Answer to Question #154303 in Statistics and Probability for Ali Ahmed

Question

A random sample of 20 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 2006. 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 sample standard deviation of the usage was 81kWh. Assuming the
that the usage is normally distributed, provide an expression for calculating a 95%
confidence interval for the mean usage in the March quarter of 2006.

Answer 1

$95\%CI=(\bar x-z_{0.025}\frac{\sigma}{\sqrt{n}},\bar x+z_{0.025}\frac{\sigma}{\sqrt{n}})=$

$=(375-1.96\frac{81}{\sqrt{20}},375+1.96\frac{81}{\sqrt{20}})=(339.50,410.50).$