Questions: The U.S. Energy information Administration claimed that U.S, residential customers used an average of 10,001 kilowatt hours (kWh) of electricity this year. A local power company believes that residents in their area use more electricity on average than EIA's reported average. To test their claim, the company chooses a random sample of 147 of their customers and calculates that these customers used an average of 10,405 kWh of electricity last year. Assuming that the population standard deviation is 2113 kWh . is there sufficient evidence to support the power company's claim at the 0.05 level of significance? Step 1 of 3: State the null and alternative hypotheses for the test. Fill in the blank below. H0: μ=10,001 Ha: μ>10,001

The U.S. Energy information Administration claimed that U.S, residential customers used an average of 10,001 kilowatt hours (kWh) of electricity this year. A local power company believes that residents in their area use more electricity on average than EIA's reported average. To test their claim, the company chooses a random sample of 147 of their customers and calculates that these customers used an average of 10,405 kWh of electricity last year. Assuming that the population standard deviation is 2113 kWh . is there sufficient evidence to support the power company's claim at the 0.05 level of significance?
Step 1 of 3: State the null and alternative hypotheses for the test. Fill in the blank below.

H0: μ=10,001
Ha: μ>10,001

Transcript text: The U.S. Energy information Administration claimed that U.S, residential customers used an average of 10,001 kilowatt hours (kWh) of electricity this year. A local power company believes that residents in their area use more electricity on average than EIA's reported average. To test their claim, the company chooses a random sample of 147 of their customers and calculates that these customers used an average of $10,405 \mathrm{kWh}$ of electricity last year. Assuming that the population standard deviation is 2113 kWh . is there sufficient evidence to support the power company's claim at the 0.05 level of significance? Step 1 of 3: State the null and alternative hypotheses for the test. Fill in the blank below. \[ \begin{array}{l} H_{0}: \mu=10,001 \\ H_{a}: \mu-10,001 \end{array} \]

Solution

Solution Steps

Step 1: State the null and alternative hypotheses

The null hypothesis $ H_0 $ and the alternative hypothesis $ H_a $ are as follows: \[ \begin{array}{l} H_{0}: \mu = 10,001 \\ H_{a}: \mu > 10,001 \end{array} \] This is a one-tailed test because the power company is testing whether the average usage is greater than the reported average.

Step 2: Calculate the test statistic

The test statistic for a one-sample $ z $-test is calculated using the formula: \[ z = \frac{\bar{x} - \mu_0}{\sigma / \sqrt{n}} \] Where:

$ \bar{x} = 10,405 $ (sample mean),
$ \mu_0 = 10,001 $ (population mean under $ H_0 $),
$ \sigma = 2,113 $ (population standard deviation),
$ n = 147 $ (sample size).

Substitute the values: \[ z = \frac{10,405 - 10,001}{2,113 / \sqrt{147}} \] First, calculate the denominator: \[ \sigma / \sqrt{n} = 2,113 / \sqrt{147} \approx 2,113 / 12.1244 \approx 174.3 \] Now, calculate the numerator: \[ \bar{x} - \mu_0 = 10,405 - 10,001 = 404 \] Finally, compute the test statistic: \[ z = \frac{404}{174.3} \approx 2.318 \]

Step 3: Determine the critical value and make a decision

At a significance level of $ \alpha = 0.05 $ for a one-tailed test, the critical $ z $-value is $ z_{\text{critical}} = 1.645 $.

Compare the test statistic $ z = 2.318 $ with the critical value: \[ 2.318 > 1.645 \] Since the test statistic exceeds the critical value, we reject the null hypothesis.

Final Answer

There is sufficient evidence to support the power company's claim that residents in their area use more electricity on average than the EIA's reported average. The test statistic is $ z \approx 2.318 $, which is greater than the critical value of $ 1.645 $. Thus, we reject $ H_0 $.

\[ \boxed{\text{Reject } H_0} \]

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