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 2.3 kWh. The mean electricity usage per family was found to be 15.7 kWh per day for a sample of 1731 families. Construct the 99% confidence interval for the mean usage of electricity. Round your answers to one decimal place.

Answer 1

Answer:

$15.7-2.58\frac{2.3}{\sqrt{1731}}=15.6$    

$15.7+2.58\frac{2.3}{\sqrt{1731}}=15.8$    

We are 99% confident that the true mean of electricity comsumption is between (15.6 and 15.8) kWh

Step-by-step explanation:

Information provided

$\bar X= 15.7$ represent the sample mean for the usage of electricity

$\mu$ population mean (variable of interest)

$\sigma= 2.3$ represent the population standard deviation

n=1731 represent the sample size  

Confidence interval

The confidence interval for the mean is given by the following formula if we know the population deviation:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$   (1)

The Confidence level provided is  0.99 or 95%, the value of significance is $\alpha=0.01$ and $\alpha/2 =0.005$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.005,0,1)".And we see that $z_{\alpha/2}=2.58$

And replacing we got:

$15.7-2.58\frac{2.3}{\sqrt{1731}}=15.6$    

$15.7+2.58\frac{2.3}{\sqrt{1731}}=15.8$    

We are 99% confident that the true mean of electricity comsumption is between (15.6 and 15.8) kWh