Question

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 pounds B. About 0.0003 pounds C. About 1.96 D. Can't be determined without knowing the population mean.

Answer 1

Answer: A. Approximately 0.017 pounds

Step-by-step explanation:

Formula to find the margin of error :

$E=z^*\dfrac{s}{\sqrt{n}}$ , where z* = critical value for confidence interval , s= standard deviation , n= sample size.

As per given , we have

s= 0.15 pound

n= 300

Critical value for 95% confidence = 1.96

Then, Margin of error for 9%% confidence interval will be :

$E=(1.96)\dfrac{0.15}{\sqrt{300}}\\\\=0.0169740979142\approx0.017$

Hence, they can expect a margin error of 0.017 pound (approximately.)

Thus , the correct answer is A. Approximately 0.017 pounds