The U.S. Post Office is interested in estimating the mean weight of packages shipped using the overnight service. They plan to sample 300 packages. A pilot sample taken last year showed that the standard deviation in weight was about 0.15 pound. If they are interested in an estimate that has 95 percent confidence, what margin of error can they expect?

Question

BlackTom AI · Accepted Answer

To assess the margin of error for a 95% confidence interval when estimating a mean, we use the formula E = z_{alpha/2} * sigma / sqrt(n).

Option 1: About 0.0003 pound. This is far smaller than the calculation would yield. Using sigma = 0.15, n = 300, sqrt(n) ≈ 17.32, and z_{0.025} = 1.96, the margin of error would be 1.96 * 0.15 / 17.32 ≈ 0.01698, not 0.0003. So this choice underestimates the true spread dramatically.

Option 2: Approximately 0.017 pound. This aligns with the standard margin-of-error computation: E = 1.96 * 0.15 / sqrt(300) ≈ 0.01698 ≈ 0.017. The numbers fit the given 95% confidence level and sample size.

Option 3: About 1.96. This value is the z-score for 95% confidence, not the margin of error itself. The margin of error scales with sigma and n, not directly to the z-score in pounds.

Option 4: Can't be determined without knowing the population mean. Not necessary here. For margins of error in confidence intervals, we use the standard deviation (sigma) and the sample size; the population mean is not required to compute E.

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