STATISTICS PROBLEM Of interest is to estimate the proporti

STATISTICS PROBLEM

Of interest is to estimate ? = the proportion of all VCU students who have eaten food from Croutons, Salads and Wraps. To estimate this proportion a 95% confidence interval will be calculated and the goal is that the margin of error will be no larger than .039. What is the minimum number of current VCU students that would need to be selected to allow the calculation of a 95% confidence interval with margin of error no greater than .039? It can be assumed for this problem only that the proportion of all current U.S. college students who have eaten food from Croutons, Salads and Wraps is .08 and this number can be used for this problem. Please circle your final answer. (5 pts)

Solution

Given a=1-0.95=0.05, Z(0.025) = 1.96 (from standard normal table)

So n=(Z/E)^2*p*(1-p)

=(1.96/0.039)^2*0.08*(1-0.08)

=185.892

Take n=186