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: The 95% confidence interval estimate of the population mean is (1760, 1956) .

Explanation:

Formula for confidence interval for population mean(
`(\mu)`) :

`\overline{x}\pm z^*(\sigma)/(√(n))`

, where n= Sample size

`\overline{x}` = sample mean.

`z^*` = Two-tailed critical z-value

`\sigma` = population standard deviation.

By considering the given information, we have

n= 81

`\sigma=450` kilowatt-hours.

`\overline{x}=1858` kilowatt-hours.

By using the z-value table ,

The critical values for 95% confidence interval :
`z^*=\pm1.960`

Now , the 95% confidence interval estimate of the population mean will be :

`1858\pm (1.960)(450)/(√(81))\\\\=1858\pm(1.960)(450)/(9)=1858\pm98\\\\=(1858-98,\ 1858+98)\\\\=(1760,\ 1956)`

Hence, the 95% confidence interval estimate of the population mean is (1760, 1956) .