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.

Respuesta :

Answer:

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

Step-by-step explanation:

We are given the following in the question:

Sample mean, [tex]\bar{x}[/tex] = 16 kWh

Sample size, n = 3394

Alpha, α = 0.02

Population standard deviation, σ = 2.5 kWh

98% Confidence interval:

[tex]\bar{x} \pm z_{critical}\dfrac{\sigma}{\sqrt{n}}[/tex]

Putting the values, we get,

[tex]z_{critical}\text{ at}~\alpha_{0.02} = 2.33[/tex]

[tex]16 \pm 2.33(\dfrac{2.5}{\sqrt{3394}} )\\\\ = 16 \pm 0.0999 = \\\\(15.9001,16.0999)\approx (15.9,16.1)[/tex]

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