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) .

Step-by-step explanation:

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

$\overline{x}\pm z^*\dfrac{\sigma}{\sqrt{n}}$

, 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 :

$1858\pm (1.960)\dfrac{450}{\sqrt{81}}\\\\=1858\pm(1.960)\dfrac{450}{9}=1858\pm98\\\\=(1858-98,\ 1858+98)\\\\=(1760,\ 1956)$

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