Answer to Question #98712 in Statistics and Probability for Juliet Beglaryan

Question

Question #7A sample of 75 students found that 55 of them had cell phones. What is the margin of errorfor a 95% confidence interval estimate for the proportion of all students with cell phones?
Question #8 You buy a package of 122 Smarties and 19 of them are red. What is a 95% confidence interval for the true proportion of red Smarties?
Question #9We want to construct a 95% confidence interval for the true proportion of all adult males who have spent time in prison, with a margin of error of 0.02. From previous studies, we believe the proportion to be somewhere around 0.07. What is the required sample size?

Answer 1

#7.

$p=\frac{55}{75}=0.7333.$

$95\%ME=1.96\sqrt{\frac{p(1-p)}{n}}=1.96\sqrt{\frac{0.733(1-0.733)}{75}}=0.1001.$

#8.

$p=\frac{19}{122}=0.1557.$

$95\%CI=(p-1.96\sqrt{\frac{p(1-p)}{n}},\;0.733+1.96\sqrt{\frac{0.733(1-0.733)}{75}})=\\ =(0.1557-1.96\sqrt{\frac{0.1557(1-0.1557)}{122}},\;0.1557+1.96\sqrt{\frac{0.1557(1-0.1557)}{122}})=\\ =(0.0914,0.2200).$

#9.

$E=1.96\sqrt{\frac{p(1-p)}{n}}.$

$n=(\frac{1.96}{E})^2p(1-p)=(\frac{1.96}{0.02})^20.07(1-0.07)=626.$