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?
A) Approximately 0.017 pound
B) About 0.0003 pound
C) About 1.96
D) Can't be determined without knowing the population mean.

Question

Quizplus · Accepted Answer

The margin of error (ME) for a sample mean can be calculated using the formula ME = z*(σ/√n), where z is the z-score corresponding to the desired confidence level (1.96 for 95% confidence), σ is the population standard deviation (0.15 pound in this case), and n is the sample size (300). Plugging these values into the formula gives ME ≈ 0.017 pound. The population mean is not required to calculate the margin of error for estimating the mean weight of packages.

The US Post Office Is Interested in Estimating the Mean Weight