First, we need to find the margin of error which is calculated using the formula:
Margin of error = Z * (standard deviation / √n)
Where Z is the Z-score corresponding to the 98% confidence level, which is 2.33. The standard deviation is the square root of the variance, which is √5.76 = 2.4 kWh.
Therefore, the margin of error is:
0.09 = 2.33 * (2.4 / √n)
Solving for n:
2.33 * (2.4 / 0.09) = √n
61.87 = √n
n = (61.87)^2
n ≈ 3834.12
Rounding up to the next integer, the sample size required to estimate the mean usage of electricity is 3835 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