A company claims that the mean monthly residential electricity consumption in a certain region is more than $880 \mathrm{kiloWatt-hours}$ (kWh). You want to test this claim. You find that a random sample of 63 residential customers has a mean monthly consumption of $920 \mathrm{kWh}$. Assume the population standard deviation is $130 \mathrm{kWh}$. At $\alpha=0.05$, can you support the claim? Complete parts (a) through (e). B. The critical values are \pm Identify the rejection region(s). Select the correct choice below. A. The rejection regions are $z<-1.64$ and $z>1.64$ B. The rejection region is $z>1.64$. C. 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.)

Step 1 ：Identify the null and alternative hypotheses. The null hypothesis is that the mean monthly residential electricity consumption is equal to 880 kWh. The alternative hypothesis, which is the claim we want to test, is that the mean monthly residential electricity consumption is more than 880 kWh.

Step 2 ：Calculate the z-score, which is the number of standard deviations a data point (in this case, the sample mean) is from the population mean. The formula for the z-score is \((sample mean - population mean) / (population standard deviation / \sqrt{sample size})\).

Step 3 ：Compare the calculated z-score to the critical value to determine whether we can reject the null hypothesis. The critical value for a one-tailed test at the 0.05 significance level is 1.64. If the z-score is greater than 1.64, we can reject the null hypothesis and support the claim that the mean monthly residential electricity consumption is more than 880 kWh.

Step 4 ：The calculated z-score is 2.44, which is greater than the critical value of 1.64. This means that the sample mean is 2.44 standard deviations above the population mean.

Step 5 ：Identify the rejection region. Since this is a one-tailed test (we are testing whether the mean is greater than a certain value), the rejection region is z > 1.64.

Step 6 ：Final Answer: The standardized test statistic is \(z=2.44\). The rejection region is \(z>1.64\). Therefore, we can reject the null hypothesis and support the claim that the mean monthly residential electricity consumption is more than 880 kWh. \(\boxed{z=2.44, z>1.64}\)