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 pounds.If they are interested in an estimate that has 95 percent confidence,what margin of error can they expect?
A)Approximately 0.017 pounds
B)About 0.0003 pounds
C)About 1.96
D)Can't be determined without knowing the population mean.

Question

Quizplus · Accepted Answer

The margin of error (ME) can be calculated as: 
ME = z*(sigma/sqrt(n)) 
where z is the z-score for the desired level of confidence (95% = 1.96), sigma is the estimated standard deviation (0.15 pounds) and n is the sample size (300). 
Substituting the values: 
ME = 1.96*(0.15/sqrt(300)) 
ME ≈ 0.017 pounds. 
Therefore, the margin of error can be expected to be approximately 0.017 pounds, which is closest to option A.

The USPost Office Is Interested in Estimating the Mean Weight of of Packages