Student Aptitude test mark (X)
x = (X - aveX)
x^2
Final mark in statistics
course y = (Y - aveY)xy
1
95
17
289
85
8
136
2
85
7
49
95
18
126
3
80
2
4
70
-7
-14
4
70
-8
64
65
-12
96
5
60
-18
324
70
-7
126
Sum
390
312
730
385
308
470
Average
78
77
Student Aptitude test mark (X)
Final mark in statistics
course Y' from the
model
e (Y-Y')
e^2
1
95
85
87.5
2.5
6.25
2
85
95
81.1
13.9
193.21
3
80
70
77.9
7.9
62.41
4
70
65
71.5
6.5
42.25
5
60
70
65.1
4.9
24.01
1. Model a linear regression equation that best predicts statistics performance, based on math aptitude scores. --report “a” to one decimal place, and “b” to two decimal places
a = 77 - (0.64)(78)
= 26.7
b= 470/730
= 0.64
Y' = 26.7- 0.64X
2. If a student made an 80 on the aptitude test, what grade would we expect her to make in statistics? Prediction Final exam mark = 26.7 + 0.64(Aptitude test mark)
= 26.7 + 0.64(80)
=77.9
78
3. By looking at the residuals of your linear regression, compute r
2
as a measure of goodness of fit
r
2
= 0.47
