(e) Interpret the decision in the context of the oniginal claim. At the 1% 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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(e) Interpret the decision in the context of the original claim.
A company claims that the mean monthly residential electricity consumption in a certain region is more than 880 kiloWatt-hours (kWh). You want to test this claim. You find that a random sample of 60 residential customers has a mean monthly
consumption of 920 kWh. Assume the population standard deviation is 129 kWh. At a=0.01, can you support the claim? Complete parts (a) through (e).
(Round to two decimal places as needed.)
O A. The critical values are t
B. The critical value is 2.33
Identify the rejection region(s). Select the correct choice below.
O A. The rejection region is z<2.33.
OB. The rejection region is z>2.33.
O C. The rejection regions are z< -2.33 andz>2.33.
(c) Find the standardized test statistic. Use technology
The standardized test statistic is z= 2.40
(Round to two decimal places as needed.)
(d) Decide whether to reject or fail to reject the null hypothesis.
O A. Fail to reject Ho because the standardized test statistic is in the rejection region.
B. Reject Ho because the standardized test statistic is in the rejection region.
O C. Reject Ho because the standardized test statistic is not in the rejection region.
O D. Fail to 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 1% 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 880 kiloWatt-hours (kWh). You want to test this claim. You find that a random sample of 60 residential customers has a mean monthly consumption of 920 kWh. Assume the population standard deviation is 129 kWh. At a=0.01, can you support the claim? Complete parts (a) through (e). (Round to two decimal places as needed.) O A. The critical values are t B. The critical value is 2.33 Identify the rejection region(s). Select the correct choice below. O A. The rejection region is z<2.33. OB. The rejection region is z>2.33. O C. The rejection regions are z< -2.33 andz>2.33. (c) Find the standardized test statistic. Use technology The standardized test statistic is z= 2.40 (Round to two decimal places as needed.) (d) Decide whether to reject or fail to reject the null hypothesis. O A. Fail to reject Ho because the standardized test statistic is in the rejection region. B. Reject Ho because the standardized test statistic is in the rejection region. O C. Reject Ho because the standardized test statistic is not in the rejection region. O D. Fail to 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 1% 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 880 kiloWatt-hours (kWh). You want to test this claim. You find that a random sample of 60 residential customers has a mean monthly
consumption of 920 kWh. Assume the population standard deviation is 129 kWh. At a =0.01, can you support the claim? Complete parts (a) through (e).
(a) Identify Ho and Ha. Choose the correct answer below.
088s1:0H 8
(Wiep) 088 <r1:"H
O A. Ho: H= 880 (claim)
088 #1:"H
O C. Ho: u> 920 (claim)
O D. Ho: uS 920
Hau>920 (claim)
0z6 s1:"H
O E. Ho: H> 880 (claim)
0z6 = rl:0H 1 O
(Wjep) oz6 ar:4
088 51:"H
(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 t
B. The critical value
2.33.
Identify the rejection region(s). Select the correct choice below.
O A. The rejection region is z<2.33.
B. The rejection region is z> 2.33.
OC. The rejection regions are z< -2.33 and z>2.33.
(c) Find the standardized test statistic. Use technology.
The standardized test statistic is z= 2.40
(Round to two decimal places as needed.)
Transcribed Image Text:A company claims that the mean monthly residential electricity consumption in a certain region is more than 880 kiloWatt-hours (kWh). You want to test this claim. You find that a random sample of 60 residential customers has a mean monthly consumption of 920 kWh. Assume the population standard deviation is 129 kWh. At a =0.01, can you support the claim? Complete parts (a) through (e). (a) Identify Ho and Ha. Choose the correct answer below. 088s1:0H 8 (Wiep) 088 <r1:"H O A. Ho: H= 880 (claim) 088 #1:"H O C. Ho: u> 920 (claim) O D. Ho: uS 920 Hau>920 (claim) 0z6 s1:"H O E. Ho: H> 880 (claim) 0z6 = rl:0H 1 O (Wjep) oz6 ar:4 088 51:"H (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 t B. The critical value 2.33. Identify the rejection region(s). Select the correct choice below. O A. The rejection region is z<2.33. B. The rejection region is z> 2.33. OC. The rejection regions are z< -2.33 and z>2.33. (c) Find the standardized test statistic. Use technology. The standardized test statistic is z= 2.40 (Round to two decimal places as needed.)
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