Part 1 of 4 (a) The best point estimate of the mean is 29 pounds. Part 2 of 4 (b) Find the \( 90 \% \) confidence interval of the mean of these pounds. Round intermediate answers to at least three decimal places. Round your 5 f to one decimal place. Correct Answer: \( 28.0<\mu<30.0 \) \( 28.0<\mu<30.1 \) Part 3 of 4 (c) Find the \( 95 \% \) confidence interval of the mean of these pounds. Round intermediate answers to at least three decimal places. Round your fi to one decimal place. Part: \( 2 / 4<\mu<\square \) P 4

We start with the point estimate given as \[ \bar{x} = 29. \] Suppose that for the 90% confidence interval the margin of error was computed as \[ E_{90} = z_{0.95} \cdot \sigma_{\bar{x}}, \] and the published interval is approximately \[ 29 \pm 1, \] so that \[ E_{90} \approx 1. \] Recall that for a 90% confidence level the standard normal critical value is \[ z_{0.95} = 1.645. \] This lets us find the standard error as \[ \sigma_{\bar{x}} = \frac{E_{90}}{1.645} \approx \frac{1}{1.645} \approx 0.608. \] For a 95% confidence interval the critical value is \[ z_{0.975} = 1.96. \] Thus the margin of error for 95% confidence is \[ E_{95} = 1.96 \cdot \sigma_{\bar{x}} \approx 1.96 \times 0.608 \approx 1.191. \] The 95% confidence interval for the mean is found by \[ \bar{x} \pm E_{95}. \] That is, \[ 29 \pm 1.191. \] Computing the endpoints gives \[ \text{Lower bound} = 29 - 1.191 \approx 27.809, \] \[ \text{Upper bound} = 29 + 1.191 \approx 30.191. \] Rounded to one decimal place, the limits are \[ 27.8 \quad \text{and} \quad 30.2. \] Thus the 95% confidence interval of the mean is \[ 27.8 < \mu < 30.2. \]