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 a = 0.10, can you support the claim? complete parts (a) through (e)
e. ( h_0: mu = 900 )
( h_a: mu
eq 900 ) (claim)
f. ( h_0: mu = 860 ) (claim)
( h_a: mu
eq 860 )
(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 ±
b. the critical value is 1.28
identify the rejection region(s). select the correct choice below
a. the rejection regions are ( z < -1.28 ) and ( z > 1.28 ).
b. the rejection region is ( z < 1.28 ).
c. the rejection region is ( z > 1.28 ).

Explanation:

Step1: Determine the type of test

The claim is that the mean monthly residential electricity consumption is more than 860 kWh. So, this is a right - tailed test.

Step2: Recall the formula for the critical value in a one - tailed z - test

For a significance level of \(\alpha = 0.10\) in a right - tailed z - test, we use the standard normal distribution \(Z\sim N(0,1)\). The critical value \(z_{\alpha}\) is such that \(P(Z>z_{\alpha})=\alpha\). Using a standard normal table or technology, for \(\alpha = 0.10\), the critical value \(z_{0.10}\) is \(1.28\).

Step3: Identify the rejection region

In a right - tailed test, the rejection region is \(z>z_{\alpha}\). Since \(z_{\alpha}=1.28\), the rejection region is \(z > 1.28\).

Answer:

C. The rejection region is \(z>1.28\)