Q:

A well-known bank credit card firm wishes to estimate the proportion of credit card holders who carry a nonzero balance at the end of the month and incur an interest charge. Assume that the desire margin of error is 0.03 at 98% confidence.a) How large a sample should be selected if it is anticipated that roughly 70% of the firm�s card holders carry a nonzero balance at the end of the month?b) How large a sample should be selected if no planning value for the proportion could be specified?

Accepted Solution

A:
Answer:a) 1263 b) 1503Step-by-step explanation:Sample size for an Interval estimate of population proportion is n = (zα/2)^2 p (1-p) / E^2 Sample size for an Interval estimate of a population mean, p is unknowmn n = (zα/2)^2 0.25 / E^2 given: proportion, p = 70 margin of error, E = 0.03 Confidence level of 98%, that means the the siginficance level α is 1 – p α = 1 – 0.98 = 0.02 Z(α /2) = Z(0.02/2) = Z (0.01) Using a Z table Z = 2.326 a) n = (Zα/2)2 p (1-p) / E2 n = 2.326^2*0.7 (1-.7)/0.03^2 n = 1262.39 = 1263 b) n = (Zα/2)2 0.25 / E2 n = 2.326^2 *0.25/0.03^2 n = 1502.85 = 1503