Free dessert: In an attempt to increase business on Monday nights, a restaurant offers a free dessert with every dinner order.Before the offer, the mean number of dinner customers on Monday was 150. Following are the numbers of diners on a randomsample of 12 days while the offer was in effect. Can you conclude that the mean number of dinners decreased while the freedessert offer was in effect? Use the a = 0.05 level of significance and the P-value method with the TI-84 Plus calculator. 175138160 121181108111120143 189161122Following is a boxplot for these data. Is it reasonable to assume that the conditions for performing a hypothesis test aresatisfied? Explain.100120140160180200220240The boxplot shows that there (Choose one)outliers.areThe boxplot shows that there are noevidence of strong skewness.We (Choose one)assume that the population is approximately normal.It (Choose one) v reasonable to assume that the conditions are satisfled.The boxplot shows that there (Choose one) outliers.The boxplot shows that there (Choose one)evidence of strong skewness.We (Choose one)assume that the population is approximately normal.canIt dcannotJeasonable to assume that the conditions are satisfied.The boxplot shows that there (Choose one) outliers.The boxplot shows that there (Choose one) evidence of strong skewness.We (Choose one) assume that the population is approximately normal.It (Choose one) reasonable to assume that the conditions are satisfled.isis notThe boxplot shows that there (Choose one) outliers.The boxplot shows that there (Choose one) evidence of strong skewness.We (Choose one) assume is nopn is approximately normal.It (Choose one) reasonable to assume that the conditions are satisfied. Part 1 of 4State the appropriate null and alternate hypotheses.HoThis hypothesis test is a (Choose one) v test.left-tailedright-talledtwo-tailedPart 2 of 4Compute the P-value. Round the P-value to at least four decimal places.P-value =Part 3 of 4Determine whether to reject H(Choose one)the null hypothesis H.RejectDo not reject

Question

Accepted Answer

The hypothesized mean is 150.