ychology professor was interested in the relationship between course grades and
ID: 3312610 • Letter: Y
Question
ychology professor was interested in the relationship between course grades and 7. An Intro to was The data below show the grades (A's, B·C-3, D-2. F-)and overall evaluation sc thigher scores are more positive evaluations) a. Calculate the Pearson r (correlation). (2 points) Course Grade Course Evaluation ssx= 6.8 SP-56 b. Interpret the meaning of this correlation in terms of the problem, stating what this suggests between students' grades and their evaluations of the course. (4 points) about the relationship d. Calculate the slope and Y-intercept for these data, write down the regressi evaluation score for a studerit who received an F in the course (X-1).(10 points) on equation, and use it to predict the ext of the study, meaning how can we interpro with respect to the variables that we measure. (4 points) 8. If an analysis of variance is used for the following data, what would be the effect of changing the value of SS 3 points)Sample Data Mi 15 M 25 ssl = 90 SS2-70 9. For a one-way independent measures ANOVA comparing four treatments, MSpeten 12. What is the valeuc SSeween? (2 points) 10. The following table shows the results of an analysis of variance comparing three treatment condit ions wit Find the missing values. (10 points), Source SS dr MS Between 20 Within TotalExplanation / Answer
Part a
Here, we have to find correlation coefficient r. Formula for Pearson correlation coefficient r is given as below:
Correlation coefficient = r = [nxy - xy]/sqrt[(nx^2 – (x)^2)*(ny^2 – (y)^2)]
Table for calculations is given as below:
No.
X
Y
X^2
Y^2
XY
1
5
6
25
36
30
2
5
8
25
64
40
3
4
9
16
81
36
4
3
4
9
16
12
5
2
5
4
25
10
Total
19
32
79
222
128
Mean
3.8
6.4
We have
X = 19
Y = 32
X^2 = 79
Y^2 = 222
XY = 128
Xbar = 3.8
Ybar = 6.4
n = 5
r = [nxy - xy]/sqrt[(nx^2 – (x)^2)*(ny^2 – (y)^2)]
r = [5*128 – 19*32]/sqrt[(5*79 – (19)^2)*(5*222 – (32)^2)]
r = [640 – 608]/sqrt[(395 – 361)*(1110 – 1024)]
r = 32/sqrt(34*86)
r = 32/ 54.07402
r = 0.591781
Correlation coefficient = r = 0.591781
Part b
The correlation coefficient between the two variables course grade and course evaluation is given as 0.5918, which means there is a considerable positive linear relationship or association exists between the given two variables course grade and course evaluation.
Part d
The formula for slope and Y-intercept is given as below:
Slope = b = (XY – n*Xbar*Ybar)/(X^2 – n*Xbar^2)
Y-intercept = a = Ybar – b*Xbar
Slope = b = (128– 5*3.8*6.4)/(79 – 5*3.8^2)
Slope = b = 0.941176
Y-intercept = a = 6.4 – 0.941176*3.8
Y-intercept = a =2.823531
Regression equation is given as below:
Y = a + b*X
Y = 2.823531 + 0.941176*X
Predicted value of Y at X = 1 is given as below:
Y = 2.823531 + 0.941176*1
Y = 3.764707
Part e
The value of slope is given as 0.9412 approximately which means there is a 0.9412 increment in the predicted value of course evaluation Y as per unit change in course grade.
No.
X
Y
X^2
Y^2
XY
1
5
6
25
36
30
2
5
8
25
64
40
3
4
9
16
81
36
4
3
4
9
16
12
5
2
5
4
25
10
Total
19
32
79
222
128
Mean
3.8
6.4
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