Essay Type (show your work in Excel) Show your hypotheses and calculated test statisticsChapter 91. The average waiting time per customer at a fast food restaurant has been 7.5 minutes. The customer waiting time has a normal distribution. The manager claims that the use of a new cashier system will significantly decrease the average customer waiting time in the store. Based on a random sample of 25 customer transactions the mean waiting time is 6.22 minutes and the standard deviation is 2.5 minutes per customer. Test the manager’s claim at 5% and 1% significance level tests.2. In an early study, researchers at an Ivy University found that 30% of the freshmen had received at least one A in their first semester. Administrators are concerned that grade inflation has caused this percentage to increase. In a more recent study, of a random sample of 400 freshmen, 136 had at least one A in their first semester Calculate the appropriate test statistic to test the hypotheses related to the concern and test at 5% and 1%.3. A microwave manufacturing company has just switched to a new automated production system. In the past the company has made a reputation by providing its customers with a completion time of less than 6 days on average. To analyze whether the mean completion time is still statistically less than 6 days, the production manager took a sample of 36 jobs and found that the sample mean completion time was 5.74 days with a sample standard deviation of 1.2 days. At significance levels of 5% and 10%, test whether the completion time is still less than 6 days. Indicate which test you are performing; show the hypotheses, the test statistic and the critical values and mention whether one-tailed or two-tailed.Chapter 104. At ? = 0.05 and 0.10, test the hypothesis that the proportion of Consumer (CON) industry companies winter quarter profit growth is more than 1percentage point greater than the proportion of Banking (BKG) companies winter quarter profit growth, given that p CON = 0.20, p BKG = 0.14, nCON = 350, nBKG=400.5. The mid-distance running coach, Zdravko Popovich, for the Olympic team of an eastern European country claims that his six-month training program significantly reduces the average time to complete a 1500-meter run. Five mid-distance runners were randomly selected before they were trained with coach Popovich’s six-month training program and their completion time of 1500-meter run was recorded (in minutes). After six months of training under coach Popovich, the same five runners’ 1500 meter run time was recorded again the results are given below.At an alpha level of .01, can we conclude that there has been a significant decrease in the mean time per mile?6. Test H0: ?1 = ?2; H1: ?1 ? ?2 at ? = 0.05 and 0.01, when 1 = 75.6, 2 = 72, s1 = 3.3, s2 = 2.1, n1 = 6, and n2 = 6. Assume equal variances. Indicate which test you are performing; show the hypotheses, the test statistic and the critical values and mention whether one-tailed or two-tailed.Chapter 127. In the past, of all the students enrolled in “Basic Business Statistics” 12% earned A’s 30% earned B’s, 22% earned C’s, 26% earned D’s and the rest either failed or withdrew from the course. Dr Johnson is a new professor teaching “Basic Business Statistics” for the first time this semester. At the conclusion of the semester, in Dr. Johnson’s class of 50 students, there were 9 A’s, 19 B’s, 14 C’s, 4 D’s and 4 W’s or F’s. Assume that Dr. Johnson’s class constitutes a random sample. Dr Johnson wants to know if there is sufficient evidence to conclude that the grade distribution of his class is different than the historical grade distribution.Use ? =.05 and .01 and perform a goodness of fit test.8. A recent national survey of hospital admissions for people between 25 and 50 years who had hospital admissions in during a two years’ period showed that 30% had 1 admission only, 25% had two admissions, 15% had 3 admissions, 12% had 4 admissions, 8 % had 5 admissions, 10% had 6 admissions or more admissions. The mayor of a small city claims that his city is much healthier than the national average. He even cites the percentages for the two extreme categories. His claim was in fact based on a sample of 300 randomly selected people in the specified age group who were interviewed by a local Newspaper. It was revealed that 114 people had only 1 admission, 84 had 2 admissions, 47 had 3 admissions, 24 had 4 admissions, 15 had 5 admissions, and 16 had 6 admissions or more admissions. Does the data support the mayor’s claim?

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