Solved: A well-known bank credit card firm wishes to estimate the proportion of credit card holders who carry a nonzero balance

ByDr. Raju Chaudhari

Oct 9, 2020

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 desired margin of error is .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?

Solution

a. It is anticipated that roughly 70% of the firm's card holders carry a nonzero balance at the end of the month.

Given that $\hat{p}=0.7$.

The formula to estimate the sample size required to estimate the proportion is
 $$n =p*(1-p)\bigg(\frac{z}{E}\bigg)^2$$
where $p$ is the proportion of success, $z$ is the $Z_{\alpha/2}$ and $E$ is the margin of error.

The margin of error is $E =0.03$. The confidence coefficient is $0.98$.

The critical value of $Z$ is $Z_{\alpha/2} = 2.3263$.

The minimum sample size required to estimate the proportion is

 \begin{aligned} n&= p(1-p)\bigg(\frac{z}{E}\bigg)^2\\ &= 0.7(1-0.7)\bigg(\frac{2.3263}{0.03}\bigg)^2\\ &=1262.7234\\ &\approx 1263. \end{aligned}

Thus, the sample of size $n=1263$ will ensure that the $98$% confidence interval for the proportion will have a margin of error $0.03$.

b. If no planning value for the proportion could be specified, then assume that $\hat{p}=0.50$.

Given that $\hat{p}=0.5$.

The formula to estimate the sample size required to estimate the proportion is

 $$n =p*(1-p)\bigg(\frac{z}{E}\bigg)^2$$

where $p$ is the proportion of success, $z$ is the $Z_{\alpha/2}$ and $E$ is the margin of error.

The margin of error is $E =0.03$. The confidence coefficient is $0.98$.

The critical value of $Z$ is $Z_{\alpha/2} = 2.3263$.

The minimum sample size required to estimate the proportion is

 \begin{aligned} n&= p(1-p)\bigg(\frac{z}{E}\bigg)^2\\ &= 0.5(1-0.5)\bigg(\frac{2.3263}{0.03}\bigg)^2\\ &=1503.2421\\ &\approx 1504. \end{aligned}

Thus, the sample of size $n=1504$ will ensure that the $98$% confidence interval for the proportion will have a margin of error $0.03$.