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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).
(a) identify ( h_0 ) and ( h_a ). choose the correct answer below.
a. ( h_0:mu>910 ) (claim)
( h_a:muleq910 )
b. ( h_0:mu>880 ) (claim)
( h_a:muleq880 )
c. ( h_0:muleq910 )
( h_a:mu>910 ) (claim)
d. ( h_0:mu = 880 ) (claim)
( h_a:mu
eq880 )
e. ( h_0:muleq880 )
( h_a:mu>880 ) (claim)
f. ( h_0:mu = 910 )
( h_a:mu
eq910 ) (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 ±

b. the critical value is

Explanation:

Step1: Determine the null and alternative hypotheses

The claim is that the mean monthly residential electricity consumption is more than 880 kWh. The null hypothesis \(H_0\) is the statement of no change or equality. The alternative hypothesis \(H_a\) is the claim we are testing. So \(H_0:\mu\leq880\) and \(H_a:\mu > 880\) (claim).

Step2: Identify the type of test and find the critical value

Since \(H_a:\mu>880\), this is a right - tailed test. For a significance level \(\alpha = 0.05\), using the standard normal distribution \(Z\) (because the population standard deviation \(\sigma\) is known), we look up the \(z\) - value. The critical value \(z_{\alpha}\) for a right - tailed test with \(\alpha=0.05\) is the \(z\) - value such that \(P(Z>z_{\alpha})=0.05\), which means \(P(Z\leq z_{\alpha})=1 - 0.05=0.95\). Using a standard normal table or technology, \(z_{\alpha}=1.64\)

Answer:

(a) E. \(H_0:\mu\leq880\), \(H_a:\mu > 880\) (claim)
(b) B. The critical value is \(1.64\)