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 \( \bar{x}=138.8, s=49.8 \), 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 \( = \) i margin of error \( = \) i The \( 90 \% \) confidence interval is

To calculate the \( 90\% \) confidence interval for the mean \( \mu \), we'll first determine the best point estimate, the margin of error, and then the confidence interval. 1. **Point Estimate**: This is simply the sample mean, which is given as \( \bar{x} = 138.8 \). 2. **Margin of Error**: We first need to find the critical value \( t^* \) for \( 90\% \) confidence with \( n - 1 = 49 \) degrees of freedom. From the \( t \)-distribution table, \( t^* \approx 1.676 \). Now, calculate the standard error (SE): \[ SE = \frac{s}{\sqrt{n}} = \frac{49.8}{\sqrt{50}} \approx 7.045 \] The margin of error (ME) is: \[ ME = t^* \times SE \approx 1.676 \times 7.045 \approx 11.83 \] 3. **Confidence Interval**: Now we can construct the confidence interval: \[ CI = \bar{x} \pm ME = 138.8 \pm 11.83 \] Which gives us: \[ \text{Lower limit} = 138.8 - 11.83 \approx 126.97 \] \[ \text{Upper limit} = 138.8 + 11.83 \approx 150.63 \] Putting it all together, we have: Point estimate \( = 138.8 \) Margin of error \( = 11.83 \) The \( 90\% \) confidence interval is \( (126.97, 150.63) \).