ExplOver e years, the mean customer satisfaction rating at a localrestaurant has been 79. The restaurant was recently remodeled, and nowthe management claims the mean customer rating, H, is not equal to 79.In a sample of 51 customers chosen at random, the mean customerrating is 79.3. Assume that the population standard deviation of customer?ratings is 2.4.Is there enough evidence to support the claim that the mean customerrating is different from 79? Perform a hypothesis test, using the 0.10 levelof significance.(a) State the null hypothesis H, and the alternative hypothesisH₁.Ho:H₁:0population mean usi(b) Perform a Z-test and find the p-value.Here is some information to help you with your Z-test.• The value of the test statistic is given byx-μ0Standard Normal DistributionStep 1: Select one-tailed or two-talled.O One-tailedOTwo-tailedStep 2: Enter the test statistic.(Round to 3 decimal places.)Step 3: Shade the area represented bythe p-value.Step 4: Enter the p-value.(Round to 3 decimal places.)• The p-value is two times the area under the curve to the right ofthe value of the test statistic.03+02+0.1+O Since the p-value is less than (or equal to) the level ofsignificance, the null hypothesis is rejected. So, there isenough evidence to support the claim that the meancustomer rating is not equal to 79.O Since the p-value is less than (or equal to) the level ofsignificance, the null hypothesis is not rejected. So, thereis not enough evidence to support the claim that themean customer rating is not equal to 79.O Since the p-value is greater than the level of significance,the null hypothesis is rejected. So, there is enoughevidence to support the claim that the mean customerrating is not equal to 79.0<0 0:0 0>0O Since the p-value is greater than the level of significance,the null hypothesis is not rejected. So, there is notenough evidence to support the claim that the meancustomer rating is not equal to 79.020 0-0 0-0XOP(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 themanagement.XX5$8$ 9:42=Ⓒ CONFIDENCE INTERVALS AND HYPOTHESIS TESTINGIntroduction to hypothesis tests for the population mean usingOver the years, the mean customer satisfaction rating at a localrestaurant has been 75. The restaurant was recently remodeled, and nowthe management claims the mean customer rating,", is not equal to 75.In a sample of 37 customers chosen at random, the mean customerrating is 72.5. Assume that the population standard deviation of customerratings is 15.6.Is there enough evidence to support the claim that the mean customerrating is different from 75? Perform a hypothesis test, using the 0.10 levelof significance.(a) State the null hypothesis H, and the alternative hypothesis H.H₁:0(b) Perform a hypothesis test. The test statistic has a normaldistribution (so the test is a "z-test"). Here is some otherinformation to help you with your test.√n• 5 is the value that cuts off an area of oos in the right tail.• The test statistic has a normal distribution and the value isgiven byStandard Normal DistributionStep 1: Select one-tailed or two-talled.O One-tailedOTwo-talledStep 2: Enter the critical value(s).(Round to 3 decimal places.)AAStep 3: Enter the test statistic(Round to 3 decimal places.)43402+artO Since the value of the test statistic lies in the rejectionregion, the null hypothesis is rejected. So, there is enoughevidence to support the claim that the mean customer ratingis not equal 75.O Since the value of the test statistic lies in the rejectionregion, the null hypothesis is not rejected. So, there is notenough evidence to support the claim that the meancustomer rating is not equal 75.H(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 themanagement.O Since the value of the test statistic doesn't lie in therejection region, the null hypothesis is rejected. So, there isenough evidence to support the claim that the meancustomer rating is not equal 75.0<0 00 00020 0-0 D-DO Since the value of the test statistic doesn't lie in therejection region, the null hypothesis is not rejected. So,there is not enough evidence to support the claim that themean customer rating is not equal 75.Xwww-awu.aleks.comX835GD田8

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Españo Over e years, the mean customer satisfaction rating at a local restaurant has been 79. The restaurant was recently remodeled, and now the management claims the mean customer rating, H. is not equal to 79. In a sample of si customers chosen at random, the mean customer rating is 79.3. Assume that the population standard deviation of customer' ratings is 2.4. Is there enough evidence to support the claim that the mean customer rating is different from 79? Perform a hypothesis test, using the 0.10 level of significance. (a) State the null hypothesis H, and the alternative hypothesis H₁. Ho: H₁:0 (b) Perform a z-test and find the p-value. Here is some information to help you with your Z-test. The value of the test statistic is given by Standard Normal Distribution Step 1: Select one-tailed or two-tailed. O One-tailed OTwo-tailed Step 2: Enter the test statistic. (Round to 3 decimal places.) Step 3: Shade the area represented by the p-value. Step 4: Enter the p-value. (Round to 3 decimal places.) • The p-value is two times the area under the curve to the right of the value of the test statistic. O 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 79. H O 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 79. O 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 79. O<0 0≤0 0>0 O 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 79. 020 0-0 0-0 X 03+ A 02- 0.1+ 8 X 20 (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. $ X I 3