Question

a company claims that the mean monthly residential electricity consumption in a certain region is more than 880 kilowatt - hours (kwh). you want to test this claim. you find that a random sample of 70 residential customers has a mean monthly consumption of 910 kwh. assume the population standard deviation is 124 kwh. at \\( \alpha = 0.05 \\), can you support the claim? complete parts (a) through (e).
(b) find the critical value(s) and identify the rejection region(s). select the correct choice below and fill in the answer box within your choice. use technology.
(round to two decimal places as needed.)
a. the critical values are \\( \pm \\).
b. the critical value is 1.64.
identify the rejection region(s). select the correct choice below.
a. the rejection regions are \\( z\lt - 1.64 \\) and \\( z\gt 1.64 \\).
b. the rejection region is \\( z\gt 1.64 \\).
c. the rejection region is \\( z\lt 1.64 \\).
(c) find the standardized test statistic. use technology.
the standardized test statistic is \\( z = \square \\).
(round to two decimal places as needed)

Explanation:

Step1: Recall the formula for the z - test statistic

The formula for the z - test statistic is \(z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\), where \(\bar{x}\) is the sample mean, \(\mu\) is the population mean, \(\sigma\) is the population standard deviation, and \(n\) is the sample size.
We are given that \(\bar{x} = 910\), \(\mu=880\), \(\sigma = 124\), and \(n = 70\).

Step2: Substitute the values into the formula

First, calculate \(\frac{\sigma}{\sqrt{n}}=\frac{124}{\sqrt{70}}\approx\frac{124}{8.3666}\approx14.82\).
Then, \(z=\frac{910 - 880}{14.82}=\frac{30}{14.82}\approx2.02\)

Answer:

\(z = 2.02\)