Stat 401 B – Lecture 12
Multiple Regression
A single numerical response
variable, Y.
 Multiple numerical explanatory
variables, X1, X2,…, Xk

1
Multiple Regression
Y = μY | x1 , x2 ,..., xk + ε
Y = β 0 + β1 x1 + β 2 x2 + ... + β k xk + ε
2
Example
Y, Response – Effectiveness
score based on experienced
teachers’ evaluations.
 Explanatory – Test 1, Test 2,
Test 3, Test 4.

3
Stat 401 B – Lecture 12
Student Eval
Teacher
1
2
3
4
5
6
Test1
489
423
507
467
340
524

23
Test2
81
68
80
107
43
129
Test3
151
156
165
149
134
163



434
76
141
Test4
46
46
76
56
49
72

54
44
45
55
43
49
50

58
4

JMP
Analyze – Fit Model
 Pick Role Variables



Y – EVAL
Construct Model Effects

Add – Test1, Test2, Test3, Test4
5
JMP

Analyze – Fit Model
Personality – Standard Least
Squares
 Emphasis – Minimal Report

6
Stat 401 B – Lecture 12
Response EVAL
Summary of Fit
RSquare
RSquare Adj
Root Mean Square Error
Mean of Response
Observations (or Sum Wgts)
0.802861
0.759052
37.53627
444.4783
23
Analysis of Variance
Source
Model
Error
C. Total
DF
4
18
22
Sum of
Squares Mean Square
103286.25
25821.6
25361.49
1409.0
128647.74
F Ratio
18.3265
Prob > F
<.0001*
Parameter Estimates
Term
Intercept
Test1
Test2
Test3
Test4
Estimate Std Error t Ratio Prob>|t|
-193.4994 125.3074
-1.54 0.1399
1.1158539 0.319746
3.49 0.0026*
2.243267 0.628449
3.57 0.0022*
-1.367001 0.563965
-2.42 0.0261*
6.0482387 1.202281
5.03 <.0001*
7
Prediction Equation

Predicted Evaluation = –193.50
+ 1.116*Test1 + 2.243*Test2
– 1.367*Test3 + 6.048*Test4
8
Conditions

The random error term,
ε , is
Independent
 Identically distributed
 Normally distributed with
standard deviation, σ .

9
Stat 401 B – Lecture 12
Estimate of Error Variance, σ 2
MSError =
SSError
df Error
 ( y − yˆ )
=
2
MSError
MSError
n − (k + 1)
25361.49
=
= 1409.0
18
10
Estimate of Error Std Dev, σ

Root Mean Square Error
RMSE = MSError
RMSE = 1409.0 = 37.54
11
Multiple R2
R2 =
SSModel
SS
= 1 − Error
SSTotal
SSTotal
R2 =
103286.25
= 0.802861
128647.74
12
Stat 401 B – Lecture 12
Interpretation

80.3% of the variation in the
evaluation scores can be
explained by the model, i.e. the
relationship with the
explanatory variables.
13
Caution

Including additional explanatory
variables in a model can only
increase the value of R2, even if
those explanatory variables
have nothing to do with the
response variable.
14
Adjusted R2
adjR 2 = 1 −
adjR 2 = 1 −
MSError
MSTotal
(25361.49 18) = 0.75905
(128647.74 22)
15
Stat 401 B – Lecture 12
Test of Model Utility

Is there any explanatory
variable in the model that is
helping to explain significant
amounts of variation in the
response?
16
Step 1: Hypotheses
H 0 : β1 = β 2 = ... = β k = 0
H A : at least one parameter is not zero
17
Step 2: Test Statistic
F=
MSModel
MSError
25821.6
= 18.3265
1409.0
P − value < 0.0001
F=
18
Stat 401 B – Lecture 12
Step 3: Decision

Reject the null hypothesis
because the P-value is so small.
19
Step 4: Conclusion
At least one of the tests is
providing statistically significant
information about the
evaluation score.
 The model is useful. Maybe not
the best, but useful.

20
Alternative Form
(R k )
2
F=
(
)
 1− R2



(
(
)
)
−
+
1
n
k


0.802861
4 = 18.3265
F=
0.197139
18
P − value < 0.0001
(
(
)
)
21