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 ) .
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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