STA113_Midterm2_HWsol

STA113_Midterm2_HWsol - 7.1 a. z / 2 = 2 . 81 suggests that...

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Unformatted text preview: 7.1 a. z / 2 = 2 . 81 suggests that / 2 = 1- (2 . 81) = 0 . 0025. Thus = 0 . 005 and the corresponding confidence level is 99 . 5%. b. z / 2 = 1 . 44 suggests that / 2 = 1- (2 . 81) = 0 . 075. Thus = 0 . 15 and the corresponding confidence level is 85%. c. 99 . 7% implies an value of 0 . 003 and consequently z / 2=0 . 0015 = 2 . 96. d. 75% implies an value of 0 . 25 and consequently z / 2=0 . 125 = 1 . 15. 7.4 a. 58 . 3 1 . 96 3 25 = [57 . 1 , 59 . 5] b. 58 . 3 1 . 96 3 100 = [57 . 7 , 58 . 9] c. 58 . 3 2 . 58 3 100 = [57 . 5 , 59 . 1] d. 82% confidence implies an value of 0 . 18 and consequently z / 2=0 . 09 = 1 . 34 and the interval based on this would be 58 . 3 1 . 34 3 100 = [57 . 9 , 58 . 7]. e. n = h 2 2 . 58 3 1 i 2 = 239 . 62 240 7.13 a. x 1 . 96 . 163 69 = [0 . 990 , 1 . 066] 7.24 8 . 17 1 . 96 1 . 42 56 = [7 . 798 , 8 . 542]. No assumptions are made on the distribution of percentage elongation. 7.34 a. A 95% lower confidence bound is 8 . 48- 1 . 771 . 79 14 = 8 . 11. With 95% confi- dence, the true mean proportional limit stress of all such joints lies in the interval [8 . 11 , ). If this interval is calculated for infinitely many samples, 95% of these intervals will contain the true mean. We must assume that the sample observations were coming from a normally distributed population.sample observations were coming from a normally distributed population....
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This note was uploaded on 01/17/2010 for the course STA 113 taught by Professor Staff during the Fall '08 term at Duke.

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STA113_Midterm2_HWsol - 7.1 a. z / 2 = 2 . 81 suggests that...

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