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).

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.
(e) interpret the decision in the context of the original claim.
at the 10% significance level, there is enough evidence to reject the claim that the mean monthly residential electricity consumption in a certain region is greater than 135 kwh.

Explanation:

Step1: Hypothesis testing basics

The null hypothesis \(H_0:\mu\leq860\) and the alternative hypothesis \(H_1:\mu > 860\) (claim). The test - statistic for a z - test (since population standard deviation \(\sigma\) is known) is \(z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\), where \(\bar{x} = 900\), \(\mu=860\), \(\sigma = 120\), \(n = 61\).

Step2: Calculate the z - statistic

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For a right - tailed test with \(\alpha=0.10\), the critical value \(z_{\alpha}\) from the standard normal table is \(z_{0.10}=1.28\).

Step3: Make a decision

Since \(z = 2.60>z_{\alpha}=1.28\) (the standardized test statistic is in the rejection region), we reject \(H_0\).

Answer:

At the \(10\%\) significance level, there is enough evidence to support the claim that the mean monthly residential electricity consumption in a certain region is greater than \(860\) kWh.