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

A sample of 3851 is required.

Explanation:

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

`\alpha = (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(\sigma)/(√(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 = √(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(\sigma)/(√(n))`

`0.09 = 2.327(2.4)/(√(n))`

`0.09√(n) = 2.327*2.4`

`√(n) = (2.327*2.4)/(0.09)`

`(√(n))^2 = ((2.327*2.4)/(0.09))^2`

`n = 3850.6`

Rounding up:

A sample of 3851 is required.