Over the years, the mean customer satisfaction rating at a local restaurant has been 85. The restaurant was recently remodeled, and now the managementclaims the mean customer rating, u, is not equal to 85. In a sample of 42 customers chosen at random, the mean customer rating is 84.1. Assume that thepopulation standard deviation of customer ratings is 5.2.Is there enough evidence to support the claim that the mean customer rating is different from 85? Perform a hypothesis test, using the 0.05 level ofsignificance.(a) State the null hypothesis H, and the alternative hypothesis H.O -3-23.(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 value of the test statistic lies in the rejection region, the null hypothesis is rejected. So, there is enough evidenceto support the claim that the mean customer rating is not equal 85.Since the value of the test statistic lies in the rejection region, the null hypothesis is not rejected. So, there is not enoughevidence to support the claim that the mean customer rating is not equal 85.Since the value of the test statistic doesn't lie 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 85.Since the value of the test statistic doesn't lie 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 85.

In order to determine whether the mean customer rating is different from 85 or not, conduct a single mean comparison hypotheses test. The hypotheses test will contain two hypotheses: null hypothesis and alternate hypothesis. The alternate hypothesis is about the claim which is to be tested. If the test statistic lies in the rejection region, then null hypothesis can be rejected and it can be concluded that test is significant.(a): Assume, X be the random variable representing the customer rating after remodeling the restaurant. μ be the population mean customer rating after remodeling the restaurant. x¯ be the sample mean customer rating after remodeling the restaurant. σ be the population standard deviation associated with customer ratings. n be the sample size. The null hypothesis for the test states that, μ is 85 and the alternate hypothesis for the test states that μ is significantly different from 85. H0: μ=85Ha: μ≠85(b): For two tailed test, the critical values of z can be given by ±zα2 which is the value of standard normal variable which corresponds to α2 proportion of area under the right tail of standard normal distribution curve. The significance level is 5%, that is, α=0.05. In this case, the critical values of z can be given by ±z0.025 which is the value of standard normal variable which corresponds to 0.025 proportion of area under the right tail of standard normal distribution curve. This gives, zα2=1.96, which implies zcritical=±1.96. The null hypothesis can be rejected if ztest statistic>1.96 or ztest statistic<-1.96. The test statistic for the test can be given as, z=x¯-μ0σn where, μ0 denotes the mean customer rating as per null hypothesis. Substitute x¯=84.1, σ=5.2, n=42 and μ0=85 in the above mentioned formula for z and calculate its value. z=84.1-855.242≈-1.122 Since, the test statistic is neither less than -1.96 nor greater than 1.96, therefore, it can be said that the test statistic does not lie in the rejection. (c): Since, the test statistic does not lie in the rejection. The null hypothesis is not rejected. At 5% significance level, there are insufficient evidence to conclude that the population mean customer rating is not equal to 85 after remodeling the restaurant. Hence, option (d) is correct. Option (a): Incorrect because the test statistic does not lie in the rejection. Option (b): Incorrect because the test statistic does not lie in the rejection. Option (c): Incorrect because the test statistic does not lie in the rejection, therefore, the null hypothesis is not rejected.