Over the years, the mean customer satisfaction rating at a local restaurant has been 75. The restaurant was recently remodeled, and now the managementclaims the mean customer rating, μ, is not equal to 75. In a sample of 37 customers chosen at random, the mean customer rating is 72.5. Assume that thepopulation 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 ofsignificance.(a) State the null hypothesis Ho and the alternative hypothesis H₁.Ho:H₁:0μXO Standard Normal DistributionStep 1: Select one-tailed or two-tailed.O One-tailedO Two-tailedStep 2: Enter the critical value(s).(Round to 3 decimal places.)Step 3: Enter the test statistic.(Round to 3 decimal places.)040.3+0.2+0.1+(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 enoughevidence 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 notenough 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 isenough evidence to support the claim that the mean customer rating is not equal 75.X

given data, n=37x¯=72.5σ=15.6claim:μ≠75α=0.10

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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