Question

Question Help A company claims that the mean monthly residential electricity consumption in a certain region is more than 870 kiloWatt-hours (kWh). You want to test this claim. You findt t sample of 64 esidential customers has a mean monthly consumption of 890 kWh with a standard deva on of 129 kWh. At α-Ο 10, can you supp he claim? Complete parts (a) through (e). a random O A. The rejection region is z < 1.28. O B. The rejection region is z > 1.28. O C. The rejection regions are z -1.28 and z> 1.28 (c) Find the test statistic. The test statistic is z Round to two decimal places as needed.) (d) Decide whether to reject or fail to reject the null hypothesis. O A. 0 B. ○ c. 0 D. Reject Ho because the test statistic is not in the rejection region Reject Ho because the test statistic is in the rejection region. Fail to reject Ho because the test statistic is not in the rejection region. Fail to reject Ho because the test statistic is in the rejection region (e) Interpret the decision in the context of the original claim. t o At the 10% significance level, there | ▼| enough evidence to the claim that the mean monthly residential electricity consumption in a certain region ▼1kWh. Click to select your answerts) and then click Check Answer

Answer 1

Solution:

Null hypothesis H0: mean<=870

Alternate hypothesis H1: mean>870

As alpha=0.1 and this is right tailed test Z =1.28

Reject H0 if test statistic is greater than 1.28 else do not reject H0.

Test stat = (890-870)/129/sqrt(64) = 20/129/8 = 1.24

Here test stat value is less than Zcritical value(1.24<1.28) so we are failed to reject H0. Because the test statistic is not in the rejection region. So its answer is C.

At the 10% significance level there are not enough evidence to support the claim that the mean monthly residential electricity consumption in a certain region is more than 870kwh.