Question

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?

Answer 1

Answer: a) 1267

b)  1509

Step-by-step explanation:

As per given , we have

Significance level : $\alpha: 1-0.98=0.02$

Using the z-value table , the critical z value for 98% confidence : $z_{\alpha/2}=2.33$

Margin of error : E= 0.03

Formula for sample size :

$n=p(1-p)(\dfrac{z_{\alpha/2}}{E})^2$ , where p is the prior estimate of population proportion.

a) p=0.70

Sample size : $n=(0.7)(1-0.7)(\dfrac{2.33}{0.03})^2$

Simplify , we get

$n=11266.74333333\approx1267$

Hence, the minimum sample size required = 1267

a) if no estimate of prior population proportion is given , we take p= 0.5

Sample size : $n=(0.5)(1-0.5)(\dfrac{2.33}{0.03})^2$

Simplify , we get

$n=1508.02777778\approx1509$

Hence, the minimum sample size required = 1509