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HW7-Solutions.pdf

HW7-Solutions.pdf - BlbTOI I01 AAA HM Solodlm Hw#7...

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Unformatted text preview: Blb§TOI I01: AAA HM) Solod‘lm Hw#7 Ht)? (002. (on [0.1+ [LS 106 (0% (/7 Exercise 12 a. The null hypothesis for this test is Ho:p=136mmHg and the alternative hypothesis is HA:p7é136mmHg. The test statistic is t I! _ ”'03 "MIA/lI 143 — 136 24-4/w/8_6 = 2.66. For a 1: distribution with 85 degrees of freedom, 2(0.0005) < p < 2(0.005) or 0.001 < p < 0.01. Therefore, we reject H0 at the 0.10 level of significance. The mean systolic blood pressure for workers who have experienced a major coronary event is not equal to the mean for workers who have not. b. The null hypothesis is H0211 = 84 mm Hg and the alternative hypothesis is H A: p. 9E 84 mm Hg. The test statistic is 51d - [‘0d 84/ J71 87 — 84 16.0/w/8—6 = 1.74. For a t distribution with 85 degrees of freedom, 2(0.025) < p < 2(0.05) or 0.05 < p < 0.10. Therefore, we again reject H0 at the 0.10 level. c. The workers who have experienced a major coronary event have a. higher mean systolic blood pressure and also a higher mean diastolic blood pressure than the workers who have not. Exercise 14 a. The probability of committing a type I error is P(typeIerror) = a 0.05. b. Since we are conducting a one-sided test at the 0.05 level of significance, the null hypothesis H02 [1. 2 24-4 mg/100 ml would be rejected for z 5 —1.645. Solving for 5:, —1.645 = : J; = 5—244 41/w/2—5 and z = 230.5. Consequently, the null hypothesis would be rejected if the sample has a mean that is less than or equal to 230.5 mg/100 ml. We must now find the proportion of the area under the curve centered at the true mean [i = 219 nag/100 ml that lies to the right of 230.5. This is the area where Ho fails to be rejected. Solving for z in this case, 230.5 — 219 41/J2‘5 = 1.40. The area to the right of z = 1.40 is 0.081. Therefore, the probability of committing a type II error is P(typeIIerror) = fl — 0.081. c. The power of the test is power ll P(reject Ho I H0 is false) 1 — I3 0.919. II II d. The power could be increased by increasing a, the probability of making a type I error; by considering a. hypothesized mean that is larger than 244 mg/ 100 ml and therefore further from 219; or by increasing the sample size n. e. Since a = 0.05, Ho would be rejected for z 3 -1.645. Writing 2 = —1.645 2‘: — 244 41/fi and solving for 5:, 41 I = 244—1.45 -— . z 6 («5) The null hypothesis would be rejected for this value. Furthermore, the value of 2 which corresponds to [i = 0.05 is 1.645; for the distribution centered at m = 219 mg/ 100 ml, 5: —- 219 41/ x/fi 1.645 = and i = 219+ 1.645 (-3). § Equating the two expressions for 5:, therefore, _ [(1.645+1.645)(41) 2 n ‘ (244-219) 1 29.1. m A sample of size 30 would be required. f. The value of 2 that corresponds to [3 = 0.10 is 1.28. In this case, therefore, _ (1.28 + 1.645)(41) ’ n _ [ (244— 219) ] ‘ = 23.0. A sample of size 23 would be required. CHAPTER 11 I ( \ 5 Exercise 5 a. The samples are paired. t b. The null hypothesis is HO: I‘corn _ Hosts = 0 and the alternative hypothesis is HA: I‘corn _ floats 7e 0' c. Since the data are paired, we begin by calculating the difierence in LDL cholesterol levels for each person in the study. Note that a? = 0.363 mmol/l and 34 = 0.406 mmol/l. Therefore, the test statistic is J — 6 Sd/x/fi 0.363 — o mos/m = 3.35. For a t distribution with 14 — 1 = 13 degrees of freedom, 0.001 < p < 0.01. We reject H0 at the 0.05 level of significance. Hxé {IX d. We conclude that the true difference in population mean cholesterol levels (or the true mean diEerence) is not equal to 0. Mean LDL cholesterol is lower when individuals are adhering to the oat bran diet. Exercise 6 a. The null hypothesis is H0: “cancer _ ”control = 0 and the alternative hypothesis is H A : ”cancer “ ”control 5e 0' Because the data are paired, the test statistic is J — 6 8.1/1/73 2.7 — 0 15.9N171 = 2.22. For a. t distribution with 171 — 1 = 170 degrees of freedom, 0.02 < p < 0.05. We reject H0 at the 0.05 level of significance and conclude that the true difierence in population mean blood levels of DDE is not equal to 0. The mean level is higher among women with breast cancer. b. Because the null hypothesis was rejected at the 0.05 level, we would not expect the 95% confidence interval to contain the value 0. Exercise 8 a. The two samples of data are independent. b. The null hypothesis is Ho: #1 = #2 and the alternative hypothesis is HA: Ill 75 I12- c. We begin by calculating the pooled estimate of the variance; note that s 2 = (in — 1):},2 + (112 — 1).sz22 m + n; — 2 P - (77 — 1)(0.026)2 + (161 — 1)(o.025)2 77 + 151 - 2 0.00064. Therefore, the test statistic is t _ (51 —52)- (#1 —l£2) ‘/sp2(1/n1 + 1/n2) (0.098 — 0.095) — 0 ,/(0.00064)[(1/77) + (1/161)] 0.86. ll A t distribution with 77 + 161 — 2 = 236 df can be approximated by the normal distribution. Therefore, p = 2(0.195) = 0.390 and we are unable to reject Ho. This study fails to provide evidence that maternal cigarette smoking has an effect on the bone mineral content of newborns. Exercise 9 a. The null hypothesis of the test is H01 #1 = #2 and the alternative hypothesis is HA3 #1 $9 #2- Since .91 = .92 = 8 mm Hg, the pooled estimate of the variance is 3P2 = 82 = 64. Furthermore, the test statistic is (5'1 ‘ 52) — (#1 — #2) 8p2[(1/n1)+(1/"»2)l (111 — 109) — o 54 [(1/23) + (1/24)] = 0.86. eo- | For a t distribution with 23 + 24 — 2 = 45 degrees of freedom, p > 0.10. Therefore, we are unable to reject H0 at the 0.01 level of significance. We do not have any evidence that mean arterial blood pressure differs for the two populations of women. b. To begin, we can approximate the t distribution with 4.5 (if by the t distribution with 40 df. In this case, 99% of the observations are enclosed by the values —2.704 and 2.704. (In fact, if df = 45, then 99% of the observations lie between -2.690 and 2.690.) Therefore, a 99% confidence interval for the true difl'erence in population means 111 — 112 is 1 1 (5:1 — 5:2) i 2.704 3,2 [— + —] n1 n; 01' 1 1 __ 1 _ _ __ (111 09)i2 704 64 [23 + 24] 01' (—4.3, 8.3). This interval does contain the value 0. Given that we were unable to reject the null hypothesis at the 0.01 level, we should have expected that it would. ...
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