Question

Calculate the mean, median, mode, variance (population) and standard deviation (sample) for the following test grades: 65, 60, 67, 71, 75, 80, 65 Use this information to answer questions #1-4. Question 1 (0.5 points) Saved The mode is: Question 1 options: 65 60 There is no modal category 75 71 Question 2 (0.5 points) Saved The median is: Question 2 options: 67 65 80 66 None of the above Question 3 (0.5 points) What is the approximate population variance for these scores? Question 3 options: 39.7 46.3 121 81.5 None of the above Question 4 (0.5 points) What is the approximate sample standard deviation for these scores? Question 4 options: 6.3 36 4 6.8 None of the above Question 5 (0.5 points) Dr. Gershon tells her class at the midpoint that the mean level of points in the class is 150 (SD=20). One student in the class (with 185 points) wants to know what percent of the class has a higher grade and what percent has a lower grade than they do. (Using the normal curve areas table) What percentage of students have a grade which is lower than this student? Question 5 options: 1.75% 96% 4% 100% None of the above Question 6 (0.5 points) Let’s say GSU is thinking about offering more online courses in the summer. We’re interested in levels of student support for more online summer courses (in the place of in-person course offerings). We conduct a survey of 1000 GSU students, finding that 51% of students would support this change (S.E. = 4%). Calculate the 95% confidence intervals for these poll results. Can we say, based on these results whether the majority of GSU students would support these changes? Which of the following is the correct interpretation of your results? Question 6 options: Based on these results, we can be 95% confident that the true approval level for these changes in the GSU student population is between approximately 43% and 59%. We cannot conclude, based on these results, that the majority of students would support these changes. Based on these results, we can be 95% confident that the true approval level for these changes in the GSU student population is between approximately 43% and 59%. We can conclude, based on these results, that the majority of students would support these changes. Based on these results, we can be 95% confident that the true approval level for these changes in the GSU student population is between approximately 41% and 61%. We cannot conclude, based on these results, that the majority of students would support these changes. Based on these results, we can be 95% confident that the true approval level for these changes in the GSU student population is between approximately 47% and 55%. We can conclude, based on these results, that the majority of students would support these changes. None of the above Question 7 (0.5 points) Let’s say that a poll by a local TV station reveals that Americans use social media to read the news an average of 10 times a week. We believe that college students will rely on social media to get their news much more frequently (due to their age) than average Americans. To test this hypothesis, we survey 1000 college students at several public universities across the country, finding that the students in our survey use social media to get the news an average of 12 times a week (sample standard deviation = 15). Please determine whether college students’ use of social media for news is significantly different than that of the average American at a p<.01 level of significance relying on the t- distribution chart. Can the null hypothesis be rejected? Question 7 options: No We cannot determine whether the null hypothesis can be rejected based on the data above Yes Question 8 (0.5 points) Calculate Lambda for Party Attachment (our dependent variable) and Race & Ethnicity (our independent variable) in this fictitious sample: African Americans Anglos Latinos Asian Americans Total Democrat 60 40 30 70 200 Republican 30 45 50 20 145 Independent 10 15 20 10 55 Total 100 100 100 100 400 Question 8 options: 200 175 .125 .143 None of the above Question 9 (0.5 points) Calculate gamma for the relationship between TV watching (our independent variable) & level of aggression in this fictitious sample of children (our dependent variable). Low amount of TV watching Medium amount of TV watching High amount of TV watching Total Low Aggression 20 25 25 70 Medium Aggression 30 35 30 95 High Aggression 40 48 58 146 Total 90 108 113 311 Question 9 options: .06 16.83 .54 .68 None of the above Question 10 (0.5 points) Pew recently ran a survey finding that Americans have become more likely to say that it is “stressful” to discuss politics with people they disagree with (https://www.people-press.org/2018/11/05/more-now-say-its-stressful-to-discuss-politics-with-people-they-disagree-with/) Let’s say that we wanted to test that relationship among college students. We believe that taking political science classes will help students effectively engage in political discourse with those they disagree with and that the experience will not be as stressful. We hypothesize that the number of political science classes students have taken will be negatively correlated with the stress they feel while engaging in discussions about politics. We sample 30 students and the correlation between number of political science classes taken and stress associated with political discussion (on a 10 point scale) is -.43. We need to determine whether we can reject the null hypothesis for the population (GSU students). Please determine whether the correlation is statistically significant at a p<.01 level of significance and interpret the results of your test. Can you reject the null hypothesis? Question 10 options: Yes No We cannot determine whether the null hypothesis can be rejected based on the data above

Answer 1

Given: 65, 60, 67, 71, 75, 80, 65

Arranging in an ascending order: 60, 65, 65, 67, 71, 75, 80

Number of test grades, n =7

Mean, =(60+65+....+80)/7 =69

Median =The middle most value =67

Mode =The most frequently occured value =65

Population variance = =39.7143

Sample standard deviation = =6.8069

Answers 1 to 4:

1.

The mode is: 65

2.

The median is: 67

3.

The approximate population variance for these scores: 39.7

4.

The approximate sample standard deviation for these scores: 6.8

Answer 5.

Mean, =150

Standard Deviation, SD= =20

X =185

Z =(X - )/ =(185 - 150)/20 =1.75

P(Z < 1.75) =0.96 =96%

Therefore, 96% of students have a grade which is lower than this student.