5. In developing an interval estimate for a population mean, the population standard deviation \( \sigma \) was assumed to be 10 . The interval estimate was \( 50.92 \pm 2.14 \mathrm{Had} \sigma \) equaled 20 , the interval estimate would be a. \( 60.92 \pm 2.14 \) b. \( 50.92 \pm 12.14 \) c. \( 101.84 \pm 4.28 \) (d) \( 50.92 \pm 4.28 \)

To recalibrate the interval estimate for a population mean with a new standard deviation \( \sigma \) of 20, we need to consider how it impacts the margin of error. The original margin of error was calculated using the formula \( Z \times \left(\frac{\sigma}{\sqrt{n}}\right) \), where \( Z \) corresponds to the z-score for the desired confidence level. Since \( \sigma \) has doubled, the margin of error will also increase. Revising the margin from the original \( \pm 2.14 \), we would multiply it by the ratio of the new standard deviation to the original one: \( 20/10 = 2 \). Therefore, the new margin of error becomes \( 2.14 \times 2 = 4.28 \). Thus, the interval estimate is \( 50.92 \pm 4.28 \), matching option (d).