A random sample of 5 fields of corn has a mean yield of 43.7 bushels per acre and standard deviation of 6.95 bushels per acre. Determine the $98 \%$ confidence interval for the true mean yield. Assume the population is approximately normal. Round your answer to one decimal place. Answer Tables Keyboard Shc How to enter your answer (opens in new window) Lower endpoint: Upper endpoint: Submit Ansh
Solution
Step 1 ：Given that a random sample of 5 fields of corn has a mean yield of 43.7 bushels per acre and standard deviation of 6.95 bushels per acre, we are to determine the $98 \%$ confidence interval for the true mean yield. We assume the population is approximately normal.
Step 2 ：The Z-score for a $98 \%$ confidence level is 2.33.
Step 3 ：We calculate the margin of error using the formula: \(Z \times \frac{s}{\sqrt{n}}\), where \(Z\) is the Z-score, \(s\) is the sample standard deviation, and \(n\) is the sample size. Substituting the given values, we get: \(2.33 \times \frac{6.95}{\sqrt{5}} \approx 7.24\).
Step 4 ：We then calculate the confidence interval using the formula: \([x_{bar} - \text{margin of error}, x_{bar} + \text{margin of error}]\), where \(x_{bar}\) is the sample mean. Substituting the given values, we get: \([43.7 - 7.24, 43.7 + 7.24] = [36.5, 50.9]\).
Step 5 ：Thus, the $98 \%$ confidence interval for the true mean yield is \(\boxed{[36.5, 50.9]}\) bushels per acre.
