The critical values for z given a two-sided test with alpha = 0.05 are +/ Since 1.73 does not exceed 1.96, do not reject the null hypothesis.

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1 Problem Answers Chapter Ho: µ = 1.0 H1: µ 1.0 This is a one-sample z-test (you are given the population standard deviation). The data are continuous, values of serum-creatinine...using equation z = ( ) / (0.4 / r12) = 1.73 The critical values for z given a two-sided test with alpha = 0.05 are +/ Since 1.73 does not exceed 1.96, do not reject the null hypothesis. 7.2 Use the logic shown in equation 7.12, use a z-table to find the exact probability of z=1.73, p= Then use the second portion of equation p = 2[ ] = 2[.0418] =.0836 Using StatCrunch (remember StatCrunch is 1-tail) A t-table is used in this problem since a sample standard deviation is used to calculate the standard error (not a population value). N=7, so look in the t- table with 6 DF and find the values that 'bracket' the given value of alpha =.90, t = alpha =.95, t = At this point, you can either interpolate, or just say that p (since it is two-sided test) is between.10 and.20 (remember it is a two-tail test and you must double the p-values found in the t- table). You can use StatCrunch to get the exact p-value of (remember StatCrunch is 1-tail). -1-

2 7.4 Ho: µ = 1.0 H1: µ 1.0 This is a one-sample t-test since you are estimating the population standard deviation using a sample. The data are continuous, values of serum-creatinine...using equation t = ( ) / (0.6 / r12) = The critical values for t given a two-sided test with alpha = 0.05 and 11 DF are +/ Since does not exceed 2.201, do not reject the null hypothesis. Look in the t-table with 11 DF and find the values that 'bracket' the given value of alpha =.85, t = alpha =.90, t = At this point, you can either interpolate, or just say that p (since it is two-sided test) is between.20 and.30 (remember it is a twotail test and you must double the p-values found in the t-table). You can use StatCrunch to get the exact p-value of (remember StatCrunch is 1-tail). 7.5 Using equation 7.31, a two-sided confidence interval is / (0.6 / r12) = (0.82, 1.58) 7.6 The answers to 7.4 and 7.5 are consistent in that in 7.4, the null hypothesis was not rejected and in 7.5, the confidence band was shown to include the population mean of

3 7.7 This is a power calculation where the following have been set... #1 two-tail test #2 alpha = 0.05 #3 difference to detect = 0.10 #4 population standard deviation = 0.54 #5 sample size = 100 You can use equation M = r100 / 0.54 = / 0.54 = = Using a z-table, 0.11 has a probability of 0.54, but since the value is negative (-0.11), calculate the area to the right of that value, or = 0.46, or 46% power. Using the Power Calculator at -3-

4 7.8 This is a power calculation where the following have been set... #1 two-tail test #2 alpha = 0.05 #3 difference to detect = 0.20 #4 population standard deviation = 0.54 #5 sample size = 100 You can use equation M = r100 / 0.54 = / 0.54 = = 1.74 Using a z-table, 1.74 has a probability of 0.96, and since 1.74 is positive, power is 96%. Using the Power Calculator at -4-

5 7.9 This is a sample size calculation where the following have been set... #1 two-tail test #2 alpha = 0.05 #3 difference to detect = 0.10 #4 population standard deviation = 0.54 #5 power = 0.80 You can use equation n = 0.54**2 ( )**2 / (0.10)**2 n = * 7.84 / 0.01 n = ~ 229 Using the Power Calculator at Ho: p =.04 H1: p >.04 Since NPQ is < 5 (50x.04x.96 = 1.92), the normal approximation is not used, rather the exact binomial is used. You must calculate the exact probability that X => 5 given N=50 and P=.04. Use STATCRUNCH on the web... Using STATCRUNCH... In STATCRUNCH, you can get the probability directly, p =.049. That value is less than.05, so you can reject the null hypothesis and conclude that there is an additional risk of breast cancer given that both a mother and sister have breast cancer. -5-

6 7.22 Ho: µ = 1/90 H1: µ 1/90 Since NPQ is > 5 (538x1/90x89/90 = 5.91), use the normal approximation. This is one-sample test for a binomial proportion with a two-sided alternative, equation z = (2/538-1/90) / r(1/90 * 89/90 / 538) z = / z = Now you can use column B of Table 3 with z = 1.64 and you will see You must multiply that number by 2 (2-sided alternative). That gives The results are NOT SIGNIFICANT. There is no evidence that older women have a higher proportion of twins. Using STATCRUNCH and the STAT menu

7 7.23 Ho: µ = 0.30 H1: µ 0.30 As in 7.22, NPQ > 5, use the normal approximation. Use equation z = (110/200) / r(0.3 * 0.7 / 200) z = 0.25 /.0324 z = 7.72 This number is so large, it is well beyond the bounds of Table 3, SIGNIFICANT. Women in the erythromicin group have nausea more often than general population of pregnant women. Using STATCRUNCH (same menus as in 7.22) Ho: µ = 0.30 H1: µ 0.30 The null hypothesis is that the proportion of smokers among the volunteers (10/100) is equal the proportion of smokers in the general population (.30). The alternative is the they are not equal (a two-sided alternative) Ho: µ = 0.30 H1: µ 0.30 NPQ > 5, use equation z = ( ) / r(0.3 * 0.7 / 100) z = / z = See the answer to SIGNIFICANT. The exact p-value can be found using equation p = 2 * [1 - M(4.36)], p <.001 Using STATCRUNCH (same menus as in 7.22)