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 2.3 kWh. The mean electricity usage per family was found to be 15.3 kWh per day for a sample of 3611 families. Construct the 99% confidence interval for the mean usage of electricity.

Answer #1

Solution :

Given that,

Sample size = n = 3611

Z_{/2}
= 2.576

Margin of error = E = Z_{/2}*
(
/n)

= 2.576 * (2.3 / 3611)

Margin of error = E = 0.1

At 99% confidence interval estimate of the population mean is,

- E < < + E

15.3 - 0.1 < < 15.3 + 0.1

15.2 < < 15.4

**(15.2 , 15.4)**

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