Question

a company claims that the mean monthly residential electricity consumption in a certain region is more than 860 kilowatt - hours (kwh). you want to test this claim. you find that a random sample of 61 residential customers has a mean monthly consumption of 900 kwh. assume the population standard deviation is 120 kwh. at \\( \alpha=0.10 \\), can you support the claim? complete parts (a) through (e)
(c) find the standardized test statistic. use technology
the standardized test statistic is \\( z = 2.60 \\)
(round to two decimal places as needed)
(d) decide whether to reject or fail to reject the null hypothesis.
a. fail to reject \\( h_{0} \\) because the standardized test statistic is not in the rejection region.
b. reject \\( h_{0} \\) because the standardized test statistic is in the rejection region.
c. fail to reject \\( h_{0} \\) because the standardized test statistic is in the rejection region.
d. reject \\( h_{0} \\) because the standardized test statistic is not in the rejection region.

Explanation:

Step1: Analyze the rejection region

The claim is that the mean is more than 860, so it is a right - tailed test. For a right - tailed test with \(\alpha = 0.10\), the critical value \(z_{c}\) is such that \(P(Z>z_{c})=0.10\), and from the standard normal table \(z_{c}=1.28\). The rejection region is \(z > 1.28\).

Step2: Compare the test statistic with the rejection region

The standardized test statistic \(z = 2.60\). Since \(2.60>1.28\) (i.e., the test statistic \(z = 2.60\) lies in the rejection region \(z>1.28\)).

Answer:

B. Reject \(H_{0}\) because the standardized test statistic is in the rejection region.