Question

Ch8 Sec1: Problem 4 Previous Problem List Next 1 point) Cora wants to determine a 99 percent confidence interval for the true proportion of high school students in the area who attend their home basketball games. How large of a sample must she have to get a margin of error less than 0.037 [To find n, use the value p-1/2 for the sample proportion and the values for z" from a z-table or t-table.] Round to the smallest integer that works.] n-

Answer 1

Given The confidence Interval = 99% it can be written as (1 - alpha) = 0.99 alpha = 0.01 implies alpha/2 = 0.005 The critical z-value for the given Interval is Z_alpha/2 = Z_0.005 = 2.58 The z-value above can be calculated from standard Normal table where probability is 0.005 we know that the Margin of Error (E) = Z_alpha/2 times standard deviation standard deviation = squareroot p(1 - p)/n Therefore E = Z_alpha/2 times squareroot p(1 - p)/n implies n = (Z_alpha/2 times squareroot p(1 - p)/E)^2 Here Assume p = 1/2 is given, Margin of error E = 0.03 is given n = (2.58 times squareroot 1/2 times 1/2/0.03)^2 = (2.58 times 1/2/0.03)^2 n = 43^2 = 1849 implies required sample size