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Over the years, the mean customer satisfaction rating at a local restaurant has been 92. The restaurant was recently remodeled, and now the management daims the mean customer rating, p, is not equal to 92. In a sample of 68 customers chosen at random, the mean customer rating is 87.4. Assume that the population standard deviation of customer ratings is 143. Is there enough evidence to support the clalm that the mean customer rating is different from 92? Perform a hypothesis test, using the 0.05 level of significance. (a) State the null hypothesis Ho and the alternative hypothesis H,. Ho: 0 H: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. F0.025 is the value that cuts off an area of 0.025 in the right tail. The test statistic has a normal distribution and the value is given by z- or two-tailed. o One-tailed o 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.) (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 evidence to support the claim that the mean customer rating is not equal 92. 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 92. 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 92. 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

Over the years, the mean customer satisfaction rating at a local restaurant has been 92. The restaurant was recently remodeled, and now the management daims the mean customer rating, p, is not equal to 92. In a sample of 68 customers chosen at random, the mean customer rating is 87.4. Assume that the population standard deviation of customer ratings is 143. Is there enough evidence to support the clalm that the mean customer rating is different from 92? Perform a hypothesis test, using the 0.05 level of significance. (a) State the null hypothesis Ho and the alternative hypothesis H,. Ho: 0 H: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. F0.025 is the value that cuts off an area of 0.025 in the right tail. The test statistic has a normal distribution and the value is given by z- or two-tailed. o One-tailed o 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.) (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 evidence to support the claim that the mean customer rating is not equal 92. 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 92. 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 92. 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

MATLAB: An Introduction with Applications
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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Over the years, the mean customer satisfaction rating at a local restaurant has been 92. The restaurant was recently remodeled, and now the management daims the mean customer rating, p, is not equal to 92. In a sample of 68 customers chosen at random, the mean customer rating is 874. Assume that the population standard deviation of customer ratings is 14.3. Is there enough evidence to support the clalm that the mean customer rating is different from 92? Perform a hypothesis test, uslng the 0.05 level of significance. (a) State the null hypothesis H, and the alternative hypothesis H. Osa a D>0 H: 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. F0.025 is the value that cuts off an area of 0.025 in the right tail. • The test statistic has a normal distribution and the value is given by z- or two-talled. o One-tailed o 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. ) (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 evidence to support the claim that the mean customer rating is not equal 92. 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 92. 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 92. 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 92
Transcribed Image Text:Over the years, the mean customer satisfaction rating at a local restaurant has been 92. The restaurant was recently remodeled, and now the management daims the mean customer rating, p, is not equal to 92. In a sample of 68 customers chosen at random, the mean customer rating is 874. Assume that the population standard deviation of customer ratings is 14.3. Is there enough evidence to support the clalm that the mean customer rating is different from 92? Perform a hypothesis test, uslng the 0.05 level of significance. (a) State the null hypothesis H, and the alternative hypothesis H. Osa a D>0 H: 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. F0.025 is the value that cuts off an area of 0.025 in the right tail. • The test statistic has a normal distribution and the value is given by z- or two-talled. o One-tailed o 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. ) (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 evidence to support the claim that the mean customer rating is not equal 92. 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 92. 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 92. 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 92
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