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 0.12 kWh. A previous study found that for an average family the standard deviation is 1.2 kWh and the mean is 17.9 kWh per day. If they are using a 90% level of confidence, how large of a sample is required to estimate the mean usage of electricity? Round your answer up to the next integer.

Answer 1

A sample size of at least 271 is required.

Explanation:

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

`\alpha = (1-0.9)/(2) = 0.05`

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*(\sigma)/(√(n))`

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

Maxium error of 0.12.

How large of a sample is required to estimate the mean usage of electricity?

We need a sample size of at least n.

n is found when
`M = 0.12, \sigma = 1.2`

So

`M = z*(\sigma)/(√(n))`

`0.12 = 1.645*(1.2)/(√(n))`

`0.12√(n) = 1.645*1.2`

`√(n) = (1.645*1.2)/(0.12)`

`(√(n))^(2) = ((1.645*1.2)/(0.12))^(2)`

`n = 270.6`

Rounding up

A sample size of at least 271 is required.