Question

Section 9.1: 8. Ginkgo for Dementia The herb ginkgo biloba is commonly used as a treatment to prevent dementia. In a study of the effectiveness of this treatment, 1545 elderly subjects were given ginkgo and 1524 elderly subjects were given a placebo. Among those in the ginkgo treatment group, 246 later developed dementia, and among those in the placebo group, 277 later developed dementia (based on data from "Ginkgo Biloba for Prevention of Dementia." by DeKosky et al., Journal of the American Medical Association, Vol. 300, No. 19). We want to use a 0.01 significance level to test the claim that ginkgo is effective in prevent ing dementia a. Test the claim using a hypothesis test. b. Test the claim by constructing an appropriate confidence interval. c. Based on the results, is ginkgo effective in preventing dementia? 14. Police Gunfire In a study of police gunfire reports during a recent year, it was found that among 540 shots fired by New York City police, 182 hit their targets; and among 283 shots fired by Los Angeles police, 77 hit their targets (based on data from the New York Times). We want to use a 0.05 significance level to test the claim that New York City police and Los Angeles police have the same proportion of hits. a. Test the claim using a hypothesis test. b. Test the claim by constructing an appropriate confidence interval. c. Based on the results, does there appear to be a difference between the hit rates of New York City police and Los Angeles police?

Answer 1

8) Answer: ---- Date: ---2/4/2020 a) Here let pj be the proportion of elderly people who developed dementia after being subject to the treatment of ginkgo and pz be the proportion of elderly people who developed demential after being given placebo. Then the null and the alternate hypothesis of our test would be: H_0:p_1\geqp_2 H_a: p_1<p_2 The test statistic would be computed as: ZA* = \frac{p_1-p_2}{SE) From the given data we have: p_1 = \frac{246}1545) = 0.1592 p_2 = \frac{277}{1524} = 0.1818 The pooled sample proportion is computed as: P = \frac{p_1*n_1+p_2*n_2Yn_1+n_2} = \frac{246+277}1545+1524} = 0.1704 Now the standard error is computed as: SE =\sqrt{ P(1-P)(\frac{1}{n_1}+\frac{1}{n_2}}} = \sqrt{0.1704(1-0.1704)(\frac{1}{1545) + \frac{1}{1524}}} = \sqrt{0.1414*0.0013} = 0.0136 Therefore the test statistic value would be: __ P1 - P20.1592 - 0.1818 SE -= -1.6618 0.0136 Therefore the p-value would be computed as: ( as it is a one tailed test)

Therefore the p-value would be computed as: ( as it is a one tailed test) p= P(Z < -1.6618) From the standard normal tables we get: p= P(Z < -1.6618) = 0.0483 Therefore as the p-value here is greater than the level of significance = 0.01, the null hypothesis cannot be rejected at 99% confidence interval here. b) The confidence interval for the true difference between the 2 population proportions is given as: P1 – P2 + 20.995 * SE From the standard normal tables we get: P(Z <2.576) = 0.995 Therefore we get: %0.995 = 2.576 Therefore the confidence interval would be: P1 – P2 + 20.995 * SE 0.1592 -0.1818 2.576 * 0.0136 -0.0226 = 0.0350 Now as clearly o lies in the above interval therefore at 99% confidence interval we cannot reject the null hypothesis here and therefore there is not much statistical significant evidence on the effectiveness of the herb ginkgo biloba. fidence interval we fan oftereiectate of the c) Clearly from both the test we see that we do not have sufficient evidence at 99% confidence interval that ginkgo effective at preventing dementia.

Date: ---2/4/2020 14) Answer: ---- Let pl be the proportion of hits for New York City police p2 be the proportion of hits for Los Angeles police (a) Ho: pl=p2 Ha: pl not equal to p2 pl=182/540 =0.337037 p2=77/283 =0.2720848 The test statistic is Z=(p1-p2)/sqrt(p1*(1-p1)/n1+p2*(1-p2)/n2) =(0.337037-0.2720848)/sqrt(0.337037*(1-0.337037)/540 + 0.2720848*(1-0.2720848)/283) -1.95 Given a=0.05, the critical values are Z(0.025)=-1.96 or 1.96 Since Z=1.95 is between -1.96 and 1.96, we do not reject Ho

(b) So 95% confidence interval is (p1-p2)+/-Z*sqrt(p1*(1-p1)/n1+p2*(1-p2)/n2) --> (0.337037-0.2720848)+/- 1.96*sqrt(0.337037*(1-0.337037)/540 + 0.2720848*(1-0.2720848)/283) --> (-0.0004549482, 0.1303593) (c) No, there does not appear to be a difference between the hit rates of New york city police and Los Angeles police