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? Round your answer up to the next integer.

Answer 1

Solution :

Given that,

variance = 2 = 5.29

Population standard deviation = = 2 = 5.29 = 2.3

Margin of error = E = 0.14

At 80% confidence level the z is,

= 1 - 80%

= 1 - 0.80 = 0.20

/2 = 0.10

Z/2 = Z0.10 = 1.282

sample size = n = [Z/2* / E] 2

n = [ 1.282 * 2.3 / 0.14 ]2

n = 443.58

Sample size = n = 444