LEAD 6341 Research Methods and Statistics Midterm Exam
LEAD 6341
Research Methods and Statistics
Midterm Exam
Part I: Calculations (Open Book)
Spring, 2013
Note: For the following computational exercises, show all steps of your work (when appropriate). Do NOT simply put an answer. Show or explain how you arrived at your answer. 1) Compute the mean, median, and mode for the following distribution: (5 pts) 1,2,2,3,3,3,3,3,3,4,4,5,6,7,7,8,8,8,8,8,8,9,9,10
The mean is what people call the average. (1+2+2+3+3+3+3+3+3+4+4+5+6+7+7+8+8+8+8+8+8+9+9+10 = 127/24 = 5.2916
Mean = 5.29 The median is the middle value. When the middle does not fall neatly in the distribution, use the following formula to identify the position. In the formula n is the sample size.
Median position = (n + 1) / 2
Median position = (24 + 1)/2
Median position = 25/2 = 12.5
Now we look for position 12 and 13. Position 12 is 5, and position 13 is 6. We take the average of the two values and that is the median. Median = (5 + 6)/2 = 11/2 = 5.5
Median = 5.5 The mode is the most frequent response in the distribution 1 – 1 6 – 1 2 – 2 7 – 2 3 – 6 8 – 6 4 – 2 9 – 2 5 – 1 10 – 1
In this example we have two modes. They are 3 and 8.
2) You have been asked to give a student two different tests; an intelligence test and a creativity test. The student scored 123 on the intelligence test and 123 on the creativity test. The mean for the intelligence test is 100, and the standard deviation is 16. The mean for the creativity test is 115, and the standard deviation is 14. What statistic would you use to compare the two scores from these two different distributions (the intelligence test distribution and the creativity test distribution)? Compute this statistic and determine on which test, if either, the student performed better. Explain your answer. (10 pts)
Intelligence test score = 123, mean = 100, standard diviation = 16
Creativity test score = 123, mean = 115, standard diviation =
Research Methods and Statistics
Midterm Exam
Part I: Calculations (Open Book)
Spring, 2013
Note: For the following computational exercises, show all steps of your work (when appropriate). Do NOT simply put an answer. Show or explain how you arrived at your answer. 1) Compute the mean, median, and mode for the following distribution: (5 pts) 1,2,2,3,3,3,3,3,3,4,4,5,6,7,7,8,8,8,8,8,8,9,9,10
The mean is what people call the average. (1+2+2+3+3+3+3+3+3+4+4+5+6+7+7+8+8+8+8+8+8+9+9+10 = 127/24 = 5.2916
Mean = 5.29 The median is the middle value. When the middle does not fall neatly in the distribution, use the following formula to identify the position. In the formula n is the sample size.
Median position = (n + 1) / 2
Median position = (24 + 1)/2
Median position = 25/2 = 12.5
Now we look for position 12 and 13. Position 12 is 5, and position 13 is 6. We take the average of the two values and that is the median. Median = (5 + 6)/2 = 11/2 = 5.5
Median = 5.5 The mode is the most frequent response in the distribution 1 – 1 6 – 1 2 – 2 7 – 2 3 – 6 8 – 6 4 – 2 9 – 2 5 – 1 10 – 1
In this example we have two modes. They are 3 and 8.
2) You have been asked to give a student two different tests; an intelligence test and a creativity test. The student scored 123 on the intelligence test and 123 on the creativity test. The mean for the intelligence test is 100, and the standard deviation is 16. The mean for the creativity test is 115, and the standard deviation is 14. What statistic would you use to compare the two scores from these two different distributions (the intelligence test distribution and the creativity test distribution)? Compute this statistic and determine on which test, if either, the student performed better. Explain your answer. (10 pts)
Intelligence test score = 123, mean = 100, standard diviation = 16
Creativity test score = 123, mean = 115, standard diviation =