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STUDY GUIDE ANSWERS Part 2

STUDY GUIDE ANSWERS Part 2 - b Xi 96.08 X = 1.5(2.7 96.08 =...

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b) 7 . 2 08 . 96 X 5 . 1 ± ² i ; X i = +1.5 (2.7) + 96.08 = 4.05 + 96.08 = 100.13 ; 7 . 2 08 . 96 X 8 . ± ± i ; X i = -.8 (2.7) + 96.08 = -2.16 + 96.08 = 93.92 2. a) area above z = -.4 = .1544 + .5 = .6544, so 65.44 area above z = .93 = .1762, so %; 17.62 % b) area beyond z = .4 is .1554, so 15.54 % Chapter 5 1. 67 . 4 54 . 52 . 2 25 7 . 2 6 . 98 08 . 96 70 . 2 , 08 . 96 X ± ± ± z V This z is larger in magnitude than the critical value (1.96) so the result is significant at the .05 level (two-tailed). 2. 33 . 9 100 7 . 2 52 . 2 ± ± z For D = .01, z crit = 2.33 (one-tailed), or 2.58 (two-tailed). In either case these results are easily significant at the .01 level. Because the sample size was 4 times larger for part b than part a, the z-score from part a could have been simply multiplied by 2 (i.e., the square-root of 4). Chapter 6 1. a) 25 7 . 2 x V ; x crit z X V P r = 96.08 ± 1.96 (.54); so, P = 96.08 ± 1.06 Therefore, the 95% CI goes from 95.02 to 97.14 (note, as a check, that the sum of these two limits is 192.16, which is exactly twice as large as the sample mean, as will always be the case). Because the proposed population mean of 98.6 is not in this 95% CI, we know that the sample mean would differ significantly from this population mean at the .05 level.
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