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

1. The sample mean is given as \( \bar{x} = 46.4 \) bushels per acre and the sample standard deviation is \( s = 2.45 \) bushels per acre, while the sample size is \( n = 16 \). 2. Since the population is approximately normal and the sample size is small, we use the \( t \)-distribution with degrees of freedom \( n-1 = 15 \). 3. The confidence level is \( 80\% \), so the significance level is \( \alpha = 0.2 \) and the tail probability is \( \frac{\alpha}{2} = 0.1 \). The critical \( t \)-value, \( t_{0.1,15} \), is approximately \( 1.341 \). 4. The margin of error (ME) is calculated by: \[ \text{ME} = t_{\frac{\alpha}{2}, n-1} \times \frac{s}{\sqrt{n}} = 1.341 \times \frac{2.45}{\sqrt{16}} = 1.341 \times \frac{2.45}{4} = 1.341 \times 0.6125 \approx 0.8. \] 5. Thus, the \(80\%\) confidence interval for the true mean yield is: \[ \bar{x} \pm \text{ME} = 46.4 \pm 0.8, \] which gives the interval: \[ (46.4 - 0.8, \, 46.4 + 0.8) = (45.6, 47.2). \] Rounded to one decimal place, the \(80\%\) confidence interval is \( (45.6, 47.2) \) bushels per acre.