Question 5 of 10, Step 2 of 2 \( \begin{array}{l}\text { A random sample of } 16 \text { fields of spring wheat has a mean yield of } 46.4 \text { bushels per acre and standard } \\ \text { deviation of } 2.45 \text { bushels per acre. Determine the } 80 \% \text { confidence interval for the true mean yield. } \\ \text { Assume the population is approximately normal. } \\ \text { Step 2 of } 2 \text { : Construct the } 80 \% \text { confidence interval. Round your answer to one decimal place. } \\ \begin{array}{l}\text { Answer } \\ \text { How to enter your answer (opens in new window) } \\ \text { Keyboard Shortcuts }\end{array} \\ \text { Previous step answers }\end{array} \)

To construct the 80% confidence interval for the true mean yield of spring wheat, we can use the formula for the confidence interval: \[ \text{CI} = \bar{x} \pm t^* \times \frac{s}{\sqrt{n}} \] Where: - \( \bar{x} = 46.4 \) (sample mean) - \( s = 2.45 \) (sample standard deviation) - \( n = 16 \) (sample size) - \( t^* \) is the t-value corresponding to 80% confidence and \( n-1 = 15 \) degrees of freedom. Looking up the t-distribution table, the t-value for 15 degrees of freedom at a 0.10 significance level (or 80% confidence) is approximately \( t^* \approx 1.341 \). Now, calculate the margin of error: \[ \text{Margin of Error} = t^* \times \frac{s}{\sqrt{n}} = 1.341 \times \frac{2.45}{\sqrt{16}} \] \[ = 1.341 \times \frac{2.45}{4} = 1.341 \times 0.6125 \approx 0.820 \] Now, construct the confidence interval: \[ \text{Lower Limit} = 46.4 - 0.820 \approx 45.6 \] \[ \text{Upper Limit} = 46.4 + 0.820 \approx 47.2 \] Thus, the 80% confidence interval for the true mean yield of spring wheat is approximately \( (45.6, 47.2) \).