Question

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 random sample of 12 days while the offer was in effect. Can you conclude that the mean number of diners changed while the free dessert offer was in effect? Use the a-0.01 level of significance and the critical value method with the Critical Values for the Student's t Distribution Table 185 148 121 130 Send data Excel Part: 0/6 Part 1 of 6 Following is a boxplot for these data. Is it reasonable to assume that the conditions for performing a hypothesis test are satisfied? Explain.

Answer 1

Solution:- Given that samples : 185,148,170,131,191,118,121,130,153,189,171,131

part 1:

Box-Plot Graph: 19 Maximum Third Quartile 185 150elian First Quartile 131 11 Minimum

part :

The provided sample mean is X = 153.1667 and the sample standard deviation is 8 = 27.2157, and the sample size is n = 12 (1) Null and Alterative Hypotheses The following null and alternative hypotheses need to be tested: | Ho H = 150 Ha: > 150 This corresponds to a right-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used. (2) Rejection Region Based on the information provided, the significance level is a = 0.01, and the critical value for a right-tailed test is t = 2.718. The rejection region for this right-tailed test is R=t:t> 2.718 (2) Test Statistics The t-statistic is computed as follows: 153.1667 – 150 27.2157/V12 = 0.403 sVn (4) Decision about the null hypothesis Since it is observed that t = 0.403 <te= 2.718, it is then concluded that the null hypothesis is not rejected. Using the P-value approach: The p-value is p = 0.3473, and since p=0.3473 > 0.01, it is concluded that the null hypothesis is not rejected. (5) Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population meani is greater than 150, at the 0.01 significance level.