In order to estimate the average electric usage per month, a sample of 81 houses was selected and the electric usage was determined. Assume a population standard deviation of 450 kilowatt-hours. If the sample mean is 1858 kWh, the 95% confidence interval estimate of the population mean is _________?

Answer:Confidence interval:  (1760,1956)Step-by-step explanation:We are given the following information in the question:Sample size, n = 81Sample mean = [tex]\bar{x} = 1858 \text{ kWh}[/tex]Population standard deviation =[tex]\sigma = 450 \text{ kilowatt-hours}[/tex]Confidence Level = 95%Significance level = 5% = 0.05Confidence interval:[tex]\bar{x} \pm z_{critical}\displaystyle\frac{\sigma}{\sqrt{n}}[/tex]Putting the values, we get,[tex]z_{critical}\text{ at}~\alpha_{0.05} = \pm 1.96[/tex][tex]1858 \pm 1.96(\displaystyle\frac{450}{\sqrt{81}} ) = 1858 \pm 98 = (1760,1956)[/tex]