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 2.4 kWh. The mean electricity usage per family was found to be 17.5 kWh per day for a sample of 796 families. Construct the 90% confidence interval for the mean usage of electricity. Round your answers to one decimal place.

Answer 1

Answer:

The 90% confidence interval for the mean usage of electricity is between 17.4 kwH and 17.6 kwH

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.9}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$

So it is z with a pvalue of 1-0.05 = 0.95, so z = 1.645

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

So

$M = 1.645*\frac{2.4}{\sqrt{796}} = 0.14$

The lower end of the interval is the mean subtracted by M. So 17.5 - 0.14 = 17.4 kwH

The upper end of the interval is M added to the mean. So 17.5 + 0.14 = 17.6 kwH

The 90% confidence interval for the mean usage of electricity is between 17.4 kwH and 17.6 kwH