Question 10 of 14, Step 1 of 1 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 0.14 kWh . A previous study found that for an average family the variance is 5.29 kWh and the mean is 19.9 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.

To determine the sample size required, we can use the formula for sample size in estimating a population mean: \[ n = \left( \frac{Z \cdot \sigma}{E} \right)^2 \] Where: - \( n \) = required sample size - \( Z \) = Z-score corresponding to the confidence level (for 80% confidence, \( Z \) is approximately 1.28) - \( \sigma \) = standard deviation (which is the square root of variance) - \( E \) = maximum error (0.14 kWh) Given that the variance is 5.29 kWh, the standard deviation (\( \sigma \)) is: \[ \sigma = \sqrt{5.29} \approx 2.3 \] Now plug the values into the formula: \[ n = \left( \frac{1.28 \cdot 2.3}{0.14} \right)^2 \] Calculating inside the parentheses: \[ \frac{1.28 \cdot 2.3}{0.14} \approx 19.2 \] Now square this result: \[ n \approx (19.2)^2 \approx 368.64 \] Since we need to round up to the next integer, the required sample size is: \[ n = 369 \] Thus, the electric cooperative needs a sample size of **369** households to estimate the mean household usage of electricity with a maximum error of 0.14 kWh at an 80% confidence level.