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Topic 5 Homework (Nonadaptive) Question 6 of 21 (1 point)|Question Attempt: 1 of Unlimited

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Question 6 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, μ, is not equal to 75. In a sample of 37 customers chosen at random, the mean customer rating is 72.5. 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 H0 and the alternative hypothesis H1.

H0:

H1:

(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. z0.05 is the value that cuts off an area of 0.05 in the right tail. The test statistic has a normal distribution and the value is given by =z−xμσn. Standard Normal Distribution Step 1: Select one-tailed or two-tailed. One-tailed Two-tailed Step 2: Enter the critical value(s). (Round to 3 decimal places.)

Step 3: Enter the test statistic. (Round to 3 decimal places.)

0.10.20.30.41−12−23−3

(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. Conclusion:

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.

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.

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.

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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