Answer to Question #329254 in Statistics and Probability for Naomi

Question

A random sample of 30 households was selected as part of a study on electricity usage, and the number of kilowatt-hours (kWh) was recorded for each household in the sample for the March quarter of 2020. The average usage was found to be 375kWh. In a very large study in the March quarter of the previous year it was found that the standard deviation of the usage was 85kWh. Assuming the standard deviation is unchanged and that the usage is normally distributed provide an expression for calculating a 99% confidence interval for the

mean usage in the March quarter of 2020

Answer 1

n=30

$\mu=375$

$\sigma=85$

z = 2.57 for 99% confidence interval

$CI=\mu \plusmn z \frac{\sigma}{\sqrt{n}}=375 \plusmn 2.57\frac{85}{\sqrt{30}}=375 \plusmn 39.9$

(335.1, 414.9)