Sample Size for Estimating the Population Proportion

If 9 out of 25 randomly selected WMU students live in Southwest Michigan, the 95% confidence interval for the true proportion is $\hat{p} \pm 2\sqrt{\hat{p}(1-\hat{p})}/\sqrt{n} = .36 \pm .19$. This result says that the true proportion can be as low as .17 or as high as .55. If we wanted to reduce the margin of error from .19 to some value M, then we set the formula for margin of error equal to M, i.e $M=2\sqrt{\hat{p}(1-\hat{p})}/\sqrt{n}$. Solving for n gives the result we need.  

   

Example: Suppose we want to reduce the margin of error for estimating the population proportion from .19 to .10. Using the estimate $\hat{p}=.36$ based on the initial sample, the sample size we need is: n=4 (.36)(.64)/(.10)²=92. To verify that this is the correct sample size, the 95% confidence interval would be computed (if the sample proportion remained at .36) as $.36 \pm 2\sqrt{(.36)(.64)}/\sqrt{92}= .36 \pm .10$.