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 66 residential customers has a mean monthly consumption of 880 kWh. Assume the population standard deviation is 127 kWh. At a = 0.05, can you support the claim? Complete parts (a) through (e). C. ine cnticai vaiue is Identify the rejection region(s). Select the correct choice below. O A. The rejection regions are z< -1.64 and z> 1.64. O 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=. (Round to two decimal places as needed.) (d) Decide whether to reject or fail to reject the null hypothesis. O A. Fail to reject H, because the standardized test statistic is not in the rejection region. O B. 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. O D. 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 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 860 kilowatt-hours (kWh). You want to test this claim. You find that a random sample of 66 residential customers has a mean monthly consumption of 880 kWh. Assume the population standard deviation is 127 kWh. At a = 0.05, can you support the claim? Complete parts (a) through (e). C. ine cnticai vaiue is Identify the rejection region(s). Select the correct choice below. O A. The rejection regions are z< -1.64 and z> 1.64. O 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=. (Round to two decimal places as needed.) (d) Decide whether to reject or fail to reject the null hypothesis. O A. Fail to reject H, because the standardized test statistic is not in the rejection region. O B. 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. O D. 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 V enough evidence to V the claim that the mean monthly residential electricity consumption in a certain region kWh.
MATLAB: An Introduction with Applications
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Chapter1: Starting With Matlab
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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 66 residential customers has a mean monthly consumption of
880 kWh. Assume the population standard deviation is 127 kWh. At a = 0.05, can you support the claim? Complete parts (a) through (e).
(a) Identify Ho and Ha. Choose the correct answer below.
Ο Α. H0: μ = 880
O B. Ho: µ> 860 (claim)
Ha: H# 880 (claim)
Ha: us 860
ОС. Но: и3860 (claim)
O D. Ho: µ> 880 (claim)
H3: uz 860
Ha: us 880
ОЕ. Но: нS860
Ha: u> 860 (claim)
O F. Ho: µs880
Ha: µ> 880 (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.)
O A. The critical values are +
O B. The critical value is
Identify the rejection region(s). Select the correct choice below.
O A. The rejection regions are z< - 1.64 and z> 1.64.
O B. The rejection region is z> 1.64.
C. The rejection region is z< 1.64.
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 66 residential customers has a mean monthly consumption of
880 kWh. Assume the population standard deviation is 127 kWh. At a = 0.05, can you support the claim? Complete parts (a) through (e).
U D.
The criticai vaiue is
Identify the rejection region(s). Select the correct choice below.
O A. The rejection regions are z< - 1.64 and z> 1.64.
O 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=
(Round to two decimal places as needed.)
(d) Decide whether to reject or fail to reject the null hypothesis.
O A. Fail to reject H, because the standardized test statistic is not in the rejection region.
O B. Reject Ho because the standardized test statistic is not in the rejection region.
OC. Reject H, because the standardized test statistic is in the rejection region.
O D. 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.
At the 5% significance level, there
enough evidence to
the claim that the mean monthly residential electricity consumption in a certain region
kWh.
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