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

We need a sample of size at least 443.

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:

Now, we have to find z in the Ztable as such z has a pvalue of .

So it is z with a pvalue of , so

Now, find the margin of error M as such

In which 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

So

Rounding up

We need a sample of size at least 443.

the equivalent expression are always the hardest ones but you look at the gcf also

step-by-step explanation:

answer: 0.64

step-by-step explanation: