nswers>Math>Statistics and Probability Question #303943 A regional hardware chain is interested in estimating the proportion of their customers who own their own homes. There is some evidence to suggest that the proportion might be around 0.70. Given this, what sample size is required if they wish a 90 percent confidence level with a margin of error of ± .025?

The critical value for $\alpha = 0.1$ is $z_c = z_{1-\alpha/2} = 1.6449.$

The corresponding confidence interval is computed as shown below:

$CI(Proportion)=(\hat{p}-z_c\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}},$

$\hat{p}+z_c\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}})$

Given $\hat{p}=0.7.$ Then

$1.6449\sqrt{\dfrac{0.7(1-0.7)}{n}}=0.025$

$n=0.21(\dfrac{1.6449}{0.025})^2\approx909$

Required sample size is 909.