and cannot be traced separately. 7.1 Definition: According to R.A. Fisher , Analysis of Variance (ANOVA) is the “ Separation of Variance ascribable to one group of causes from the variance ascribable to other group”. By this technique the total variation in the sample data is expressed as the sum of its nonnegative components where each of these components is a measure of the variation due to some specific independent source or factor or cause. 186 7.2 Assumptions: For the validity of the F-test in ANOVA the following assumptions are made. (i) The observations are independent (ii) Parent population from which observations are taken is normal and (iii) Various treatment and environmental effects are additive in nature. 7.3 One way Classification: Let us suppose that N observations xij , i = 1, 2, … k ; … j = 1,2… i) of a random variable X are grouped on some basis, .n k into k classes of sizes n1, n2 , … k respectively ( N = ∑ n i ) as ..n i =1 exhibited below x11 x12 x22 … x2n2 . . . xi1 . . . xi2 . . . xk1 . . . xk2 … . . . xini Total T1. x2 . . . … x1n1 x21 Mean x1 . T2. . xi . . . . xk . . . . … xknk … . . . Ti. . . . Tk. G The total variation in the observation xij can be spilit into the following two components : (i) The variation between the classes or the variation due to different bases of classification, commonly known as treatments. 187 (ii) The variation within the classes i.e., the inherent variation of the random variable within the observations of a class. The first type of variation is due to assignable causes which can be detected and controlled by human endeavour and the second type of variation due to chance causes which are beyond the control of human hand. In particular, let us consider the effect of k different rations on the yield in milk of N cows (of the same breed and stock) divided k into k classes of sizes n1, n2 , …nk respectively. N = ∑ n i . Hence .. i =1 the sources of variation are (i) Effect of the rations (ii) Error due to chance causes produced by numerous causes that they are not detected and identified. 7.4 Test Procedure: The steps involved in carrying out the analysis are: 1) Null Hypothesis: The first step is to set up of a null hypothesis H0: µ1 = µ2 = … µk = Alternative hypothesis H1: all µi ‘ s are not equal (i = 1,2,… ,k) 2) Level of significance : Let α : 0.05 3) Test statistic: Various sum of squares are obtained as follows. a) Find the sum of values of all the (N) items of the given data. Let this grand total represented by ‘ G’ . G2 Then correction factor (C.F) = N b) Find the sum of squares of all the individual items (xij) and then the Total sum of squares (TSS) is TSS = Σ xi2j – C.F c) Find the sum of squares of all the class totals (or each treatment total) Ti (i:1,2,… and then the sum of .k) squares between the classes or between the treatments (SST) is 188 Ti 2 ∑ n - C.F i =1 i Where ni (i: 1,2,… is the number of observations in ..k) the ith class or number of observations received by ith treatment d) Find the sum of squares within the class or sum of squares due to error (SSE) by subtraction. SSE = TSS - SST 4) Degrees of freedom (d.f): The degrees of freedom for total sum of squares (TSS) is (N−1). The degrees of freedom for SST is (k−1) and the degrees of freedom for SSE is (N−k) 5) Mean sum of squares: SST The mean sum of squares for treatments is and mean k −1 SSE sum of squares for error is N−k 6) ANOVA Table The above sum of squares together with their respective degrees of freedom and mean sum of squares will be summarized in the following table. k SST = ANOVA Table for one-way classification Sources of variation Between treatments Error Total N−1 d.f S.S M.S.S F ratio K−1 SST MST = FT MSE N−k SSE SST = MST k −1 SSE =MSE N−k Calculation of variance ratio: Variance ratio of F is the ratio between greater variance and smaller variance, thus 189 Variance between the treatments Variance within the treatment MST = MSE If variance within the treatment is more than the variance between the treatments, then numerator and denominator should be interchanged and degrees of freedom adjusted accordingly. 7) Critical value of F or Table value of F: The Critical value of F or table value of F is obtained from F table for (k-1, N-k) d.f at 5% level of significance. 8) Inference: If calculated F value is less than table value of F, we may accept our null hypothesis H0 and say that there is no significant difference between treatments. If calculated F value is greater than table value of F, we reject our H0 and say that the difference between treatments is significant. F= Example 1: Three processes outputs are equivalent. made: A 10 12 B 9 11 C 11 10 A, B and C are tested to see whether their The following observations of outputs are 13 10 15 11 12 14 10 13 12 14 15 13 13 Carry out the analysis of variance and state your conclusion. Solution: To carry out the analysis of variance, we form the following tables Total Squares A 10 12 13 11 10 14 15 13 98 9604 B 9 11 10 12 13 55 3025 C 11 10 15 14 12 13 75 5625 G = 228 190 Squares: A B C 100 144 169 121 100 196 225 169 81 121 100 144 169 121 100 225 196 144 169 Total = 2794 Test Procedure: Null Hypothesis: H0: µ1 = µ2 = µ3 i.e., There is no significant difference between the three processes. Alternative Hypothesis H1: µ1 ≠ µ2 ≠ µ3 Level of significance : Let α : 0.05 Test statistic G2 Correct factor (c.f) = N 2282 = 19 51984 = 19 = 2736 Total sum of squares (TSS) = xij2 − C. F = 2794 – 2736 = 58 Sum of squares due to processes = (SST) 3 ∑T . 2 i − C.F ni 9604 3025 5625 = + + − 2736 8 5 6 = (1200.5 + 605 + 937.5) − 2736 = 2743 − 2736 =7 Sum of squares due to error (SSE) = TSS – SST = 58 − 7 = 51 = 191 i =1 ANOVA Table Sources of variation Between Processes Error Total d.f S.S M.S.S F ratio 3 −1 = 2 7 3.5 = 1.097 3.19 16 51 7 = 3.50 2 51 = 3.19 16 19 −1 = 18 Table Value: Table value of Fe for (2,16) d.f at 5% level of significance is 3.63 Inference: Since calculated F0 is less than table value of Fe, we may accept our H0 and say that there is no significant difference between the three processes. Example 2: A test was given to five students taken at random from the fifth class of three schools of a town. The individual scores are School I 9 7 6 5 8 School II 7 4 5 4 5 School III 6 5 6 7 6 Carry out the analysis of variance Solution: To carry out the analysis of variance, we form the following tables. School I School II School III 9 7 6 7 4 5 6 5 6 5 4 7 192 Total Squares 8 35 1225 5 25 625 6 30 900 Total G=90 2750 Squares: School I School II School III 81 49 36 49 16 25 36 25 36 25 64 16 25 49 36 Total = 568 Test Procedure : Null Hypothesis: H0: µ1 = µ2 = µ3 i.e., There is no significant difference between the performance of schools. Alternative Hypothesis: H1: µ1 ≠ µ2 ≠ µ3 Level of significance: Let α :0.05 Test Statistic: G2 Correct factor (c.f) = N 902 = 15 8100 = = 540 15 Total sum of squares (TSS) = xij2 − C. F = 568 – 540 = 28 ΣTi 2 Sum of squares between schools = − C.F ni 2750 = − 540 5 = 550 – 540 = 10 Sum of squares due to error (SSE) = TSS – SST = 28-10 = 18 ANOVA TABLE: Source of d.f S.S M.S.S F ratio variation Between 3-1 = 2 10 10 5 = 5.0 = 3.33 Schools 2 1.5 Error 12 18 18 = 1.5 12 Total 15 -1 = 14 193 Table Value: Table value of Fe for (2,12) d.f at 5% level of significance is 3.8853 Inference: Since calculated F0 is less than table value of Fe, we may accept our H0 and say that there is no significant difference between the performance of schools 7.5 Two way classification: Let us consider the case when there are two factors which may affect the variate values xij, e.g the yield of milk may be affected by difference in treatments i.e., rations as well as the difference in variety i.e., breed and stock of the cows. Let us now suppose that the N cows are divided into h different groups or classes according to their breed and stock, each group containing k cows and then let us consider the effect of k treatments (i.e., rations given at random to cows in each group) on the yield of milk. Let the suffix i refer to the treatments (rations) and j refer to the varieties (breed of the cow), then the yields of milk xij (i:1,2, … j:1,2… of N = h × k cows furnish the data for the ..k; .h) comparison of the treatments (rations). The yields may be expressed as variate values in the following k × h two way table. Mean Total T1. x11 x12 x1j … 1h x x1 . T2. x21 x22 x2j … 2h x x. 2 . . . . xi1 . . xi2 xk1 . . . xk2 . . . . . . xkj … kh x . . . . . . . xi . . . . xk . xij … ih x Mean x .1 . x .2 …x .j …x .h . Total T.1 T.2. … T.j… h .. .T. . . Ti. . . . Tk. x G 194 The total variation in the observation xij can be split into the following three components: (i) The variation between the treatments (rations) (ii) The variation between the varieties (breed and stock) (iii) The inherent variation within the observations of treatments and within the observations of varieties. The first two types of variations are due to assignable causes which can be detected and controlled by human endevour and the third type of variation due to chance causes which are beyond the control of human hand. 7.6 Test procedure for Two - way analysis: The steps involved in carrying out the analysis are: 1. Null hypothesis: The first step is to setting up a null hypothesis H0 Ho : µ1. = µ2. = … µk. = µ … Ho : µ . 1 = µ . 2 = … . h = µ µ i.e., there is no significant difference between rations (treatments) and there is no significant difference between varieties ( breed and stock) 2.Level of significance: Let α : 0.05 3.Test Statistic: Various sums of squares are obtained as follows: a) Find the sum of values of all the N (k×h) items of the given data. Let this grand total represented by G2 ‘ G’ Then correction factor (C.F) = N b) Find the sum of squares of all the individual items (xij) and then the total sum of squares (TSS) k k i −1 j−1 ∑∑ x 2ij − C.F c) Find the sum of squares of all the treatment (rations) totals, i.e., sum of squares of row totals in the h × k two-way table. Then the sum of squares between treatments or sum of squares between rows is 195 Ti . 2 SST = SSR = ∑ − C.F h i −1 where h is the number of observations in each row d) Find the sum of squares of all the varieties (breed and stock) totals, in the h × k two - way table. Then the sum of squares between varieties or sum of squares between columns is k k ∑T 2 .j j−1 − C.F where k is the number k of observations in each column. e) Find the sum of squares due to error by subtraction: i.e., SSE = TSS – SSR − SSC 4. Degrees of freedom: (i) The degrees of freedom for total sum of squares is N−1 = hk−1 (ii) The degrees of freedom for sum of squares between treatments is k−1 (iii) The degree of freedom for sum of squares between varieties is h – 1 (iv) The degrees of freedom for error sum of squares is (k−1) (h−1) 5. Mean sum of squares (MSS) SST (i) Mean sum of squares for treatments (MST) is k −1 SSV (ii) Mean sum of squares for varieties (MSV) is h −1 SSE (iii) Mean sum of squares for error (MSE) is (h − 1)(k − 1) 6. ANOVA TABLE The above sum of squares together with their respective degrees of freedom and mean sum of squares will be summarized in the following table. SSV = SSC = 196 ANOVA Table for Two-way classification Sources of d.f SS variation Between k-1 SST Treatments Between h-1 SSV Varieties Error (h-1) (k-1) SSE Total N-1 MSS F0 - ratio MST MST = FR MSE MSV = Fc MSE MSV MSE 7. Critical values Fe or Table values of F: (i) The critical value or table value of ‘ F’ for between treatments is obtained from F table for [(k−1, (k−1) (h−1)] d.f at 5% level of significance. (ii) The critical value or table value of Fe for between varieties is obtained from F table for [ (h−1), (k−1) (h−1)] d.f at 5% level of significance. 8. Inference: (i) If calculated F0 value is less than or greater than the table value of Fe for between treatments (rows) H0 may be accepted or rejected accordingly. (ii) If calculated F0 value is less than or greater than the table value of Fe for between varieties (column), H0 may be accepted or rejected accordingly. Example 3: Three varieties of coal were analysed by four chemists and the ash-content in the varieties was found to be as under. Chemists Varieties 1 2 A 8 5 B 7 6 C 3 6 Carry out the analysis of variance. 197 3 5 4 5 4 7 4 4 Solution: To carry out the analysis of variance we form the following tables Varieties 1 2 A 8 5 B 7 C Chemists 3 4 Total 5 7 25 625 6 4 4 21 441 3 6 5 4 18 324 Total 18 17 14 15 G = 64 1390 Squares 324 289 196 225 1034 Varieties A B C 1 64 49 9 Individual squares Chemists 2 25 36 36 3 25 16 25 Squares 4 49 16 16 Total = 366 Test Procedure: Null hypothesis: H0 : µ1. = µ2. = µ3. = µ H0 : µ.1 = µ.2 = µ.3 = µ.4 = µ (i) i.e., there is no significant difference between varieties (rows) (ii) i.e., there is no significant difference between chemists (columns) Alternative hypothesis H1: (i) not all µi. ’ s equal (ii) not all µ.j’ s equal Level of significance : Let α : 0.05 198 Test statistic: G2 G2 = N h×k 2 (64) (64)2 = = 3× 4 12 4096 = = 341.33 12 Correction factor (c.f) = k k i −1 Total sum of squares (TSS) = j−1 ∑∑ x 2ij − C.F = 366 – 341.33 = 24.67 Sum of squares between varieties (Rows) 2 ∑ Ti. = − C.F 4 1390 = − 341.33 4 = 347.5 – 341.33 = 6.17 Sum of squares between chemists (columns) 2 ∑ T. j = − C.F 3 1034 = 341.33 3 = 344.67 − 341.33 = 3.34 Sum of square due to error (SSE) = TSS – SSR – SSC = 24.67 – 6.17 – 3.34 = 24.67 – 9.51 = 15.16 199 ANOVA TABLE Sources of d.f SS MSS F ratio variation 6.17 3.085 Between 3.085 3 −1 = 2 = 1.22 Varieties 2.527 3.34 1.113 Between 2.527 4 −1 = 3 = 2.27 Chemists 1.113 Error 6 15.16 2.527 Total 12−1 = 11 Table value : (i) Table value of Fe for (2,6) d.f at 5% level of significance is 5.14 (ii) Table value of Fe for (6,3) d.f at 5% level of significance is 8.94 Inference: (i) Since calculated F0 is less than table value of Fe, we may accept our H0 for between varieties and say that there is no significant difference between varieties. (ii) Since calculated F0 is less than the table value of Fe for chemists, we may accept our Ho and say that there is no significant difference between chemists. Exercise – 7 I. Choose the best answers: 1. Equality of several normal population means can be tested by (a). Bartlet’ s test (b) F - test (c) χ2 -test (d) t- test 2. Analysis of variance technique was developed by (a) S. D. Poisson (b) Karl - Pearson (c) R.A. Fisher (d) W. S. Gosset 3. Analysis of variance technique originated in the field of (a) Agriculture (b) Industry (c) Biology (d) Genetics 4. One of the assumption of analysis of variance is that the population from which the samples are drawn is (a) Binomial (b) Poisson (c) Chi-square (d) Normal 200 5. In the case of one-way classification the total variation can be split into (a) Two components (b) Three components (c) Four components (d) Only one component 6. In the case of one-way classification with N observations and t treatments, the error degrees of freedom is (a) N−1 (b) t −1 (c) N − t (d) Nt 7. In the case of one-way classification with t treatments, the mean sum of squares for treatment is (a) SST/N−1 (b) SST/ t−1 (c) SST/N−t (d) SST/t 8. In the case of two-way classification with r rows and c columns, the degrees of freedom for error is (a) (rc) – 1 (b) (r-1).c (c) (r-1) (c-1) (d) (c-1).r 9. In the case of two-way classification, the total variation (TSS) equals. (a) SSR + SSC + SSE (b) SSR − SSC + SSE (c) SSR + SSC – SSE (d) SSR + SSC. 10. With 90, 35, 25 as TSS, SSR and SSC respectively in case of two way classification, SSE is (a) 50 (b) 40 (c) 30 (d) 20 I. Fill in the blanks 11. The technique of analysis of variance was developed by ____________ 12. One of the assumptions of Analysis of variance is: observations are ______________ 13. Total variation in two – way classification can be split into ____________ components. 14. In the case of one way classification with 30 observations and 5 treatment, the degrees freedom for SSE is _____________ 15. In the case of two-way classification with 120, 54, 45 respectively as TSS, SSC, SSE, the SSR is _____________ III. Answer the following: 16. What is analysis of variance? 17. Distinguish between t-test for difference between means and ANOVA. 201 18. 19. 20. 21. 22. 23. 24. 25. 26. State all the assumptions involved in analysis of variance technique. Explain the structure for one-way classification. Write down the ANOVA table for one-way classification. Distinguish between one - way classification and two-way classification. Explain the structure of two-way classification data. Explain the procedure of obtaining various sums of squares in one-way classification. Write down ANOVA table for two-way classification. Explain the procedure of obtaining various sums of squares in two-way classification. A test was given to a number of students taken at random from the eighth class from each of the 5 schools. The Individual Scores are: Schools I II III IV V 8 9 12 10 12 9 7 14 11 11 10 11 15 9 10 7 12 12 12 9 8 13 11 10 13 Carry out the analysis of variance and give your conclusions. 27. The following figures relate to production in kg of three varieties A, B and C of wheat shown in 12 plots. A: 20 18 19 B: 17 16 19 18 C: 20 21 20 19 18 Is there any significant difference in the production of the three varieties 202 28. 29. A special type of fertilizer was used in four agricultural fields A,B,C and D each field was divided into four beds and the fertilizer was applied over them. The respective yields of the beds of four fields are given below. Find whether the difference in mean yields of fields is significant or not? Plot yield A B C D 8 9 3 3 12 4 8 7 1 7 2 8 9 1 5 2 The following table gives the retail prices of a commodity in (Rs. Per Kg) in some shops selected at random in four cities. A 22 24 20 21 CITY B 20 19 21 22 C 19 17 21 18 D 20 22 21 22 Analysis the data to test the significance of the differences between the price of commodity in four cities. 30. For experiments determine the moisture content of sample of a powder, each man taking a sample from each of six consignments Their assessments are: Observer 1 2 3 4 1 9 12 11 12 Consignment 2 3 10 9 11 9 10 10 13 11 4 10 11 12 14 5 11 10 11 12 6 11 10 10 10 Perform an analysis of variance of these data and discuss if there is any significant difference between consignments or between observers. 203 31. The following are the defective pieces produced by four operators working in turn, on four different machines: Operator Machine I II III A 3 2 3 B 3 2 3 C 2 3 4 D 3 4 3 IV 2 4 3 2 Perform analysis of variance at 5% level of significance to ascertain whether variability in production is due to variability in operator’ s performance or variability in machine’ s performance. 32. Apply the technique of Analysis of variance to the following data relating to yields of 4 varieties of wheat in 3 blocks. Blocks Varieties 1 2 3 I 10 9 8 II 7 7 6 III 8 5 4 IV 5 4 4 33. Four Varieties of potato are planted, each on five plots of ground of the same size and type and each variety is treated with five different fertilizers. The yields in tons are as follows. Varieties V1 V2 V3 V4 F1 1.9 2.5 1.7 2.1 Fertilizers F2 F3 2.2 2.6 1.9 2.2 1.9 2.2 1.8 2.5 F4 1.8 2.6 2.0 2.2 Perform an analysis of variance and test whether there is any significant difference between yields of different varieties and fertilizers. 204 F5 2.1 2.2 2.1 2.5 34. In an experiment on the effects of temperature conditions in human performance, 8 persons were given a test on 4 temperature conditions. The scores in the test are shown in the following table. Temperature 1 2 3 4 1 70 70 75 65 2 80 80 85 75 Persons 3 4 70 90 80 90 80 95 70 85 5 80 80 75 80 6 100 100 85 90 7 90 90 95 80 8 80 80 75 75 Perform the analysis of variance and state whether there is any significant difference between persons and temperature conditions. 35. The following table gives the number of refrigerators sold by 4 salesmen in three months may, June and July Sales Man Months A B C D May 50 40 48 39 June 46 48 50 45 July 39 44 40 39 Carry out the analysis Answers I. 1. b 2. c 3. a 4. d 5. a 6. c 7. b 8. c 9. a 10.c II. 11. R.A. Fisher 12. Independent 14. 25 15. 21 205 13. Three III. 26. Calculated F = 4.56, Table value of F (4,20) = 2.87 27. Calculated F = 9.11, Table value of F (9,2) = 19.39 28. Calculated F = 1.76, Table value of F (12,3) = 8.74 29. Calculated F = 3.29, Table value of F (3,12) = 3.49 30. Calculated FR = 5.03, Table value of F (3,15) = 3.29 Calculated FC = 2.23, Table value of F (5,15) = 2.90 31. Calculated FR = 2.76, FC Table value of F (9,3) = 8.81 32. Calculated FR = 18.23, Table value of F (3,6) = 4.77 Calculated FC = 6.4, Table value of F (2,6) = 5.15 33. Calculated FR = 1.59, Table value of F (3,12) = 3.49 Calculated FC = 3.53, Table value of F (4,12) = 3.25 34. Calculated FR = 3.56, Table value of F (3,21) = 3.07 Calculated FC = 14.79, Table value of F (7,21) = 2.49 35. Calculated FR = 3.33, Table value of F (2,6) = 5.15 Calculated FC = 1.02, Table value of F (3,6) = 4.77 206 8. TIME SERIES 8.0 Introduction: Arrangement of statistical data in chronological order ie., in accordance with occurrence of time, is known as “Time Series”. Such series have a unique important place in the field of Economic and Business statistics. An economist is interested in estimating the likely population in the coming year so that proper planning can be carried out with regard to food supply, job for the people etc. Similarly, a business man is interested in finding out his likely sales in the near future, so that the businessman could adjust his production accordingly and avoid the possibility of inadequate production to meet the demand. In this connection one usually deal with statistical data, which are collected, observed or recorded at successive intervals of time. Such data are generally referred to as ‘ time series’ . 8.1 Definition: According to Mooris Hamburg “A time series is a set of statistical observations arranged in chronological order” Ya-Lun- chou defining the time series as “A time series may be defined as a collection of readings belonging to different time periods, of some economic variable or composite of variables. A time series is a set of observations of a variable usually at equal intervals of time. Here time may be yearly, monthly, weekly, daily or even hourly usually at equal intervals of time. Hourly temperature reading, daily sales, monthly production are examples of time series. Number of factors affect the observations of time series continuously, some with equal intervals of time and others are erratic studying, interpreting analyzing the factors is called Analysis of Time Series. The Primary purpose of the analysis of time series is to discover and measure all types of variations which characterise a time series. The central objective is to decompose the various elements present in a time series and to use them in business decision making. 207 8.2 Components of Time series: The components of a time series are the various elements which can be segregated from the observed data. The following are the broad classification of these components. Components Long Term Secular Trend Short Term Cyclical Seasonal Irregular (or) Erratic Regular In time series analysis, it is assumed that there is a multiplicative relationship between these four components. Symbolically, Y=T×S×C×I Where Y denotes the result of the four elements; T = Trend ; S = Seasonal component; C = Cyclical components; I = Irregular component In the multiplicative model it is assumed that the four components are due to different causes but they are not necessarily independent and they can affect one another. Another approach is to treat each observation of a time series as the sum of these four components. Symbolically Y = T + S+ C+ I The additive model assumes that all the components of the time series are independent of one another. 1) Secular Trend or Long - Term movement or simply Trend 2) Seasonal Variation 3) Cyclical Variations 4) Irregular or erratic or random movements(fluctuations) 8.2.1 Secular Trend: It is a long term movement in Time series. The general tendency of the time series is to increase or decrease or stagnate 208 during a long period of time is called the secular trend or simply trend. Population growth, improved technological progress, changes in consumers taste are the various factors of upward trend. We may notice downward trend relating to deaths, epidemics, due to improved medical facilities and sanitations. Thus a time series shows fluctuations in the upward or downward direction in the long run. 8.2.2 Methods of Measuring Trend: Trend is measured by the following mathematical methods. 1. Graphical method 2. Method of Semi-averages 3. Method of moving averages 4. Method of Least Squares Graphical Method: This is the easiest and simplest method of measuring trend. In this method, given data must be plotted on the graph, taking time on the horizontal axis and values on the vertical axis. Draw a smooth curve which will show the direction of the trend. While fitting a trend line the following important points should be noted to get a perfect trend line. (i) The curve should be smooth. (ii) As far as possible there must be equal number of points above and below the trend line. (iii) The sum of the squares of the vertical deviations from the trend should be as small as possible. (iv) If there are cycles, equal number of cycles should be above or below the trend line. (v) In case of cyclical data, the area of the cycles above and below should be nearly equal. Example 1: Fit a trend line to the following data by graphical method. Year Sales (in Rs ‘ 000) 1996 1997 1998 1999 2000 2001 2002 60 72 75 65 80 85 95 209 Solution: 100 Sales 80 60 40 20 0 1996 1997 1998 1999 2000 2001 2002 2003 Year The dotted lines refers trend line Merits: 1. It is the simplest and easiest method. It saves time and labour. 2. It can be used to describe all kinds of trends. 3. This can be used widely in application. 4. It helps to understand the character of time series and to select appropriate trend. Demerits: 1. It is highly subjective. Different trend curves will be obtained by different persons for the same set of data. 2. It is dangerous to use freehand trend for forecasting purposes. 3. It does not enable us to measure trend in precise quantitative terms. Method of semi averages: In this method, the given data is divided into two parts, preferably with the same number of years. For example, if we are given data from 1981 to 1998 i.e., over a period of 18 years, the two equal parts will be first nine years, i.e., 1981 to 1989 and from 1990 to 1998. In case of odd number of years like 5,7,9,11 etc, two equal parts can be made simply by omitting the middle year. For example, if the data are given for 7 years from 1991 to 1997, the two equal parts would be from 1991 to 1993 and from 1995 to 1997, the middle year 1994 will be omitted. 210 After the data have been divided into two parts, an average of each parts is obtained. Thus we get two points. Each point is plotted at the mid-point of the class interval covered by respective part and then the two points are joined by a straight line which gives us the required trend line. The line can be extended downwards and upwards to get intermediate values or to predict future values. Example 2: Draw a trend line by the method of semi-averages. Year 1991 1992 1993 1994 1995 Sales Rs in 60 75 81 110 106 (1000) 1996 117 Solution: Divide the two parts by taking 3 values in each part. Year Sales (Rs) Semi total Semi Trend average values 59 60 1991 72 216 72 75 1992 85 81 1993 98 110 1994 111 333 111 106 1995 124 117 1996 Difference in middle periods = 1995 –1992 = 3 years Difference in semi averages = 111 –72 = 39 ∴ Annual increase in trend = 39/3 = 13 Trend of 1991 = Trend of 1992 -13 = 72-13 = 59 Trend of 1993 = Trend of 1992 +13 = 72 + 13 = 85 Similarly, we can find all the values The following graph will show clearly the trend line. 211 140 120 Sales 100 80 60 40 20 0 1991 1992 1993 1994 1995 1996 Year Example 3 : Calculate the trend value to the following data by the method of semi- averages. Year 1995 1996 1997 1998 1999 2000 2001 Expenditure (Rs in 1.5 1.8 2.0 2.3 2.4 2.6 3.0 Lakhs) Solution: Year 1995 1996 1997 1998 1999 2000 2001 Expenditure (Rs) 1.5 1.8 2.0 2.3 2.4 2.6 3.0 Semi total Semi average 5.3 1.77 8.0 2.67 Difference between middle periods = 2000 – 1996 = 4 years Difference between semi-averages = 2.67 - 1.77 = 0.9 212 Trend values 1.545 1.770 1.995 2.220 2.445 2.670 2.895 0.9 4 = 0.225 Trend of 1995 = Trend of 1996 – 0.225 = 1.77 – 0.225 = 1.545 Trend of 1996 = 1.77 Trend of 1997 = 1.77 + 0.225 = 1.995 Similarly we can find all the trend values ∴ Annual trend values = 3.5 Expenditure 3 2.5 2 1.5 1 0.5 0 1995 1996 1997 1998 1999 2000 2001 Year Merits: 1. It is simple and easy to calculate 2. By this method every one getting same trend line. 3. Since the line can be extended in both ways, we can find the later and earlier estimates. Demerits: 1. This method assumes the presence of linear trend to the values of time series which may not exist. 2. The trend values and the predicted values obtained by this method are not very reliable. Method of Moving Averages: This method is very simple. It is based on Arithmetic mean. Theses means are calculated from overlapping groups of successive 213 time series data. Each moving average is based on values covering a fixed time interval, called “period of moving average” and is shown against the center of the interval. The method of ‘ odd period of moving average is as follows. a+b+c ( 3 or 5) . The moving averages for three years is , 3 b+c +d c+d+e , etc 3 3 a +b+c+d+e The formula for five yearly moving average is , 5 b+c+d+e+f c+d+e+f +g , etc. 5 5 Steps for calculating odd number of years. 1. Find the value of three years total, place the value against the second year. 2. Leave the first value and add the next three years value (ie 2nd, 3rd and 4th years value) and put it against 3rd year. 3. Continue this process until the last year’ s value taken. 4. Each total is divided by three and placed in the next column. These are the trend values by the method of moving averages Example 4 : Calculate the three yearly average of the following data. Year 1975 1976 1977 1978 1979 1980 Production in (tones) 50 36 43 45 Year 1981 1982 1983 1984 Production 33 42 41 34 in (tones) 214 39 38 Solution: Year Production (in tones) 3 years moving total 3 years moving average as Trend values 43.0 41.3 42.3 40.7 36.7 37.7 38.7 39.0 - 50 1975 129 36 1976 124 43 1977 127 45 1978 122 39 1979 110 38 1980 113 33 1981 116 42 1982 117 41 1983 34 1984 Even Period of Moving Averages: When the moving period is even, the middle period of each set of values lies between the two time points. So we must center the moving averages. The steps are 1. Find the total for first 4 years and place it against the middle of the 2nd and 3rd year in the third column. 2. Leave the first year value, and find the total of next four-year and place it between the 3rd and 4th year. 3. Continue this process until the last value is taken. 4. Next, compute the total of the first two four year totals and place it against the 3rd year in the fourth column. 5. Leave the first four years total and find the total of the next two four years’ totals and place it against the fourth year. 6. This process is continued till the last two four years’ total is taken into account. 7. Divide this total by 8 (Since it is the total of 8 years) and put it in the fifth column. These are the trend values. Example 5 : The production of Tea in India is given as follows. Calculate the Four-yearly moving averages 215 Year 1993 1994 1995 1996 1997 1998 Production (tones) 464 518 502 540 Year 1999 2000 2001 2002 Production (tones) 557 586 515 571 467 612 Solution: Year Production (in tones) 1993 1994 Trend Values - 3966 495.8 4029 503.6 4093 511.6 4236 529.5 4424 553.0 4580 464 Total of Two four years - 4 years Moving total 572.5 515 1964 1995 518 2002 1996 467 2027 1997 502 2066 1998 540 1999 557 2000 571 2170 2254 2326 2001 586 - 2002 612 216 Merits: 1. The method is simple to understand and easy to adopt as compared to other methods. 2. It is very flexible in the sense that the addition of a few more figures to the data, the entire calculations are not changed. We only get some more trend values. 3. Regular cyclical variations can be completely eliminated by a period of moving average equal to the period of cycles. 4. It is particularly effective if the trend of a series is very irregular. Demerits: 1. It cannot be used for forecasting or predicting future trend, which is the main objective of trend analysis. 2. The choice of the period of moving average is sometimes subjective. 3. Moving averages are generally affected by extreme values of items. 4. It cannot eliminate irregular variations completely. 8.3 Method of Least Square: This method is widely used. It plays an important role in finding the trend values of economic and business time series. It helps for forecasting and predicting the future values. The trend line by this method is called the line of best fit. The equation of the trend line is y = a + bx, where the constants a and b are to be estimated so as to minimize the sum of the squares of the difference between the given values of y and the estimate values of y by using the equation. The constants can be obtained by solving two normal equations. y = na + b x …… (1) …. xy = a x + b x2 …… … (2) Here x represent time point and y are observed values. ‘ n’ is the number of pair- values. When odd number of years are given Step 1: Writing given years in column 1 and the corresponding sales or production etc in column 2. 217 Step 2: Write in column 3 start with 0, 1, 2 .. against column 1 and denote it as X Step 3: Take the middle value of X as A Step 4: Find the deviations u = X A and write in column 4 Step 5: Find u2 values and write in column 5. Step 6: Column 6 gives the product uy Now the normal equations become y = na + b u (1) where u = X-A uy = a u + b u2 (2) Since u = 0 , From equation (1) Σy a= n From equation (2) uy = b u2 uy ∴b= Σu 2 ∴ The fitted straight line is y = a + bu = a + b ( X - A) Example 6: For the following data, find the trend values by using the method of Least squares Year 1990 1991 1992 1993 1994 Production 50 55 45 (in tones) Estimate the production for the year 1996 Solution: 52 54 u2 uy Trend values -2 4 -100 50.2 1 -1 1 -55 50.7 45 2A 0 0 0 51.2 1993 52 3 1 1 52 51.7 1994 54 4 2 4 108 52.2 Total 256 10 5 X= x -1990 u = X-A Year (x) Production (y) 1990 50 0 1991 55 1992 = X-2 218 Where A is an assumed value The equation of straight line is Y = a + bX = a + bu , where u = X - 2 the normal equations are ….(1) Σy = na+ bΣu…… ..(2) Σuy = aΣu +bΣu2 … since Σu = 0 from(1) Σy = na Σy 256 = a= = 51.2 n 5 From equation (2) Σuy = bΣu2 5 = 10b 5 = 0.5 b= 10 The fitted straight line is y = a+ bu y = 51.2 + 0.5 (X-2) y = 51.2 + 0.5X –1.0 y = 50.2 + 0.5X Trend values are, 50.2, 50.7, 51.2, 51.7, 52.2 The estimate production in 1996 is put X = x – 1990 X = 1996 –1990 = 6 Y = 50.2 + 0.5X = 50.2 +0.5(6) = 50.2 +3.0 = 53.2 tonnes. 60 production 50 40 30 20 10 0 1990 1991 1992 Year 219 1993 1994 When even number of years are given Here we take the mean of middle two values of X as A X−A Then u = = 2 (X-A). The other steps are as given in the 1/ 2 odd number of years. Example 7: Fit a straight line trend by the method of least squares for the following data. Year Sales (Rs. in lakhs) 1983 1984 1985 1986 1987 1988 3 8 7 9 11 14 Also estimate the sales for the year 1991 Solution: Sales X = Year u (x) (y) x-1983 =2X-5 1983 3 0 -5 1984 8 1 -3 1985 7 2 -1 1986 9 3 1 1987 11 4 3 1988 14 5 5 Total 52 0 X−A u= 1/ 2 = 2 (X − 2.5) = 2X − 5 The straight line equation is y = a + bX = a + bu The normal equations are … Σy = na ….(1) 2 … Σuy = bΣu …(2) From (1) 52 = 6a 220 u2 25 9 1 1 9 25 70 uy -15 -24 -7 9 33 70 66 Trend values 3.97 5.85 7.73 9.61 11.49 13.37 52 6 = 8.67 From (2) 66 = 70 b 66 b= 70 = 0.94 The fitted straight line equation is y = a+bu y = 8.67+0.94(2X-5) y = 8.67 + 1.88X - 4.7 y = 3.97 + 1.88X -----------(3) The trend values are Put X = 0, y = 3.97 X = 1, y = 5.85 X = 2, y = 7.73 X = 3, y = 9.61 X = 4, y = 11.49 X = 5, y = 13.37 The estimated sale for the year 1991 is; put X = x –1983 = 1991 – 1983 = 8 y = 3.97 + 1.88 × 8 = 19.01 lakhs The following graph will show clearly the trend line. Sales a= 16 14 12 10 8 6 4 2 0 1983 1984 1985 1986 Year 221 1987 1988 Merits: 1. Since it is a mathematical method, it is not subjective so it eliminates personal bias of the investigator. 2. By this method we can estimate the future values. As well as intermediate values of the time series. 3. By this method we can find all the trend values. Demerits: 1. It is a difficult method. Addition of new observations makes recalculations. 2. Assumption of straight line may sometimes be misleading since economics and business time series are not linear. 3. It ignores cyclical, seasonal and irregular fluctuations. 4. The trend can estimate only for immediate future and not for distant future. 8.4 Seasonal Variations: Seasonal Variations are fluctuations within a year during the season. The factors that cause seasonal variation are i) Climate and weather condition. ii) Customs and traditional habits. For example the sale of ice-creams increase in summer, the umbrella sales increase in rainy season, sales of woolen clothes increase in winter season and agricultural production depends upon the monsoon etc., Secondly in marriage season the price of gold will increase, sale of crackers and new clothes increase in festival times. So seasonal variations are of great importance to businessmen, producers and sellers for planning the future. The main objective of the measurement of seasonal variations is to study their effect and isolate them from the trend. Measurement of seasonal variation: The following are some of the methods more popularly used for measuring the seasonal variations. 1. Method of simple averages. 2. Ratio to trend method. 3. Ratio to moving average method. 4. Link relative method 222 Among the above four methods the method of simple averages is easy to compute seasonal variations. 8.4.1 Method of simple averages The steps for calculations: i) Arrange the data season wise ii) Compute the average for each season. iii) Calculate the grand average, which is the average of seasonal averages. iv) Obtain the seasonal indices by expressing each season as percentage of Grand average The total of these indices would be 100n where ‘ n’ is the number of seasons in the year. Example 8: Find the seasonal variations by simple average method for the data given below. Quarter Year I II III IV 34 36 40 30 1989 44 50 52 34 1990 48 54 58 40 1991 62 68 76 54 1992 82 86 92 80 1993 Solution: Year 1989 1990 1991 1992 1993 Total Average Seasonal Indices I 30 34 40 54 80 238 47.6 85 Quarter II 40 52 58 76 92 318 63.6 113.6 223 III 36 50 54 68 86 294 58.8 IV 34 44 48 62 82 270 54 105 96.4 47.6 + 63.6 + 58.8 + 54 4 224 = = 56 4 Seasonal Index for First quarterly Average I quarter = × 100 Grand Average 47.6 = × 100 = 85 56 Seasonal Index for Second quarterly Average II quarter = × 100 Grand Average 63.6 = × 100 = 113.6 56 Seasonal Index for Third quarterly Average III quarter = × 100 Grand Average 58.8 = × 100 = 105 56 Seasonal Index for Fourth quarterly Average IV quarter = × 100 Grand Average 54 = ×100 = 96.4 56 Example 9: Calculate the seasonal indices from the following data using simple average method. Year Quarter 1974 1975 1976 1977 1978 I 72 76 74 76 74 II 68 70 66 74 74 III 80 82 84 84 86 IV 70 74 80 78 82 Grand average = 224 Solution: Year 1974 1975 1976 1977 1978 Total Average Seasonal Indices I 72 76 74 76 74 372 74.4 97.6 Quarter II 68 70 66 74 74 352 70.4 92.4 III 80 82 84 84 86 416 83.2 109.2 74.4 + 70.4 + 83.2 + 76.8 4 304.8 = 76.2 = 4 Seasonal Index for First quarterly Average I quarter = × 100 Grand Average 74.4 = × 100 76.2 = 97.6 Seasonal Index for Second quarterly Average II quarter = × 100 Grand Average 70.4 = × 100 76.2 = 92.4 Seasonal Index for Third quarterly Average III quarter = × 100 Grand Average 83.2 = × 100 76.2 = 109.2 Grand Average = 225 IV 70 74 80 78 82 384 76.8 100.8 Seasonal Index for Fourth quarterly Average IV quarter = × 100 Grand Average 76.8 = ×100 76.2 = 100.8 The total of seasonal indices calculated must be equal to 400 here we have = 97.6 + 92.4 + 109.2 + 100.8 = 400 hence verified. Cyclical variations: The term cycle refers to the recurrent variations in time series, that extend over longer period of time, usually two or more years. Most of the time series relating to economic and business show some kind of cyclic variation. A business cycle consists of the recurrence of the up and down movement of business activity. It is a four-phase cycle namely. 1. Prosperity 2. Decline 3. Depression 4. Recovery Each phase changes gradually into the following phase. The following diagram illustrates a business cycle. The study of cyclical variation is extremely useful in framing suitable policies for stabilising the level of business activities. Businessmen can take timely steps in maintaining business during booms and depression. Irregular variation: Irregular variations are also called erratic. These variations are not regular and which do not repeat in a definite pattern. 226 These variations are caused by war, earthquakes, strikes flood, revolution etc. This variation is short-term one, but it affect all the components of series. There is no statistical techniques for measuring or isolating erratic fluctuation. Therefore the residual that remains after eliminating systematic components is taken as representing irregular variations. FORECASTING 8.5 Introduction: A very important use of time series data is towards forecasting the likely value of variable in future. In most cases it is the projection of trend fitted into the values regarding a variable over a sufficiently long period by any of the methods discussed latter. Adjustments for seasonal and cyclical character introduce further improvement in the forecasts based on the simple projection of the trend. The importance of forecasting in business and economic fields lies on account of its role in planning and evaluation. If suitably interpreted, after consideration of other forces, say political, social governmental policies etc., this statistical technique can be of immense help in decision making. The success of any business depends on its future estimates. On the basis of these estimates a business man plans his production stocks, selling market, arrangement of additional funds etc. Forecasting is different from predictions and projections. Regression analysis, time series analysis, Index numbers are some of the techniques through which the predictions and projections are made. Where as forecasting is a method of foretelling the course of business activity based on the analysis of past and present data mixed with the consideration of ensuring economic policies and circumstances. In particularly forecasting means fore-warning. Forecasts based on statistical analysis are much reliable than a guess work. According to T.S.Levis and and R.A. Fox, “ Forecasting is using the knowledge we have at one time to estimate what will happen at some future movement of time”. 227 8.5.1 Methods of Business forecasting: There are three methods of forecasting 1. Naive method 2. Barometric methods 3. Analytical Methods 1. Naive method : It contains only the economic rhythm theory. 2. Barometric methods: It covers i) Specific historical analogy ii) Lead- Lag relationship iii) Diffusion method iv) Action –reaction theory 3. Analytical Methods: It contains i) The factor listing method ii) Cross-cut analysis theory iii) Exponential smoothing iv) Econometric methods The economic rhythm theory: In this method the manufactures analysis the time-series data of his own firm and forecasts on the basis of projections so obtained. This method is applicable only for the individual firm for which the data are analysed, The forecasts under this method are not very reliable as no subjective matters are being considered. Diffusion method of Business forecasting The diffusion index method is based on the principle that different factors, affecting business, do not attain their peaks and troughs simultaneously. There is always time-log between them. This method has the convenience that one has not to identify which series has a lead and which has a lag. The diffusion index depicts the movement of broad group of series as a whole without bothering about the individual series. The diffusion index shows the percentage of a given set of series as expanding in a time period. It should be carefully noted that the peaks and troughs of diffusion index are not the peaks troughs of the business cycles. All 228 series do not expand or contract concurrently. Hence if more than 50% are expanding at a given time, it is taken that the business is in the process of booming and vice - versa. The graphic method is usually employed to work out the diffusion index. The diffusion index can be constructed for a group of business variables like prices, investments, profits etc. Cross cut analysis theory of Business forecasting: In this method a thorough analysis of all the factors under present situations has to be done and an estimate of the composite effect of all the factors is being made. This method takes into account the views of managerial staff, economists, consumers etc. prior to the forecasting. The forecasts about the future state of the business is made on the basis of over all assessment of the effect of all the factors. Exercise – 8 I. Choose the best answer: 1. A time series consists of a) Two components b) Three Components c) Four components d) Five Components 2. Salient features responsible for the seasonal variation are a) Weather b) Social customers c) Festivals d) All the above 3. Simple average method is used to calculate a) Trend Values b) Cyclic Variations c) Seasonal indices d) None of these 4. Irregular variations are a) Regular b) Cyclic c) Episodic d) None of the above 5. If the slope of the trend line is positive it shows a) Rising Trend b) Declining trend c) Stagnation d) None of the above 6. The sales of a departmental store on Diwali are associated with the component of time-series a) Secular trend b) Seasonal variation c) Irregular variation d) All the above 229 7. The component of time-series attached to long term variation is termed as a) Secular Trend b) Seasonal Variation c) Irregular variation d) Cyclic variation 8. Business forecasts are made on the basis of a) Present Data b) Past data c) Polices and circumstances d) All the above 9. Econometric methods involve a) Economics and mathematics b) Economics and Statistics c) Economics, Statistics and Mathematics d) None of the above 10. The economic rhythm theory comes under the category of a) Analytical methods b) Naive method c) Barometric methods d) None of the above II. Fill in the blanks: 11. A time series in a set of values arranged in ________ order 12. Quarterly fluctuations observed in a time series represent _______ variation 13. Periodic changes in a business time series are called _________ 14. A complete cycle passes through _____________stages of phenomenon. 15. An overall tendency of rise and fall in a time series represents ___________ 16. The trend line obtained by the method of least square is known as the ___________ 17. Forecasting is different from ________ and _________. 18. No statistical techniques measuring or isolating _________ III. Answer the following questions 19. What is a time series? 20. What are the components of time series. 21. Write briefly about seasonal variation. 22. What is cyclic variation. 23. Give the names of different methods of measuring trend. 24. What are the merits and demerits of the semi-average method. 25. Discuss the mathematical models for a time series analysis. 230 26. Discuss irregular variation in the context of time series. 27. What do you understand by business fore-casting 28. Give the names of different methods of fore casting. 29. Write briefly about any one method of forecasting? 30. In what sense forecasting differ from prediction and projection. IV. Problems 31. With the help of graph paper obtain the trend values. Year 1996 1997 1998 1999 2000 2001 Value 65 85 95 75 100 80 2002 130 32. Using graphical method, fit a trend-line to the following data. Year 1982 1983 1984 1985 1986 1987 Value 24 22 25 26 27 26 33. Draw a trend line by the method of semi-averages. 98 99 Year 1993 94 95 96 97 Sales 210 200 215 205 220 235 210 34. The following figures are given relating to factory. Draw a trend-line with the help of averages. Year 1996 1997 1998 1999 2000 Output 600 800 1000 800 1200 35. 2000 235 the output in a method of semi2001 2002 1000 1400 Calculate three yearly moving average of the following data Year No of students 91 92 93 94 95 96 97 98 99 00 15 18 17 20 23 25 29 33 36 40 36. The following figures relating to the profits of a commercial concern for 8 years. Find the 3-yearly moving averages. Years Profits Years Profits 1995 1996 1997 1998 15,420 14,470 15,520 21,020 1999 2000 2001 2002 231 26,120 31,950 35,370 35,670 37. Construct a four yearly centered moving average from the following data. Year 1940 1950 1960 1970 1980 1990 2000 Imported 129 131 106 91 95 84 93 cotton consumption (‘ 000) 38. From the following data calculate the 4-yearly moving average and determine the trend values. Find the short-term fluctuations. Plot the original data and the trend on a graph. Year Value 93 50 94 36.5 95 43 96 44.5 97 38.9 98 38.1 99 32.6 00 41.7 01 41.1 02 33.8 39. Calculate trend value by taking 5 yearly period of moving average from the data given below Year 1987 Production 4 in tones Year 95 96 Production 8 7 in tones 88 89 5 90 91 92 93 94 7 9 6 5 7 6 97 98 99 2000 01 02 6 8 9 10 7 9 40. Fit a straight line trend by the method of least squares to the following data and calculate trend values. Year 1996 1997 Sales of TV Sets 4 6 (Rs ‘ 000 ) Estimate the sales for the year 2005 232 1998 7 1999 8 2000 10 41. Below are given the figures of production in ‘ 000 quintals of a sugar factory. Year 1994 95 96 97 98 99 2000 Production 80 90 92 83 94 99 92 in tones 42. Fit a straight line trend by the method of least square to the following data. Year 1996 97 98 99 2000 2001 Profit 300 700 600 800 900 700 43. Fit a straight line trend by the method of least squares to the following data. Estimate the earnings for the year 2002. Year 1993 94 95 96 97 98 99 2000 Earnings 38 40 65 72 69 60 87 95 44. Compute the average seasonal movement for the following series. Year Ist quarter 1999 3.5 2000 3.5 2001 3.5 2002 4.0 2003 4.1 IInd quarter 3.9 4.1 3.9 4.6 4.4 IIIrd quarter 3.4 3.7 3.7 3.8 4.2 IVth quarter 3.6 4.0 4.2 4.5 4.5 45. Obtain seasonal fluctuations from the following time-series Quarterly output of coal for four years. Year I II III IV 2000 65 58 56 61 2001 58 63 63 67 233 2002 70 59 56 52 2003 60 55 51 58 Answers I. 1. (c) 7. (a) 2.(d) 8.(d) 3. (c) 9. (c) 4. (c) 10.(b) 5. (a) 6. (b) II. 11. Chronological 14. four 16. line of best fit 18. Erratic Fluctuation. 12. Seasonal 13. Cycles 15. secular trend 17. Prediction, projection IV. 33. Trend values are 200.94, 205.31, 209.69, 214.06, 218.43, 222.80, 227.19, 231.56 34. 700, 800, 900, 1000, 1100, 1200, 1300 35. 16.7, 18.3, 20, 22.7, 25.7, 29, 32.7, 36.3 36. 15137, 17003, 20363, 26363, 31.147, 34330 37. 110.0, 99.88, 92.38 38. 42.1, 40.9, 39.8, 38.2, 38.1, 37.8, 39. 6.2, 6.6, 6.6, 6.8, 7.0, 6.6, 6.6, 7.2, 7.6, 8.0, 8.0 40. Trend values are 4.2, 5.6, 7; 8.4, 9.8 41. 84, 86, 88, 90, 92, 94, 96 42. 446.67, 546.67, 626.67, 706.67, 786.67, 866.67 43. 40.06, 47.40, 54.74, 62.08, 69.42, 76.76, 84.10, 91.44 44. 94.18, 105.82, 95.19, 105.32 45. 106.4, 98.7, 94.9,100 234 9. THEORY OF ATTRIBUTES 9.0 Introduction: Generally statistics deal with quantitative data only. But in behavioural sciences, one often deals with the variable which are not quantitatively measurable. Literally an attribute means a quality on characteristic which are not related to quantitative measurements. Examples of attributes are health, honesty, blindness etc. They cannot be measured directly. The observer may find the presence or absence of these attributes. Statistics of attributes based on descriptive character. 9.1 Notations: Association of attribute is studied by the presence or absence of a particular attribute. If only one attribute is studied, the population is divided into two classes according to its presence or absence and such classification is termed as division by dichotomy. If a class is divided into more than two scale-classes, such classification is called manifold classification. Positive class which denotes the presence of attribute is generally denoted by Roman letters generally A,B,… and the .etc negative class denoting the absence of the attribute and it is denoted by the Greek letters α, β… For example, A represents .etc the attribute ‘ Literacy’ and B represents ‘ Criminal’ . α and β represents the ‘ Illiteracy’ and ‘ Not Criminal’ respectively. 9.2 Classes and Class frequencies: Different attributes, their sub-groups and combinations are called different classes and the number of observations assigned to them are called their class frequencies. If two attributes are studied the number of classes will be 9. (i.e.,) (A) , (α), (B), (β), (A β) (α β), (α B) and N. 235 The chart given below illustrate it clearly. N (α) (A) (AB) (Aβ) (αB) (αβ) The number of observations or units belonging to class is known as its frequency are denoted within bracket. Thus (A) stands for the frequency of A and (AB) stands for the number objects possessing the attribute both A and B. The contingency table of order (2×2) for two attributes A and B can be displayed as given below A B β Total α Total (AB) (Aβ) (A) (αB) (αβ) (α) (B) (β ) N Relationship between the class frequencies: The frequency of a lower order class can always be expressed in terms of the higher order class frequencies. i.e., N = ( A ) + (α ) = (B) + (β) (A) = (AB) + (Aβ) (α) = (αB) + (αβ) (B) = (AB) + (α B) (β) = (Aβ) + (α β) If the number of attributes is n, then there will be 3n classes and we have 2n cell frequencies. 236 9.3 Consistency of the data: In order to find out whether the given data are consistent or not we have to apply a very simple test. The test is to find out whether any or more of the ultimate class-frequencies is negative or not. If none of the class frequencies is negative we can safely calculate that the given data are consistent (i.e the frequencies do not conflict in any way each other). On the other hand, if any of the ultimate class frequencies comes to be negative the given data are inconsistent. Example 1: Given N = 2500, (A) = 420, (AB) = 85 and Find the missing values. Solution: We know N = (A) +(α) = (B) + (β) (A) = (AB) + (Aβ) (α) = (αB) + (αβ) (B) = (AB) + (α B) (β) = (Aβ) + (α β) From (2) 420 = 85 + (Aβ) ∴ (Aβ) = 420 –85 (A β) = 335 From (4) 670 = 85 + (αB) ∴ (αB) = 670 − 85 (αB) = 585 From (1) 2500 = 420 + (α) ∴ (α) = 2500 − 420 (α) = 2080 From (1) (β) = 2500 −670 (β) = 1830 From (3) = 2080 = 585 + (αβ) ∴(αβ) = 1495 237 (B) = 670. Example 2: Test the consistency of the following data with the symbols having their usual meaning. N = 1000 (A) = 600 (B) = 500 (AB) = 50 Solution: A α Total B 50 450 500 β 550 -50 500 Total 600 400 1000 Since (αβ)) = −50, the given data is inconsistent. Example 3: Examine the consistency of the given data. N = 60 (A) = 51 (B) = 32 (AB) = 25 Solution: A α Total B 25 7 32 β 26 2 28 Total 51 9 60 Since all the frequencies are positive, it can be concluded that the given data are consistent. 9.4 Independence of Attributes: If the attributes are said to be independent the presence or absence of one attribute does not affect the presence or absence of the other. For example, the attributes skin colour and intelligence of persons are independent. 238 If two attributes A and B are independent then the actual frequency is equal to the expected frequency (A).(B) (AB) = N (α).(β) Similarly (α β ) = N 9.4.1 Association of attributes: Two attributes A and B are said to be associated if they are not independent but they are related with each other in some way or other. The attributes A and B are said to be positively associated if (A).(B) (AB) > N (A).(B) If (AB) < ,. then they are said to be negatively associated. N Example 4: Show that whether A and B are independent, positively associated or negatively associated. (AB) = 128, (αB) = 384, (Aβ) = 24 and (αβ) = 72 Solution: (A) = (AB) + (Aβ) = 128 + 24 (A) = 152 (B) = (AB) + (αB) = 128 +384 (B) = 512 (α) = (αB) + (αβ) = 384 + 72 ∴(α) = 456 (N) = (A) + (α) = 152 + 456 = 608 239 (A) × (B) 152 × 512 = N 608 = 128 (AB) = 128 (A) × (B) ∴ (AB) = N Hence A and B are independent Example 5: From the following data, find out the types of association of A and B. 1) N = 200 (A) = 30 (B) = 100 (AB) = 15 2) N = 400 (A) = 50 (B) = 160 (AB) = 20 3) N = 800 (A) = 160 (B) = 300 (AB) = 50 Solution: (A).(B) N (30)(100) = = 15 200 Since the actual frequency is equal to the expected frequency, ie 15 = 15, therefore A and B are independent. 1. Expected frequency of (AB) = (A).(B) N (50)(160) = = 20 400 Since the actual frequency is greater than expected frequency. i.e., 25 > 20, therefore A and B are positively associated. 2. Expected frequency of (AB) = 3. Expected frequency of (AB) = (A).(B) (160)(300) = = 60 N 800 Since Actual frequency is less than expected frequency i.e., 50 < 60 therefore A and B are negatively associated. 240 9.5 Yules’ co-efficient of association: The above example gives a rough idea about association but not the degree of association. For this Prof. G. Undy Yule has suggested a formula to measure the degree of association. It is a relative measure of association between two attributes A and B. If (AB), (αB), (Aβ) and (αβ) are the four distinct combination of A, B, α and β then Yules’ co-efficient of association is (AB)(αβ) − (Aβ).(αB) Q= (AB)(αβ) + (Aβ).(αB) Note: I. If Q = +1 there is perfect positive association If Q = -1 there is perfect negative association If Q = 0 there is no association (ie) A and B are independent 1. For rememberance of the above formula , we use the table below A α B AB αB β Aβ αβ Example 6: Investigate the association between darkness of eye colour in father and son from the following data. Fathers’ with dark eyes and sons’ with dark eyes = 50 Fathers’ with dark eyes an sons’ with no dark eyes = 79 Fathers’ with no dark eyes and sons with dark eyes = 89 Neither son nor father having dark eyes = 782 Solution: Let A denote the dark eye colour of father and B denote dark eye colour of son. A Total α B 50 89 139 79 782 861 β Total 129 871 1000 241 Yules’ co-efficient of association is (AB)(αβ) − (Aβ).(αB) Q= (AB)(αβ) + (Aβ).(αB) 50 × 782 − 79 × 89 = 50 × 782 + 79 × 89 32069 = = 0.69 46131 ∴ there is a positive association between the eye colour of fathers’ and sons’ . Example 7 : Can vaccination be regarded as a preventive measure of small pox from the data given below. Of 1482 persons in a locality, exposed to small pox, 368 in all were attacked, among the 1482 persons 343 had been vaccinated among these only 35 were attacked. Solution: Let A denote the attribute of vaccination and B denote that of attacked. A Total α B 35 333 368 308 806 1114 β Total 343 1139 1482 Yules’ co-efficient of association is (AB)(αβ) − (Aβ).(αB) Q= (AB)(αβ) + (Aβ).(αB) 35 × 806 − 308 × 333 = 35 × 806 + 308 × 333 −74354 = = − 0.57 130774 i.e., there is a negative association between attacked and vaccinated. In other words there is a positive association between not attacked and vaccinated. Hence vaccination can be regarded as a preventive measure for small pox. 242 Example 8: In a co-educational institution, out of 200 students, 150 were boys. They took an examination and it was found that 120 passed, 10 girls failed. Is there any association between sex and success in the examination. Solution: Let A denote boys and α denote girls. Let B denote those who passed the examination and β denote those who failed. We have given N = 200 (A) = 150 (AB) = 120 (αβ) = 10 Other frequencies can be obtained from the following table α 40 10 50 A 120 30 150 B β Total Total 160 40 200 Yule’ s co-efficient of association is (AB)(αβ) − (Aβ).(αB) Q= (AB)(αβ) + (Aβ).(αB) 120 ×10 − 30 × 40 = =0 120 × 10 + 30 × 40 Therefore, there is no association between sex and success in the examination. Recall (A) (B) denote positive attributes (α) (β) denote negative attributes 2 ×2 contingency table. X A Total α B (AB) (αB) (B) β (Aβ) (αβ) (β ) Total (A) (α) N 243 Vertical Total (AB) + (Aβ) = (A) (αB) + (αβ) = (α) (A) + (α) = N Types of Association Horizontal Total (AB) + (αB) = B (Aβ) + (αβ) = β (B) + (β) = N (A).(B) N (A).(B) Negative Association if (AB) < N (A).(B) Independent if (AB) = N Yule’ s co-efficient of Association (AB)(αβ) − (Aβ).(αB) Q= (AB)(αβ) + (Aβ).(αB) Positive Association if (AB) > Exercise – 9 I. Choose the best answer: 1. Measures of association in usually deal with (a) Attributes (b) Quantitative factors (c) Variables (d) Numbers 2. The frequency of class can always be expressed as a sum of frequencies of (a) Lower order classes (b) Higher order classes (c) Zero order classes (d) None of the above 3. With the two attributes the total number of class frequencies is (a) Two (b) Four (c) Eight (d) Nine (A).(B) 4. If for two the attributes are A and B, (AB) > the N attributes are (a) Independent (b) Positively associated (c) Negatively associated (d) No conclusion 244 5. In case of two attributes A and B the class frequency (AB) = 0 the value of Q is (a) 1 (b) − 1 (c) 0 (d) −1 ≤ Q ≤ 1 II. Fill in the blanks: 6. If an attribute has two classes it is said to be ____________ 7. In case of consistent data, no class frequency can be _________ 8. If A and B are independent Yule’ s co-efficient is equal to ________ 9. If A and B are negatively associated then __________ 10. If N = 500, (A) = 300, (B) = 250 and (AB) = 40 the data are ________ III. Answer the following: 11. Give a brief idea of notations used in classification of attributes 12. How can the frequencies for various attributes be displayed in contingency table 13. What do you understand by consistency of data. 14. Write briefly about association of attributes. 15. Give Yule’ s co-efficient of association IV. Problems 16. For two attributes A and B, we have (AB) = 35, (A) = 55; N=100 and (B) = 65. Calculate the missing values. 17. From the following ultimate class frequencies, find the frequencies of positive and negative classes and the total number of observations. (AB) = 9, (Aβ) = 14, (αB) = 4 and (αβ) = 37 18. Verify whether the given data N = 100, (A) = 75, (B) = 60 and (AB) = 15 are consistent. 19. Find whether A and B are independent in the following data (AB) = 256 (αB) = 768 (Aβ) = 48 (αβ) = 144 20. In a report on consumer’ s preference it was given that out of 500 persons surveyed 410 preferred variety A 380 preferred 245 variety B and 270 persons linked both. Are the data consistent? 21. For two attributes A and B, we have (AB) = 35, (A) = 55, N=100, (αβ) = 20. Calculate the Yule’ s co-efficient of association. 22. Given N = 1500, (A) = 383, (B) = 360 and (AB) = 35. Prepare 2 × 2 contingency table and compute Yule’ s coefficient of association and interpret the result. 23. In an experiment on immunization of cattle from tuberculosis the following results were obtained. Affected Unaffected Inoculated 12 26 Not inoculated 16 6 By calculating Yule’ s co-efficient of association, examine the effect of vaccine is in controlling the disease. 24. Calculate the co-efficient of association between the intelligence of fathers and sons from the following data Intelligent fathers with intelligent sons = 300 Intelligent fathers with dull sons = 100 Dull fathers with intelligent sons = 50 Dull fathers with dull sons = 500 25. Out of 3000 unskilled workers of a factory, 2000 come from rural area and out of 1200 skilled workers 300 come from rural area. Determine the association between skill and residence 26. In an anti-malarial campaign in a certain area, quinine was administrated to 812 persons out of a total population of 3428. The number of fever cases is shown below: Treatment Quinine No quinine Fever 20 220 No Fever 792 2216 Examine the effect of quinine on controlling malaria. 27. 1500 candidates appeared for competitive examinations 425 were successful. 250 had attended a coaching class and of 246 these 150 came out successful. Estimate the utility of the coaching class. 28. In an examination at which 600 candidates appeared of them 348 were boys. Number of passed candidates exceeded the number of failed candidates by 310. Boys failing in the examination numbered 88. Find the coefficient of association between male sex and success in examination. 29. Following data relate to literacy and unemployment in a group of 500 persons. Calculate Yule’ s co-efficient of association between literacy and unemployment and interpret it Literate unemployed = 220 Literate employed = 20 Illiterate Employed = 180 30. In a group of 400 students, the number of married is 160. Out of 120 students who failed 48 belonged to the married group. Find out whether the attributes of marriage and failure are independent. Answers I. 1. (a) 2. (b) 3. (d) II. 6. Dichotomy (A).(B) 9. AB < N 4. (b) 7. Negative 8. 0 10. Inconsistent IV. 16 B β Total A 35 20 55 α 30 15 45 247 5. (b) Total 65 35 100 17. A α B 9 4 14 37 β Total 23 41 Total No of observations = 64 Total 13 51 64 18. Inconsistent 19. A and B are independent 20. Inconsistent 21. 0.167 22. – 0.606, Negative association 23. − 0.705, Vaccine is effective 24. + 0.935 25. Negative association between skill and residence. 26. – 0.59. Negative association ∴quinine is effective. 27. + 0.68. Coaching class are useful 28. – 0.07 29. 0.92 Positive association between literacy and unemployment 30. Q = 0, Marriage and failure are independent. 248 10. DECISION THEORY 10. 0 Introduction: Decision theory is primarily concerned with helping people and organizations in making decisions. It provides a meaningful conceptual frame work for important decision making. The decision making refers to the selection of an act from amongst various alternatives, the one which is judged to be the best under given circumstances. The management has to consider phases like planning, organization, direction, command and control. While performing so many activities, the management has to face many situations from which the best choice is to be taken. This choice making is technically termed as “decision making” or decision taking. A decision is simply a selection from two or more courses of action. Decision making may be defined as - “ a process of best selection from a set of alternative courses of action, that course of action which is supposed to meet objectives upto satisfaction of the decision maker. The knowledge of statistical techniques helps to select the best action. The statistical decision theory refers to an optimal choice under condition of uncertainty. In this case probability theory has a vital role, as such, this probability theory will be used more frequently in the decision making theory under uncertainty and risk. The statistical decision theory tries to reveal the logical structure of the problem into alternative action, states of nature, possible outcomes and likely pay-offs from each such outcome. Let us explain the concepts associated with the decision theory approach to problem solving. The decision maker: The decision maker refers to individual or a group of individual responsible for making the choice of an appropriate course of action amongst the available courses of action. 249 Acts (or courses of action): Decision making problems deals with the selection of a single act from a set of alternative acts. If two or more alternative courses of action occur in a problem, then decision making is necessary to select only one course of action. Let the acts or action be a1, a2, a3,…then the totality of all these actions is known as action space denoted by A. For three actions a1, a2 a3; A = action space = (a1, a2, a3) or A = (A1, A2, A3). Acts may be also represented in the following matrix form i.e., either in row or column was Acts A1 Acts A1 A2 … An A2 . . An In a tree diagram the acts or actions are shown as A1 Start A2 A3 Events (or States of nature): The events identify the occurrences, which are outside of the decision maker’ s control and which determine the level of success for a given act. These events are often called ‘ States of nature’ or outcomes. An example of an event or states of nature is the level of market demand for a particular item during a stipulated time period. A set of states of nature may be represented in any one of the following ways: 250 S = {S1, S2, … n} ,S or E = {E1, E2, … n} ,E or = {θ1, θ2, θ3} For example, if a washing powder is marketed, it may be highly liked by outcomes (outcome θ1) or it may not appeal at all (outcome θ2) or it may satisfy only a small fraction, say 25% (outcome θ3) ∴ = { θ1 , θ2, θ3} In a tree diagram the places are next to acts. We may also get another act on the happening of events as follows: Acts Events E1 A1 E2 E1 A2 E2 E1 A3 E2 In matrix form, they may be represented as either of the two ways: States of nature S1 Acts A1 A2 OR 251 S2 Acts A1 A2,… n ,A States of nature S1 S2 10.1 Pay-off: The result of combinations of an act with each of the states of nature is the outcome and momentary gain or loss of each such outcome is the pay-off. This means that the expression pay-off should be in quantitative form. Pay -off may be also in terms of cost saving or time saving. In general, if there are k alternatives and n states of nature, there will be k × n outcomes or pay-offs. These k × n pay-offs can be very conveniently represented in the form of a k × n pay -off table. States of nature E1 E2 . . . En A1 a11 a21 . . . an1 Decision alternative A2 ……… Ak …… a12 ……… a1k …… a22 ……… a2k …… . ……… . …… . ……… . …… . .……… . …… an2 ……… ank …… where aij = conditional outcome (pay-off) of the ith event when jth alternative is chosen. The above pay-off table is called pay-off matrix. For example, A farmer can raise any one of three crops on his field. The yields of each crop depend on weather conditions. We have to show pay –off in each case, if prices of the three products are as indicated in the last column of yield matrix. 252 Dry (E1) Damp (E3) Price Rs.per . kg Paddy (A1) 500 1700 4500 1.25 Gound nut (A2) 800 1200 1000 4.00 Tobacco (A3) Yield in kg per hectare Weather Moderate (E2) 100 300 200 15.00 Solution: Pay - off Table E1 E2 E3 A1 500 ×1.25 = 625 1700 ×1.25= 2125 4500 ×1.25 = 5625 A2 800 × 4 = 3200 1200 ×4 = 4800 1000 × 4 A3 100 × 15 = 1500 300 × 15 = 4500 200 × 15 = 3000 = 4000 10.1.1 Regret (or Opportunity Loss): The difference between the highest possible profit for a state of nature and the actual profit obtained for the particular action taken is known as opportunity loss. That is an opportunity loss is the loss incurred due to failure of not adopting the best possible course of action. Opportunity losses are calculated separately for each state of nature. For a given state of nature the opportunity loss of possible course of action is the difference between the pay-off value for that course of action and the pay-off for the best possible course of action that could have been selected. Let the pay-off of the outcomes in the 1st row be P11,P12…… 1n and similarly for the other rows. …P 253 Pay-off table States of nature S1 S2 ……Sn … P11 P12 …… 1n …P P21 P22 ……P2n … . . . Acts A1 A2 . . . . . . . . . Am Pm1 Pm2 …… mn …P Consider a fixed state of nature Si. The pay-off corresponding to the n strategies are given by Pi1, Pi2,… in. Suppose ,P Mi is the maximum of these quantities. The Pi1 if A1 is used by the decision maker there is loss of opportunity of M1 – Pi1, and so on Then a table showing opportunity loss can be computed as follows: Regret (or opportunity loss table) States of nature Acts S1 S2 … Sn A1 M1- P11 M2 - P12 …Mn - P1n M1- P21 M2 - P22 … Mn - P2n A2 . . . . . . . . . . . . Am M1-Pm1 M2 - Pm2 … Mn - Pmn Types of decision making: Decisions are made based upon the information data available about the occurrence of events as well as the decision situation. There are three types of decision making situations: certainty , uncertainty and risk. Decision making under certainty: In this case the decision maker has the complete knowledge of consequence of every decision choice with certainty. In this 254 decision model, assumed certainty means that only one possible state of nature exists. Example 1: A canteen prepares a food at a total average cost of Rs. 4 per plate and sells it at a price of Rs 6. The food is prepared in the morning and is sold during the same day. Unsold food during the same day is spoiled and is to be thrown away. According to the past sale, number of plates prepared is not less than 50 or greater than 53. You are to formulate the (i) action space (ii) states of nature space (iii) pay-off table (iv) loss table Solution: (i) The canteen will not prepare less than 50 plates or more than 53 plates. Thus the acts or courses of action open to him are a1 = prepare 50 plates a2 = prepare 51 plates a3 = prepare 52 plates a4 = prepare 53 plates Thus the action space is A = {a1, a2 , a3, a4} (ii) The state of nature is daily demand for food plates. Then are four possible state of nature ie S1 = demand is 50 plates S2 = demand is 51 plates S3 = demand is 52 plates S4 = demand is 53 plates Hence the state of nature space, S= {S1, S2, S3, S4} iii) The uncertainty element in the given problem is the daily demand. The profit of the canteen is subject to the daily demand. Let n = quantity demanded m = quantity produced For n ≥ m, profit = (Cost price – Selling price) x m = (6 – 4) x m = 2m For m > n, 255 profit ={(Cost price – Selling price) x n}- {Cost price x (m-n) } = 2n – 4 (m-n) = 6n – 4m Supply (m) ( a1 ) ( a2 ) ( a3 ) ( a4 ) 50 51 52 53 (S1) 50 100 96 92 88 Pay-off table Demand (n) (S2) (S3) 51 52 100 100 102 102 98 104 94 100 (S4) 53 100 102 104 106 (iv) To calculate the opportunity loss we first determine the maximum pay-off in each state of nature. In this state First maximum pay-off = 100 Second maximum pay-off = 102 Third maximum pay-off = 104 Fourth maximum pay-off = 106 Loss table corresponding to the above pay-off table Demand (n) (S1) (S2) (S3) (S4) Supply 50 51 52 53 (m) (a1) 50 100 -100 = 0 102-100 = 2 104-100 = 4 106 -100 = 6 (a2) 51 100 - 96 = 4 102-102 = 0 104-102 = 2 106 -102 = 4 (a3) 52 100 - 92 = 8 102 - 98 = 4 104-104 = 0 106 -104 = 2 (a4) 53 100 - 88 =12 102 - 94 = 8 104-100 = 4 106 -106 = 0 10.2 Decision making under uncertainty (without probability): Under conditions of uncertainty, only pay-offs are known and nothing is known about the lilkelihood of each state of nature. Such situations arise when a new product is introduced in the 256 market or a new plant is set up. The number of different decision criteria available under the condition of uncertainty is given below. Certain of optimism (Maximax ): The maximax criterion finds the course of action or alternative strategy that maximizes the maximum pay-off. Since this decision criterion locates the alternative with the highest possible gain, it has also been called an optimistic decision criterion. The working method is (i) Determine the best outcome for each alternative. (ii) Select the alternative associated with the best of these. Expected Monetary value (EMV): The expected monetary value is widely used to evaluate the alternative course of action (or act). The EMV for given course of action is just sum of possible pay-off of the alternative each weighted by the probability of that pay-off occurring. The criteria of pessimism or Maximin: This criterion is the decision to take the course of action which maximizes the minimum possible pay-off. Since this decision criterion locates the alternative strategy that has the least possible loss, it is also known as a pessimistic decision criterion. The working method is: 1) Determine the lowest outcome for each alternative. 2) Choose the alternative associated with the best of these. Minimax Regret Criterion (Savage Criterion): This criterion is also known as opportunity loss decision criterion because decision maker feels regret after adopting a wrong course of action (or alternative) resulting in an opportunity loss of pay-off. Thus he always intends to minimize this regret. The working method is (a) Form the given pay-off matrix, develop an opportunity loss (or regret) matrix. (i) find the best pay-off corresponding to each state of nature and (ii) subtract all other entries (pay-off values) in that row from this value. 257 (b) Identify the maximum opportunity loss for each alternatives. (c) Select the alternative associated with the lowest of these. Equally likely decision (Baye’ s or Laplace)Criterion: Since the probabilities of states of nature are not known, it is assumed that all states of nature will occur with equal probability. ie., each state of nature is assigned an equal probability. As states of nature are mutually exclusive and collectively exhaustive, so the probability of each these must be 1 /(number of states of nature). The working method is (a) Assign equal probability value to each state of nature by using the formula: 1/(number of states of nature) (b) Compute the expected (or average) value for each alternative by multiplying each outcome by its probability and then summing. (c) Select the best expected pay-off value (maximum for profit and minimum for loss) This criterion is also known as the criterion of insufficient reason because, expect in a few cases, some information of the likelihood of occurrence of states of nature is available. Criterion of Realism (Hurwicz Criterion): This criterion is a compromise between an optimistic and pessimistic decision criterion. To start with a co-efficient of optimism α (0 ≤ α ≤ 1) is selected. When α is close to one, the decision maker is optimistic about the future and when α is close to zero, the decision maker is pessimistic about the future. According to Hurwicz , select strategy which maximizes H = α (maximum pay-off in row) + (1 - α) minimum pay-off in row. Example 2: Consider the following pay-off (profit) matrix 258 States Action (S1) (S2) (S3) (S4) A1 5 10 18 25 A2 8 7 8 23 A3 21 18 12 21 A4 30 22 19 15 No Probabilities are known for the occurrence of the nature states . Compare the solutions obtained by each of the following criteria: (i) Maximin (ii) Laplace (iii) Hurwicz (assume that α = 0.5) Solution: i) Maximin Criterion: A1: 5 10 A2: 8 7 A3: 21 18 A4: 30 22 Best action is A4 ii) Laplace criterion 18 8 12 19 Minimum 5 7 12 15 maximum 25 23 21 15 E(A1) = 1/4 [5 +10+18+25] = 14.5 E(A2) = 1/4 [8 +7+8+23] = 11.5 E(A3) = 1/4 [21 +18+12+21] = 18.0 E(A4) = 1/4 [30 +22+19+15] = 21.5 maximum E(A4) is maximum. So the best action is A4 iv) Hurwicz Criterion (with α = 0.5) Minimum Maximum α (max) + (1-α) min A1 5 25 0.5(25) + 0.5(5) A2 7 23 0.5(7) + 0.5 (23) = 15 A3 12 21 0.5(12) + 0.5 (21) = 16.5 A4 15 30 0.5(15) + 0.5 (30) = 22.5 maximum Best action is A4 259 = 15 Example 3: Suppose that a decision maker faced with three decision alternatives and two states of nature. Apply (i) Maximin and (ii) Minimax regret approach to the following pay-off table to recommend the decisions. States of Nature Act A1 A2 A3 S1 S2 10 20 30 15 12 11 Solution: (i) Maximin Act Minimum A1 10 12 maximum A2 A3 11 Act A2 is recommended ii) Minimax regret States of Nature S1 S2 Maximum Regret Act A1 30-10 = 20 15-15 = 0 20 A2 30-20 = 10 15-12 = 3 10 A3 30-30 = 0 15-11 = 4 4 Minimum of the maximum regrets is 4 which corresponds to the act A3. So the act A3 is recommended Example 4: A business man has to select three alternatives open to him each of which can be followed by any of the four possible events. The conditional pay-off (in Rs) for each action event combination are given below: 260 Alternative X Y Z A 8 -4 14 Pay-offs conditional events B C 0 -10 12 18 6 0 D 6 -2 8 Determine which alternative should the businessman choose, if he adopts the a) Maximin criterion b) Maximax criterion c) Hurwicz criterion with degree of optimism is 0.7 d) Minimax regret Criterion e) Laplace criterion Solution: For the given pay-off martrix, the maximum assured and minimum possible pay-off for each alternative are as given below. Minimum (α =0.7) Alternative Maximum pay-off pay-off H = α (maximum pay-off) (Rs) (Rs) + (1- α)(minimum pay-off) X 8 -10 2.6 Y 18 -4 11.4 Z 14 0 9.8 a) Since z yields the maximum of the minimum pay-off, under maximin criterion, alternative z would be chosen. b) Under maximax criterion, the businessman would choose the alternative Y. c) It will be optimal to choose Y under Hurwicz Criterion. d) For the given pay-off matrix, we determine the regrets as shown below, when the regret payoffs amounts when event A occurs, are computed by the relation Regret pay-off = maximum pay-offs from A – pay-off. Similarly for the other events. 261 Alternative X Y Z Maximum pay-off Pay- off amount AB C 8 0 -10 - 4 12 18 14 6 0 14 12 18 D Regret pay-off amount A BCD 66 - 2 18 8 0 8 12 28 00 6 18 2 10 0 Maximum Regret 28 18 18 Since alternative Y and Z both corresponding to the minimal of the maximum possible regrets, the decision maker would choose either of these two (e) Laplace Criterion In this method assigning equal probabilities to the pay-off of each strategy, results in the following expected pay-off. Pay-off Expected pay-off value Alternative A B CD P=1/4 P=1/4 P=1/4 P=1/4 X ¼[ 8 + 0 –10 + 6] = 1 8 0 −10 6 Y - 4 12 18 - 2 ¼[- 4 + 12 +18 -2] = 6 Z 14 6 0 8 ¼[ 14 + 6 +0 + 8] = 7 Since the expected pay-off value for z is the maximum the businessman would choose alternative z 10.3 Decision making under risk (with probability): Here the decision maker faces many states of nature. As such, he is supposed to believe authentic information, knowledge, past experience or happenings to enable him to assign probability values to the likelihood of occurrence of each state of nature. Sometimes with reference to past records, experience or information, probabilities to future events could be allotted. On the basis of probability distribution of the states of nature, one may select the best course of action having the highest expected pay-off value. 262 Example 5: The pay-off table for three courses of action (A) with three states of nature (E) (or events) with their respective probabilities (p) is given. Find the best course of action. Events Probability Acts A1 A2 A3 E1 E2 E3 0.2 0.5 0.3 2 3 4 1 2 2 -1 0 1 The expected value for each act is A1 : 2(0.2) + 1(0.5) - 1(0.3) = 0.6 A2 : 3(0.2) + 2(0.5) + 0(0.3) = 1.6 A3 : 4(0.2) + 2(0.5) + 1(0.3) = 2.1 The expected monetary value for the act 3 is maximum. Therefore the best course of action is A3. Example 6: Given the following pay-off of 3 acts: A1, A2, A3 and their events E1, E2, E3. Act A1 A2 A3 States of Nature E1 E2 E3 35 200 550 -10 240 640 -150 200 750 The probabilities of the states of nature are respectively 0.3, 0.4 and 0.3. Calculate and tabulate EMV and conclude which of the acts can be chosen as the best. 263 Solution: Events Prob. A1 A2 A3 35 × 0.3 = 10.5 -10 × 0.3 = -3 -150 × 0.3 = - 45 E1 0.3 200 × 0.4 = 80.0 240 × 0.4 = 96 200 × 0.4 = 80 E2 0.4 E3 0.3 550 ×0.3 = 165.0 640 × 0.3 = 192 750 × 0.3 = 225 285 260 255.5 EMV The EMV of A2 is maximum, therefore to choose A2 Example 7: A shop keeper has the facility to store a large number of perishable items. He buys them at a rate of Rs.3 per item and sells at the rate of Rs.5 per item. If an item is not sold at the end of the day then there is a loss of Rs.3 per item. The daily demand has the following probability distribution. Number of 3 4 5 6 Items demanded Probability : 0.2 0.3 0.3 0.2 How many items should he stored so that his daily expected profit is maximum? Solution: Let m = number of items stocked daily n = number of items demanded daily Now, for n ≥ m, profit = 2m And for m > n, profit = 2n – 3(m-n) = 2n – 3m + 3n = 5n – 3m Stock (m) 3 4 5 6 Probability Pay - off table Demand (n) 3 4 5 6 3 0 −3 0.2 6 8 5 2 0.3 264 6 8 10 7 0.3 6 6 8 10 12 0.2 Stock(m) 3 4 5 6 Thus stocked. So, daily. Expected gain 6 × 0.2 + 6 × 0.3 + 6 × 0.3 + 6 × 0.2 = Rs. 6.00 3 × 0.2 + 8 × 0.3 + 8 × 0.3 + 8 × 0.2 = Rs. 7.00 0 × 0.2 + 5 × 0.3 + 10 × 0.3 + 10 × 0.2 = Rs. 6.50 −3 × 0.2 + 2 × 0.3 + 7 × 0.3 + 12 × 0.2 = Rs. 4.50 the highest expected gain is Rs 7.00 when 4 units he can store 4 items to get maximum expected profit Example 8: A magazine distributor assigns probabilities to the demand for a magazine as follows: Copies demanded : 2 3 4 5 Probability : 0.4 0.3 0.2 0.1 A copy of magazine which he sells at Rs.8 costs Rs6. How many should he stock to get the maximum possible expected profit if the distributor can return back unsold copies for Rs.5 each? Solution: Let m = no of magazines stocked daily n = no of magazines demanded Now, For n ≥ m, profit = Rs 2m and for m > n, profit = 8n –6m +5(m-n) = 8n –6m +5m – 5n = 3n –m Stock (m) Pay-off table Demand (n) 3 4 4 4 2 2 4 5 4 3 3 6 6 6 4 2 5 8 8 5 1 4 265 7 10 Probability Stock 2 3 4 5 0.4 0.3 0.2 0.1 Expected Profit (in Rs) 4 × 0.4 + 4 × 0.3 + 4 × 0.2 + 4 × 0.1 = 4.0 3 × 0.4 + 6 × 0.3 + 6 × 0.2 + 6 × 0.1 = 4.8 2 × 0.4 + 5 × 0.3 + 8 × 0.2 + 8 × 0.1 = 4.7 1 × 0.4 + 4 × 0.3 + 7 × 0.2 + 10 ×0.1= 4.0 Thus the highest expected profit is Rs. 4.8, when 3 magazines stocked. So, the distributor can stock 3 magazines to get the maximum possible expected profit. 10.4 Decision Tree Analysis: A decision problem may also be represented with the help of a diagram. It shows all the possible courses of action, states of nature, and the probabilities associated with the states of nature. The ‘ decision diagram’ looks very much like a drawing of a tree, therefore also called ‘ decision tree’ . A decision tree consists of nodes, branches, probability estimates and pay-offs. Nodes are of two types, decision node (designated as a square) and chance node (designated as a circle). Alternative courses of action originate from decision node as the main branches (decision branches). Now at the terminal point of decision node, chance node exists from where chance nodes, emanate as sub-branches. The respective pay-offs and the probabilities associated with alternative courses, and the chance events are shown alongside the chance branches. At the terminal of the chance branches are shown the expected pay-off values of the outcome. There are basically two types of decision trees-deterministic and probabilistic. These can further be divided into single stage and multistage trees. A single stage deterministic decision tree involves making only one decision under conditions of certainty (no chance events). In a multistage deterministic tree a sequence or chain of decisions are to be made, The optimal path (strategy) is one that corresponds to the maximum EMV. 266 In drawing a decision tree, one must follow certain basic rules and conventions as stated below: 1. Identify all decisions (and their alternatives) to be made and the order in which they must be made. 2. Identify the chance events or state of nature that might occur as a result of each decision alternative. 3. Develop a tree diagram showing the sequence of decisions and chance events. The tree is constructed starting from left and moving towards right. The square box denotes a decision point at which the available courses of action are considered. The circle O represents the chance node or event, the various states of nature or outcomes emanate from this chance event. 4. Estimate probabilities that possible events or states of nature will occur as a result of the decision alternatives. 5. Obtain outcomes (usually expressed in economic terms) of the possible interactions among decision alternatives and events. 6. Calculate the expected value of all possible decision alternatives. 7. Select the decision alternative (or course of action) offering the most attractive expected value Advantages of decision tree: 1. By drawing of decision tree, the decision maker will be in a position to visualise the entire complex of the problem. 2. Enable the decision - maker to see the various elements of his problem in content and in a systematic way. 3. Multi-dimensional decision sequences can be strung on a decision tree without conceptual difficulties. 4. Decision tree model can be applied in various fields such as introduction of a new product, marketing strategy etc… Example 9: A manufacturing company has to select one of the two products A or B for manufacturing. Product A requires investment 267 of Rs.20,000 and product B Rs 40,000. Market research survey shows high, medium and low demands with corresponding probabilities and returns from sales in Rs. Thousand for the two products in the following table. Market demand High Medium Low Probability A 0.4 0.3 0.3 B 0.3 0.5 0.2 Return from sales A B 50 80 30 60 10 50 Construct an appropriate decision tree. What decision the company should take? Market demand High Medium Low Total A X(‘ 000) P 50 0.4 30 0.3 10 0.3 268 PX X(‘ 000) 20 80 9 60 3 50 32 B P 0.3 0.5 0.2 PX 24 30 10 64 Product Return (Rs) Investment(Rs) Profit (Rs) A 32,000 20,000 12,000 B 64,000 40,000 24,000 Since the profit is high in case of product B, so the company’ s decision in favour of B. Example 10: A farm owner is considering drilling a farm well. In the past, only 70% of wells drilled were successful at 20 metres of depth in that area. Moreover, on finding no water at 20 metres, some person drilled it further up to 25 metres but only 20% struck water at 25 metres. The prevailing cost of drilling is Rs.500 per metres. The farm owner has estimated that in case he does not get his own well, he will have to pay Rs.15000 over the next 10 years to buy water from the neighbour. Draw an appropriate decision tree and determine the farm owner’ s strategy under EMV approach. Solution: The given data is represented by the following decision tree diagram. Decision Event 1. Drill upto 25 metres Water struck No water struck Probability Cash out flows Decision at point D2 0.2 Rs.12500 0.8 Expected cash out flow Rs 2500 Rs 22000 EMV(out flows) 269 Rs27500 Rs.24500 2. Do not drill EMV (out flow) = Rs.25000 The decision at D2 is : Drill up to 25 metres Decision at point D1 0.7 Rs.10000 Rs 7000 Water 1. Drill upto 20 struck metres 0.3 Rs.24500 Rs.7350 No water struck EMV(out Rs.14,350 flows) 2. Do not EMV (out flow) = Rs.15000 drill The decision at D1 is : Drill up to 20 metres. Thus the optimal strategy for the farm-owner is to drill the well up to 20 metres. Exercise – 10 I. Choose the correct answers: 1. Decision theory is concerned with (a) The amount of information that is available (b) Criteria for measuring the ‘ goodness’ of a decision (c) Selecting optimal decisions in sequential problems (d) All of the above 2. Which of the following criteria does not apply to decision – making under uncertainly (a) Maximin return (b) Maximax return (c) Minimax return (d) Maximize expected return 3. Maximin return, maximax return and minimax regret are criteria that (a) Lead to the same optimal decision. (b) Cannot be used with probabilities (c) Both a and b (d) None of the above 270 4. Which of the following does not apply to a decision tree? (a) A square node is a point at which a decision must be made. (b) A circular node represents an encounter with uncertainty. (c) One chooses a sequence of decisions which have the greatest probability of success. (d) One attempts to maximize expected return. 5. The criterion which selects the action for which maximum pay-off is lowest is known as (a) Max-min criterion (b) Min-max criterion (c) Max –max criterion (d) None of these II. Fill in the blanks: 6. Decision trees involve ____________ of decisions and random outcomes. 7. One way to deal with decision making in the ‘ uncertainity’ context is to treat all states of nature as ___________ and maximize expected return. 8. Maximizing expected net rupee return always yields the same optimal policy as ___________ expected regret. 9. The different criteria for making decisions under risk always yields the same _________ choice. 10. In decision under uncertainty, the Laplace criterion is the least conservative while the ___________ criterion is the most conservative. III. Answer the following: 11. Explain the meaning of ‘ statistical decision theory’ 12. What techniques are used to solve decision making problems under uncertainty? 13. Write a note on decision tree. 14. What is a pay-off matrix? 15. Describe how you would determine the best decision using the EMV criterion with a decision tree. 271 IV. Problems: 16. The pay-off table for three courses of action (A) with three states of nature (E) (or events) with their respective probabilities (P) are given. Find the best course of action. Events Acts A1 A2 A3 Probability E1 E2 E3 2.5 4.0 3.0 0.2 2.0 2.6 1.8 0.6 −1 0 1 0.2 17. Calculate EMV and thus select the best act for the following pay-off table: States of nature X Y Z Probability Pay-off (Rs) by the player A B C 0.3 20 −2 −5 0.4 20 −10 −5 0.3 40 60 30 18. Consider the pay-off matrix States of nature Probability High demand Medium demand Low demand 0.4 Act A1 do Act A2 Expand not 200 units expand 2500 3500 Act A3 Expand 400 units 5000 0.4 2500 3500 2500 0.2 2500 1500 1000 272 Using EMV criterion decide the best act. 19. Apply (i) maximin (ii) minimax regret to the following pay-off matrix to recommended the decisions without any knowledge of probability. States of nature Act S1 S2 S3 a1 14 8 10 a2 11 10 7 a3 9 12 13 20. A shop keeper of some highly perishable type of fruits sees that the daily demand X of this fruit in his area the following probability distribution. Daily Demand (in Dozen) : 6 7 8 9 Probability : 0.1 0.3 0.4 0.2 He sells for Rs10.00 a dozen while he buys each dozen at Rs4.00. Unsold fruits in a day are traded on the next day at Rs.2.00 per dozen, assuming that the stocks the fruits in dozen, how many should he stock so that his expected profit will be maximum? [Hint: profit = 6m for n ≥ m = 10n – 4m +2(m-n) = 8n – 2m for n< m] 21. A florist, in order to satisfy the needs of a number of regular and sophisticated customers, stocks a highly perishable flowers. A dozen flowers cost Rs 3 and sell at Rs10.00 Any flower not sold on the day are worthless. Demand distribution in dozen of flowers is as follows: Demand 1 2 3 4 Probability 0.2 0.3 0.3 0.2 How many flowers should he stock daily in order to maximize his expected net profit? 273 22. A florist stock highly perishable flower. A dozen of flower costs Rs3.00 and sells for Rs.10.00 Any flower not sold the day are worthless. Demand in dozen of flowers is as follows: Demand in 0 1 2 3 4 dozen Probability 0.1 0.2 0.4 0.2 0.1 Assuming that failure to satisfy any one customer’ s request will result in future lost profit amounting to Rs.5.00, in addition to the lost profit on the immediate sale, how many flowers should the florist stock to expect maximum profit? 23. A newspaper agent’ s experience shows that the daily demand x of newspaper in his area has the following probability distribution Daily Demand(x) Probability 300 400 500 0.1 0.3 0.4 600 0.1 700 0.1 He sales the newspapers for Rs.2.00 each while he buys each at Rs.1.00. Unsold copies are treated as scrap and each such copy fetches 10 paisa. Assuming that he stocks the news papers in multiple of 100 only. How many should he stock so that his expected profit is maximum? 24. Suppose that a decision maker faced with three decision alternatives and four states of nature. Given the following profit pay-off table. States of nature Acts S1 S2 S3 S4 a1 16 10 12 7 a2 13 12 9 9 a3 11 14 15 14 Assuming that he has no knowledge of the probabilities of occurrence of the states of nature, find the decisions to be recommended under each of the following criteria. 274 (i) maximin (ii) maximax (iii) minimax Regret 25. Pay-off of three acts A, B and C and states of nature X, Y and Z are given below Pay-off (in Rs) Acts States of A B C nature X 2000 − 20 −50 Y 200 −100 −50 Z 400 600 300 The probabilities of the states of nature are 0.3, 0.4 and 0.3. Calculate the EMV for the above and select the best art. Answers: I. 1. (d) II. 6. Sequence 9. Optimal 2. (d) 3. (b) 7. equally likely 10. Minimax 4. (c) 5. (a) 8. minimizing IV. 16. A2 is the best 17. Select A with the highest EMV Rs.194 18. EMV: 3200, decide, Act A3, expand 400 units 19. (i) maximin : Act a3 (ii) minimax regret Act a1 20. So the shop keeper should stock 8 dozen of fruits to get maximum expected profit. 21. He should stock 3 dozen of flowers to get maximum expected net profit. 22. He stocks 3 dozen of flowers to expect maximum profit Rs.9.50 23. To stock 405 copies so that his expected profit is maximum 24. (i) Act a3 is recommended 275 (ii) Act a1 is recommended (iii) Act a3 is recommended 25. EMA for A is highest. So the best act is A is selected 276 277 278 279 280 281 282 283 284 285 286 287 288 289 ...
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This note was uploaded on 12/09/2011 for the course BBA ms 08 taught by Professor Sharma during the Spring '06 term at Amity University.

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