Over the years, the mean customer satisfaction rating at a local restaurant has been 85. The restaurant was recently remodeled, and now the management claims the mean customer rating, u, is not equal to 85. In a sample of 42 customers chosen at random, the mean customer rating is 83.1. Assume that the population standard deviation of customer ratings is 5.3. 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 of significance. (a) State the null hypothesis H, and the alternative hypothesis H. р x B H: OSO D> H: 0 . 20 O=D Х $ ? (b) Perform a Z-test and find the p-value. Here some information to help you with your Z-test. X • The value of the test statistic is given by 0 The p-value is two times the area under the curve to the left of the value of the test statistic. . Standard Normal Distribution 0. Step 1: Select one-tailed or two-tailed. 1-. O One-tailed Two-tailed 0.3 + Step 2: Enter the test statistic. (Round to 3 decimal places.) 0.2 Step 3: Shade the area represented by : the p-value 0.1 Step 4: Enter the p-value. (Round to 3 decimal places.) S ? (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 85. х 5 ? 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 85. 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 85. Since the p-value is greater than 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 85.