Question

Suppose that shopping times for customers at a local mall are normally distributed with a
known population standard deviation of 20 minutes. A random sample of 64 shoppers in the local
grocery store had a mean time of 75 minutes. Find the standard error, margin of error, and the
upper and lower confidence limit of a 95% confidence interval for the population mean (

Answer 1

$\sigma=20 \\ n=64 \\ \bar{x}=75$

The value of z at 95% confidence level is 1.96.

The standard error is

$SE = \frac{\sigma}{\sqrt{n}} \\ = \frac{20}{\sqrt{64}}=2.5$

The margin of error is

$ME = Z \times SE \\ = 1.96 \times 2.5 = 4.9$

And the 95% confidence interval is

$\bar{x}±ME = 75 ± 4.9 \\ =(70.1, 79,9)$