Question

Exercise 6.12 presents the results of a poll where 48% of 331 Americans who decide to not go to college do so because they cannot a ord it. (a) Calculate a 90% con dence interval for the proportion of Americans who decide to not go to college because they cannot a ord it, and interpret the interval in context. (b) Suppose we wanted the margin of error for the 90% con dence level to be about 1.5%. How large of a survey would you recommend?

Answer 1

Answer with explanation:

As per given , we have

$\hat{p}=0.48$   , n=331

Critical value for 90% Confidence interval : $z_{\alpha/2}=1.645$

a) Confidence interval :

$\hat{p}\pm z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

$0.48\pm (1.645)\sqrt{\dfrac{0.48(1-0.48)}{331}}\\\\\approx 0.48\pm0.02746\\\\=(0.48-0.02746,\ 0.48+0.02746)\\\\=(0.45254,\ 0.50746)$

Hence, a 90% confidence interval for the proportion of Americans who decide to not go to college because they cannot a ord it : $(0.45254,\ 0.50746)$

b) Margin of error : E=1.5%=0.015

Formula for sample size : $n=p(1-p)(\dfrac{z_{\alpha/2}}{E})^2$

For p =0.48 , we  have

$n=0.48(1-0.48)(\dfrac{1.645}{0.015})^2=3001.88373333\approx3002$

Hence, the required sample size to survey = 3002