410Hw08ans.pdf - STAT 410 Spring 2019 A Stepanov Homework#8(due Wednesday May 1 by 4:30 p.m Please include your name with your last name underlined your

410Hw08ans.pdf - STAT 410 Spring 2019 A Stepanov...

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STAT 410 Homework #8 (due Wednesday, May 1, by 4:30 p.m.) Spring 2019 A. Stepanov Please include your name ( with your last name underlined ) , your NetID, and your section number at the top of the first page. No credit will be given without supporting work. 1. A certain STAT 410 instructor believes that at most 30% of the students start working on their homework before Thursday ( for a homework that is due on Friday ) . Let p denote the overall proportion of students who start working on their homework before Thursday. We wish to test H 0 : p ≤ 0.30 vs. H 1 : p > 0.30. A random sample of size n = 25 students is obtained. Let X denote the number of students in the sample who start working on their homework before Thursday. a) Find the “best” Rejection Region with the significance level closest to 0.10. Hint 1: Reject H 0 if X c . Hint 2: X ~ Binomial ( n = 25, p ) . Hint 3: Want c such that 0.10 = P ( Reject H 0 | H 0 is true ) = P ( X c | p = 0.30 ) . = P ( Reject H 0 | H 0 is true ) = P ( X c | p = 0.30 ) = 1 – CDF ( c – 1 | p = 0.30 ) . Want ( 1 – CDF ( c – 1 | p = 0.30 ) ) 0.10, CDF ( c – 1 | p = 0.30 ) 0.90. CDF ( 10 | p = 0.30 ) = 0.902, c – 1 = 10, c = 11. Reject H 0 if X 11 . = 0.098 . b) What is the power of the Rejection Region obtained in part (a) if p = 0.40 ? If p = 0.50 ? Hint: Power ( p ) = P ( Reject H 0 | H 0 is NOT true ) = P ( X c | p ) . Power ( p ) = P ( Reject H 0 | p ) = P ( X 11 | p ) = 1 – CDF ( 10 | p ) .
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Power ( 0.40 ) = P ( Reject H 0 | p = 0.40 ) = P ( X 11 | p = 0.40 ) = 1 – P ( X 10 | p = 0.40 ) = 1 – 0.586 = 0.414 . Power ( 0.50 ) = P ( Reject H 0 | p = 0.50 ) = P ( X 11 | p = 0.50 ) = 1 – P ( X 10 | p = 0.50 ) = 1 – 0.212 = 0.788 . c) Suppose that 10 out of 25 students in the sample start working on their homework before Thursday. Find the p-value of the test. p-value = P ( value of X as extreme or more extreme than X = 10 | H 0 true ) = P ( X 10 | p = 0.30 ) = 1 – CDF ( 9 | p = 0.30 ) = 1 – 0.811 = 0.189 .
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2 – 3. Bert and Ernie noticed that the following are satisfied when Cookie Monster eats cookies: (a) the number of cookies eaten during non-overlapping time intervals are independent; (b) the probability of exactly one cookie eaten in a sufficiently short interval of length h is approximately h ; (c) the probability of two or more cookies eaten in a sufficiently short interval is essentially zero. Therefore, X t , the number of cookies eaten by Cookie Monster by time t , is a Poisson process, and for any t > 0, the distribution of X t is Poisson ( t ). However, Bert and Ernie could not agree on the value of , the average number of cookies that Cookie Monster eats per minute. Bert claimed that it equals 1.5, but Ernie insisted that it has been less than 1.5 ever since Cookie Monster was forced to eat broccoli and carrots. Thus, the two friends decided to test H 0 : = 1.5 vs. H 1 : < 1.5. Bert decided to count the number of cookies Cookie Monster would eat in 7 minutes, X 7 , and then Reject H 0 if X 7 is too small. Ernie, who was the less patient of the two, decided to note how much time Cookie Monster needs to eat the first 4 cookies, T 4 , and then Reject H 0 if T 4 is too large.
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  • Spring '08
  • AlexeiStepanov

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