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

Confidence interval: (1760,1956)

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(\sigma)/(√(n))`

Putting the values, we get,

`z_(critical)\text{ at}~\alpha_(0.05) = \pm 1.96`

`1858 \pm 1.96(\displaystyle(450)/(√(81)) ) = 1858 \pm 98 = (1760,1956)`