Exam 10: Categorical Data Analysis

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A coffeehouse wishes to see if customers have any preference among 5 different brands of coffee. A sample of 200 customers provided the data below. Find the rejection region used to test the claim that the probabilities show no preference. Use α=0.01\alpha = 0.01 . Brand 1 2 3 4 5 Customers 55 65 18 32 30

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A professor chose a random sample of 50 recent graduates of an MBA program and recorded the gender of each graduate (M or F) and whether the graduate chose to complete his or her degree requirements by completing a research project (RP) or by taking comprehensive exams (CE). The results are shown below. M, CE M, RP F, RP M, CE M, CE F, CE F, RP, M, CE M, RP F, RP F, CE M, RP F, RP F, CE M, RP F, RP M, RP F, CE M, CE M, CE M, RP F, CE F, RP M, CE M, CE F, CE M, RP F, RP M,CE M, CE F, CE F, RP, M, CE M, RP F, RP M, CE M, RP F, RP F, CE M, RP F, RP M, RP M, CE M,CE M, CE M, RP F, CE F, RP M,CE F, CE a. Create a contingency table for the data. b. Perform a χ2\chi ^ { 2 } -test to determine if there is any evidence that gender and choice of research project or comprehensive exams are not independent. Use α=0.05\alpha = 0.05 .

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A sports researcher is interested in determining if there is a relationship between the number of home team and visiting team wins and different sports. A random sample of 526 games is selected and the results are given below. Calculate the chi-square test statistic χ2\chi ^ { 2 } used to test the claim that the number of home team and visiting team wins is independent of the sport. Use α=0.01\alpha = 0.01 . Football Basketball Soccer Baseball Home team wins 39 156 25 83 Visiting team wins 31 98 19 75

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The contingency table below shows the results of a random sample of 200 state representatives that was conducted to see whether their opinions on a bill are related to their party affiliations. Opinion Party Approve Disapprove No Opinion Republican 42 20 14 Democrat 50 24 18 Independent 10 16 6 Test the claim of independence. Use \alpha=.05 .

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A multinomial experiment with k=4k = 4 cells and n=300n = 300 produced the data shown in the following table. Cell 1 2 3 4 65 69 80 86 Do these data provide sufficient evidence to contradict the null hypothesis that p1=.20,p2=.20,p3=.30p _ { 1 } = .20 , p _ { 2 } = .20 , p _ { 3 } = .30 , and p4=.30p _ { 4 } = .30 ? Test using α=.05\alpha = .05 .

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A multinomial experiment with k=3k = 3 cells and n=30n = 30 has been conducted and the results are shown in the table. Cell 1 2 3 16 12 2 Explain why the sample size is not large enough to test whether p1=.55,p2=.40p _ { 1 } = .55 , p _ { 2 } = .40 , and p3=.05p _ { 3 } = .05 .

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A sports researcher is interested in determining if there is a relationship between the number of home team and visiting team wins and different sports. A random sample of 526 games is selected and the results are given below. Find the rejection region used to test the claim that the number of home team and visiting team wins is independent of the sport. Use α=0.01\alpha = 0.01 Football Basketball Soccer Baseball Home team wins 39 156 25 83 Visiting team wins 31 98 19 75

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The data below show the age and favorite type of music of 779 randomly selected people. Test the claim that age and preferred music type are independent. Use α=0.05\alpha = 0.05 Age Country Rock Pop Classical 15-21 21 45 90 33 21-30 68 55 42 48 30-40 65 47 31 57 40-50 60 39 25 53

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Inc. Technology reported the results of consumer survey in which 300 Internet users indicated their level of agreement with the following statement: "The government needs to be able to scan Internet messages and user communications to prevent fraud and other crimes." The possible responses were "agree strongly", "agree somewhat", "disagree somewhat", and "disagree strongly". The number of Internet users in each category is summarized in the table. RESPONSE NUMBER Agree Strongly 60 Agree Somewhat 110 Disagree Somewhat 80 Disagree Strongly 50 In order to determine whether the true proportions of Internet users in each response category differ, a one-way chi-square analysis should be conducted. Use the chi-square distribution to determine the rejection region when testing at α=.05\alpha = .05 .

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Economists at USF are researching the problem of absenteeism at U.S. firms. A random sample of 100 U.S. organizations was selected to participate in a 1-year study. As part of the study, the economists had collected data on the following two variables for each company: shiftwork available (Yes or No), and union-management relationship (Good or Poor). As part of their analyses, the economists wanted to determine whether or not a company makes shiftwork available depends on the relationship between union and management. The collected data are Shown below: Relation Shiftwork Good Bad Total No 11 22 33 Yes 25 42 67 Total 36 64 100 Use the chi-square distribution to determine the rejection region for this test when testing at α=\alpha = 0.050.05 .

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Many track runners believe that they have a better chance of winning if they start in the inside lane that is closest to the field. For the data below, the lane closest to the field is Lane 1, the next lane is Lane 2, and so on until the outermost lane, Lane 6. The table displays the starting positions for the winners of 240 competitions. Test the claim that the probability of winning is the same regardless of starting position. Use α=0.05\alpha = 0.05 . The results are based on 240 wins. Starting Position 1 2 3 4 5 6 Number of Wins 36 33 44 50 32 45

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Many track runners believe that they have a better chance of winning if they start in the inside lane that is closest to the field. For the data below, the lane closest to the field is Lane 1 , the next lane is Lane 2, and so on until the outermost lane, Lane 6. The table displays the starting positions for the winners of 240 competitions. Find the rejection region used to test the claim that the probability of winning is the same regardless of starting position. Use α=0.05\alpha = 0.05 . The results are based on 240 wins. Starting Position 1 2 3 4 5 6 Number of Wins 45 32 36 44 50 33

(Multiple Choice)
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A random sample of 160 car accidents are selected and categorized by the age of the driver determined to be at fault. The results are listed below. The age distribution of drivers for the given categories is 18% for the under 26 group, 39% for the 26-45 group, 31% for the 46-65 group, and 12% for the group over 65. Test the claim that all ages have crash rates proportional to their number of drivers. Use α=0.05\alpha = 0.05 Age Under 26 26-45 46-65 Over 65 Drivers 66 39 25 30

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Inc. Technology reported the results of consumer survey in which 300 Internet users indicated their level of agreement with the following statement: "The government needs to be able to scan Internet messages and user communications to prevent fraud and other crimes." The possible responses were "agree strongly", "agree somewhat", "disagree somewhat", and "disagree strongly". The number of Internet users in each category is summarized in the table. RESPONSE NUMBER Agree Strongly 60 Agree Somewhat 110 Disagree Somewhat 80 Disagree Strongly 50 In order to determine whether the true proportions of Internet users in each response category differ, a one-way chi-square analysis should be conducted. Calculate the value of the test statistic for the desired analysis.

(Multiple Choice)
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A multinomial experiment with k=4k = 4 cells and n=400n = 400 produced the data shown in the following table. Cell 1 2 3 4 50 213 104 33 Previous studies in this area have shown that p1=p2=p3=p4=.25p _ { 1 } = p _ { 2 } = p _ { 3 } = p _ { 4 } = .25 . Construct a 95%95 \% confidence interval for the multinomial probability associated with cell 2

(Multiple Choice)
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Use the appropriate table to find the following  chi-square value: χ.052 for df=3\text { chi-square value: } \chi _ { .05 } ^ { 2 } \text { for } \mathrm { df } = 3

(Multiple Choice)
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Inc. Technology reported the results of consumer survey in which 300 Internet users indicated their level of agreement with the following statement: "The government needs to be able to scan Internet messages and user communications to prevent fraud and other crimes." The possible responses were "agree strongly", "agree somewhat", "disagree somewhat", and "disagree strongly". The number of Internet users in each category is summarized in the table. RESPONSE NUMBER Agree Strongly 60 Agree Somewhat 110 Disagree Somewhat 80 Disagree Strongly 50 In order to determine whether the true proportions of Internet users in each response category differ, a one-way chi-square analysis should be conducted. As part of that analysis, a 90% confidence interval for the multinomial probability associated with the "Disagree Somewhat" response was desired. Which of the following confidence intervals should be used?

(Multiple Choice)
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A drug company developed a honey-based liquid medicine designed to calm a child's cough at night. To test the drug, 105 children who were ill with an upper respiratory tract infection were randomly selected to participate in a clinical trial. The children were randomly divided into three groups - one group was given a dosage of the honey drug, the second was given a dosage of liquid DM (an over-the-counter cough medicine), and the third (control group) received a liquid placebo (no dosage at all). After administering the medicine to their coughing child, parents rated their children's cough diagnosis as either better or worse. The results are shown in the table below: Diagnosis Treatment Better Worse Total Control 4 33 37 DM 12 21 33 Honey 24 11 35 Total 40 65 105 In order to determine whether the treatment group is independent of the coughing diagnosis, a two-way chi-square test was conducted. Use the chi-square distribution to determine the rejection region for this test when testing at α=0.025\alpha = 0.025 .

(Multiple Choice)
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A teacher finds that final grades in the statistics department are distributed as: A, 25%; B, 25%; C, 40%; D, 5%; F, 5%. At the end of a randomly selected semester, the following grades were recorded. Determine if the grade distribution for the department is different than expected. Use α=0.01\alpha = 0.01 Grade A B C D F Number 36 42 60 14 8

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