Unformatted text preview: Lessons in Business Statistics
Prepared By
P.K. Viswanathan Chapter 8: Hypothesis Testing Introduction
Managers have to make decisions with
minimum risk in an environment
characterized by uncertainty. Acceptance
or rejection of a decision depends on
acceptance or rejection of a hypothesis. For
example, a marketing manager is facing a
decision whether to introduce a new
product in the market or not. If his
company could get a market share of 15
percent or more, then the new product
would be introduced in the market. A
suitable hypothesis formulation and testing
it would help the manager take the right
decision. This chapter covers the various
tests of hypothesis that are useful in making
sound decisions. 1) Statistical HypothesisA Conceptual
Framework
What is a Statistical Hypothesis?
A statistical hypothesis is a statement about
a population parameter. It may or may not
be true. The manager has to ascertain the
truth of the hypothesis. 1) Statistical HypothesisA Conceptual
FrameworkContinues
What is a Statistical Hypothesis?Example
Statement 1: Not more than 20% of the adults watch children’s
program in the television.
Statement 2: More than 20% of the adults watch children’s program
in the television.
First, it should be noted that these two hypotheses cannot be
simultaneously true. Only one of them will be true. Likewise, these
two hypotheses cannot be simultaneously false. Only one of them will
be false. The acceptance or rejection of a particular hypothesis leads
to the acceptance or rejection of a particular decision.
Statement 1 given above is called the Null Hypothesis H0
.
Statement 2 is called the Alternative Hypothesis H1 . 1) Statistical HypothesisA Conceptual
Framework
Type I Error and Type II Error
Null Hypothesis Reject Accept True
Type I
Error
()
No
Error False
No
Error
Type II
Error (
) 1) Statistical HypothesisA Conceptual
Framework
The probability of making a Type I error is called the
level of significance of the test. It is designated by the
Greek letter alpha (1 is the confidence level that
.
)
says that you are right in your assessment (1 of the
)%
times. If you set 0.05 and happen to reject the null
=
hypothesis at this level, there is a 5% probability that you
have rejected the null hypothesis when in fact it is true.
This also means that you are 95% confident that you have
accepted the null hypothesis when it is true. 1) Statistical HypothesisA Conceptual
Framework
The probability of making a Type II error is symbolized by the
Greek letter 1is called the power of the test. The power of
.
the test is the probability of rejecting the null hypothesis when in
fact it is false. Suppose you keep =10%, it means that the
power of the test is 90%. That is, the probability of rejecting the
null hypothesis when it is false is 90% and only 10% of the time
you commit the error of accepting the null hypothesis when it is
false. It is desirable to keep both and at minimum level.
However, a decrease in will lead to an increase in and an in
,
increase in will lead to a decrease in It is a general practice
.
to fix and let vary. It is convention to set = 0.05 that
,
corresponds to the confidence level of 95%. Some practitioners
at times also keep = 0.01 1) Statistical HypothesisA Conceptual
Framework 1) Statistical HypothesisA Conceptual
Framework
A hypothesis could be directional or nondirectional.
A directional hypothesis is one in which the
population parameter is structured to be greater than
or equal to or less than or equal to a specified value.
This is known as a onetailed test(onesided test ) in
the parlance of statistical hypothesis. A nondirectional hypothesis is one in which the population
parameter is structured to be equal to a specified
value. This is known as a twotailed test(twosided
test). 2) Hypothesis Testing–Univariate Case
(One Sample) Hypothesis Test for a Single Mean
Primary Purpose of the Test :
To test hypotheses that compare the
population mean of interest to a
specified value
2) Hypothesis Testing–Univariate Case
(One Sample)
Illustration Is the average waiting time for the
customers of Smart Supermarket
at the checkouts greater than 15 minutes? 2) Hypothesis Testing–Univariate Case
Hypothesis Structure for the Illustration H0: H1: 15
> 15 2) Hypothesis Testing–Univariate Case
Population Mean
Example
The marketing manager of a large restaurant has been asked to
conduct a survey of its customers belonging to a particular income
class. The president of the restaurant is interested in the mean
income of its customers. He is further interested in comparing this
mean income with that of a recently concluded census study by the
government. The government study shows a mean income of
Rs.300000 per year for this class of customers with a standard
deviation of Rs. 30000. The president is desirous of finding out
whether the population mean of its customers in this category is Rs.
300000 per year or not. The marketing manager has picked up a
random sample of 100 customers of this class from the customer
database. The sample data show a mean income of Rs. 293000 per
year. Perform a comprehensive statistical hypothesis testing
procedure and state your conclusions. 2) Hypothesis Testing–Univariate Case
Population Mean
Solution to the Example
Step 1: Formulate the null and the alternative hypothesis.
H0: 300000 (the mean income of the population is equal to
=
Rs.300000)
H1: 300000 (the mean income of the population is not equal to
Rs.300000)
Step 2: Select the right test statistic. The correct test statistic to be
used here is the Z test. Why? Because, the sample size is large.
The formula for the Z test is given below: Please note that Z
follows a standard normal distribution with mean 0 and standard
deviation 1. 2) Hypothesis Testing–Univariate Case
Population Mean
Solution the Example Continues
Step 3: Decide on the level of significance When the value of
.
the level of significance is not specified in a problem, it is a
convention to set the value equal to 0.05. What we’re saying is that
only 5% of the time we make the mistake of rejecting the null
hypothesis when it is true.
Step 4: Compute the test statistic based on sample data. The
formula to be used is X μ Z
σ n 2) Hypothesis Testing–Univariate Case
Population Mean
Solution the Example Continues
Under the assumption of the null hypothesis being true, you can
substitute the value Rs. 300000 in the place of μ
. X σ =293000. n =100.
= 30000.
Upon substitution of these values in the formula, we have 293000 00000 = 2.33
3 Z
30000 100 2) Hypothesis Testing–Univariate Case
Population Mean
Solution the Example Continues
Step 5: Determine the Critical Value for the chosen level of
significance. Here, =0.05. The critical value corresponding to the
twotailed test where each tail contains an area of can be easily
/2
worked out by using Microsoft Excel. The methodology is already
covered in the previous chapters. Here, =0.025. The critical
/2
value of Z =1.96 for positive Z and –1.96 for negative results.
Since the normal distribution is symmetrical, we can ignore the
sign of Z and just take the positive value of Z. That is the critical
value of Z is 1.96. Incidentally, if you choose =0.01, then the
critical value of Z =2.58. 2) Hypothesis Testing–Univariate Case
Population Mean
Solution the Example Continues Step 6: Compare the computed test statistic with the critical value.
Here, computed Z = 2.33. Since the normal distribution is
symmetrical, take the positive value of Z and compare it with the
Critical Z =1.96. If you take the negative computed Z, then
compare it with –1.96. Take the positive Z. Simple is best. Why
bother?
Step 7: Decision. If the computed Z is greater than the table Z,
reject the null hypothesis H0 and accept H1. Else accept H0. This is
same as finding out whether the computed Z falls in the acceptance
region or the rejected region. In our case, computed value of Z
(take just the positive value) 2.33 is greater than the critical value
of Z =1.96. Hence, it falls in the rejection region. Reject H0 and
accept H1. (See picture in next slide) 2) Hypothesis Testing–Univariate Case
Population Mean 2) Hypothesis Testing–Univariate Case
Population Mean
Interpretation of the Results for our example:
We have rejected H0 and accepted H1. What does
this mean? This means that the population mean
income of the category of interest to the president of
the restaurant is not equal to Rs. 300000 per year at
5% level of significance. Is it more than or less than
Rs. 300000? The sample mean suggests that it may
be less than Rs. 300000 per year. Can you do this
exercise now as a onetailed hypothesis test? This is
a Progressive Test Question for you. 2) Hypothesis Testing–Univariate Case
(One Sample)Population Proportion
Test for a Single Proportion
Primary Purpose of the Test :
To test hypotheses that compare the
population proportion of Interest to
a specified value 2) Hypothesis Testing–Univariate Case
(One Sample)Illustration Is the proportion of households
owning Color TVs in Chennai less
than 0.4? 2)Hypothesis Testing–Univariate Case
(One Sample)
Hypothesis Structure for the Illustration
Statement of the Null and
Alternative hypothesis H0 : H1 : P 0.40
P 0.40 2)Hypothesis Testing–Univariate Case
(One Sample)
Example Problem
A marketing manager of an enterprise is facing a
decision whether to introduce a new product into
the market or not. Consumer acceptance measured
in a blind comparison test is agreed upon as an
appropriate basis for evaluation. Marketing of the
new product will be pursued only if the acceptance
rate exceeds 30%. Otherwise, the new product will
not be introduced in the market. A random sample
of 200 consumers reveals that the acceptance rate is
32%. Using a level of significance of 0.05, perform
the hypothesis testing and recommend your action. 2)Hypothesis Testing–Univariate Case
(One Sample)
Solution: This is a classical hypothesistesting problem of the
population proportion. This is also a onetailed test. This
involves a large sample in which n =200. The Z test for the
proportion is the appropriate test.
H0: P 0.30 (The population proportion of consumer acceptance
is less than or equal to 0.30)
H1: P > 0.30 (The population proportion of consumer acceptance
is greater than 0.30)
The Z test for proportion is to be used here. It is given by
Z p
P
P(1 .
P)
n Please note that Z follows a standard normal with mean 0 and
standard deviation 1. 2)Hypothesis Testing–Univariate Case
(One Sample)
Under the null hypothesis being true, P=0.30. p=0.32. Substituting, we have
Z 0.32 0.30 0.30(1 0.30)
200 = 0.62. The critical value of Z for a one  tailed test is 1.65. Since, the com puted Z is
less than critical Z, accept H 0 . W hat do you conclude?
W e have no evidence to reject the null hypothesis based on the sample data
at 5% level of significance. In this case even at 1% level of significance, we
cannot reject H 0 . This im plies that you accept H 0 and conclude that the
population proportion of consum er accepta nce is less than or equal to30% .
Hence, the m anager should not introduce the new product in the m arket.
You m ay wonder how com e when the sam ple proportion is 32% , you say
that you should not introduce the new product? Is not 32% better than the
30% stipula ted? Yes, but you see statistically speaking, 32% sample
proportion has arisen due to chance and not a real one. This is why you say
statistically not significant. As long as statistical significance does not take
place, you cannot reject the null hypothes is. This is a real beauty of testing
of hypothesis. Unless the m anager wants to gamble, he should not venture to
introduce the new product. 2)Hypothesis Testing–Univariate
Case (Small Sample) Test Statistic (X )
μ
t
S
n 2)Hypothesis Testing–Univariate Case
(Small Sample)
Example:
An investigator took a random sample of eight
pieces of aluminum diecastings and observed the
sample mean strength to be 31.5. Before taking
the measurement, the investigator knew that the
population mean strength for an older type of
aluminum diecasting was 33. The standard
deviation of the sample measurements was 1.3.
The investigator would like to know whether the
population mean strength of the aluminum diecasting is 33. Setup the null and the alternative
hypothesis, perform the test, and comment on the
results. 2)Hypothesis Testing–Univariate Case
(Small Sample)Solution to Example This is a small sample case with unknown
population standard deviation. The appropriate
test is the t test. Please note also from the
wording of the problem, you need to perform a
twotailed test. Just like the normal distribution,
t is also symmetrical and it is enough if you
compare the positive value of the computed t
with the critical t for n1d.f at 5% level of
significance. 2)Hypothesis Testing–Univariate Case
(Small Sample)Solution to ExampleContinues
H0: =33 H1: 33 (X ) =
μ
t
S
n (31.5 33)
1.3
8 The positive value of the computed t =3.26. The critical t value for
7d.f(n1 =81) at 5% level of significance from Excel paste function
is 2.36(2 places of decimal). Since the calculated t value is greater
than the critical t, reject H0 an accept H1. The conclusion is that the
mean strength of aluminum die casting of the population is not 33
at 5% level of significance. 3)Hypothesis Testing–Bivariate Case
Population Mean
Test of Two Population Means
Primary Purpose of the Test:
To test hypotheses that compare the Population
Mean of interest for two separate populations.
(Samples are independent).
Example: Is the average expenditure per household
on eating out significantly higher in Bangalore
than in Calcutta? 3)Hypothesis Testing–Bivariate Case
Difference Between Means(Large
Sample)
Example: A test in computer course was conducted for a group of students,
consisting of 70 boys and 60 girls. The marks scored by the students are
given below: Boys Girls n1 = 70 n2 = 60 X1 = 70 X2 = 65 (
X 1 X 1 ) 2 = 7,500 (
X 2 X 2 ) 2 = 7,800 Is there a significant difference between the performance of the boys and the
girls? 3)Hypothesis Testing–Bivariate Case
Difference Between Means(Large
Sample)
Solution: From the wording of the question, it is clear
that the problem could be structured as a twotailed test.
The sample sizes for both the populations are large and
hence the Z test is appropriate to use. Let us take a level
of significance of 5%.
H0: = ( The population mean score of boys
1
2
=The population mean score of girls)
H1: ( The population mean score of boys 1
2
The population mean score of girls) 3)Hypothesis Testing–Bivariate Case
Difference Between Means(Large
Sample)
X 1 X 2 Z= 1
2
n1
n2
2 2 . The standard deviation of the two populations are not given. We can use the sample standard deviations in the place of the population standard deviations.
2
2
2
2
S1
S2 1 2 . In the data for the problem, we are
That is use
in the place of
n1
n2
n1
n2 ( X
(
X X 1 ) 2 =7500 and X 2 X 2 ) 2 = 7800. By definition
(
(
X2
1 1)
2 X 2 X 2 ) 2
S2 and
. Substituting and simplifying we have
n1 1
n2 1 given
S1 2 1 S1 2 (7500/69) = 108.70 S 2 2 (7800/59) =132.20. Z= X 1 X 2
2 2 S1
S
2
n1
n2 70 65
=
=2.58.
108.70 132.20 70
60 3)Hypothesis Testing–Bivariate Case
Difference Between Means(Large
Sample)
Conclusion:
In our example, calculated Z(2.58) is greater than the
table Z(1.96) at 5% level of significance. Reject H0
and accept H1. That is the performance of the boys
and the girls are not identical. The null hypothesis of
equally good performance is rejected. The difference
in the mean scores between boys and girls is
significant. 3)Hypothesis Testing–Bivariate Case
Difference Between Means(Small
Sample)
Example:Aptitude test was conducted for two groups of
executivesgroup1 consists of engineers and group2 consists of
accountants. The scores obtained by the candidates are given
below:
Engineers 125
Accountants 112 115
98 119
109 85
96 97
77 107
70 125
114 125
100 118 Do you find any significant difference between the scores of
these two groups? 3)Hypothesis Testing–Bivariate Case
Difference Between Means(Small Sample) 3)Hypothesis Testing–Bivariate Case
Difference Between Means(Small Sample)
Now compute t = we have t = X 1 X 2 1
S2 n
1
112.89 97 . Substituting the values form the spreadsheet above ,
1 n2 = 2.18. As the calculated t value exceeds the critical t for 15
1 1
224.73 9 8
degrees of freedom at 5% level of significance (2.13 from the t table in Appendix E),
Reject the null hypothesis and accept the alternative hypothesis. 3)Hypothesis Testing–Bivariate Case
Difference Between Means(Small Sample
Paired)
Test of Two means:Small sample
Primary Purpose of the Test
To test hypotheses that compare
the Population Mean of the
same variable and the data are
collected based on before and after situation
scenario ( dependent sample). 3)Hypothesis Testing–Bivariate Case
Difference Between Means(Small Sample
Paired)Structure of the Null and Alternative
Hypothesis H0 : H1 :  = 0
1
2  = 0
1
2 Test Statistic
t=D
where D = XA  XB
S/
n 3)Hypothesis Testing–Bivariate Case
Difference Between Means(Small Sample Paired)
Example
A company conducted a promotional campaign in 10 randomly
chosen retail outlets. Monthly sales in 1000 units are shown before
and after the campaign. Is there any significant difference in sales
before and after the campaign?
Outlet
No
1
2
3
4
5
6
7
8
9
10 Before
the
Campaign
240
225
250
280
200
150
165
100
130
170 After the
Campaign
270
245
260
290
190
160
160
130
135
175 3)Hypothesis Testing–Bivariate Case
Difference Between Means(Small Sample
Paired)
Solution
This is a twotailed hypothesis problem. Let us take a level of
significance of 0.05. The appropriate test statistic is a paired t for
the dependent sample.
H0: = (The means sales are same before and after the
1
2
campaign)
H1: (The mean sales are not same before and after the
1
2
campaign) 3)Hypothesis Testing–Bivariate Case
Difference Between Means(Small Sample
Paired)
Performing Calculation for the Paired t Test Outlet No Before the
Campaign(XB) After the
Campaign(XA) D = X AX B D D 1
2
3
4
5
6
7
8
9
10 240
225
250
280
200
150
165
100
130
170 270
245
260
290
190
160
160
130
135
175 30 19.5 20 9.5 10 0.5 10 0.5 10 20.5 S= 174.7222 10 0.5 S= 13.21825 5 15.5 30 19.5 5 5.5 5 5.5 D 10.5 D
t 2.5119745
(S / n ) 2 3)Hypothesis Testing–Bivariate Case
Difference Between Means(Small
Sample Paired)
The table value of t from Appendix E for 9
degrees of freedom at 5% level of significance
is 2.26. Since the calculated t is greater than
critical t, reject the null hypothesis and accept
the alternative. 3)Hypothesis Testing–Bivariate Case
Difference Between Proportions
Test of Two Proportions
Primary Purpose of the Test:
To test hypotheses that compare the
Population Proportion of Interest for two
separate populations 3)Hypothesis Testing–Bivariate Case
Difference Between Proportions
Example: Two random sample surveys, conducted with two
months gap between the two, assessed public opinions on the
outcome: The question that was posed was “If the general election
was going to take place tomorrow, would you cast your vote for
or against the ruling party?”
The results of the two surveys are tabulated below:
1st Survey
Sample size…………………
1000
For the ruling party……………
520
Against the ruling party………
480 2nd Survey
800
380
420 Set up the appropriate hypotheses, test and draw your conclusions. 3)Hypothesis Testing–Bivariate Case
Difference Between Proportions
Solution: This requires a structuring of the null hypothesis as no change in pattern of
voting between the two months by the public. Symbolically
H 0: P1 =P2 (The population proportions favoring the ruling party in the two months gap
is the same)
H 1: P1 P2 (No. It is not the same).
Let us take a level of significance of 5%. It is a two tailed test and therefore critical value
of Z =1.96.
The test statistic Z p1 p 2 1
1
p (1 p) n 1 n2 n p 2 p2
n
p1 1
. First let us calculate p.
n1 2
n n p 2 p2
n
1000(520 / 1000) (380 / 800)
800
p1 1
p
=
=0.50.
n1 2
n
1000 800 3)Hypothesis Testing–Bivariate Case
Difference Between Proportions
Z p1 p 2 0.52 .475
0
=Z =1.90
1 1
1
1
0.5(1 .5)
0 p (1 p ) 1000 800 n
1 n2 The calculated value of Z is less than the table value of Z. Accept H 0 . The inference is
that the population proportion of the public favoring the ruling party in the two months
gap has not changed at a level of significance of 5%. ...
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