hw10sol - STAT401: Homework 10 Solutions (Spring 2009)...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: STAT401: Homework 10 Solutions (Spring 2009) 9.5.3. 1. p-value is 1- (1 . 61) = 1- . 9463 = 0 . 0537 9.6.4. 1. (a) ( a ) = - a / n + z Here, = 1, a = 2, = . 05, = 4 . 1, Z . 05 = 1 . 645 The probability of type II error is: 1- 2 4 . 1 / 36 + 1 . 645 = (0 . 1815854) = 0 . 5720459 . (b) Note that H a : > 1, we need to use formulas below: n = ( z . 05 + z . 1 ) - a 2 Since our = 0 . 1, Z = 1 . 28 , = 0 . 05, Z = 1 . 645 that n = ( z . 05 + z . 1 ) - a 2 = 4 . 1(1 . 645 + 1 . 28) 1- 2 2 = 143 . 82 . The number of sample we need is n = 144. 10.4.3. 1. i. H : 1- 2 = 0 vs H a : 1- 2 6 = 0 ii. TS: Z = X 1- X 2 r S 2 1 n 1 + S 2 2 n 2 = 160592- 159778 q 3954 2 36 + 15533 2 42 = 0 . 33275 RR: | Z | z / 2 iii. No assumption for the distribution iv. z / 2 = z . 025 = 1 . 96 Since Z = 0 . 33 < 1 . 96 we fail to reject the null hypothesis....
View Full Document

Page1 / 2

hw10sol - STAT401: Homework 10 Solutions (Spring 2009)...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online