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 0.14 kWh. A previous study found that for an average family the variance is 5.29 kWh and the mean is 19.9 kWh per day. If they are using a 80% level of confidence, how large of a sample is required to estimate the mean usage of electricity

Answer 1

`n=((1.28(2.3))/(0.14))^2 =442.2 \approx 443`

So the answer for this case would be n=443 rounded up to the nearest integer

Explanation:

first we need to find a quantile for the normal distribution associated to the confidence level given.

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 80% of confidence, our significance level would be given by
`\alpha=1-0.80=0.2` and
`\alpha/2 =0.1`. And the critical value would be given by:

`z_(\alpha/2)=-1.28, z_(1-\alpha/2)=1.28`

The margin of error is given by this formula:

`ME=z_(\alpha/2)(\sigma)/(√(n))` (a)

And on this case we have that ME =0.14 the desired error and we are interested in order to find the value of n, if we solve n from equation (a) we got:

`n=((z_(\alpha/2) \sigma)/(ME))^2` (b)

The standard deviation would be
`\sigma =√(5.29) = 2.3`. If we replace into formula (b) we got:

`n=((1.28(2.3))/(0.14))^2 =442.2 \approx 443`

So the answer for this case would be n=443 rounded up to the nearest integer

Answer 2

We need a sample of size at least 443.

Explanation:

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

`\alpha = (1-0.8)/(2) = 0.1`

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
`1-0.1 = 0.9`, so
`z = 1.28`

Now, find the margin of error M as such

`M = z*(\sigma)/(√(n))`

In which
`\sigma` is the standard deviation of the population(square root of the variance) and n is the size of the sample.

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

We need a sample of size at least n.

n is found when
`M = 0.14, \sigma = √(5.29) = 2.3`

So

`M = z*(\sigma)/(√(n))`

`0.14 = 1.28*(2.3)/(√(n))`

`0.14√(n) = 1.28*2.3`

`√(n) = (1.28*2.3)/(0.14)`

`(√(n))^(2) = ((1.28*2.3)/(0.14))^(2)`

`n = 442.2`

Rounding up

We need a sample of size at least 443.