Answer to Question #215881 in Statistics and Probability for Hasibul

Question #215881
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 (
1
Expert's answer
2021-07-16T03:26:32-0400

σ=20n=64xˉ=75\sigma=20 \\ n=64 \\ \bar{x}=75

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

The standard error is

SE=σn=2064=2.5SE = \frac{\sigma}{\sqrt{n}} \\ = \frac{20}{\sqrt{64}}=2.5

The margin of error is

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

And the 95% confidence interval is

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


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