Many people think that a national lobby's successful fight against gun control legislation is reflecting the will of a minority of Americans. A previous random sample of 4000 citizens yielded 2250 who are in favor of gun control legislation. How many citizens would need to be sampled if a 99% confidence interval was desired to estimate the true proportion to within 5%? 611 690 653 664

Answer:416Step-by-step explanation:Given that:A previous random sample = 4000Sample mean = 2250[tex]\hat p = \dfrac{x}{n}[/tex][tex]\hat p = \dfrac{2250}{4000}[/tex][tex]\hat p = 0.5625[/tex]The confidence interval level = 90%Desired Margin of error E = 0.04The level of significance at 90% C.I  = 1 - 0.90 = 0.10The critical value [tex]Z_{\alpha/2} = Z_{0.10/2}[/tex][tex]\implies Z_{0.05} = 1.645[/tex]   ( From the  z tables)[tex]n = \hat p \times (1- \hat p ) \bigg ( \dfrac{z_{0.05}}{E} \bigg)^2[/tex][tex]n = 0.5625 \times (1- 0.5625) \bigg ( \dfrac{1.645}{0.04} \bigg)^2[/tex][tex]n = 0.5625 \times (0.4375) \bigg ( 41.125 \bigg)^2[/tex]n = 416.21 Thus, the number of citizens required to be sampled is [tex]\simeq[/tex] 416