This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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....
View
Full
Document
This note was uploaded on 01/17/2010 for the course STA 113 taught by Professor Staff during the Fall '08 term at Duke.
 Fall '08
 Staff

Click to edit the document details