Question

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 1

Answer:

Confidence interval:  (1760,1956)

Step-by-step explanation:

We are given the following information in the question:

Sample size, n = 81

Sample mean =

$\bar{x} = 1858 \text{ kWh}$

Population standard deviation =

$\sigma = 450 \text{ kilowatt-hours}$

Confidence Level = 95%

Significance level = 5% = 0.05

Confidence interval:

$\bar{x} \pm z_{critical}\displaystyle\frac{\sigma}{\sqrt{n}}$

Putting the values, we get,

$z_{critical}\text{ at}~\alpha_{0.05} = \pm 1.96$

$1858 \pm 1.96(\displaystyle\frac{450}{\sqrt{81}} ) = 1858 \pm 98 = (1760,1956)$