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a company claims that the mean monthly residential electricity consumpt…

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 α = 0.05, can you support the claim? complete parts (a) through (e). c. ( h_{0}: mu leq 910 ) ( h_{a}: mu>910 ) (claim) d. ( h_{0}: mu = 880 ) (claim) ( h_{a}: mu
eq 880 ) e. ( h_{0}: mu leq 880 ) ( h_{a}: mu>880 ) (claim) f. ( h_{0}: mu = 910 ) ( h_{a}: mu
eq 910 ) (claim) (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<-1.64 ) and ( z>1.64 ). b. the rejection region is ( z>1.64 ). c. the rejection region is ( z<1.64 ).

Explanation:

Step1: Determine the type of test

Since the alternative hypothesis \(H_{a}:\mu > 880\) (claim), this is a right - tailed test.

Step2: Find the critical value

For a right - tailed test with \(\alpha = 0.05\), using the standard normal distribution \(Z\), we look up the \(z\) - value such that \(P(Z>z_{\alpha})=\alpha\). From the standard normal table or using technology, \(z_{\alpha}=1.64\) (rounded to two decimal places).

Step3: Identify the rejection region

In a right - tailed test, the rejection region is where the test statistic \(z\) is greater than the critical value. So the rejection region is \(z > 1.64\).

Answer:

B. The critical value is \(1.64\).
B. The rejection region is \(z>1.64\).