(1 point) In a sample of credit card holders the mean monthly value of credit card purchases was $\$ 244$ and the sample variance was 57 ( $\$$ squared). Assume that the population distribution is normal. Answer the following, rounding your answers to two decimal places where appropriate. (a) Suppose the sample results were obtained from a random sample of 12 credit card holders. Find a $95 \%$ confidence interval for the mean monthly value of credit card purchases of all card holders. (ㅁ) (b) Suppose the sample results were obtained from a random sample of 19 credit card holders. Find a $95 \%$ confidence interval for the mean monthly value of credit card purchases of all card holders. (1) (1) Before you answer (b) consider whether the confidence interval will be wider than or narrower than the confidence interval found for (a). The check that your answer verifies this.

Step 1 ：Given values are mean = \$244, variance = 57, and z_score = 1.96. The standard deviation is calculated as the square root of the variance, which is approximately 7.55.

Step 2 ：For part (a), the sample size is 12. The confidence interval is calculated as \(mean \pm z\_score \times \frac{std\_dev}{\sqrt{n}}\), which gives us the interval (239.73, 248.27).

Step 3 ：For part (b), the sample size is 19. The confidence interval is calculated in the same way as in part (a), which gives us the interval (240.61, 247.39).

Step 4 ：As expected, the confidence interval for the larger sample size is narrower than for the smaller sample size. This is because the larger sample size gives us a more precise estimate of the population mean.

Step 5 ：Final Answer: The $95 \%$ confidence interval for the mean monthly value of credit card purchases of all card holders for a sample size of 12 is approximately \(\boxed{(239.73, 248.27)}\) and for a sample size of 19 is approximately \(\boxed{(240.61, 247.39)}\).