Questions: 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 70 residential customers has a mean monthly consumption of 930 kWh. Assume the population standard deviation is 128 kWh. At α=0.05, can you support the claim? Complete parts (a) through (e). (a) Identify H0 and Ha. Choose the correct answer below. A. H0: μ>890 (claim) Ha: μ ≤ 890 B. H0: μ>930 (claim) C. H0: μ ≤ 930 Ha: μ ≤ 930 Ha: μ>930 (claim) D. E. H0: μ=890 (claim) Ha: μ ≠ 890 H0: μ=930 Ha: μ ≠ 930 (claim) H0: μ ≤ 890 Ha: μ>890 (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.) A. The critical values are ± B. The critical value is

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 70 residential customers has a mean monthly consumption of 930 kWh. Assume the population standard deviation is 128 kWh. At α=0.05, can you support the claim? Complete parts (a) through (e). (a) Identify H0 and Ha. Choose the correct answer below. A. H0: μ>890 (claim) Ha: μ ≤ 890 B. H0: μ>930 (claim) C. H0: μ ≤ 930 Ha: μ ≤ 930 Ha: μ>930 (claim) D. E. H0: μ=890 (claim) Ha: μ ≠ 890 H0: μ=930 Ha: μ ≠ 930 (claim) H0: μ ≤ 890 Ha: μ>890 (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.) A. The critical values are ± B. The critical value is
Transcript 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 70 residential customers has a mean monthly consumption of 930 kWh. Assume the population standard deviation is 128 kWh. At $\alpha=0.05$, can you support the claim? Complete parts (a) through (e). (a) Identify $\mathrm{H}_{0}$ and $\mathrm{H}_{\mathrm{a}}$. Choose the correct answer below. A. $\mathrm{H}_{0}: \mu>890$ (claim) $\mathrm{H}_{\mathrm{a}}: \mu \leq 890$ B. $\mathrm{H}_{0}: \mu>930$ (claim) C. $\mathrm{H}_{0}: \mu \leq 930$ $H_{a}: \mu \leq 930$ $H_{a}: \mu>930$ (claim) D. E. \[ \begin{array}{l} \mathrm{H}_{0}: \mu=890 \text { (claim) } \\ \mathrm{H}_{\mathrm{a}}: \mu \neq 890 \end{array} \] $\mathrm{H}_{0}: \mu=930$ $H_{a}: \mu \neq 930$ (claim) $\mathrm{H}_{0}: \mu \leq 890$ $H_{a}: \mu>890$ (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.) A. The critical values are $\pm$ $\square$ B. The critical value is $\square$
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Solution

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Solution Steps

Step 1: Hypotheses

We are testing the claim that the mean monthly residential electricity consumption in a certain region is more than 890 890 kWh. Therefore, we set up our hypotheses as follows:

H0:μ≤890(null hypothesis) H_0: \mu \leq 890 \quad \text{(null hypothesis)} Ha:μ>890(alternative hypothesis) H_a: \mu > 890 \quad \text{(alternative hypothesis)}

Step 2: Calculate the Standard Error

The standard error SE SE is calculated using the formula:

SE=σn=12870≈15.2989 SE = \frac{\sigma}{\sqrt{n}} = \frac{128}{\sqrt{70}} \approx 15.2989

Step 3: Calculate the Test Statistic

The test statistic Z Z is calculated using the formula:

Z=xˉ−μ0SE=930−89015.2989≈2.6146 Z = \frac{\bar{x} - \mu_0}{SE} = \frac{930 - 890}{15.2989} \approx 2.6146

Step 4: Calculate the P-value

For a right-tailed test, the P-value is calculated as:

P=1−Φ(Z)≈0.0045 P = 1 - \Phi(Z) \approx 0.0045

Step 5: Determine the Critical Value and Rejection Region

For a significance level α=0.05 \alpha = 0.05 in a right-tailed test, the critical value Zcritical Z_{critical} is:

Zcritical=1.645 Z_{critical} = 1.645

The rejection region is defined as:

Z>1.645 Z > 1.645

Step 6: Conclusion

Since the calculated test statistic Z≈2.6146 Z \approx 2.6146 falls within the rejection region Z>1.645 Z > 1.645 and the P-value 0.0045 0.0045 is less than α=0.05 \alpha = 0.05 , we reject the null hypothesis.

Final Answer

We can support the claim that the mean monthly residential electricity consumption is more than 890 890 kWh. Thus, the answer is:

Reject H0 \boxed{\text{Reject } H_0}

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