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A company claims that the mean monthly residential electricity consumption in a certain region is more than 890 kiloWatt-hours (kWh). You want to test this claim. You find that a random sample of 68 residential customers has a mean monthly consumption of 930 kWh. Assume the population standard deviation is 123 kWh. At a = 0.10, can you support the claim? Complete parts (a) through (e). (Round to two decimal places as needed.) CA. The critical value is 1.28 O B. The critical values are t Identify the rejection region(s). Select the correct choice below. O A. The rejection regions are z< - 1.28 and z>1.28. YB. The rejection region isz>1.28. O C. The rejection region is z<1.28. (c) Find the standardized test statistic. Use technology. The standardized test statistic isz= 2.68 (Round to two decimal places as needed.) (d) Decide whether to reject or fail to reject the null hypothesis. CA. Reject Ho because the standardized test statistic is in the rejection region. O B. Fail to reject Ho because the standardized test statistic is not in the rejection region. O C. Fail to reject Ho because the standardized test statistic is in the rejection region. O D. Reject Ho 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 V enough evidence to v the claim that the mean monthly residential electricity consumption in a certain region kWh.

A company claims that the mean monthly residential electricity consumption in a certain region is more than 890 kiloWatt-hours (kWh). You want to test this claim. You find that a random sample of 68 residential customers has a mean monthly consumption of 930 kWh. Assume the population standard deviation is 123 kWh. At a = 0.10, can you support the claim? Complete parts (a) through (e). (Round to two decimal places as needed.) CA. The critical value is 1.28 O B. The critical values are t Identify the rejection region(s). Select the correct choice below. O A. The rejection regions are z< - 1.28 and z>1.28. YB. The rejection region isz>1.28. O C. The rejection region is z<1.28. (c) Find the standardized test statistic. Use technology. The standardized test statistic isz= 2.68 (Round to two decimal places as needed.) (d) Decide whether to reject or fail to reject the null hypothesis. CA. Reject Ho because the standardized test statistic is in the rejection region. O B. Fail to reject Ho because the standardized test statistic is not in the rejection region. O C. Fail to reject Ho because the standardized test statistic is in the rejection region. O D. Reject Ho 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 V enough evidence to v the claim that the mean monthly residential electricity consumption in a certain region kWh.

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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A company claims that the mean monthly residential electricity consumption in a certain region is more than 890 kiloWatt-hours (kWh). You want to test this claim. You find that a random sample of 68 residential customers has a mean monthly consumption of 930 kWh. Assume the population standard deviation is 123 kWh. At a = 0.10, can you support the claim? Complete parts (a) through (e). (Round to two decimal places as needed.) C A. The critical value is 1.28 O B. The critical values are t Identify the rejection region(s). Select the correct choice below. O A. The rejection regions are z< -1.28 and z>1.28. YB. The rejection region is z>1.28. O C. The rejection region is z<1.28. (c) Find the standardized test statistic. Use technology. The standardized test statistic is z= 2.68 (Round to two decimal places as needed.) (d) Decide whether to reject or fail to reject the null hypothesis. C A. Reject Ho because the standardized test statistic is in the rejection region. O B. Fail to reject Ho because the standardized test statistic is not in the rejection region. O C. Fail to reject Ho because the standardized test statistic is in the rejection region. O D. Reject H, 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 V enough evidence to V the claim that the mean monthly residential electricity consumption in a certain region kWh.
Transcribed Image Text:A company claims that the mean monthly residential electricity consumption in a certain region is more than 890 kiloWatt-hours (kWh). You want to test this claim. You find that a random sample of 68 residential customers has a mean monthly consumption of 930 kWh. Assume the population standard deviation is 123 kWh. At a = 0.10, can you support the claim? Complete parts (a) through (e). (Round to two decimal places as needed.) C A. The critical value is 1.28 O B. The critical values are t Identify the rejection region(s). Select the correct choice below. O A. The rejection regions are z< -1.28 and z>1.28. YB. The rejection region is z>1.28. O C. The rejection region is z<1.28. (c) Find the standardized test statistic. Use technology. The standardized test statistic is z= 2.68 (Round to two decimal places as needed.) (d) Decide whether to reject or fail to reject the null hypothesis. C A. Reject Ho because the standardized test statistic is in the rejection region. O B. Fail to reject Ho because the standardized test statistic is not in the rejection region. O C. Fail to reject Ho because the standardized test statistic is in the rejection region. O D. Reject H, 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 V enough evidence to V the claim that the mean monthly residential electricity consumption in a certain region kWh.
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