Question

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 2.52.5 kWh. The mean electricity usage per family was found to be 1616 kWh per day for a sample of 33943394 families. Construct the 98%98% confidence interval for the mean usage of electricity. Round your answers to one decimal place.

Answer 1

(15.9,16.1) is the required 98% confidence interval for mean usage of electricity.

Explanation:

We are given the following in the question:

Sample mean,
`\bar{x}` = 16 kWh

Sample size, n = 3394

Alpha, α = 0.02

Population standard deviation, σ = 2.5 kWh

98% Confidence interval:

`\bar{x} \pm z_(critical)(\sigma)/(√(n))`

Putting the values, we get,

`z_(critical)\text{ at}~\alpha_(0.02) = 2.33`

`16 \pm 2.33((2.5)/(√(3394)) )\\\\ = 16 \pm 0.0999 = \\\\(15.9001,16.0999)\approx (15.9,16.1)`

(15.9,16.1) is the required 98% confidence interval for mean usage of electricity.