The electric cooperative needs to know the mean household usage of electricity by its non-commercial customers in kWh per day. Assume that the population standard deviation is 1.8 kWh. The mean electricity usage per family was found to be 15.8 kWh per day for a sample of 872 families. Construct the 98% confidence interval for the mean usage of electricity. Round your answers to one decimal place.

Answer: (14.4, 17.2)

Step-by-step explanation: We are to construct a 98% confidence interval for mean household usage of electricity.

We have been given that

Sample size (n) = 872

Sample mean (x) = 15.8

Population standard deviation (σ) = 1.8

The formulae that defines the 98% confidence interval for mean is given below as

u = x + Zα/2 * σ/√n ... For the upper limit

u = x - Zα/2 * σ/√n ... For the lower limit

Zα/2 = critical value for a 2% level of significance in a two tailed test = 2.33

By substituting the parameters we have that

For upper tailed

u = 15.8 + 2.33 * (1.8/√872)

u = 15.8 + 2.33 (0.6096)

u = 15.8 + 1.4203

u = 17.2

For lower tailed

u = 15.8 - 2.33 * (1.8/√872)

u = 15.8 - 2.33 (0.6096)

u = 15.8 - 1.4203

u = 14.4

Hence the 98% confidence interval for population mean usage of electricity is (14.4kwh, 17.2kwh)