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13 Pages

### 409Hw03ans

Course: STATISTICS stat 410, Spring 2011
School: University of Illinois,...
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Word Count: 2132

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409 Fall STAT 2009 Homework #3 (due Friday, September 18, by 4:00 p.m.) 1. Let X 1 , X 2 , , X n be a random sample of size probability density function f X (x) = f X ( x ; ) = ( 1 ) 2 a) ln x x n from the distribution with x &gt; 1, , &gt; 1. Find the sufficient statistic Y = u ( X 1 , X 2 , , X n ) for . n f ( x 1; ) f ( x 2; ) f ( x n; ) = n = ( 1 ) 2 n xi i =1...

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409 Fall STAT 2009 Homework #3 (due Friday, September 18, by 4:00 p.m.) 1. Let X 1 , X 2 , , X n be a random sample of size probability density function f X (x) = f X ( x ; ) = ( 1 ) 2 a) ln x x n from the distribution with x > 1, , > 1. Find the sufficient statistic Y = u ( X 1 , X 2 , , X n ) for . n f ( x 1; ) f ( x 2; ) f ( x n; ) = n = ( 1 ) 2 n xi i =1 n Y1 = Xi i =1 ( 1 ) 2 ln xi i =1 xi n ln xi . i =1 is a sufficient statistic for . n n Y 2 = ln Y 1 = ln X i = ln X i is also a sufficient statistic for . i =1 i =1 OR f X ( x; ) = ( 1 ) 2 Y2 = x = exp { ln x + ln ln x + 2 ln ( 1 ) } K ( x ) = ln x . ln x n ln X i i =1 Y1 = e Y2 n = is a sufficient statistic for . Xi i =1 is also a sufficient statistic for . b) What is the probability distribution of W = ln X ? Using the change-of-variable technique ( Section 3.5, pp. 173 174 ): X continuous r.v. with p.d.f. f X ( x ). Y = u(X) X= u ( x ) one-to-one, differentiable ( strictly increasing or strictly decreasing ) u 1( Y ) = v(Y) f Y ( y ) = f X ( v( y ) ) | v ' ( y ) | W = ln X X = eW = v(W ) v'( w) = e w fW( w ) = ( 1 )2 w e w e w = ( 1 )2 w e w ( 1) ( 1 ) 2 w 2 1 e w ( 1 ), ( 2 ) w > 0. W has Gamma ( = 2, usual = 1 ) distribution. 1 = n c) What is the probability distribution of ln X i ? i =1 Suppose X and Y are independent, X is Gamma ( 1 , ), Y is Gamma ( 2 , ). If random variables X and Y are independent, then M X + Y ( t ) = M X ( t ) M Y ( t ). 1 1 M X + Y( t ) = 1 X + Y is Gamma ( 1 + 2 , ); n ln X i = i =1 ( 1 t ) 1 ( 1 t ) 2 n Wi i =1 = ( 1 t ) 1 + 2 has Gamma ( = 2 n, usual = , t< 1 ) distribution. 1 1 . 2. Let X 1 , X 2 , , X n be a random sample from the distribution with probability density function f X (x) = f X ( x ; ) = a) x 2 e x > 0. x>0 Find the sufficient statistic Y = u ( X 1 , X 2 , , X n ) for . f ( x 1, x 2, x n; ) = f ( x 1; ) f ( x 2; ) f ( x n; ) n = 2n in 1 x i n = e 1 . i =1 x i n By Factorization Theorem, Y = i =1 X i is a sufficient statistic for . OR f ( x ; ) = exp x + ln ln 2 ln x . 1 2 n Y= i =1 b) K( x ) = x. X i is a sufficient statistic for . What is the probability distribution of W = X? FX( x ) = P( X x ) = 1 e x , x > 0. FW( w ) = P( W w ) = P( X w ) = P( X w2 ) = FX( w2 ) = 1 e w, w > 0. OR W= X = W 2 = v(W) X fW(w ) = 2w e w 2 w = e w, v'( w) = 2 w w > 0. 1 W has Exponential ( usual = ) = Gamma ( = 1, usual = 1 ) distribution. n c) What is the probability distribution of i =1 Xi ? Suppose X and Y are independent, X is Gamma ( 1 , ), Y is Gamma ( 2 , ). If random variables X and Y are independent, then M X + Y ( t ) = M X ( t ) M Y ( t ). 1 1 M X + Y( t ) = 1 X + Y is Gamma ( 1 + 2 , ); n ln X i = i =1 ( 1 t ) 1 ( 1 t ) 2 n Wi i =1 = ( 1 t ) 1 + 2 has Gamma ( = n, usual = 1 , ) distribution. t< 1 . 3. Let X 1 , X 2 , , X n be a random sample of size n from a shifted Exponential ( 1 ) distribution with probability density function f X (x) = f X ( x ; ) = e ( x ) , x > , R. Find the sufficient statistic Y = u ( X 1 , X 2 , , X n ) for . Define 1 if A is true I{A} = . 0 if A is false n f ( x 1; ) f ( x 2; ) f ( x n; ) = e ( xi i =1 n e i =1 xi e n I { x i > } n = )I{x i =1 = [e n ] nx I { min x i > } e i =1 i . Y = min X i is a sufficient statistic for . i > } 4. Let > 0 and let X 1 , X 2 , , X n be a random sample of size double exponential distribution. That is, f X ( x ) = f ( x; ) = a) 2 e x n from a < x < . , Find the sufficient statistic Y = u ( X 1 , X 2 , , X n ) for . f ( x 1, x 2, x n; ) = f ( x 1; ) f ( x 2; ) f ( x n; ) = n By Factorization Theorem, Y = i =1 Xi n n is a sufficient statistic for . OR f ( x ; ) = exp{ x + ln ln 2 }. n Y= K( Xi ) = i =1 b) n i =1 K( x ) = x . is a sufficient statistic for . Xi What is the probability distribution of Y? x P(|X| > x) = |X| 2 e y dy + x2 has Exponential ( = n i =1 Xi 1 e y dy = e x , ) = Gamma ( = 1, = has Gamma ( = n, = 1 ) distribution. exp xi . 2n i =1 1 x > 0. ) distribution. 5. A machine fastens plastic screw-on caps onto containers of motor oil. If the machine applies more torque than the cap can withstand, the cap will break. Both the and torque applied the strength of the caps vary. The capping machine torque has the normal distribution with mean 7.9 inch-pounds and standard deviation 0.9 inch-pounds. The cap strength (the torque that would break the cap) has the normal distribution with mean 10 inch-pounds and standard deviation 1.2 inch-pounds. The cap strength and the torque applied by the machine are independent. What is the probability that a cap will break while being fastened by the capping machine? That is, what is the probability P( Strength < Torque ) ? Need P( Strength Torque < 0 ) = ? E( Strength Torque ) = 10 7.9 = 2.1. Var( Strength Torque ) = 1.2 2 + 0.9 2 = 2.25. SD( Strength Torque ) = 1.5. ( Strength Torque ) is normally distributed. 0 2 .1 = P( Z < 1.40 ) = ( 1.40 ) = 0.0808. P( Strength Torque < 0 ) = P Z < 1 .5 6. The weight of an almond varies with mean 0.051 ounce and standard deviation 0.018 ounce. What is the probability (approximately) that the total weight (of a random sample) of 60 almonds is less than 3 ounces? E( Total ) = 60 0.051 = 3.06. Var( Total ) = 60 0.018 2 = 0.01944. n = 60 large. SD( Total ) 0.1394. Central Limit Theorem: Total n n Z. 3 3.06 P( Total < 3 ) P Z < = P( Z < 0.43 ) = ( 0.43 ) = 0.3336. 0.1394 OR Total < 3 n = 60 large. Average < 3 = 0.05. 60 Central Limit Theorem: Need P( X < 0.05 ) = ? X Z. n 0.05 0.051 P( X < 0.05 ) P Z < = P( Z < 0.43 ) = ( 0.43 ) = 0.3336. 0.018 60 From the textbook: 6.3-3 (show work) N ( 0, 2 ) a) ( ) n 1 L 2 ; x 1 , x 2 ,..., x n = i =1 2 2 1 2 2 exp x i2 n n exp 1 x i2 . 2 2 i =1 2 2 = 1 By Factorization Theorem, n X i2 Y= i =1 is a sufficient statistic for 2. OR 1 f(x) = 2 2 1 2 2 exp K ( x ) = x 2. n Y= X i2 i =1 b) () ln L 2 = d d d d 2 n 2 () ln ( 2 ) ln L 2 = () L 2 =0 2 n 2 () ln 2 n 1 + 22 2 2 () ( 1 2 ) x 2 ln 2 2 , < x < . is a sufficient statistic for 2. n 1 2 2 x i2 . i =1 n 2 x i2 . i =1 1n Y 2 = X i2 = . n i =1 n OR ln L ( ) = d d d d n ln ( 2 ) n ln ( ) 2 ln L ( ) = n x i2 . 2 2 i =1 n 1n2 + x . 3 i =1 i L( ) = 0 2 = 1n X i2 n i =1 2 E ( X i ) = Var ( X i ) + [ E ( X i ) ] 2 = 2 + 2 = 2, E( 2 ) = . n 2 = 1 X i2 = Y . n i =1 n c) 1 E ( X i2 ) n 1n = i =1 1 n ( n 2 ) = 2. i = 1, 2, , n. 2 is unbiased for 2. OR (X i ) 2 is 2 2 ( n ); here = 0. n X i2 i =1 2 E ( ( n ) ) = n. 2 = n2 2 is ( n ) 2 E ( 2 ) = 2. n2 E = n. 2 2 is unbiased for 2. 6.3-4 x 1 0 < x <1 otherwise f (x ) = n a) fX (X i ; ) i =1 0 n = n X i i =1 0 < < . 1 . n Y1 = By Factorization Theorem, Xi i =1 is sufficient statistic for . n n Y 2 = ln X i = ln X i is also a sufficient statistic for . i =1 i =1 OR f ( x ) = exp { ( 1 ) ln x + ln }, 0 < x < 1. K ( x ) = ln x . n By Theorem 6.3-1, b) Y1 = e Y2 Y2 = n = Xi i =1 ln X i i =1 is a sufficient statistic for . is also a sufficient statistic for . Likelihood function: L() = n fX (X i ; ) i =1 n = n X i i =1 1 . n ln L() = n ln + ( 1 ) ln X i . i =1 ( ( ) ) = n + ln X i d ln L d n i =1 = 0. = n n ln X i i =1 = n n ln X i i =1 . c) n n = = is also a sufficient statistic for , since any single-valued Y2 ln Y1 function of a sufficient statistic not involving but with a single-valued inverse is also a sufficient statistic for . 6.3-6 Gamma ( , ) distribution: f ( x; , ) = 1 f x i ; , = ( ) i =1 n a) ( ) 1 ( ) n x 1 e x n x i =1 i 1 n By Factorization Theorem, Y = Xi i =1 b) , 0 < x < . 1n exp x i i =1 . is a sufficient statistic for . n n 1 L( ) = f X i ; , = X i 1 e X i . i =1 i =1 ( ) ( ) n 1n ln L( ) = n ln ( ) n ln + ( 1 ) ln X i X i . i =1 i =1 n d ln L( ) n 1 n = + X = 0. d 2 i =1 i Xi Y X = i =1 = =. n n E ( X ) = = . E ( ) = and is an unbiased estimator for . 6.4-2 x = 85, 2 = 72, n = 8. 85 z 2 72 8 a) z 0.005 = 2.576, 85 7.728 b) z 0.025 = 1.96, 85 5.88 c) z 0.05 = 1.645, 85 4.935 or ( 80.065, 89.935 ). d) z 0.10 = 1.282, 85 3.846 or ( 81.154, 88.846 ). or or ( 77.272, 92.728 ). ( 79.12, 90.88 ). 6.4-4 2 = 4, = 2. a) x = 56.8; b) z 0.025 = 1.96, 56.8 1.96 2 10 or 56.8 1.24 or ( 55.56, 58.04 ); 52 56.8 c) P ( X < 52 ) P Z < = P ( Z < 2.4 ) = 0.0082. 2
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University of Illinois, Urbana Champaign - STATISTICS - stat 410
STAT 409Fall 2005Homework #4(due Friday, September 25, by 4:00 p.m.)From the textbook:6.4-86.4-106.8-26.5-26.4-126.8-46.4-186.8-66.5-4Hint: Equal variances10.The National Security Agency (NSA) hires you to compute a 95% confidenceinterval
University of Illinois, Urbana Champaign - STATISTICS - stat 410
STAT 409Fall 2005Homework #4(due Friday, September 25, by 4:00 p.m.)From the textbook:6.4-8a) x = 46.42;b) s 2 = 41.682, s 6.456, t 0.05 ( 4 ) = 2.132,46.42 2.132 6.4565or46.42 6.156or( 40.264, 52.576 ).6.4-10t 0.10 ( 27 ) = 1.314,( 21.45
University of Illinois, Urbana Champaign - STATISTICS - stat 410
STAT 409Fall 2009Homework #5(due Thursday, October 1, by 4:00 p.m.)From the textbook:6.6-26.6-46.6-86.6-106.7-26.7-46.7-126.7-186.8-12 if we believe the seed distributors claim;b)10.a) if we ignore the seed distributors claim.Let X have
University of Illinois, Urbana Champaign - STATISTICS - stat 410
STAT 409Homework #5(due Thursday, October 1, by 4:00 p.m.)From the textbook:6.6-2For these 9 weights, x = 20.90, s = 1.858.a)A point estimate for is s = 1.858.b)8 1.858 2,17.548 1.858 2 = [ 1.255, 3.559 ].2.180 c)8 1.858 2,15.518 1.858
University of Illinois, Urbana Champaign - STATISTICS - stat 410
STAT 409Fall 2009Homework #6(due Friday, October 9, by 4:00 p.m.)1.Let a &gt; 0, &gt; 0 and let X 1 , X 2 , , X n be a random sample of sizedistribution with probability density functionf X (x) = f X ( x ; ) =aa x a 1 ,nfrom anfrom a0 &lt; x &lt; .Sup
University of Illinois, Urbana Champaign - STATISTICS - stat 410
University of Illinois, Urbana Champaign - STATISTICS - stat 410
STAT 409Fall 2009Homework #7 (Answers)8.1-6H 0 : p = 1/6 vs. H 1 : p &lt; 1/6orH 0 : p 1/6 vs. H 1 : p &lt; 1/6Left tailed.n = 8000.a)The test statistic isz= = 0.05.b)y py1n0=80006.p 0 (1 p 0 )1 1 166n8000(The critical (rejection) reg
University of Illinois, Urbana Champaign - STATISTICS - stat 410
STAT 409Homework #8Fall 2009(due Friday, October 23, by 4:00 p.m.)1.In a random sample of 100 Hawk &amp; Hummingbird Airline (HHA) direct flightsfrom New York to Boston, the average number of passengers was 56.3, withsample standard deviation 11.33.a)
University of Illinois, Urbana Champaign - STATISTICS - stat 410
STAT 409Fall 2009Homework #8(due Friday, October 23, by 4:00 p.m.)1.In a random sample of 100 Hawk &amp; Hummingbird Airline (HHA) direct flightsfrom New York to Boston, the average number of passengers was 56.3, withsample standard deviation 11.33.X
University of Illinois, Urbana Champaign - STATISTICS - stat 410
STAT 409Fall 2009Homework #9(due Friday, October 30, by 4:00 p.m.)From the textbook:8.2-169.1-49.1-69.1-109.1-79.1-127.Let X 1 , X 2 , , X 25 be a random sample from a N ( , 100 ) population,and suppose the null hypothesis H 0 : = 100 is to b
University of Illinois, Urbana Champaign - STATISTICS - stat 410
STAT 409Fall 2009Homework #9(due Friday, October 30, by 4:00 p.m.)From the textbook:8.2-16H 0 : 2 = 30 vs. H 1 : 2 = 80.Right tailed.Recall: If X 1 , X 2 , , X n are i.i.d. N ( , 2 ), thena)Test Statistic:2 =2is 2 ( n 1 ).(n 1) s 2 = 18 s 2
University of Illinois, Urbana Champaign - STATISTICS - stat 410
STAT 409Homework #10Fall 2009(due Thursday, November 5, by 4:00 p.m.)1 2.Bert and Ernie noticed that thefollowing are satisfied whenCookie Monster eats cookies:(a)the number of cookies eaten duringnon-overlapping time intervals areindependent;
University of Illinois, Urbana Champaign - STATISTICS - stat 410
STAT 409Homework #10Fall 2009(due Thursday, November 5, by 4:00 p.m.)1 2.Bert and Ernie noticed that thefollowing are satisfied whenCookie Monster eats cookies:(a)the number of cookies eaten duringnon-overlapping time intervals areindependent;
University of Illinois, Urbana Champaign - STATISTICS - stat 410
STAT 409Homework #11Fall 2009(due Wednesday, November 13, by 4:00 p.m.)1.Let &gt; 0 and let X 1 , X 2 , , X n be independent random variables, each withthe probability density function +1f(x) = x0x 1.x &lt;1We wish to test H 0 : = 1 vs. H 1 : &gt; 1.
University of Illinois, Urbana Champaign - STATISTICS - stat 410
STAT 409Fall 2009Homework #11(due Friday, November 13, by 4:00 p.m.)1.Let &gt; 0 and let X 1 , X 2 , , X n be independent random variables, each withthe probability density function +1f(x) = x0x 1.x &lt;1We wish to test H 0 : = 1 vs. H 1 : &gt; 1.a)
University of Illinois, Urbana Champaign - STATISTICS - stat 410
STAT 409Fall 2009Homework #12(due Friday, November 20, by 4:00 p.m.)8.6-195(O E )3237.268.423.446.2360.038710.09883264318322859.2E45532210.432883E7188.8O3624.845.615.630.8300240.246154 0.031169 0.4444442000.649324 0
University of Illinois, Urbana Champaign - STATISTICS - stat 410
STAT 409Fall 2009Homework #13(due Friday, December 4, by 4:00 p.m.)1.Seventy percent of the light aircraft that disappear while in flight in Neverlandare subsequently discovered. Of the aircraft that are discovered, 60% have anemergency locator, wh
University of Illinois, Urbana Champaign - STATISTICS - stat 410
STAT 409Fall 2009Homework #13(due Friday, December 4, by 4:00 p.m.)1.Seventy percent of the light aircraft that disappear while in flight in Neverlandare subsequently discovered. Of the aircraft that are discovered, 60% have anemergency locator, wh
University of Illinois, Urbana Champaign - STATISTICS - stat 410
STAT 409Homework #14Fall 2009(due Wednesday, December 9, by 4:30 p.m.)1.When correctly adjusted, a machine that makes widgets operates with a 5% defectiverate. However, there is a 10% chance that a disgruntled employee kicks the machine,in which ca
University of Illinois, Urbana Champaign - STATISTICS - stat 410
University of Illinois, Urbana Champaign - STATISTICS - stat 410
1.Let X 1 , X 2 , , X n be a random sample from the distribution with probabilitydensity functionf (x ) = 4 x 3 e x4 &gt; 0.x&gt;0a)Find the sufficient statistic Y = u ( X 1 , X 2 , , X n ) for .b)Obtain the maximum likelihood estimator of ,c)Isa c
University of Illinois, Urbana Champaign - STATISTICS - stat 410
Practice Problems1.The label on 1-gallon can of paint states that the amount of paint in the can issufficient to paint at least 400 square feet (on average). Suppose the amount ofcoverage is approximately normally distributed, and the overall standard
University of Illinois, Urbana Champaign - STATISTICS - stat 410
STAT 409Fall 2009Name _Quiz 1(3 points)Be sure to show all your work, your partial credit might depend on it.No credit will be given without supporting work.1.Metaltech Industries manufactures carbide drill tips used in drilling oil wells.The lif
University of Illinois, Urbana Champaign - STATISTICS - stat 410
STAT 409Fall 2009NameANSWERS.Quiz 1(3 points)Be sure to show all your work, your partial credit might depend on it.No credit will be given without supporting work.1.Metaltech Industries manufactures carbide drill tips used in drilling oil wells.
University of Illinois, Urbana Champaign - STATISTICS - stat 410
STAT 409Fall 2009Name _Quiz 2(3 points)Be sure to show all your work, your partial credit might depend on it.No credit will be given without supporting work.1.At Anytown State University, the population standard deviation of the SAT scoresof ente
University of Illinois, Urbana Champaign - STATISTICS - stat 410
STAT 409Fall 2009NameANSWERS.Quiz 2(3 points)Be sure to show all your work, your partial credit might depend on it.No credit will be given without supporting work.1.At Anytown State University, the population standard deviation of the SAT scores
University of Illinois, Urbana Champaign - STATISTICS - stat 410
STAT 409Fall 2009Name _Version AQuiz 3(3 points)Be sure to show all your work, your partial credit might depend on it.No credit will be given without supporting work.a) (1) Suppose we Do NOT Reject H 0 for = 0.05. Then for = 0.10, we _.( Circle o
University of Illinois, Urbana Champaign - STATISTICS - stat 410
STAT 409Fall 2009NameANSWERS.Version AQuiz 3(3 points)Be sure to show all your work, your partial credit might depend on it.No credit will be given without supporting work.a) (1) Suppose we Do NOT Reject H 0 for = 0.05. Then for = 0.10, we _.(
University of Illinois, Urbana Champaign - STATISTICS - stat 410
STAT 409Fall 2009Name _Version BQuiz 3(3 points)Be sure to show all your work, your partial credit might depend on it.No credit will be given without supporting work.a) (1) Suppose we Do NOT Reject H 0 for = 0.05. Then for = 0.01, we _.( Circle o
University of Illinois, Urbana Champaign - STATISTICS - stat 410
STAT 409Fall 2009NameANSWERS.Version BQuiz 3(3 points)Be sure to show all your work, your partial credit might depend on it.No credit will be given without supporting work.a) (1) Suppose we Do NOT Reject H 0 for = 0.05. Then for = 0.01, we _.(
University of Illinois, Urbana Champaign - STATISTICS - stat 410
STAT 409Fall 2009Name _Version AQuiz 4(5 points)Be sure to show all your work, your partial credit might depend on it.No credit will be given without supporting work.1.At Anytown College, the administration would like the students grade distribut
University of Illinois, Urbana Champaign - STATISTICS - stat 410
STAT 409Fall 2009NameANSWERS.Version AQuiz 4(5 points)Be sure to show all your work, your partial credit might depend on it.No credit will be given without supporting work.1.At Anytown College, the administration would like the students grade d
University of Illinois, Urbana Champaign - STATISTICS - stat 410
STAT 409Fall 2009Name _Version BQuiz 4(5 points)Be sure to show all your work, your partial credit might depend on it.No credit will be given without supporting work.1.At Anytown College, the administration would like the students grade distribut
University of Illinois, Urbana Champaign - STATISTICS - stat 410
STAT 409Fall 2009NameANSWERS.Version BQuiz 4(5 points)Be sure to show all your work, your partial credit might depend on it.No credit will be given without supporting work.1.At Anytown College, the administration would like the students grade d
University of Illinois, Urbana Champaign - STATISTICS - stat 410
STAT 409Fall 2009Name _Version AQuiz 5(5 points)Be sure to show all your work, your partial credit might depend on it.No credit will be given without supporting work.1.A travel agent randomly sampled individuals in her target market to determine
University of Illinois, Urbana Champaign - STATISTICS - stat 410
STAT 409Fall 2009NameANSWERS.Version AQuiz 5(5 points)Be sure to show all your work, your partial credit might depend on it.No credit will be given without supporting work.1.A travel agent randomly sampled individuals in her target market to de
University of Illinois, Urbana Champaign - STATISTICS - stat 410
STAT 409Fall 2009Name _Version BQuiz 5(5 points)Be sure to show all your work, your partial credit might depend on it.No credit will be given without supporting work.1.A travel agent randomly sampled individuals in her target market to determine
University of Illinois, Urbana Champaign - STATISTICS - stat 410
STAT 409Fall 2009NameANSWERS.Version BQuiz 5(5 points)Be sure to show all your work, your partial credit might depend on it.No credit will be given without supporting work.1.A travel agent randomly sampled individuals in her target market to de
University of Illinois, Urbana Champaign - STATISTICS - stat 410
Cumulative Binomial Probabilities p n2Cumulative Binomial Probabilities p0.60 0.160 0.640 0.064 0.352 0.784 0.026 0.179 0.525 0.870 0.010 0.087 0.317 0.663 0.922 0.004 0.041 0.179 0.456 0.767 0.953 0.002 0.019 0.096 0.290 0.580 0.841 0.972 0.001 0.009
University of Illinois, Urbana Champaign - STATISTICS - stat 410
Left tailed testH 0 : p = p0vs.H 1 : p &lt; p0If H 0 is TRUE :Use p 0 .Reject H 0Do NOT Reject H 0Type I ErrorCorrect decisiona0a+1nRejection Rule for a Left tailed test:Find a such that P( Y a ) = CDF @ a .( using Binomial ( n , p 0 ) tables
University of Illinois, Urbana Champaign - STATISTICS - stat 410
Chapter 6Estimation6.1Sample Characteristics6.12 (a) x =(b) s2 =4= 1.333;388= 1.275.696.14 (a) x = 1.711, s = 0.486;(b) and (d) graphs.0.80.70.60.50.40.30.20.10.3 0.6 0.9 1.2 1.5 1.8 2.1 2.4 2.7123456Figure 6.14: (b) Female un
University of Illinois, Urbana Champaign - STATISTICS - stat 410
Chapter 8Tests of Statistical Hypotheses8.1Tests about Proportions8.12 (a) C = cfw_x : x = 0, 1, 2;(b) = P (X = 0, 1, 2; p = 0.6)= (0.4)4 + 4(0.6)(0.4)3 + 6(0.6)2 (0.4)2 = 0.5248;= P (X = 3, 4; p = 0.4)= 4(0.4)3 (0.6) + (0.4)4 = 0.1792.OR(a ) C
University of Illinois, Urbana Champaign - STATISTICS - stat 410
Consider a coin being tossed 20 times.There are 2 20 = 1,048,576 possible outcomes (samples).Suppose we wish to test whether the coin is fair or not.We can rank all these outcomes (samples) from 20 C 10 = 184,756 best outcomes(samples) with 10 Hs and
University of Illinois, Urbana Champaign - STATISTICS - stat 410
STAT 409 / MATH 409Monday, Wednesday, FridayInstructor:Office:E-mail:Office hours:Text:Fall 20092:00 2:50 p.m.103 Talbot LaboratoryAlexey Stepanov101-A Illini Hallstepanov@illinois.eduph.:265-6550Monday 3:30 4:30 p.m., Thursday 1:30 3:00 p.
University of Illinois, Urbana Champaign - STATISTICS - stat 410
William Gosset(1876-1937)Thet DistributionEXCEL: = TINV ( , v ) = TDIST ( t , v , 1 ) = TDIST ( t , v , 2 ) gives gives givest2for t distribution with v degrees of freedom the upper tail probability for t distribution with v degrees of freedom, P (
Bilkent University - DSB - 243
BLECK NVERSTES REKTRLNDENniversitemiz Sosyal Bilimler Enstitsne 2011-2012 Eitim retim Yl Gz Yarylnda TezliYksek Lisans Programlarna renci alnacaktr.BAVURU ARTLARI1.2.ALES snavndan; bavurulan programn n grd puan trnde yksek lisans iin en az 55 puana
University of South Africa - MNG 2016 - 2016
University of South Africa - MNG 2016 - 2016
9/19/2011MNG 2016General ManagementPresenting Lecturer:MRS K STANDERstandk@unisa.ac.zaOBJECTIVES OF THE GROUP DISCUSSION CLASS1. AN OVERVIEW OF THE CONTENT OF THIS MODULE.2. PREPARATION FOR THE EXAMINATION.3. THE FORMAT OF THE EXAMINATION PAPER.
University of South Africa - MNG 2016 - 2016
University of South Africa - MNG 2016 - 2016
University of South Africa - MNG 2016 - 2016
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University of South Africa - MNG 2016 - 2016
University of South Africa - MNG 2016 - 2016
University of South Africa - MNG 2016 - 2016
University of South Africa - MNE202 - 202
10APPENDIX A:MNE 202-V/101ASSIGNMENTS FOR 2007ASSIGNMENT 01SELF-EVALUATIONThis assignment covers topics 1 and 2 of your study guide. Itcontains short and long questions.The aim of this assignment is to familiarise you with thecontent of the work
University of South Africa - MNE202 - 202
9MNE 202-V/101ANNEXURE A: ASSIGNMENTS FOR 2008ASSIGNMENT 01SELF-EVALUATIONThis assignment covers topics 1 and 2 of your study guide. Itcontains short and long questions.The aim of this assignment is to familiarise you with thecontent of the work a
University of South Africa - MNE202 - 202
33MNE 202-V/201/2/2008FEEDBACK ON ASSIGNMENT 02NOTE: Question 14 is from chapter 16 which is not prescribed in the syllabusfor this course. This question will be ignored and your assignment will bemarked out of 24 marks.Question 1The correct answe
University of South Africa - MNE202 - 202
33MNE202V/201/1/2009FEEDBACK ON ASSIGNMENT 01NOTE: Question 20 is based on chapter 16, which is not part of the syllabusfor this course. This question will be ignored and your assignment will bemarked out of 24 marks.Question 1The correct answer i
University of South Africa - MNE202 - 202
University of South Africa - MNE202 - 202
University of South Africa - MNE202 - 202
MNE202V/201/3/20103FEEDBACK ON ASSIGNMENT 01Question 1The correct answer is 3. See page 15 in the prescribed book.Entrepreneurial success factors include the following: creativity andinnovation; risk orientation; leadership; good human relations; a