stats 363 HW-Section-10-6

stats 363 HW-Section-10-6 - Section 10.6 Exercise 10.49 ....

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: Section 10.6 Exercise 10.49 . It is given that n = 15 and s 2 = 0 : 3214. A 90% con&dence interval for & 2 is ( n & 1) s 2 2 &= 2 < & 2 < ( n & 1) s 2 2 1 & &= 2 : With = 0 : 10 and df = n & 1 = 14, we read from Table 5 that 2 &= 2 = 2 : 05 = 23 : 6848 and 2 1 & &= 2 = 2 : 95 = 6 : 57063 and the desired 90% con&dence interval for & 2 is (15 & 1) (0 : 3214) 23 : 6848 < & 2 < (15 & 1) (0 : 3214) 6 : 57063 or : 190 < & 2 < : 685 : Exercise 10.51 . The sample variance s 2 is calculated as follows: n X i =1 x i = 353 + 351 + 351 + 355 = 1 ; 410 n X i =1 x 2 i = 353 2 + 351 2 + 351 2 + 355 2 = 497 ; 036 and s 2 = 1 n & 1 n X i =1 ( x i & x ) 2 = 1 n & 1 2 4 n X i =1 x 2 i & 1 n n X i =1 x i ! 2 3 5 = 1 4 & 1 & 497 ; 036 & 1 4 (1 ; 410) 2 = 11 3 : The hypotheses to be tested are H : & = 0 : 7 versus H a : & > : 7 : The test statistic is 2 = ( n & 1) s 2 & 2 = (4 & 1) 11 3 (0 : 7) 2 = 22 : 449 : With = 0 : 05 and df = n & 1 = 3, we read from Table 5 that 2 & = 2 : 05 = 7...
View Full Document

This note was uploaded on 04/20/2008 for the course MTH 1363 taught by Professor Cheng during the Spring '08 term at Creighton.

Page1 / 4

stats 363 HW-Section-10-6 - Section 10.6 Exercise 10.49 ....

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