**Hypothesis Testing of Mean Monthly Residential Electricity Consumption**A company claims that the mean monthly residential electricity consumption in a certain region is more than 860 kilowatt-hours (kWh). You want to test this claim. You find that a random sample of 66 residential customers has a mean monthly consumption of 900 kWh. Assume the population standard deviation is 123 kWh. At a significance level (\(\alpha\)) of 0.10, can you support the claim? Complete parts (a) through (e).---**Part (a):**Define the null and alternative hypotheses:- **Option A**: \[ H_0: \mu \neq 860 \quad (\text{claim}) \\ H_a: \mu \neq 860 \]- **Option B**: \[ H_0: \mu \geq 900 \quad (\text{claim}) \\ H_a: \mu < 900 \]- **Option C**: \[ H_0: \mu \neq 900 \\ H_a: \mu \neq 900 \quad (\text{claim}) \]- **Option D**: \[ H_0: \mu \leq 860 \\ H_a: \mu > 860 \quad (\text{claim}) \quad \text{(Selected)} \]- **Option E**: \[ H_0: \mu \geq 860 \quad (\text{claim}) \\ H_a: \mu < 860 \]**Part (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.)- **Option A**: The critical values are \(\_\).- **Option B**: The critical value is \(\_\).*Explanation:*This section allows you to calculate the critical value for a one-tailed test using a standard normal distribution, given a significance level of 0.10. Use statistical software or a z-table to find this value.

From the given information, Sample size n =66 Sample mean, x̅=900 kWh. Population standard…

A company claims that the mean monthly residential electricity consumption in a certain region is more than 860 kiloWatt-hours (kWh). You want to test this claim. You find that a random sample of 66 residential customers has a mean monthly consumption of 900 kWh. Assume the population standard deviation is 123 kWh. At a= 0.10, can you support the claim? Complete parts (a) through (e). O A. Ho u= 860 (claim) H u# 860 O B. Ho: u> 900 (claim) H us 900 O C. Ho: = 900 Ha u# 900 (claim) YD. Ho: us860 H3 p> 860 (claim) OE Ho us 900 H p> 900 (claim) OF. Ho H> 860 (claim) H us 860 (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.) O A. The critical values are + OB. The critical value is

A company claims that the mean monthly residential electricity consumption in a certain region is more than 860 kiloWatt-hours (kWh). You want to test this claim. You find that a random sample of 66 residential customers has a mean monthly consumption of 900 kWh. Assume the population standard deviation is 123 kWh. At a= 0.10, can you support the claim? Complete parts (a) through (e). O A. Ho u= 860 (claim) H u# 860 O B. Ho: u> 900 (claim) H us 900 O C. Ho: = 900 Ha u# 900 (claim) YD. Ho: us860 H3 p> 860 (claim) OE Ho us 900 H p> 900 (claim) OF. Ho H> 860 (claim) H us 860 (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.) O A. The critical values are + OB. The critical value is

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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