Estimate the population mean by finding a $95 \%$ confidence interval given a sample of size 24 , with a mean of 49.3 and a standard deviation of 7.7. Preliminary: \[ \text { Is } n \geq 30 \text { ? } \] No Yes Confidence Interval: What is the $95 \%$ confidence interval to estimate the population mean? Enter the tri-linear inequality values accurate to one decimal place. Question Help: VIDEO

Step 1 ：Given a sample size of 24, a sample mean of 49.3, and a standard deviation of 7.7, we are asked to find the 95% confidence interval for the population mean.

Step 2 ：Since the sample size is less than 30, we should use the t-distribution instead of the z-distribution.

Step 3 ：The formula for a confidence interval is \(\bar{x} \pm t \frac{s}{\sqrt{n}}\), where \(\bar{x}\) is the sample mean, \(t\) is the t-score corresponding to the desired confidence level, \(s\) is the standard deviation of the sample, and \(n\) is the sample size.

Step 4 ：The t-score for a 95% confidence interval with 23 degrees of freedom (n-1) can be found from a t-distribution table or calculated using a statistical function. In this case, the t-score is approximately 2.07.

Step 5 ：Substituting the given values into the formula, we get the lower bound of the confidence interval as \(49.3 - 2.07 \times \frac{7.7}{\sqrt{24}} \approx 46.0\) and the upper bound as \(49.3 + 2.07 \times \frac{7.7}{\sqrt{24}} \approx 52.6\).

Step 6 ：Thus, the 95% confidence interval to estimate the population mean is \(\boxed{[46.0, 52.6]}\).