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 0.1 kWh. A previous study found that for an average family the variance is 4 kWh and the mean is 19.6 kWh per day. If they are using a 80% 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

To estimate the mean electricity usage with an 80% confidence level and maximum error of 0.1 kWh, a sample size of 656 households is required.

Step-by-step explanation:

To answer how large of a sample is required to estimate the mean usage of electricity with a maximum error of 0.1 kWh and an 80% level of confidence, we need to use the formula for sample size in estimating a population mean. The formula we use is n = (Z*σ/E)^2, where Z is the Z-value for the confidence level, σ is the population standard deviation, and E is the maximum error allowed.

The variance is given as 4 kWh, so the standard deviation σ is the square root of variance, which is 2 kWh. The maximum error E is 0.1 kWh. For an 80% confidence level, the Z-value (from the standard normal distribution table) is approximately 1.28. Plugging these values into the formula, we get:

n = (1.28*2/0.1)^2

n = (25.6)^2

n = 655.36

Since we need a whole number of samples and we round up to ensure the maximum error does not exceed 0.1 kWh, we round up to the next integer, therefore, the sample size needed is 656.