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.

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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.