Two researchers are examining academic performance. They use the same set of 11
ID: 3048411 • Letter: T
Question
Two researchers are examining academic performance. They use the same set of 11 college student participants, but they each measure academic performance differently. Researcher 1 measures academic performance by asking the participants to self-report which letter grade is most typical for them to receive on an assignment on the standard A through F scale. Researcher 2 measures academic performance by accessing the volunteer’s academic record and retrieving their overall grade point average from the college system. The participants’ scores are shown below.
Researcher 1: Letter Grade
Researcher 2: Overall GPA
B
3.25
A
3.31
C
2.45
B
3.09
A
4.00
C
1.90
D
.60
B
2.85
A
2.96
B
3.02
C
1.54
FOR QUESTIONS 20-24: Use the sample mean and standard deviations in your calculations.
How many standard deviations away from the mean would someone be who had an overall GPA of 2.25? (3 pts)
How many standard deviations away from the mean would someone be who had an overall GPA of 1.50? (3 pts)
If someone had a Z-score of 1.85, what would their overall GPA be? (3 pts) .
If someone had an overall GPA that was 1.60 standard deviations below the mean, what would their raw score be? (3 pts)
Would someone with a GPA of 2.25 be considered “extreme” or “typical”? Provide an explanation why with your answer. (3 pts)
Researcher 1: Letter Grade
Researcher 2: Overall GPA
B
3.25
A
3.31
C
2.45
B
3.09
A
4.00
C
1.90
D
.60
B
2.85
A
2.96
B
3.02
C
1.54
Explanation / Answer
ans) The overall GPA is been related to researcher 2. So we need to calculate z-scores based on data of researcher 2. The z-score measures how many standard deviation is the raw score is away from the mean. If it is negative then the raw score is some standard deviation below the mean and if it is positive then the raw score is some standard deviation above the mean. Here we will use sample mean and sample standard deviation. The z-score = (raw score - sample mean)/ ( sample standard deviation). The sample mean here is 2.633636 and sample standard deviation is 0.9549269.
a) Given GPA is 2.25 therefore the z-score is (2.25 - 2.633636) / 0.9549269 = -0.4017442 . So it is 0.4017442 standard deviation below the sample mean.
b) Given GPA is 1.50 therefore the z-score is (1.50 - 2.633636) / 0.9549269 = -1.187145 . So it is 1.187145 standard deviation below the sample mean.
c) If z-score is 1.85 then GPA is 2.633636)+(1.85*0.9549269) = 4.400251
d) If GPA is 1.60 sd below the mean so z-score is -1.60 . So the GPA is 2.633636 +(-1.60*0.9549269) = 1.105753
e) If someone has GPA of 2.25 then first draw the histogram. From the histogram it can be noted that the value is the mid value of the middle class . Then calculate the mean , median and mode of the data . The values are 2.633636, 2.96 and 3.042467 respectively. So 2.25 is not a typical value. It is also not the minimum or maximum value. So it is not extreme.
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