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.
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.
1 answer
1 answer