Question

a simple random of 256 college students showed that 142 of them agree with a certain law. construct the 95% confidence interval estimate of the proportion of students who agree with this law

a. state the critical value

b. compute the margin of error

c. state the confidence interval

d. compute the minimum sample size that will keep the margin of error within 2%. use the information given in this problem

Answer 1

Given na 256, x= 142, q=0.05 P = 142 = 0.5546875 @ Critical value, 2x = 20.025 = 1.959964 ® Margin of eroors 2g * PCI-P) = 1-959964 *, (0:5546875)*(-0.5546875) = 0.060881415 © Confidence interval for the proportion of students who agree with the law SE § £ margin of error SE 0.5546875 $ 0.060881415 256 ie - 0.493806084 < s < 0.615568915 - ② Given, margin of error = 0.02 FPC1P 20.025 n = 0.02 1959964 * (0.5546875) (1-0.5546875) η. 0.02 1.959 964 * 0.247009277 = 0.02

(1.9599642 (0.247009277) (0.02)² n = 2372.189952 in a 2872