Question

"In order to estimate the average electric usage per month, a sample of 33 houses were selected and the electric usage determined. The sample mean is 2,000 KWH. Assume a population standard deviation of 106 kilowatt hours. At 99% confidence, compute the upper bound of the interval estimate for the population mean."

Answer 1

Answer:

2,157.89 KWH is the upper bound of the interval estimate for the population mean.

Step-by-step explanation:

We are given the following in the question:

Sample mean, $\bar{x}$ = 2,000 KWH

Sample size, n = 33

Alpha, α = 0.01

Population standard deviation, σ = 106 KWH

99% Confidence interval:

$\mu \pm z_{critical}\dfrac{\sigma}{\sqrt{n}}$

Putting the values, we get,

$z_{critical}\text{ at}~\alpha_{0.01} = 2.58$

$2000 \pm 2.58(\dfrac{106}{\sqrt{3}} )\\\\ = 2000 \pm 157.89 \\\\= (1842.11,2157.89)$

2,157.89 KWH is the upper bound of the interval estimate for the population mean