Answer:
A 68 percent confidence interval for the proportion of persons who work less than 40 hours per week is (0.251, 0.351), or equivalently (25.1%, 35.1%)
Step-by-step explanation:
We have a large sample size of n = 83 respondents. Let p be the true proportion of persons who work less than 40 hours per week. A point estimate of p is [tex]\hat{p} = 0.301[/tex] because about 30.1 percent of the sample work less than 40 hours per week. We can estimate the standard deviation of [tex]\hat{p}[/tex] as [tex]\sqrt{\hat{p}(1-\hat{p})/n}=\sqrt{0.301(1-0.301)/83} = 0.0503[/tex]. A [tex]100(1-\alpha)%[/tex] confidence interval is given by [tex]\hat{p}\pm z_{\alpha/2}\sqrt{\hat{p}(1-\hat{p})/n[/tex], then, a 68% confidence interval is [tex]0.301\pm z_{0.32/2}0.0503[/tex], i.e., [tex]0.301\pm (0.9944)(0.0503)[/tex], i.e., (0.251, 0.351). [tex]z_{0.16} = 0.9944[/tex] is the value that satisfies that there is an area of 0.16 above this and under the standard normal curve.
