1. KNNL Problem 2.26. To change the axis on a plot, add an “order” statement to the
axis definition. For example, to make the vertical axis go from -30 to 30 in
increments of 10, do
axis3 label=(angle=90 ‘Residuals’) order=(-30 to 30 by 10);
Then use vaxis=axis3 as an option to the plot statement, along with haxis and vref=0.
The mean value of Y appears as Dependent Mean in the output from proc reg.
2.26: Refer to Plastic hardness Problem 1.22.
(a) Set up the ANOVA table.
Analysis of Variance
Source
DF
Sum of
Mean
Squares
Square
F Value Pr > F
Model
1 5297.51250 5297.51250
Error
14
146.42500
10.45893
Corrected Total
15 5443.93750
506.51
<.0001
(b) Test by means of an F test whether or not there is a linear association between the
hardness of the plastic and the elapsed time. Use a = .01. State the alternatives,
decision rule, and conclusion.
Testing H0: β1 = 0 vs Ha: β1  0. Reject H0 if P-value < 0.01. Since P-value < 0.0001,
we reject H0 and conclude that there is a significant linear relationship between hardness
and time.
(c) Plot the deviations Yi  Yˆi against Xi on a graph. Plot the deviations Yˆi  Y
against Xi on another graph, using the same scales as for the first graph. From
your two graphs, does SSE or SSR (SSM) appear to be the larger component of
SSTO (SST)? What does this imply about the magnitude of r2?
Re s i d u a l s
( Ob s e r v e d -
Pr e d i c t e d )
30
20
10
0
- 10
- 20
- 30
10
20
30
40
t i me ( h o u r s )
De v i a t i o n s
( Pr e d i c t e d -
Me a n )
30
20
10
0
- 10
- 20
- 30
10
20
30
40
t i me ( h o u r s )
On the graphs, it is apparent that the residuals Yi  Yˆi are much smaller than the
deviations Yˆ  Y of the predictors from the mean. Thus the SSM should be much larger
i
than the SSE, and we expect r 2 
SSM
to be close to 1.
SST
(d) Calculate r2 and r.
The values of r2 and r can be obtained from the SAS output: r2 = 0.9731 and so r =
0.9865. (Choose the positive square root because b1 > 0).
Root MSE
3.23403 R-Square 0.9731
Dependent Mean
225.56250 Adj R-Sq 0.9712
Coeff Var
1.43376
2. Problem 2.31
a)
Source
Regression
Error
Total
SS
93,462,942
455,273,165
548,736,107
df
MS
1 93,462,942
82 5,552,112
83
b). H0 : β1 = 0, Ha : β1 = 0. F ∗ = 93, 462, 942/5, 552, 112 = 16.8338, F (.99; 1,
82) =6.9544. If F ∗ ≤ 6.9544 conclude H0 , otherwise Ha. Conclude Ha . (t∗ )2
= (−4.102895)2 = 16.8338 = F ∗ . [t(.995; 82)]2 = (2.63712)2 = 6.9544 = F (.99; 1,
82)
so the two tests are, indeed, equivalent
c) SSR = 93, 462, 942 which is 17.03% or 0.1703
d) -0.4127
3. Given that R2 = SSM/SST, it can be shown that R2 / (1 – R2) = SSM / SSE. If you
have n = 28 cases and R2 = 0.35, what is the F statistic for the test that the slope
is equal to zero?
dfR = 1 and dfE = n – 2 = 26. The F-statistic is
MSM SSM / df M SSM df E
R 2 26 0.35  26
F




 14
MSE
SSE / df E
SSE df M
0.65
1  R2 1


The α = 0.05 critical value for (1,24) df is 4.26 (from page 1342) and the P-value using
that df is 0.001. The results for (1,26) df will be similar. We reject H0: β1 = 0 and
conclude that the slope is non-zero.
4. Problem 2.54 – Yes to the first question and no to the second one.
5. KNNL Problem 2.55
2
_
_ 
_
 ^ _  


SSR   Y i Y



   b 0 b1X i Y    b 0 b1X i b 0 b1X  b   X i  X







2
2
n
n
n
i 1
i 1
i 1
6. KNNL Problem 2.57 (The degrees of freedom dfR referred to in the problem are
error degrees of freedom for the reduced model, i.e. dfE(R)).: The normal error
regression model (2.1) is assumed to be applicable.
2
1
n
i 1
(a) When testing H0: β1 = 5 versus Ha: β1  5 by means of a general linear test, what
is the reduced model? What are the degrees of freedom dfR?
The reduced model is Y  0  5 X   . Since there is only one model parameter to
estimate, the error df for the reduced model is n-1.
(b) When testing H0: β0 = 2, β1 = 5 versus Ha: not both β0 = 2 and β1 = 5 by means
of a general linear test, what is the reduced model? What are the degrees of
freedom dfR?
The reduced model is Y  2  5 X   . Since there are no model parameters to estimate,
the error df for the reduced model is n.
For the next 3 questions use the grade point average data described in the text with
KNNL problem 1.19.
7. Describe the distribution of the explanatory variable. Show the plots and output
that were helpful in learning about this variable.
Using PROC UNIVARIATE we see there are 20 observations ranging between 3.9 and
6.3 with a mean of 5 and median of 4.85; their standard deviation is 0.69. There are no
extreme observations (i.e., ones far away from the others) as shown in the histogram plot
below. The distribution appears to be reasonably symmetric but not completely so; it has
a slight skew to the right.
The UNIVARIATE Procedure
Variable: testscore
Moments
N
20
Sum Weights
Mean
5
Sum Observations
Std Deviation
0.69282032
Variance
Skewness
0.49695171
Kurtosis
Uncorrected SS
509.12
Corrected SS
Coeff Variation
13.8564065
Std Error Mean
Basic Statistical Measures
Location
Variability
Mean
5.000000
Std Deviation
Median
4.850000
Variance
Mode
4.700000
Range
Interquartile Range
Extreme Observations
----Lowest-------Highest--Value
Obs
Value
Obs
3.9
4
5.5
1
4.1
19
5.9
18
4.3
16
6.0
7
4.3
10
6.2
6
4.5
5
6.3
14
Stem
6
5
5
4
4
3
Leaf
023
59
0024
5677789
133
9
----+----+----+----+
#
3
2
4
7
3
1
20
100
0.48
-0.595083
9.12
0.15491933
0.69282
0.48000
2.40000
0.90000
Boxplot
|
|
+--+--+
*-----*
|
|
35
30
25
P
e
r
c
e
n
t
20
15
10
5
0
2. 7
3. 3
3. 9
4. 5
5. 1
5. 7
6. 3
6. 9
7. 5
t est scor e
6. 5
6. 0
5. 5
t
e
s
t
s
c
o
r
e
5. 0
4. 5
4. 0
3. 5
- 2
- 1
0
No r ma l
1
2
Qu a n t i l e s
8. Run the linear regression to predict GPA from the entrance test score and obtain
the residuals (do not include a list of the residuals in your solution).
a. Verify that the sum of the residuals is zero by running PROC
UNIVARIATE with the output from the regression.
The UNIVARIATE Procedure
Variable: resid (Residual)
Moments
N
Mean
Std Deviation
Skewness
Uncorrected SS
Coeff Variation
20
0
0.4234117
0.05677081
3.40627193
.
Sum Weights
Sum Observations
Variance
Kurtosis
Corrected SS
Std Error Mean
20
0
0.17927747
-1.0342174
3.40627193
0.09467773
b. Plot the residuals versus the explanatory variable and briefly describe the
plot noting any unusual patterns or points.
r esi dual
pl ot
0. 8
0. 7
0. 6
0. 5
0. 4
0. 3
0. 2
0. 1
0. 0
- 0. 1
- 0. 2
- 0. 3
- 0. 4
- 0. 5
- 0. 6
- 0. 7
- 0. 8
3
4
5
Tes t
6
7
Sc o r e
There does not appear to be any obvious pattern or outlier in this residual plot. It looks
like a random scatter of points and the variance is reasonably constant.
c. Plot the residuals versus the order in which the data appear in the data
file and briefly describe the plot noting any unusual patterns or points.
r esi dual
pl ot
0. 8
0. 7
0. 6
0. 5
0. 4
0. 3
0. 2
0. 1
0. 0
- 0. 1
- 0. 2
- 0. 3
- 0. 4
- 0. 5
- 0. 6
- 0. 7
- 0. 8
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Ob s e r v a t i o n S e q u e n c e
There is some suggestion that the residuals may become more negative over time but it
does not appear to be a big problem.
d. Examine the distribution of the residuals by getting a histogram and a
normal probability plot of the residuals by using the histogram and
qqplot statements in PROC UNIVARIATE. What do you conclude?
30
25
20
P
e
r
c
e
n
t
15
10
5
0
- 1. 65
- 1. 35
- 1. 05
- 0. 75
- 0. 45
- 0. 15
0. 15
0. 45
0. 75
1. 05
1. 35
1. 65
Re s i d u a l
1. 00
0. 75
0. 50
R
e
s
i
d
u
a
l
0. 25
0
- 0. 25
- 0. 50
- 0. 75
- 2
- 1
0
No r ma l
1
2
Qu a n t i l e s
The residuals appear reasonably normal and symmetric since the histogram appears fairly
normal and the qqplot is fairly linear. There is some suggestion of an S-shape to the qqplot and a flattening of the histogram, but it is not too bad.