We are interested in conducting a study to determine the percentage of voters of a state would vote for the incumbent governor. what is the minimum sample size needed to estimate the population proportion with a margin of error of .05 or less at 95% confidence? hint: use a planning proportion of p*=0.50a. 200.b. 100.c. 385.d. 58.

Answer:[tex]n=\frac{0.5(1-0.5)}{(\frac{0.05}{1.96})^2}=384.16[/tex]  And rounded up we have that n=385 Step-by-step explanation:In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2 =0.025[/tex]. And the critical value would be given by: [tex]z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96[/tex] The margin of error for the proportion interval is given by this formula:  [tex] ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]    (a)  And on this case we have that [tex]ME =\pm 0.05[/tex] and we are interested in order to find the value of n, if we solve n from equation (a) we got:  [tex]n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}[/tex]   (b)  The estimated proportion for this case is [tex]\hat p=0.5[/tex]. And replacing into equation (b) the values from part a we got: [tex]n=\frac{0.5(1-0.5)}{(\frac{0.05}{1.96})^2}=384.16[/tex]  And rounded up we have that n=385