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 0.09 kWh. A previous study found that for an average family the standard deviation is 2.1 kWh and the mean is 16.2 kWh per day. If they are using a 99% 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.

Please elaborate on how to obtain the Z a/2. That is the part I am struggling with the most. Thank you. :)

Answer 1

Solution :

Given that,

Population standard deviation = = 2.1

Margin of error = E = 0.09

At 99% confidence level the z is,

= 1 - 99%

= 1 - 0.99 = 0.01

/2 = 0.005

Z_/2 = Z_0.005 = 2.576

sample size = n = [Z_/2* / E] ²

n = [ 2.576 * 2.1 / 0.09 ]²

n = 3612.81

Sample size = n = 3613