**Hypothesis Testing Example****Scenario:**A company claims that the mean monthly residential electricity consumption in a certain region exceeds 860 kiloWatt-hours (kWh). You want to test this claim. You have a random sample of 66 residential customers, with a mean monthly consumption of 900 kWh. Assume the population standard deviation is 123 kWh. With a significance level (\( \alpha \)) of 0.10, can you support the claim? Follow steps (a) through (e).**(a) Critical Value:**- **Chosen Option:** B. The critical value is 1.28**(b) Identify the Rejection Region(s):**- **Chosen Option:** B. The rejection region is \( z > 1.28 \)**(c) Standardized Test Statistic:**- The standardized test statistic \( z \) needs to be calculated.- The problem prompt instructs rounding the value to two decimal places once computed. To solve this, use the formula for the standardized test statistic:\[z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}\]Where:- \( \bar{x} = 900 \) (sample mean)- \( \mu = 860 \) (population mean under the null hypothesis)- \( \sigma = 123 \) (population standard deviation)- \( n = 66 \) (sample size)Compute the test statistic to decide whether to reject the null hypothesis in favor of the alternative claim that the mean consumption is indeed greater than 860 kWh.

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Description: **Hypothesis Testing Example****Scenario:**A company claims that the mean monthly residential electricity consumption in a certain region exceeds 860 kiloWatt-hours (kWh). You want to test this claim. You have a random sample of 66 residential custo

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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 cust 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) B. The critical value is 1.28 Identify the rejection region(s). Select the correct choice below. O A. The rejection regions are z< -1.28 and z> 1.28. YB. The rejection region is z> 1.28 OC. The rejection region is z <1.28. (c) Find the standardized test statistic. Use technology. The standardized test statistic is z=

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 cust 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) B. The critical value is 1.28 Identify the rejection region(s). Select the correct choice below. O A. The rejection regions are z< -1.28 and z> 1.28. YB. The rejection region is z> 1.28 OC. The rejection region is z <1.28. (c) Find the standardized test statistic. Use technology. The standardized test statistic is z=

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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