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 0.13 kWh. A previous study found that for an average family the standard deviation is 2.1 kWh and the mean is 15.8 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 = =2.1

Margin of error = E = 0.13

At 99% confidence level the z is,

= 1 - 99%

= 1 - 0.99 = 0.01

/2 = 0.005

Z_/2 = 2.58 ( Using z table ( see the 0.005 value in standard normal (z) table corresponding z value is 2.58 )  

sample size = n = [Z_/2* / E] ²

n = ( 2.58* 2.1/ 0.13 )²

n =1737

Sample size = n =1737

( when you have z value 3 decimal than Z_/2 = 2.576 n=1732)