In a survey sample of 83 respondents, about 30.1 percent of the sample work less

Question

In a survey sample of 83 respondents, about 30.1 percent of the sample work less than 40 hours per week. What is the estimated standard error of the proportion for this group? In a survey sample of 83 respondents, about 30.1 percent of the sample work less than 40 hours per week. What is the estimated standard error for the group of respondents who work hours or more per week? In a survey sample of 83 respondents, about 30.1 percent of the sample work less than 40 hours per week. Calculate a 68 percent confidence interval for the proportion of persons who work less than 40 hours per week. In a survey sample of 83 respondents, about 30.1 percent of the sample work less than 40 hours per week. Calculate a 68 percent confidence interval for the proportion of persons who work 40 hours or more per week.

Explanation / Answer

(28) let p=proportion who work less than 40 hours=p=0.301

sample size=n=83

standard error of proportion p=SE(p)=sqrt(p(1-p)/n)=sqrt(0.301*(1-0.301)/83)=0.0503

(29) proportion who work 40 or more hours=q=1-proportion who work less than 40 hours=1-p=1-0.301=0.699

sample size=n=83

standard error of proportion q=SE(q)=sqrt(q(1-q)/n)=sqrt(0.699*(1-0.699)/83)=0.0503

(30) (1-alpha)*100% confidence interval for p=p± z(alpha/2)*SE(p)

68% confidence interval for p=0.301±z(0.32/2)*0.0503=0.301±0.9945*0.0503=0.301±0.05=(0.251,0.351)

(31) (1-alpha)*100% confidence interval for q=q± z(alpha/2)*SE(q)

68% confidence interval for q=0.699±z(0.32/2)*0.0503=0.699±0.9945*0.0503=0.699±0.050=(0.649,0.749)

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