Question 6 of 10, Step 1 of 2 A random sample of 6 fields of corn has a mean yield of 23.8 bushels per acre and standard deviation of 4.46 bushels per acre. Determine the \( 95 \% \) confidence interval for the true mean yield. Assume the population is approximately normal. Step 1 of 2 : Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.

To find the critical value for a 95% confidence interval using a t-distribution (since the sample size is small and we don't know the population standard deviation), you will be looking for the t-score that corresponds to a cumulative area of 0.975 (because we need to account for both tails of the distribution). With a sample size of \( n = 6 \), the degrees of freedom will be \( n - 1 = 5 \). Using a t-table or calculator, the critical t-value for 5 degrees of freedom at the 0.975 level is approximately 2.571. So, rounding to three decimal places, the answer is 2.571. Now you can proceed to construct the confidence interval using this critical value!