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stats 363 HW-Section-10-6

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

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

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