Question 6 of 10, Step 2 of 2 A random sample of 10 fields of durum wheat has a mean yield of 39.8 bushels per acre and standard deviation of 3.87 bushels per acre. Determine the \( 90 \% \) confidence interval for the true mean yield. Assume the population is approximately normal. Step 2 of 2 : Construct the \( 90 \% \) confidence interval. Round your answer to one decimal place. Tables Keypad Answer

To construct the \( 90 \% \) confidence interval for the mean yield, we use the formula: \[ \text{Confidence Interval} = \bar{x} \pm t \left( \frac{s}{\sqrt{n}} \right) \] Where \( \bar{x} = 39.8 \) (sample mean), \( s = 3.87 \) (sample standard deviation), and \( n = 10 \) (sample size). For a \( 90 \% \) confidence interval with \( n - 1 = 9 \) degrees of freedom, the critical t-value is approximately \( 1.833 \). 1. Calculate the standard error (SE): \[ SE = \frac{s}{\sqrt{n}} = \frac{3.87}{\sqrt{10}} \approx 1.222 \] 2. Calculate the margin of error (ME): \[ ME = t \times SE = 1.833 \times 1.222 \approx 2.24 \] 3. Construct the confidence interval: \[ \text{Lower limit} = 39.8 - 2.24 \approx 37.6 \] \[ \text{Upper limit} = 39.8 + 2.24 \approx 42.0 \] Rounding to one decimal place, the \( 90 \% \) confidence interval for the true mean yield is approximately \( (37.6, 42.0) \).