Overweight Men For a random sample of 50 overweight men, the mean of the number of pounds that they were overweight was 29 . The standard deviation of the population is 4.5 pounds. Part 1 of 4 (a) The best point estimate of the mean is 29 pounds. Part: \( \mathbf{1} / 4 \quad 2 \) of 4 Part 2 (b) Find the \( 90 \% \) confidence interval of the mean of these pounds. Round intermediate answers to at least three decimal places. Round your final answers to one decimal place. < \( \mu<\square \)

**Step 1. Identify the appropriate formula** For a known population standard deviation and a normally distributed population, the \(100(1-\alpha)\%\) confidence interval for the mean is given by \[ \bar{x} \pm z_{\alpha/2} \times \frac{\sigma}{\sqrt{n}}, \] where \(\bar{x}\) is the sample mean, \(z_{\alpha/2}\) is the critical value for the confidence level, \(\sigma\) is the population standard deviation, and \(n\) is the sample size. --- **Step 2. Identify the given values** \[ \begin{aligned} \bar{x} &= 29 \text{ pounds}, \\ \sigma &= 4.5 \text{ pounds}, \\ n &= 50, \\ \text{Confidence level} &= 90\%. \end{aligned} \] For a \(90\%\) confidence interval, the level of significance is \(\alpha = 0.10\) so that \(\alpha/2 = 0.05\). --- **Step 3. Determine the critical value \(z_{\alpha/2}\)** From the standard normal table, the critical value corresponding to \(z_{0.05}\) is approximately \[ z_{0.05} \approx 1.645. \] --- **Step 4. Calculate the standard error** The standard error (SE) is calculated as: \[ \text{SE} = \frac{\sigma}{\sqrt{n}} = \frac{4.5}{\sqrt{50}}. \] Calculating \(\sqrt{50} \approx 7.071\), we have \[ \text{SE} \approx \frac{4.5}{7.071} \approx 0.636. \] --- **Step 5. Compute the margin of error** The margin of error (ME) is: \[ \text{ME} = z_{\alpha/2} \times \text{SE} \approx 1.645 \times 0.636 \approx 1.046. \] --- **Step 6. Construct the confidence interval** The \(90\%\) confidence interval for the mean is: \[ \bar{x} \pm \text{ME} = 29 \pm 1.046. \] This gives the interval: \[ (29 - 1.046,\; 29 + 1.046) \approx (27.954,\; 30.046). \] Rounding the final answers to one decimal place, we obtain: \[ (28.0,\; 30.0). \] --- **Final Answer:** \[ 28.0 < \mu < 30.0. \]