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

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

n = ( 2.58* 2.1 / 0.13 )2

n =1736.96

Sample size = n =1737

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