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 61 residential customers has a mean monthly consumption of 920 kWh. Assume the population standard deviation is 127 kWh. At a = 0.01, can you support the claim? Complete parts (a) through (e). (a) Identify Ho and H₂. Choose the correct answer below. OA. Ho: H=890 (claim) H₂:μ*890 OC. Ho: >890 (claim) H₂:μ≤890 OE. Ho: 920 H₂:μ#920 (claim) OB. Ho: 5920 OD. Ho: >920 (claim) H₂HS920 Ha: > 920 (claim) OF. Ho: 5890 (Round to two decimal places as needed.) OA. The critical values are t OB. The critical value is Identify the rejection region(s). Select the correct choice below. At the 1% significance level, there electricity consumption in a certain region (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 region is z <2.33. OB. The rejection regions are z< -2.33 and z>2.33. OC. The rejection region is z>2.33. (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. H₂:μ> 890 (claim) OA. Fail to reject Ho because the standardized test statistic is not in the rejection region. OB. Reject H, because the standardized test statistic is not in the rejection region. OC. Reject H, because the standardized test statistic is in the rejection region. OD. Fail to reject H, because the standardized test statistic is in the rejection region. (e) Interpret the decision in the context of the original claim. enough evidence to 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 890 kilowatt-hours (kWh). You want to test this claim. You find that a random sample of 61 residential customers has a mean monthly consumption of 920 kWh. Assume the population standard deviation is 127 kWh. At a = 0.01, can you support the claim? Complete parts (a) through (e). (a) Identify Ho and H₂. Choose the correct answer below. OA. Ho: H=890 (claim) H₂:μ*890 OC. Ho: >890 (claim) H₂:μ≤890 OE. Ho: 920 H₂:μ#920 (claim) OB. Ho: 5920 OD. Ho: >920 (claim) H₂HS920 Ha: > 920 (claim) OF. Ho: 5890 (Round to two decimal places as needed.) OA. The critical values are t OB. The critical value is Identify the rejection region(s). Select the correct choice below. At the 1% significance level, there electricity consumption in a certain region (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 region is z <2.33. OB. The rejection regions are z< -2.33 and z>2.33. OC. The rejection region is z>2.33. (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. H₂:μ> 890 (claim) OA. Fail to reject Ho because the standardized test statistic is not in the rejection region. OB. Reject H, because the standardized test statistic is not in the rejection region. OC. Reject H, because the standardized test statistic is in the rejection region. OD. Fail to reject H, because the standardized test statistic is in the rejection region. (e) Interpret the decision in the context of the original claim. enough evidence to 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 890
kilowatt-hours (kWh). You want to test this claim. You find that a random sample of 61 residential customers has a
mean monthly consumption of 920 kWh. Assume the population standard deviation is 127 kWh. At a = 0.01, can you
support the claim? Complete parts (a) through (e).
(a) Identify Ho and H₂. Choose the correct answer below.
OA. Ho: H=890 (claim)
H₂: *890
OC. Ho: >890 (claim)
Ha: μ≤ 890
OE. Ho: 920
Ha: #920 (claim)
OB. Ho: ≤920
Ha: >920 (claim)
O D. Ho: > 920 (claim)
Ha: μS920
At the 1% significance level, there
electricity consumption in a certain region
OF. Ho: s890
OA. The rejection region is z <2.33.
B. The rejection regions are z< -2.33 and z>2.33.
OC. The rejection region is z>2.33.
(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.
(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 values are
.
OB. The critical value is.
Identify the rejection region(s). Select the correct choice below.
Ha: μ> 890 (claim)
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 not in the rejection region.
OC. Reject Ho because the standardized test statistic is 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.
enough evidence to
kWh.
the claim that the mean monthly residential
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