Current Attempt in Progress Use the t-distribution to find a confidence interval for a mean $\mu$ given the relevant sample results. Give the best point estimate for $\mu$. the margin of error, and the confidence interval. Assume the results come from a random sample from a population that is approximately normally distributed. A $90 \%$ confidence interval for $\mu$ using the sample results $\bar{x}=135.8, s=57.1$, and $n=50$ Round your answer for the point estimate to one decimal place, and your answers for the margin of error and the confidence interval to two decimal places. point estimate $=\mathrm{i}$ margin of error $=\mathbf{i}$ The $90 \%$ confidence interval is to i

Step 1 ：The problem provides us with the sample mean (\(\bar{x}\)) which is 135.8, the sample standard deviation (\(s\)) which is 57.1, and the sample size (\(n\)) which is 50. These values will be used to estimate the population mean (\(\mu\)).

Step 2 ：The point estimate for \(\mu\) is simply the sample mean, which is \(\boxed{135.8}\).

Step 3 ：The margin of error for a confidence interval is calculated using the formula: Margin of Error = \(t \times \frac{s}{\sqrt{n}}\), where \(t\) is the t-score for the desired level of confidence (in this case, 90%), \(s\) is the sample standard deviation, and \(n\) is the sample size.

Step 4 ：Using the given values, the t-score is approximately 1.6765508919142629 and the margin of error is approximately 13.538415763410537. Therefore, the margin of error is \(\boxed{13.54}\).

Step 5 ：The confidence interval is then calculated as: Confidence Interval = \(\bar{x} \pm\) Margin of Error. Using the given values, the confidence interval is approximately (122.26158423658947, 149.33841576341055).

Step 6 ：Therefore, the $90\%$ confidence interval for \(\mu\) is \(\boxed{(122.26, 149.34)}\).