Over the years, the mean customer satisfaction rating at a local restaurant has been 80. The restaurant was recently remodeled, and now the management claims the mean customer rating, μ, is not equal to 80. In a sample of 54 customers chosen at random, the mean customer rating is 75.4. Assume that the population standard deviation of customer ratings is 21.2.Is there enough evidence to support the claim that the mean customer rating is different from 80? Perform a hypothesis test, using the 0.05 level of significance.(a) State the null hypothesis $H_0$ and the alternative hypothesis $H_1$:- $H_0: \mu = 80$- $H_1: \mu \neq 80$(b) Perform a Z-test and find the p-value.Here is some information to help you with your Z-test:- The value of the test statistic is given by $\frac{\bar{X} - \mu}{\frac{\sigma}{\sqrt{n}}}$.- The p-value is two times the area under the curve to the left of the value of the test statistic.Graph: Standard Normal Distribution 1. Select one-tailed or two-tailed. - Two-tailed2. Enter the test statistic. (Round to 3 decimal places.)3. Shade the area represented by the p-value.4. Enter the p-value. (Round to 3 decimal places.)(c) Based on your answer to part (b), choose what can be concluded, at the 0.05 level of significance, about the claim made by the management.- [ ] Since the p-value is less than (or equal to) the level of significance, the null hypothesis is rejected. So, there is enough evidence to support the claim that the mean customer rating is not equal to 80.- [ ] Since the p-value is less than (or equal to) the level of significance, the null hypothesis is not rejected. So, there is not enough evidence to support the claim that the mean customer rating is not equal to 80.- [ ] Since the p-value is greater than the level of significance, the null hypothesis is rejected. So, there is enough evidence to support the claim that the mean customer rating is not equal to 80.- [ ] Since the p-value is greater than the level of significance, the null

Here test the two tailed claim whether the mean is 80 or not n = 54 X bar = 75.4 Sigma =21.2

Answered: Over the years, the mean customer…

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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(b) Perform a Z-test and find the p-value. Here is some information to help you with your Z-test. • The value of the test statistic is given by √n The p-value is the area under the curve to the right of the value of the test statistic. O One-tailed o Two-tailed Step 2: Enter the test statistic. (Round to 3 decimal places.) Step 3: Shade the area represented by the p- value. Step 4: Enter the p- value. (Round to 3 decimal places.) 0.3 0.2 0.1 (c) Based on your answer to part (b), choose what can be concluded, at the 0.05 level of significance, about the claim made by the studio. o Since the p-value is less than (or equal to) the level of significance, the null hypothesis is rejected. So, there is enough evidence to support the claim that the mean player rating is higher than 90. Since the p-value is less than (or equal to) the level of significance, the null hypothesis is not rejected. So, there is not enough evidence to support the claim that the mean player rating is higher than 90.…
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Almost all medical schools in the United States require applicants to take the Medical College Admission Test (MCAT). On one exam, the scores of all applicants on the biological sciences part of the MCAT were approximately Normal with mean 9.4 and standard deviation 2.6. For applicants who actually entered medical school, the mean score was 10.6 and the standard deviation was 1.7. (a) What percent of all applicants had scores higher than 11? (b) What percent of those who entered medical school had scores between 8 and 12?

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