Help me solve this Question HelpA physical therapist wants to determine the difference in the proportion of men and women whoparticipate in regular sustained physical activity. What sample size should be obtained if she wishes theestimate to be within two percentage points with 90% confidence, assuming that(a) she uses the estimates of 21.5% male and 19.8% female from a previous year?(b) she does not use any prior estimates?= (Round up to the nearest whole number.)(b) n=(Round up to the nearest whole number.)

Name: Help me solve this Question HelpA physical therapist wants to determin
Uploaded: 2024-03-05T11:52:21.000Z
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Description: Help me solve this Question HelpA physical therapist wants to determine the difference in the proportion of men and women whoparticipate in regular sustained physical activity. What sample size should be obtained if she wishes theestimate to be withi

Given: A physical therapist wants to determine the difference in the proportion of men and women who participate in the regular sustained physical activity. The level of confidence is 95%, which implies: The level of significance(α)=1-0.95 =0.05 The margin of error( M.E) =2%=0.02. (a) Let p^1 be the sample proportion of men who participated in regular sustained physical activity (p^1=21.5%). Let p^2 be the sample proportion of women who participated in regular sustained physical activity(p^2=19.8%). Let n represents the size of the sample. As the sample proportions are known, the sample size is given as: n=zcM.E2p^1(1-p^1)+p^2(1-p^2) =1.960.022(0.215×(1-0.215)+(0.198×(1-0.198) =3145.99 ~3146 Therefore, the sample size when the therapist estimate of 21.5% men and 19.8% women from a previous year is 3146.(b) When the therapist does not use any prior estimate, the sample size is given as: n=12zcM.E2 =121.960.022 =4802 Therefore, the sample size when the therapist does not use any prior estimate is 4802. Note: From the z table, the critical value at 0.05 level of significance is zc=1.96.