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

Answer:

A sample size of at least 271 is required.

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:

$\alpha = \frac{1-0.9}{2} = 0.05$

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

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

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

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

$0.12 = 1.645*\frac{1.2}{\sqrt{n}}$

$0.12\sqrt{n} = 1.645*1.2$

$\sqrt{n} = \frac{1.645*1.2}{0.12}$

$(\sqrt{n})^{2} = (\frac{1.645*1.2}{0.12})^{2}$

$n = 270.6$

Rounding up

A sample size of at least 271 is required.