Download Homework # 3 Key - Illinois State University Department of Psychology

Homework # 3 Key
 Chapter 4: 12, 15, 16, plus a fourth question not from the book
80 total points
4.12 Predicted therapy satisfaction = 1 + (.4)*depression score (2pts each)
a) 1 + .4(5) = 3
b) 1 + .4(6) = 3.4
c) 1 + .4(7) = 3.8
d) 1 + .4(8) = 4.2
e) 1 + .4(9) = 4.6
f) 1 + .4(10) = 5
4.15 They may do this one with z-scores or using raw scores. (This is also the data from
one of the problems on the last homework, so they may have “reused” some of their old
computations (r = -0.985). The problem also gives some information (SStotal = 84)
X
Y
1
1
2
4
Sum 8
Mean 2
SS
6
10
8
4
-2
20
5
84
stdev
4.58
1.22
Using Z-scores
ZX
ZY
ZX ZY
-0.82
-0.82
0.0
1.64
1.09
0.66
-0.22
-1.53
-0.89
-0.54
0.0
-2.51
-3.94
r = -0.985
(X  X)
-1
-1
0
2
Using raw scores
(Y  Y ) (X  X) (Y  Y )
5
-5
3
-3
-1
0
-7
-14
SP = -22
SP
22
r

SSX SSY
6 * 84
r = -0.985
(a) Determine the raw-score prediction formula for predicting anxiety from
dexterity (10 pts)
Standardized Beta = r = -0.985
SP 22
1  slope 

 3.67
SSX
6
ZYˆ  (0.985)Z X
  intercept  Y   X
0
computing intercept
(what does Y = when X = 0?)
(X  X) (0  2)
ZX 0 

 1.64
sX
1.22
ZYˆ  (0.985)(1.64)  1.615
 
Y  sY  ZY)  Y
YX  0  4.58 1.615   5  12.3
Computing the raw score for slope
1

 5.0  3.67 2.0 
 12.34
Yˆ  12.34  (3.67)X
(X  X) (1  2)

 0.82
sX
1.22
ZYˆ  (0.985)(0.82)  0.81
Z X 1 
 
Y  sY  ZY)  Y
YX 1  4.58 0.81  5  8.71
)
)
slope  YX 1  YX 0  8.71  12.3  3.6
Yˆ  12.3  (3.6)X
b) (specific numbers will vary with slight round off differences, give credit if they’re in
the right ballpark) (8 pts)
Y  12.3  3.67 1  8.63
)
(Y  Y )
10  8.63  1.37
)
(Y  Y )2
1.8
8
Y  12.3  3.67 1  8.63
8  8.63  0.63
0.4
2
4
Y  12.3  3.67 2   4.96
4  4.96  0.96
0.9
4
-2
2  2.38   0.38
0.14
X
Y
1
10
1
Y  0  1 X
Y  12.3  3.67 4   2.38
SSerror = 3.2
c) plot the scatterplot with the regression line (4 pts)
10
8
6
4
2
1
2
3
4
5
-2
d) figure the error and squared error for each of the four predictions (see table in b) (4pts)
e) find the proportionate reduction in error using SSerror and SStotal. (4 pts)
r2 
SStotal  SSerror 84  3.2

 0.96
SStotal
84
f) take the square root of (e) and compare it to the correlation coefficient . (4 pts)
r 2  0.96  0.98
They are essentially the same (except you lose the direction information because squaring
a negative number leads to a positive number [and square rooting a positive leads to a
positive]). To regain the direction of the relationship you need to examine the sign of
your slope
g) answers will vary here. They should try to explain what regression is good for
(making predictions) and how it works. . (4 pts)
4.16 Same data as in 4.15, but now they switch the X and Y variables (predict dexterity
from anxiety). They may do this one with z-scores or using raw scores. (This is also the
data from one of the problems on the last homework, so they may have “reused” some of
their old computations (r = -0.985). The problem also gives some information (SStotal =
6)
X
Y
10
8
4
-2
Sum 20
Mean 5
SS
84
1
1
2
4
8
2
6
stdev
1.22
4.58
Using Z-scores
ZX
ZY
ZX ZY
1.09
0.66
-0.22
-1.53
-0.82
-0.82
0.0
1.64
-0.89
-0.54
0.0
-2.51
-3.94
r = -0.985
(X  X)
5
3
-1
-7
Using raw scores
(Y  Y ) (X  X) (Y  Y )
-1
-5
-1
-3
0
0
2
-14
SP = -22
SP
22
r

SSX SSY
6 * 84
r = -0.985
(a) Determine the raw-score prediction formula for predicting anxiety from
dexterity (12pts)
Standardized Beta = r = -0.985
SP 22
1  slope 

 0.26
SSX
84
ZYˆ  (0.985)Z X
  intercept  Y   X
0
computing intercept
(what does Y = when X = 0?)
(X  X) (0  5)
ZX 0 

 1.09
sX
4.58
ZYˆ  (0.985)(1.09)  1.075
 
Y  sY  ZY)  Y
YX  0  1.22 1.075   2  3.31
1

 2.0  0.26 5.0 
 3.31
Yˆ  3.31  (0.26)X
Computing the raw score for slope
(X  X) (1  5)
Z X 1 

 0.87
sX
4.58
ZYˆ  (0.985)(0.87)  0.86
 
Y  sY  ZY)  Y
YX 1  1.22 0.86   2  3.05
)
)
slope  YX 1  YX 0  3.05  3.31  0.26
Yˆ  3.31  (0.26)X
b) (specific numbers will vary with slight round off differences, give credit if they’re in
the right ballpark) (8pts)
Y  3.31  0.26 10   0.71
)
(Y  Y )
1 0.71  0.29
)
(Y  Y )2
0.08
1
Y  3.31  0.26 8   1.23
1 1.23  0.23
0.05
4
2
Y  3.31  0.26 4   2.27
2  2.27  0.27
0.07
-2
4
Y  3.31  0.26 2   3.83
4  3.83  0.17
0.03
X
Y
10
1
8
Y  0  1 X
SSerror = 0.23
c) plot the scatterplot with the regression line (4pts)
5
4
3
2
1
-2
2
4
6
8
10
d) figure the error and squared error for each of the four predictions (see table in b) (4pts)
e) find the proportionate reduction in error using SSerror and SStotal. (4pts)
r2 
SStotal  SSerror 6  0.23

 0.96
SStotal
6
f) take the square root of (e) and compare it to the correlation coefficient (4pts)
r 2  0.96  0.98
They are essentially the same (except you lose the direction information because squaring
a negative number leads to a positive number [and square rooting a positive leads to a
positive]). To regain the direction of the relationship you need to examine the sign of
your slope
g) answers will vary here. They should try to explain what regression is good for
(making predictions) and how it works. (4pts)
Fourth question) Additionally answer the following questions based on the data
contained in height.sav. Note: This is an SPSS datafile. You'll need to use SPSS to do
these questions (so don't try to open the file on a computer that doesn't have SPSS on it. It
won't work.)
Compute the following regression models predicting HEIGHT:
(a) predict height with average height of parents
(b) predict height with average calcium intake
(c) predict height with weight
(d) predict height with average height of parents and weight
(e) predict height with average height of parents and average calcium intake
(f) predict height with all three variables (avg height of parents, avg calium, weight)
Based on the SPSS output (you don't need to print it out and include it), report the r2 for
each model. Which model do you think is the "best?" Explain why. (12 pts)
Model
1
2
3
4
5
6
Variables
average height of parents
average calcium intake
weight
average height of parents
weight
average height of parents
average calcium intake
average height of parents
average calcium intake
weight
Unstandardized ß
0.8
0.41
0.79
0.49
0.47
0.81
-0.01
0.47
0.03
0.48
Model r2
0.641
0.168
0.631
0.767
0.621
0.768
Model 6 accounts for slightly more variance than model 4. However, model 4 has fewer
variables in it. Since model 4 is has fewer variables and accounts for nearly the same
amount of variance (as model 6) that should