Direction: This test is worth 150 points. You are required to complete this test within 55 minutes.

Transcription

1 Term Test 3 (Part A) November 1, 004 Name Math 6 Studet Number Directio: This test is worth 10 poits. You are required to complete this test withi miutes. I order to receive full credit, aswer each problem completely ad must show all work. Please make sure that you have all the 7 pages. GOOD LUCK! 1

2 1. ( poits) I samplig from a oormal distributio with a variace of, how large must the sample size be so that the legth of a 9% cofidece iterval for the mea is 1.96? Aswer: The cofidece iterval whe the sample is take from a ormal populatio with a variace of is [ ( ) ( ) ] σ σ x z α, x + z α. Thus the legth of the cofidece iterval is l = z α σ = z 0.0 = (1.96). But we are give that the legth of the cofidece iterval is l = Thus 1.96 = (1.96) = 10 = 100. Hece, the sample size must be 100 so that the legth of the 9% cofidece iterval will be 1.96.

3 . ( poits) Let X 1, X,..., X be a radom sample from a ormal populatio X with kow mea µ ad ukow variace σ. Fid a pivot for σ ad the usig this pivot costruct a 100(1 α)% cofidece iterval for σ. Aswer: We first fid a pivotal quatity for σ. Sice X N ( µ, σ ), we have X i N ( µ, σ ), ( ) Xi µ N(0, 1), σ ( ) Xi µ χ (1). σ ( ) Xi µ χ (). σ We defie the pivotal quatity Q(X 1, X,..., X, σ ) as Q(X 1, X,..., X, σ ) = ( Xi µ which has a chi-square distributio with degrees of freedom. Hece 1 α = P (a Q b) ( ( ) Xi µ = P a b) σ ( ) 1 = P a σ (X i µ) 1 b ( = P (X i µ) σ (X i µ) ) a b ( = P (X i µ) σ (X i µ) ) b a ( = P (X i µ) ) χ σ (X i µ) 1 () χ α α () Therefore, the (1 α)% cofidece iterval for σ whe mea is kow is give by [ (X i µ) χ, 1 () α σ ) ] (X i µ) χ. α () 3

4 3. ( poits) If X 1, X,..., X is a radom sample from a populatio with desity θx θ 1 if 0 < x < 1 f(x; θ) = 0 otherwise, where θ > 0 is a ukow parameter, what is a 100(1 α)% cofidece iterval for θ? Aswer: To costruct a cofidece iterval for θ, we eed a pivotal quatity. That is, we eed a radom variable which is a fuctio of the sample ad the parameter, ad whose probability distributio is kow but does ot ivolve θ. We use the radom variable Q = θ l X i χ () as the pivotal quatity. The 100(1 α)% cofidece iterval for θ ca be costructed from ( ) 1 α = P χ α () Q χ 1 α () ( ) = P χ α () θ l X i χ 1 α () = P χ α () θ l X i χ 1 α (). l X i Hece, 100(1 α)% cofidece iterval for θ is give by χ α (), l X i χ 1 () α l X i. 4

5 4. ( poits) Let X 1, X,..., X be a radom sample from a populatio X with probability desity fuctio { p x (1 p) (1 x) if x = 0, 1 f(x; p) = 0 otherwise. What is a 100(1 α)% approximate cofidece iterval for the parameter p? Aswer: The likelihood fuctio of the sample is give by L(p) = p xi (1 p) (1 xi). Takig the logarithm of the likelihood fuctio, we get l L(p) = [x i l p + (1 x i ) l(1 p)]. Differetiatig, the above expressio, we get d l L(p) dp = 1 p Settig this equals to zero ad solvig for p, we get x i 1 1 p x p x 1 p = 0, (1 x i ). that is (1 p) x = p ( x), which is x p x = p p x. Hece p = x. Therefore, the maximum likelihood estimator of p is give by p = X. The variace of X is V ar ( X ) = σ. Sice X Ber(p), the variace σ = p(1 p), ad V ar ( p) = V ar ( X ) = p(1 p). Sice V ar ( p) is ot free of the parameter p, we replave p by p i the expressio of V ar ( p) to get V ar ( p) p (1 p). The 100(1 α)% approximate cofidece iterval for the parameter p is give by X (1 X) X (1 X) X z α, X + z α.

6 . ( poits) Let p represet the proportio of defectives i a maufacturig process. To test H o : p 1 4 versus H a : p > 1 4, a radom sample of size is take from the process. If the umber of defectives is 4 or more, the ull hypothesis is rejected. What is the probability of rejectig H o if p = 1? Aswer: Let X deote the umber of defectives out of a radom sample of size. The X is a biomial radom variable with = ad p = 1. Hece, the probability of rejectig H o is give by α = P (Reject H o / H o is true) = P (X 4 / H o is true) ( / = P X 4 p = 1 ) ( / = P X = 4 p = 1 ) ( + P X = ( ) ( ) = p 4 (1 p) 1 + p (1 p) 0 4 ( ) 4 ( ) ( ) = + ( ) 1 = [0 + 1] = Hece the probability of rejectig the ull hypothesis H o is / p = 1 ) 6

7 6. ( poits) Let X be the umber of idepedet trails required to obtai a success where p is the probability of success o each trial. The hypothesis H o : p = 0.1 is to be tested agaist the alterative H a : p = 0.3. The hypothesis is rejected if X 4. What is the power of the test if H a is true? Aswer: The power fuctio is give by P (Type I Error) if p = 0.1 π(p) = 1 P (Type II Error) if p = 0.3. Hece, we have α = 1 P (Accept H o / H o is false) = P (Reject H o / H a is true) = P (X 4 / H a is true) = P (X 4 / p = 0.3) 4 = P (X = k /p = 0.3) = = k=1 4 (1 p) k 1 p (where p = 0.3) k=1 4 (0.7) k 1 (0.3) k=1 = (0.7) k 1 k=1 = 1 (0.7) 4 = Hece, the power of the test at the alterative is

8 Term Test 3 (Part B) November 1, 004 Name Math 6 Studet Number Directio: This test is worth 100 poits. You are required to retur this test o Moday o or before 10:00 am. Please do these four problems by yourself. You may use the text book ad/or your class otes for these problems. Please make sure that you have all the pages. GOOD LUCK! 8

9 7. ( poits) Let X 1, X,..., X be a radom sample from a distributio with desity fuctio 1 β f(x) = e (x 4) β for x > 4 0 otherwise, where β > 0. What is a 100(1 α)% approximate cofidece iterval for θ if the sample size is large? Aswer: First we calculate the expected value E(X) of the populatio X. E(X) = The likelihood fuctio of the sample is give by 4 L(β) = x 1 β e (x 4) β dx = β β e (xi 4) β. Thus l L(β) = l β 1 β Differetiatig with respect to β, we see that (x i 4). d dβ l L(β) = β + 1 β (x i 4). Settig this derivative to zero ad solvig for β, we obtai β = X 4. Next we see that d dβ l L(β) = β β 3 (x i 4) = β (X 4). β3 Hece ( ) d E l L(β) = dβ β β 3 ( E( X ) 4) = β β 3 β = β. Therefore V ar( β) = β, where β = X 4. The approximate (1 α)100% cofidece iterval for β is give by [ X 4 z α X 4, X 4 + z α ] X 4. 9

10 8. ( poits) Let X 1, X,..., X be a radom sample from a populatio with desity fuctio e (x θ) if θ < x < f(x; θ) = 0 otherwise, where θ R is a ukow parameter. The show that Q = X (1) θ is a pivotal quatity. Usig this pivotal quatity fid a 100(1 α)% cofidece iterval for θ. Aswer: The cumulative probability desity (cdf) of the populatio X is F (x) = x θ e t e θ dt = 1 e (x θ). The probability desity g(y) of the radom variable Y = X (1) is give by g(y) = ( ) f(y) [ 1 F (y) ] 1 = e (y θ), θ < y <. 1, 1 Suppose W = Y θ. The the cdf H(w) of W is give by H(w) = P (W w) = P (Y θ w) = P (Y w + θ) = w+θ θ e (w θ) dw. Differetiatig H(w) with respect to w, we get the pdf h(w) of the radom variable W. Thus h(w) = d dw H(w) = d dw w+θ θ e (w θ) dw = e w, 0 < w <. Hece the radom variable X (1) θ Exp ( 1 ). Therefore the distributio of X(1) θ is idepedet of θ. Hece it is a pivot for θ. Thus (1 α)100% cofidece iterval for θ ca be obtaied from α = P ( a X (1) θ b ) that is where a ad b satisfy α = a Solvig the last two equatios, we get a = 1 l ( cofidece iterval for θ is 0 α = ( X (1) b θ X (1) a ), e w dw ad 1 α b = e w dw. α [ X (1) 1 ( ) l, X (1) 1 ( )] α l, α ) 0 ad b = 1 l ( α). Therefore a (1 α)100% 10

11 9. ( poits) Let X 1, X,..., X 6 be a radom sample from a distributio with desity fuctio { θ x θ 1 for 0 < x < 1 where θ > 0 f(x) = 0 otherwise. The ull hypothesis H o : θ = 1 is to be rejected i favor of the alterative H a : θ > 1 if ad oly if at least of the sample observatios are larger tha 0.7. What is the sigificace level of the test? Aswer: Let X be the radom variable associated with the give populatio. Thus P (X > 0.7) = θ x θ 1 dx = 1 (0.7) θ. Hece the probability of each X i > 0.7 is 1 (0.7) θ. The probability of each X i > 0.7 whe θ = 1 is 1 (0.7) 1 = 0.3. α = P (Reject H o / H a is true) = P (at least five X i > 0.7 / H a is true) = ( ) 6 (0.3) (0.7) 1 + ( ) 6 (0.3) 6 (0.7) 0 6 =

12 10. ( poits) A researcher wats to test H o : θ = 0 versus H a : θ = 1, where θ is a parameter of a populatio of iterest. The statistic W, based o a radom sample of the populatio, is used to test the hypothesis. Suppose that uder H o, W has a ormal distributio with mea 0 ad variace 1, ad uder H a, W has a ormal distributio with mea 4 ad variace 1. If H o is rejected whe W > 1.0, the what are the probabilities of a Type I or Type II error respectively? Aswer: The probability of type I error is give by α = P (Reject H o / H a is true) = P (W > 1. / W N(0, 1) ) = 1 P (W 1.) = = Similarly, the probability of type II error is give by β = 1 P (Reject H o / H a is true) = 1 P (W > 1. / W N(4, 1) ) = P (W 1. / W N(4, 1) ) = P (W ) = P (Z.) =