Statistics II Exam Key

STATISTICS SECTION II Part A Questions 1-5 Spend about 65 minutes on this part of the exam. Percent of Section II score-75 Directions: Show all your work. Indicate clearly the methods you use, because you will be scored on the correctness of your methods as well as on the accuracy and completeness of your results and explanations. 1. An administrator at a large university is interested in determining whether the residential status of a studeaf is associated with level of participation in extracurricular activities. Residential status is categorized as on campus for students living in university housing and off campus otherwise. A simple random sample of 100 students in the university was taken, and each student was asked the following two questions. • Are you an on campus student or an off campus student? In how many extracurricular activities do you participate? The responses of the 100 students are summarized in the frequency table shown. Residential Status Level of Participation in Extracurricular Activities On campus Off campus Total No activities 9 30 39 One activity 17 25 42 Two or more activities 7 12 19 Total 33 67 100 (a) Calculate the proportion of on campus students in the sample who participate in at least one extracurricular activity and the proportion of off campus students in the sample who participate in at least one extracurricular activity. On campus proportion: Off campus proportion: 24 332.7272 67 x.5522 The responses of the 100 students are summarized in the segmented bar graph shown. None On campus Off campus One Two or more 0% 20% 40% 60% 80% 100% (b) Write a few sentences summarizing what the graph reveals about the association between residential status and level of participation in extracurricular activities among the 100 students in the sample. To participating in at extracurricular is roughly the same for both on/off "and also more Campus. "On campus" has fewer "none's" than "off campus" a Students who participate in one activity. GO ON TO THE NEXT PAGE. -6- (c) After verifying that the conditions for inference were satisfied, the administrator performed a chi-square test of the following hypotheses. Ha: There is no association between residential status and level of participation in extracurricular activities among the students at the university. H: There is an association between residential status and level of participation in extracurricular activities among the students at the university. The test resulted in a p-value of 0.23. Based on the p-value, what conclusion should the administrator make? Fail to reject Ho bla .232.05. There is not enough evidence to suggest there is an association b/w residential status & level of part. 2. Nine sales representatives, 6 men Jad 3 women, dat a small company wanted to attend a national convention. There were only enough travel funds to send 3 people. The manager selected 3 people to attend and stated that the people were selected at random. The 3 people selected were women. There were concerns that no men were selected to attend the convention. (a) Calculate the probability that cradonly selecting 3 people from a group of 6 men and 3 women will result in selecting 3 women 3 987 # 504= 84x.0119 (b ) Based on your answer to part (a), is there reason to doubt the manager's claim that the 3 people were selected at random? Explain. Yes, ble the chance of selecting 3 women is 1.27%, which is highly unlikely if the drawing is truly random. (c) An alternative to calculating the exact probability is to conduct a sinaulation to estimate the probability. A proposed simulation process is described below. aman Each trial in the simulation consists of rolling three fair, siz-Sided dice, que die for ench of the convention attendees. For each die, rolling a1, 2, 3, or 4 represents selecting a man, rolling a 5 or 6 represents selecting a woman. After 1,000 trials, the number of times the dice indicate selecting 3 women is recorded. Does the proposed process correctly simulate the random selection of 3 women from a group of 9 people consisting of 6 men and 3 women? Explain why or why not. No, there is always a 2/3 prob. of getting a man and 113 prob. of a woman. Since the selection does not involve replacement, Leponding on cement, the ratio changes who was previously selected. 3. Schools in a certain ställe receive funding based on the number of students who attend the school. To determine the number of students who attend a school, one school day is selected at random and the number of students in attendance that day is counted and used for funding purposes. The daily number of absences at High School A in the state is approximately normally distributed with mean of 120 students and standard deviation of 10.5 students. (a) If more than 140 students are absent on the day the attendance count is taken for funding purposes, the school will lose some of its state funding in the subsequent year. Approximately what is the probability that High School A will lose some state funding? 20 140 2140-120 10.5. P/27 1.90476) (10284 1,90476 9/27기.. (b) The principals association in the state suggests that instead of choosing one day at randonı, the state should choose 3 days at random. With the suggested plan, High School A would lose some of its state funding in the subsequent year if the mean gumber of students absent for the 3 days is greater than 140. Would High School A be more likely, less likely, or equally likely to lose funding using the suggested plan compared to the plan described in part (a)? Justify your choice. der of the sampling dist. is now Less likely, St.dev 10.5/53 5/53, so there is less Variability & the schools average of the 3 days should be closer to 120. (c) A typical school week consists of the days Monday, Tuesday, Wednesday, Thursday, and Friday. The principal at High School A believes that the muumber of absences tends to be greater on Mondays and Fridays, and there is concern that the school will lose state funding if the attendance count occurs on a Monday or Friday. If one school day is chosen at random from each of 3 typical school weeks, what is the probability that none of the 3 days chosen is a Tuesday, Wednesday, or Thursday? weekl 8 125 6064 repeat for each of 3 weeks GO ON TO THE NEXT PAGE. -8- 4. As part of its twenty-fifth reunion celebration, the class of 1988 (students who graduated in 1988) at a state university held a reception on campus. In an informal survey, the director of alumni development asked 50 of the attendees about their incomes. The director computed the mean income of the 50 attendees to be $189,952 In a news release, the director aunounced, "The members of our class of 1988 enjoyed resounding success. Last year's mean income of its members was $189,95217 X (a) What would be a statistical advantage of using the median of the reported incomes, rather than the mean, as the estimate of the typical income? The median is not affected by outliers, and since salary dist. tend to be skewed right, the median better measures the center. (b) The director felt the members who attended the reception may be different from the class as a whole. A more detailed survey of the class was planned to find a better estimate of the inconue as well as other facts about the alumni. The staff developed two methods based on the available funds to carry out the survey Method 1: Send out an e-mail to all 6,826 members of the class asking them to complete an online form The staff estimates that at least 600 members will respond. Metikai 2: Select a simple random sample of members of the class and contact the selected members directly by phone. Follow up to ensure that all responses are obtamed) Because method 2 will require more time than method T, the staff estimates that only 100 members of the class could be contacted using method 2. Which of the two methods would you select for estimating the average yearly income of all 6,826 members of the class of 1988? Explain your reasoning by comparing the two methods and the effect of each method on the estimate. Method 2. Method 1. is a voluntary response sample and is likely to be blased and overestimate the true income since those with higher incomes will be more likely to respond since they can "brag" about their Income level. method 2 is an SRS & while the sample size is smaller, it is more likely to better estimate the mean because the sample is being randomly selected and should better represent the whole population,