Saisiaction Fating at a löcal restaurant has been 75. The restaurant was recently remodeled, and now the mana the mean customer rating, µ, is not equal to 75. In a sample of 37 customers chosen at random, the mean customer rating is 80.5. Assume t tion standard deviation of customer ratings is 15.6. re enough evidence to support the claim that the mean customer rating is different from 75? Perform a hypothesis test, using the 0.10 level of cance. tate the null hypothesis H, and the alternative hypothesis H,. OSO H: I D=0 Perform a hypothesis test. The test statistic has a normal distribution (so the test is a "Z-test"). Here is some other information to help you with your • Zn os is the value that cuts off an area of 0.05 in the right tail. Ix
Saisiaction Fating at a löcal restaurant has been 75. The restaurant was recently remodeled, and now the mana the mean customer rating, µ, is not equal to 75. In a sample of 37 customers chosen at random, the mean customer rating is 80.5. Assume t tion standard deviation of customer ratings is 15.6. re enough evidence to support the claim that the mean customer rating is different from 75? Perform a hypothesis test, using the 0.10 level of cance. tate the null hypothesis H, and the alternative hypothesis H,. OSO H: I D=0 Perform a hypothesis test. The test statistic has a normal distribution (so the test is a "Z-test"). Here is some other information to help you with your • Zn os is the value that cuts off an area of 0.05 in the right tail. Ix
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
Related questions
Question

Transcribed Image Text:Over the years, the mean customer satisfaction rating at a local restaurant has been 75. The restaurant was recently remodeled, and now the management
claims the mean customer rating, u, is not equal to 75. In a sample of 37 customers chosen at random, the mean customer rating is 80.5. Assume that the
population standard deviation of customer ratings is 15.6.
Is there enough evidence to support the claim that the mean customer rating is different from 75? Perform a hypothesis test, using the 0.10 level of
significance.
(a) State the null hypothesis H, and the alternative hypothesis H,.
Ho:
O<O O<O
H: 0
D=D D>0
(b) Perform a hypothesis test. The test statistic has a normal distribution (so the test is a "Z-test"). Here is some other information to help you with your test.
Znns is the value that cuts off an area of 0.05 in the right tail.
Z0.05

Transcribed Image Text:(c) Based on your answer to part (b), choose what can be concluded, at the 0.10 level of significance, about the claim made by the management.
O Since the value of the test statistic lies in the rejection region, the null hypothesis is rejected. So, there is enough evidence
to support the claim that the mean customer rating is not equal 75.
O Since the value of the test statistic lies in the rejection region, the null hypothesis is not rejected. So, there is not enough
evidence to support the claim that the mean customer rating is not equal 75.
O Since the value of the test statistic doesn't lie in the rejection region, the null hypothesis is rejected. So, there is enough
evidence to support the claim that the mean customer rating is not equal 75.
O Since the value of the test statistic doesn't lie in the rejection region, the null hypothesis is not rejected. So, there is not
enough evidence to support the claim that the mean customer rating is not equal 75.
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