1. Chapter 12. Repulation mean (20 points) The article "americans' big Debt burden growing, not cvenly Disinbuted" (www-gallup com, retrieved December 14, 2016) reperted that for a representative sample of Americans born between 1965 and 1971 (known as Generation X), the sample mean number of credit cards owned was 4.5 . Suppose that the sample standard deviation (which was not reported) was 1.0 and that the sample size was \( n-300 \). Construct a \( 95 \% \) confidenes interval.
a. Define \( A \)
b. Check
c. Calculate
d. Intermreation of Confidence Interyal
e. Interpretation of Confidenee level

Step 1 ：a. Define \( A \) as the \( 95 \% \) confidence interval for the population mean number of credit cards owned by Americans born between 1965 and 1971.\\\

Step 2 ：b. Check the given conditions:\\ n = 300, \bar{x} = 4.5, s = 1.0. Since n > 30, we can use the z-distribution to construct the confidence interval.\\\

Step 3 ：c. Calculate the confidence interval using the formula: \( A = \bar{x} \pm z * \frac{s}{\sqrt{n}} \)\\ Substitute the given values: \( A = 4.5 \pm 1.96 * \frac{1}{\sqrt{300}} \)\\ Calculate the margin of error: \( 1.96 * \frac{1}{\sqrt{300}} \approx 0.1131 \)\\ The confidence interval: \( A = (4.5 - 0.1131, 4.5 + 0.1131) = (4.3869, 4.6131) \)\\\

Step 4 ：d. Interpretation of Confidence Interval: We are \( 95 \% \) confident that the true population mean number of credit cards owned by Americans born between 1965 and 1971 lies between 4.3869 and 4.6131.\\\

Step 5 ：e. Interpretation of Confidence Level: If we were to take 100 different representative samples of Americans born between 1965 and 1971 and construct \( 95 \% \) confidence intervals for each, we would expect about 95 of those intervals to contain the true population mean number of credit cards owned.