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 variance is 5.76
kWh and the mean is 16.6
kWh per day. If they are using a 98%
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.
kWh. A previous study found that for an average family the variance is 5.76
kWh and the mean is 16.6
kWh per day. If they are using a 98%
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.
Answers
GPT 3.5
To calculate the sample size required, we can use the formula:
n = (Z^2 * σ^2) / E^2
Where:
n = sample size
Z = Z-score corresponding to the confidence level (98% in this case, which corresponds to a Z-score of 2.33)
σ = standard deviation of the population (sqrt(5.76) = 2.4 kWh)
E = maximum error allowed (0.09 kWh)
Plugging in the values:
n = (2.33^2 * 5.76) / 0.09^2
n = (5.4289 * 5.76) / 0.0081
n = 31.2464 / 0.0081
n = 3865.679
Rounding up to the next whole number, the sample size required is 3866 households.
n = (Z^2 * σ^2) / E^2
Where:
n = sample size
Z = Z-score corresponding to the confidence level (98% in this case, which corresponds to a Z-score of 2.33)
σ = standard deviation of the population (sqrt(5.76) = 2.4 kWh)
E = maximum error allowed (0.09 kWh)
Plugging in the values:
n = (2.33^2 * 5.76) / 0.09^2
n = (5.4289 * 5.76) / 0.0081
n = 31.2464 / 0.0081
n = 3865.679
Rounding up to the next whole number, the sample size required is 3866 households.