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 find that a random sample of 69 residential customers has a mean monthly consumption of 900 kWh. Assume the population standard deviation is 122 kWh. At a = 0.05, can you support the claim? Complete parts (a) through (e). (a) Identify Ho and H. Choose the correct answer below. OA. Ho: $900 Ha: >900 (claim) OC. Ho: >870 (claim) H: 870 OE. Ho: >900 (claim) Ha: μs 900 B. Ho: H=870 (claim) H₂H870 OD. Ho: μ = 900 H₂:μ*900 (claim) F. Ho: 870 (Round to two decimal places as needed.) A. The critical values are OB. The critical value is Identify the rejection region(s). Select the correct choice below. OA. The rejection regions are z< -1.64 and z> 1.64. B. The rejection region is z> 1.64. OC. The rejection region is z < 1.64. (c) Find the standardized test statistic. Use technology. The standardized test statistic is z=2.04. (Round to two decimal places as needed.) (d) Decide whether to reject or fail to reject the null hypothesis. H₂:μ> 870 (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. OA. Reject Ho because the standardized test statistic is in the rejection region. B. Reject Ho because the standardized test statistic is not in the rejection region. OC. Fail to reject Ho because the standardized test statistic is not in the rejection region. OD. Fail to reject Ho because the standardized test statistic is in the rejection region. (e) Interpret the decision in the context of the original claim. At the 5% significance level, there enough evidence to support the claim that the mean monthly residential electricity consumption in a certain region is greater than 870 kWh.
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 find that a random sample of 69 residential customers has a mean monthly consumption of 900 kWh. Assume the population standard deviation is 122 kWh. At a = 0.05, can you support the claim? Complete parts (a) through (e). (a) Identify Ho and H. Choose the correct answer below. OA. Ho: $900 Ha: >900 (claim) OC. Ho: >870 (claim) H: 870 OE. Ho: >900 (claim) Ha: μs 900 B. Ho: H=870 (claim) H₂H870 OD. Ho: μ = 900 H₂:μ*900 (claim) F. Ho: 870 (Round to two decimal places as needed.) A. The critical values are OB. The critical value is Identify the rejection region(s). Select the correct choice below. OA. The rejection regions are z< -1.64 and z> 1.64. B. The rejection region is z> 1.64. OC. The rejection region is z < 1.64. (c) Find the standardized test statistic. Use technology. The standardized test statistic is z=2.04. (Round to two decimal places as needed.) (d) Decide whether to reject or fail to reject the null hypothesis. H₂:μ> 870 (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. OA. Reject Ho because the standardized test statistic is in the rejection region. B. Reject Ho because the standardized test statistic is not in the rejection region. OC. Fail to reject Ho because the standardized test statistic is not in the rejection region. OD. Fail to reject Ho because the standardized test statistic is in the rejection region. (e) Interpret the decision in the context of the original claim. At the 5% significance level, there enough evidence to support the claim that the mean monthly residential electricity consumption in a certain region is greater than 870 kWh.
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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Transcribed Image Text: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 find that a random sample of 69 residential customers has a
mean monthly consumption of 900 kWh. Assume the population standard deviation is 122 kWh. At a = 0.05, can you
support the claim? Complete parts (a) through (e).
(a) Identify Ho and H₂. Choose the correct answer below.
OA. Ho: 5900
Ha: >900 (claim)
OC. Ho: >870 (claim)
Η:15 870
O E. Ho: >900 (claim)
H₂900
OB. Ho: H=870 (claim)
H₂H870
O D. Ho: H=900
OF. Ho: H≤ 870
H: 900 (claim)
(Round to two decimal places as needed.)
ⒸA. The critical values are t
OB. The critical value is.
Identify the rejection region(s). Select the correct choice below.
(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.
OA. The rejection regions are z<- 1.64 and z> 1.64.
B. The rejection region is z> 1.64.
OC. The rejection region is z < 1.64.
(c) Find the standardized test statistic. Use technology.
The standardized test statistic is z=2.04.
(Round to two decimal places as needed.)
(d) Decide whether to reject or fail to reject the null hypothesis.
H₂: >870 (claim)
O A. Reject Ho because the standardized test statistic is in the rejection region.
B. Reject Ho because the standardized test statistic is not in the rejection region.
OC. Fail to reject Ho because the standardized test statistic is not in the rejection region.
OD. Fail to reject Ho because the standardized test statistic is in the rejection region.
(e) Interpret the decision in the context of the original claim.
At the 5% significance level, there is enough evidence to support the claim that the mean monthly residential
electricity consumption in a certain region is greater than 870 kWh.
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