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mont4e_sm_ch08_sec04 - Section 8-4 8-38 02.05,10 = 18.31...

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Section 8-4 8-38 31 . 18 2 10 , 05 . 0 = χ 49 . 27 2 15 , 025 . 0 = χ 22 . 26 2 12 , 01 . 0 = χ 93 . 46 2 25 , 005 . 0 = χ 85 . 10 2 20 , 95 . 0 = χ 01 . 7 2 18 , 99 . 0 = χ 14 . 5 2 16 , 995 . 0 = χ 8-39 99% lower confidence bound for σ 2 For α = 0.01 and n = 15, 29.14 = 2 1 , n α χ = 2 14 , 01 . 0 χ 2 2 2 00003075 . 0 14 . 29 ) 008 . 0 ( 14 σ σ 8-40 95% two sided confidence interval for σ 8 . 4 10 = = s n 02 . 19 2 9 , 025 . 0 2 1 , 2 / = = χ χ α n and 70 . 2 2 9 , 975 . 0 2 1 , 2 / 1 = = χ χ α n 76 . 8 30 . 3 80 . 76 90 . 10 70 . 2 ) 8 . 4 ( 9 02 . 19 ) 8 . 4 ( 9 2 2 2 2 < < σ σ σ 8-41 95% confidence interval for σ : given n = 51, s = 0.37 First find the confidence interval for σ 2 : For α = 0.05 and n = 51, 71.42 and 32.36 χ α / , 2 1 2 n = χ 0 025 50 2 . , = χ α 1 2 1 2 = / ,n χ 0 975 50 2 . , = 36 . 32 ) 37 . 0 ( 50 42 . 71 ) 37 . 0 ( 50 2 2 2 σ 0.096 σ 2 0.2115 Taking the square root of the endpoints of this interval we obtain, 0.31 < σ < 0.46 8-42 95% confidence interval for σ 09 . 0 17 = = s n 85 . 28 2 16 , 025 . 0 2 1 , 2 / = = χ χ α n and 91 . 6 2 16 , 975 . 0 2 1 , 2 / 1 = = χ χ α n 137 . 0 067 . 0 0188 . 0 0045 . 0 91 . 6 ) 09 . 0 ( 16 85 . 28 ) 09 . 0 ( 16 2 2 2 2 < < σ σ σ 8-14
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8-43 The data appear to be normally distributed based on examination of the normal probability plot below. Therefore, there is evidence to support that the mean temperature is normally distributed.
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