DSRT 734 - Week 6 Quiz

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School

Texas State University *

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Course

116

Subject

Marketing

Date

Feb 20, 2024

Type

pdf

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Uploaded by talkmath2me

2/14/24, 6:32 PM Take Test: Week 6 Quiz – 2024 Spring - Infer Stats in ... https://ucumberlands.blackboard.com/webapps/assessment/take/launch.jsp?course_assessment_id=_257027_1&course_id=_162343_1&content_id=_5732234_1&ste… 1/9 Take Test: Week 6 Quiz Test Information Description Instructions Timed Test This test has a time limit of 1 hour.This test will save and submit automatically when the time expires. Warnings appear when half the time , 5 minutes , 1 minute , and 30 seconds remain. Multiple Attempts Not allowed. This test can only be taken once. Force Completion This test can be saved and resumed at any point until time has expired. The timer will continue to run if you leave the test. Your answers are saved automatically. This exam cover the Spearman's r s and Chi-Square tests. Please select the best answer. For the toolbar, press ALT+F10 (PC) or ALT+FN+F10 (Mac). QUESTION 1 6A A marketing team conducted a survey of 300 people to see if a person’s age affected their brand choice. People under the age of twenty-five made the following choices: Brand A (25), Brand B (42), Brand C (33). People over 65 made the following selections: Brand A (20), Brand B (38), Brand C (42). The group between those ages made these choices: Brand A (44), Brand B(40), Brand C (16). Form your hypotheses, run the appropriate test, and provide a conclusion to the marketing team. 15 points View Rubric Saved Question Completion Status: Remaining Time: 34 minutes, 21 seconds. Click Save and Submit to save and submit. Click Save All Answers to save all answers.

2/14/24, 6:32 PM Take Test: Week 6 Quiz – 2024 Spring - Infer Stats in ... https://ucumberlands.blackboard.com/webapps/assessment/take/launch.jsp?course_assessment_id=_257027_1&course_id=_162343_1&content_id=_5732234_1&ste… 2/9 We can use a chi-square test for independence to analyze whether a person's age affects their brand choice. This test will help us determine if there is a significant association between age group and brand choice. Null Hypothesis (H0): No association between age group and brand choice exists. Alternative Hypothesis (H1): Age group and brand choice are associated. We will use a significance level of 0.05. Brand A Brand B Brand C Total Under 25 25 42 33 100 25-65 44 40 16 100 Over 65 20 38 42 100 Total 89 120 91 300 Expected Frequencies brand age A B C Total <25 29.667 40 30.333 100 >65 29.667 40 30.333 100 (25, 65) 29.667 40 30.333 100 Total 89 120 91 300 (fo-fe)^2/fe 0.734082 0.1 0.234432 3.149813 0.1 4.487179 QUESTION 2 Remaining Time: 34 minutes, 21 seconds. Click Save and Submit to save and submit. Click Save All Answers to save all answers.

2/14/24, 6:32 PM Take Test: Week 6 Quiz – 2024 Spring - Infer Stats in ... https://ucumberlands.blackboard.com/webapps/assessment/take/launch.jsp?course_assessment_id=_257027_1&course_id=_162343_1&content_id=_5732234_1&ste… 3/9 For the toolbar, press ALT+F10 (PC) or ALT+FN+F10 (Mac). QUESTION 2 6B A social psychologist hypothesized that a factor in juvenile delinquency was the presence or absence of a strong father-figure in the home. He examined the folders of 100 inmates in the federal reformatory and found that only 50 of these young men grew up with a strong father- figure in the home. He also examined the records of 100 randomly selected male college students and found that 70 of them had strong father-figures in their boyhood homes. Form your hypotheses, run the appropriate test, and provide a conclusion. We can use a chi-square test for independence to analyze whether the presence or absence of a strong father figure in the home is associated with juvenile delinquency. This test will help us determine if there is a significant association between the two variables: the presence of a strong father figure and being an inmate or a college student. Null Hypothesis (H0): There is no association between the presence of a strong father figure in the home and being an inmate or a college student. Alternative Hypothesis (H1): There is an association between the presence of a strong father figure in the home and being an inmate or a college student. We will use a significance level of 0.05. First, let us organize the data into a contingency table: Strong Father-Figure No Strong Father-Figure Inmates 50 50 College Students 70 30 Now, let us calculate the expected frequencies for each cell, assuming there is no association between a strong father figure in the home and being an inmate or a college student. Next we will compute the chi-square statistic: 15 points View Rubric Saved Remaining Time: 34 minutes, 21 seconds. Click Save and Submit to save and submit. Click Save All Answers to save all answers.

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2/14/24, 6:32 PM Take Test: Week 6 Quiz – 2024 Spring - Infer Stats in ... https://ucumberlands.blackboard.com/webapps/assessment/take/launch.jsp?course_assessment_id=_257027_1&course_id=_162343_1&content_id=_5732234_1&ste… 4/9 Next, we will compute the chi-square statistic: χ2=∑Ei(Oi−Ei)2 Where: Oi = Observed frequency in cell i Ei = Expected frequency in cell i We will compare this statistic to the critical value from the chi-square distribution with 1 degree of freedom. We reject the null hypothesis if the calculated chi-square value exceeds the critical value. After calculating the chi-square statistic and comparing it to the critical value, we will conclude whether there is a significant association between a strong father figure in the home and being an inmate or a college student. Let us perform these calculations. First, let us calculate the expected frequencies for each cell: For the toolbar, press ALT+F10 (PC) or ALT+FN+F10 (Mac). QUESTION 3 6C In an activity wheel, experimental animals were either isolated or put together and maintained on a restricted feeding schedule. After 25 days, the animals were either alive with stabilized weights or they were dead. Form your hypotheses, run the appropriate test, and provide a conclusion. Isolated Together Alive 9 8 Dead 1 12 15 points View Rubric Saved Remaining Time: 34 minutes, 21 seconds. Click Save and Submit to save and submit. Click Save All Answers to save all answers.

2/14/24, 6:32 PM Take Test: Week 6 Quiz – 2024 Spring - Infer Stats in ... https://ucumberlands.blackboard.com/webapps/assessment/take/launch.jsp?course_assessment_id=_257027_1&course_id=_162343_1&content_id=_5732234_1&ste… 5/9 We can use Fisher's exact test to analyze whether isolation or being together in experimental animals affects their survival rate after 25 days. This test is appropriate for comparing two groups (isolated and together) with small sample sizes in each contingency table cell. Null Hypothesis (H0): There is no association between isolation/togetherness and survival of experimental animals after 25 days. Alternative Hypothesis (H1): There is an association between isolation/togetherness and survival of experimental animals after 25 days. We will use a significance level of 0.05. First, let us organize the data into a contingency table: Isolated Together Total Alive 9 (5.66) 8 (11.333) 17 Dead 1 (4.333) 12(8.66667) 13 Total 10 20 30 QUESTION 4 7A An aerobics teacher believed that physical conditioning is an important factor in the ability to perform skills, even if the skill does not require strength or endurance. To test this hypothesis, she selected 10 high BMI (body mass index) women enrolled in her aerobics course. Before the first class, she tested them on a button-sorting task. After a full semester of regular exercise, she tested the women on the same button-sorting task. Scores represent the number of seconds required to sort the buttons. Analyze the data with a nonparametric test and comment on the results. Subject Pretest Score Rank Posttest Score Rank 1 1 2 2 2 1 3 3 4 4 4 3 5 5 6 6 6 5 7 7 9 15 points View Rubric Saved Remaining Time: 34 minutes, 21 seconds. Click Save and Submit to save and submit. Click Save All Answers to save all answers.

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2/14/24, 6:32 PM Take Test: Week 6 Quiz – 2024 Spring - Infer Stats in ... https://ucumberlands.blackboard.com/webapps/assessment/take/launch.jsp?course_assessment_id=_257027_1&course_id=_162343_1&content_id=_5732234_1&ste… 7/9 positive and negative differences: Subject Difference Absolute Difference Rank 1 -1 8 2 1 4 3 -1 8 4 1 4 5 -1 8 6 1 4 7 -2 10 8 0 - 9 -1 8 10 3 1 Now, we sum the ranks of the positive and negative differences: Sum of ranks of positive differences (W+): 4 + 4 + 4 + 1 = 13 The sum of ranks of negative differences (W-): 8 + 8 + 8 + 10 = 34 For the toolbar, press ALT+F10 (PC) or ALT+FN+F10 (Mac). QUESTION 5 7B Analyze the data with a nonparametric test and comment on the results and write a conclusion. Rank in Weight Rank in Dominance Hierarchy 1 3 2 4 3 1 4 8 5 7 6 2 7 5 8 6 9 9 15 points View Rubric Saved Remaining Time: 34 minutes, 21 seconds. Click Save and Submit to save and submit. Click Save All Answers to save all answers.

2/14/24, 6:32 PM Take Test: Week 6 Quiz – 2024 Spring - Infer Stats in ... https://ucumberlands.blackboard.com/webapps/assessment/take/launch.jsp?course_assessment_id=_257027_1&course_id=_162343_1&content_id=_5732234_1&ste… 8/9 We can use the Spearman rank correlation coefficient to analyze the data provided. This nonparametric test assesses the strength and direction of association between two ranked variables. Null Hypothesis (H0): No significant correlation exists between rank in weight and rank in dominance hierarchy. Alternative Hypothesis (H1): a significant correlation exists between rank in weight and rank in dominance hierarchy. We will use a significance level of 0.05. The Spearman rank correlation coefficient, denoted by \( \rho \), ranges from -1 to 1. A positive value indicates a positive association, meaning that higher ranks in one variable correspond to higher ranks in the other. Conversely, a negative value indicates a negative association, meaning that higher ranks in one variable correspond to lower ones in the other. A value close to 0 suggests little to no association. Let us calculate the Spearman rank correlation coefficient using the provided data: Rank in Weight Rank in Dominance Hierarchy 1 3 2 4 3 1 4 8 5 7 6 2 7 5 8 6 9 9 First, let us calculate the differences between ranks and their squares: Subject Rank in Weight Rank in Dominance Hierarchy Difference (d) d^2 1 1 3 -2 4 2 2 4 -2 4 3 3 1 2 4 4 4 8 -4 16 5 5 7 -2 4 6 6 2 4 16 7 7 5 2 4 8 8 6 2 4 9 9 9 0 0 Now, let us calculate the Spearman rank correlation coefficient (\( \rho \)): \[ \rho = 1 - \frac{6 \sum d^2}{n(n^2 - 1)} \] Where: Remaining Time: 34 minutes, 21 seconds. Click Save and Submit to save and submit. Click Save All Answers to save all answers.

2/14/24, 6:32 PM Take Test: Week 6 Quiz – 2024 Spring - Infer Stats in ... https://ucumberlands.blackboard.com/webapps/assessment/take/launch.jsp?course_assessment_id=_257027_1&course_id=_162343_1&content_id=_5732234_1&ste… 9/9 Where: ∑d^2 is the sum of the squared differences n is the number of subjects \[ \rho = 1 - \frac{6 \times (4 + 4 + 4 + 16 + 4 + 16 + 4 + 4)}{9(9^2 - 1)} \] \[ \rho = 1 - \frac{6 \times 56}{9(80)} \] \[ \rho = 1 - \frac{336}{720} \] \[ \rho = 1 - 0.4667 Remaining Time: 34 minutes, 21 seconds. Click Save and Submit to save and submit. Click Save All Answers to save all answers.

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DSRT 734 - Week 6 Quiz

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