Suppose that the neighboring cities of Tweed and Ledee are long-term rivals. Nea

Question

Suppose that the neighboring cities of Tweed and Ledee are long-term rivals. Neal, who was born and raised in Tweed, is confident that Tweed residents are more concerned about the environment than the residents of Ledee. He knows that the average electricity consumption of Tweed households last February was 854.11 kWh and decides to test if Ledee residents used more electricity that month, on average. He collects data from 65 Ledee households and calculates the average electricity consumption to be 879.28 kWh with a standard deviation of 133.29 kWh. There are no outliers in his sample data.

Neal does not know the population standard deviation nor the population distribution. He uses a one-sample t-test with a significance level of = 0.05 to test the null hypothesis, H0:=854.11, against the alternative hypothesis, H1:>854.11, where is the average electricity consumption of Ledee households last February. Neal calculates a tstatistic of 1.522 and a p-value of 0.066.

Based on these results, complete the following sentences to state the decision and conclusion of the test.

Neal's decision is to the Op 0.066). There is evidence to the claim that the average electricity consumption of is Answer Bank Some all Ledee households 5% 0066 sample proportion null hypothesis equal to not equal to reject greater than fail to accept sufficient fail to reject alternative hypothesi prove the Ledee household sample 0.05 less than 854.11 kWh Support 1.5222 insufficient 879.28 kWh one-sample t-test no 133.29 kWh disprove 65 accept

Explanation / Answer

Solution:

For the given scenario, we are given the following information:

t Test for Hypothesis of the Mean

Data

Null Hypothesis                m=

854.11

Level of Significance

0.05

Sample Size

65

Sample Mean

879.28

Sample Standard Deviation

133.29

Intermediate Calculations

Standard Error of the Mean

16.5326

Degrees of Freedom

64

t Test Statistic

1.5224

Upper-Tail Test

Upper Critical Value

1.6690

p-Value

0.0664

Do not reject the null hypothesis

Here, P-value = 0.0664 > Alpha value = 0.05, so we do not reject the null hypothesis.

Neal’s decision is to fail to reject the null hypothesis (p = 0.066). There is insufficient evidence to prove the claim that the average electricity consumption of all Ledee households is greater than 854.11 kWh.

t Test for Hypothesis of the Mean

Data

Null Hypothesis                m=

854.11

Level of Significance

0.05

Sample Size

65

Sample Mean

879.28

Sample Standard Deviation

133.29

Intermediate Calculations

Standard Error of the Mean

16.5326

Degrees of Freedom

64

t Test Statistic

1.5224

Upper-Tail Test

Upper Critical Value

1.6690

p-Value

0.0664

Do not reject the null hypothesis

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