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 65 residential customers has a mean monthly consumption of $920 \mathrm{kWh}$. Assume the population standard deviation is $126 \mathrm{kWh}$. At $\alpha=0.05$, can you support the claim? Complete parts (a) through (e). (c) Find the standardized test statistic. Use technology. The standardized test statistic is $z=1.92$. (Round to two decimal places as needed.) (d) Decide whether to reject or fail to reject the null hypothesis. A. Reject $\mathrm{H}_{0}$ because the standardized test statistic is in the rejection region. B. Fail to reject $\mathrm{H}_{0}$ because the standardized test statistic is in the rejection region. C. Fail to reject $\mathrm{H}_{0}$ because the standardized test statistic is not in the rejection region. D. Reject $\mathrm{H}_{0}$ because the standardized test statistic is not in the rejection region. (e) Interpret the decision in the context of the original claim. At the $5 \%$ significance level, there enough evidence to the claim that the mean monthly residential electricity consumption in a certain region kWh.

Step 1 ：The null hypothesis is that the mean monthly residential electricity consumption is 890 kWh, and the alternative hypothesis is that the mean is more than 890 kWh.

Step 2 ：We are given a sample mean of 920 kWh, a population standard deviation of 126 kWh, and a sample size of 65.

Step 3 ：We calculate the standardized test statistic, which is a z-score. The z-score tells us how many standard deviations the sample mean is from the population mean. The calculated z-score is approximately 1.92.

Step 4 ：We decide whether to reject or fail to reject the null hypothesis by comparing the z-score to the critical value for a one-tailed test at the 0.05 significance level. The critical value for a one-tailed test at the 0.05 significance level is approximately 1.645.

Step 5 ：The calculated z-score is greater than the critical value, so we reject the null hypothesis. This means that we have enough evidence to support the claim that the mean monthly residential electricity consumption is more than 890 kWh.

Step 6 ：Final Answer: The standardized test statistic is \( \boxed{1.92} \). We should \( \boxed{\text{Reject } H_0} \) because the standardized test statistic is in the rejection region. At the 5% significance level, there is \( \boxed{\text{enough evidence to support}} \) the claim that the mean monthly residential electricity consumption in a certain region is more than 890 kWh.