05exS_p1

05exS_p1 - √ n 1 . 176 > z ∗ 3 √ 70 z 6 1...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
ISOM111 Business Statistics (Fall 2010) Solutions for Tutorial Exercise 5 1. Let X be the number of recovery days by new therapy and μ be its mean. (a) x =12 σ =3 n =70 As n =70 > 30 , by CLT, X N ( μ, σ 2 70 ) The 95% con f dence interval for the mean recovery days by new therapy is : x ± z 0 . 025 ( σ n )=12 ± 1 . 96( 3 70 )=(11 . 2972 , 12 . 7028) (b) x =12 σ =3 n =25 As we don’t know the distribution of X n =25 < 30 assume X N X N ( μ, σ 2 70 ) The 95% con f dence interval for the mean recovery days by new therapy is : x ± z 0 . 025 ( σ n )=12 ± 1 . 96( 3 25 )=(10 . 824 , 13 . 176) (c) B =0 . 5 B> n n> ( 2 . 576 3 0 . 5 ) 2 > 238 . 8879 Take z =2 . 576 ,n> 238 . 8879 n =239 Take z =2 . 575 ,n> 238 . 7025 n =239 Take z =2 . 58 ,n> 239 . 6304 n =240 (d) From part (b), B =1 . 176 B >
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: √ n 1 . 176 &gt; z ∗ 3 √ 70 z 6 1 . 176 ∗ √ 70 3 6 3 . 28 With z = 3 . 28 , con f dence level = (0 . 49948 ∗ 2) ∗ 100% = 99 . 89% (e) 92% C.I. for the di f erence in mean recovery time is : Take z = 1 . 75 ¡ X 1 − X 2 ¢ ± Z . 04 ∗ r σ 2 1 n 1 + s 2 2 n 2 = (12 − 14) ± 1 . 75 ∗ r 3 2 70 + 2 2 60 = ( − 2 . 773251 , − 1 . 226749) Take z = 1 . 76 ¡ X 1 − X 2 ¢ ± Z . 04 ∗ r σ 2 1 n 1 + s 2 2 n 2 = (12 − 14) ± 1 . 76 ∗ r 3 2 70 + 2 2 60 = ( − 2 . 777669 , −− 1 . 222331) 1...
View Full Document

This note was uploaded on 01/06/2011 for the course ISOM 111 taught by Professor Hu,inchi during the Fall '10 term at HKUST.

Ask a homework question - tutors are online