Question

The electric cooperative needs to know the mean household usage of electricity by its non-commercial customers in kWh per day. Assume that the population standard deviation is 2.5 kWh. The mean electricity usage per family was found to be 16.5 kWh per day for a sample of 3861 families. Construct the 98% confidence interval for the mean usage of electricity. Round your answers to one decimal place.

Answer 1

Answer: $(16.4,\ 16.6)$

Step-by-step explanation:

Given : Sample size : n= 3861

Significance level :  $\alpha=1-0.98=0.02$

Critical value for significance level of $\alpha=0.02$ : $z_{\alpha/2}= 2.33$

Sample mean : $\overline{x}=16.5$

Standard deviation : $\sigma= 2.5$

The formula to find  the confidence interval for population mean is given by :-

$\overline{x}\pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}$

i.e $16.5\pm (2.33)\dfrac{2.5}{\sqrt{3861}}$

$=16.5\pm0.0937445500445\\\\\approx16.5\pm0.1=(16.5-0.1,\ 16.5+0.1)=(16.4,\ 16.6)$

Hence, the 98% confidence interval for the mean usage of electricity :

$(16.4,\ 16.6)$