A company claims that the mean monthly residential electricity consumption in a certain region is more than 860 kilowatt-hours (kWh). You want to test this claim. You find that a random sample of 63 residential customers has a mean monthly consumption of 900 kWh. Assume the population standard deviation is 128 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: μ>900 (claim) OC. Ho: μ=860 (claim) H₂:μ*860 OE. Ho: H>860 (claim) H₂H860 OB. Ho: μ≤ 860 Ha: μ> 860 (claim) OD. Ho: >900 (claim) H₂:μ≤900 OF. Ho: H=900 H₂:μ*900 (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.) OA. The critical value is OB. The critical values are ± Identify the rejection region(s). Select the correct choice below. OA. The rejection region is z < 1.64. B. The rejection regions are z< -1.64 and 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 = (Round to two decimal places as needed.) (d) Decide whether to reject or fail to reject the null hypothesis. OA. Fail to reject Ho because the standardized test statistic is not in the rejection region. OB. Reject Ho because the standardized test statistic is in the rejection region. OC. Fail to reject Ho because the standardized test statistic is in the rejection region. OD. Reject H, because the standardized test statistic is not in the rejection region. (e) Interpret the decision in the context of the original claim. enough evidence to At the 5% significance level, there electricity consumption in a certain region kWh. the claim that the mean monthly residential
A company claims that the mean monthly residential electricity consumption in a certain region is more than 860 kilowatt-hours (kWh). You want to test this claim. You find that a random sample of 63 residential customers has a mean monthly consumption of 900 kWh. Assume the population standard deviation is 128 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: μ>900 (claim) OC. Ho: μ=860 (claim) H₂:μ*860 OE. Ho: H>860 (claim) H₂H860 OB. Ho: μ≤ 860 Ha: μ> 860 (claim) OD. Ho: >900 (claim) H₂:μ≤900 OF. Ho: H=900 H₂:μ*900 (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.) OA. The critical value is OB. The critical values are ± Identify the rejection region(s). Select the correct choice below. OA. The rejection region is z < 1.64. B. The rejection regions are z< -1.64 and 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 = (Round to two decimal places as needed.) (d) Decide whether to reject or fail to reject the null hypothesis. OA. Fail to reject Ho because the standardized test statistic is not in the rejection region. OB. Reject Ho because the standardized test statistic is in the rejection region. OC. Fail to reject Ho because the standardized test statistic is in the rejection region. OD. Reject H, because the standardized test statistic is not in the rejection region. (e) Interpret the decision in the context of the original claim. enough evidence to At the 5% significance level, there electricity consumption in a certain region kWh. the claim that the mean monthly residential
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 860
kilowatt-hours (kWh). You want to test this claim. You find that a random sample of 63 residential customers has a
mean monthly consumption of 900 kWh. Assume the population standard deviation is 128 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.
O A. Ho: ≤900
Ha: μ>900 (claim)
OC. Ho: μ-860 (claim)
H₂: H860
CD
O E. Ho: >860 (claim)
H₂: H≤860
OB. Ho: S860
OD. Ho: >900 (claim)
Η:με 900
OF. Ho: 900
Ha: >860 (claim)
OB. The critical values are +
t
Identify the rejection region(s). Select the correct choice below.
OA. The rejection region is z <1.64.
OB. The rejection regions are z< -1.64 and z> 1.64.
OC. The rejection region is z> 1.64.
(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.)
OA. The critical value is
(c) Find the standardized test statistic. Use technology.
The standardized test statistic is z =.
(Round to two decimal places as needed.)
(d) Decide whether to reject or fail to reject the null hypothesis.
Ha: μ#900 (claim)
(e) Interpret the decision in the context of the original claim.
At the 5% significance level, there
enough evidence to
electricity consumption in a certain region
O A. Fail to reject Ho because the standardized test statistic is not in the rejection region.
OB. Reject Ho because the standardized test statistic is in the rejection region.
OC. Fail to reject Ho because the standardized test statistic is in the rejection region.
OD. Reject Ho because the standardized test statistic is not in the rejection region.
kWh.
the claim that the mean monthly residential
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