Question

The electric cooperative needs to know the mean household usage of electricity by its non-commercial customers in kWh per day. They would like the estimate to have a maximum error of 0.09 kWh. A previous study found that for an average family the variance is 5.76 kWh and the mean is 16.6 kWh per day. If they are using a 98% level of confidence, how large of a sample is required to estimate the mean usage of electricity

Answer 1

Answer:

A sample of 3851 is required.

Step-by-step explanation:

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

$\alpha = \frac{1 - 0.98}{2} = 0.01$

Now, we have to find z in the Z-table as such z has a p-value of .

That is z with a pvalue of , so Z = 2.327.

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.

Variance is 5.76 kWh

This means that $\sigma = \sqrt{5.76} = 2.4$

They would like the estimate to have a maximum error of 0.09 kWh. How large of a sample is required to estimate the mean usage of electricity?

This is n for which M = 0.09. So

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

$0.09 = 2.327\frac{2.4}{\sqrt{n}}$

$0.09\sqrt{n} = 2.327*2.4$

$\sqrt{n} = \frac{2.327*2.4}{0.09}$

$(\sqrt{n})^2 = (\frac{2.327*2.4}{0.09})^2$

$n = 3850.6$

Rounding up:

A sample of 3851 is required.