Question

Q-8 (1 of 1) 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 standard deviation is 2.5 kWh and the mean is 17 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.

Answer 1

Solution :

Given that,

standard deviation = σ =2.5

Margin of error = E = 0.14

At 99% confidence level the z is,

= 1 - 99%

= 1 - 0.99 = 0.01

/2 = 0.005

Z/2  = 2.58

sample size = n = [ Z/2 *  σ/ E]2

n = ( 2.58* 2.5 /0.14 )2

n =2122.57

Sample size = n =2123