10 Topic Handout: ANCOVA & RCBD Designs
Learning Goals for this Activity: (1) Experience in the use of ANCOVA and RCBD Models; (2) An
understanding of why such models are worthwhile.
10.1
Output for Example II (from notes) is given below on pages 1-2. Complete the analysis of this example
(using the analysis of Example I completed in class as a guide). Make sure to consider the following:
a. Assumptions Assessment.
b. Analysis with the covariate included.
c. Compare to previous analysis (without the covariate). What changes? What stays the same?
ANOVA MODEL
The GLM Procedure
Dependent Variable: y
Source
trt
Error
Corrected Total
R-Square
0.013384
DF
2
6
8
Coeff Var
37.83138
Sum of
Squares
1.55555556
114.6666667
116.2222222
Root MSE
4.371626
Mean Square
0.77777778
19.1111111
F Value
0.04
Pr > F
0.9604
y Mean
11.55556
t Tests (LSD) for y
NOTE: This test controls the Type I comparisonwise error rate, not the experimentwise error rate.
Alpha
0.05
Error Degrees of Freedom
6
Error Mean Square
19.11111
Critical Value of t
2.44691
Least Significant Difference
8.734
16
1
15
3
14
Means with the same letter are not significantly different.
2
13
1
2
3
12
GRP
A
A
A
Mean
12.000
11.667
11.000
N
3
3
3
trt
1
3
2
11
10
9
8
7
ANCOVA Interaction Model
Source
Square
F Value
x
trt
x*trt
Error
Corrected Total
R-Square
0.873088
Pr > F
Coeff Var
19.18864
1
3
6
2
1
DF
Sum of
Squares
1
2
2
3
8
9.66896458
87.46642860
4.33686046
14.7499686
116.2222222
Root MSE
2.217353
y Mean
11.55556
2
3
4
5
Di s e a s e
trt
Mean
9.66896458
43.73321430
2.16843023
4.9166562
1.97
8.89
0.44
0.2554
0.0548
0.6793
1
6
7
St age
2
3
8
9
r esi d
3
3
2
2
1
r
e
s
i
d
1
0
0
-1
-1
-2
-2
-3
-3
- 1. 5
-1
- 0. 5
0
N o r ma l
0. 5
ANCOVA MODEL
1. 5
6
7
8
9
10
11
12
13
14
15
16
17
pr ed
Sum of
Squares
97.1353932
19.0868290
116.2222222
Source
Model
Error
Corrected Total
DF
3
5
8
R-Square
0.835773
Root MSE
1.953808
Coeff Var
16.90795
1
Qu a n t i l e s
Mean Square
32.3784644
3.8173658
F Value
8.48
Pr > F
0.0209
y Mean
11.55556
Source
x
trt
DF
1
2
Type I SS
9.66896458
87.46642860
Mean Square
9.66896458
43.73321430
F Value
2.53
11.46
Pr > F
0.1724
0.0136
Source
x
trt
DF
1
2
Type III SS
95.57983762
87.46642860
Mean Square
95.57983762
43.73321430
F Value
25.04
11.46
Pr > F
0.0041
0.0136
Least Squares Means
Adjustment for Multiple Comparisons: Tukey-Kramer
trt
1
2
3
y LSMEAN
-1.1648624
10.3148399
25.5166892
Standard
Error
2.8625918
1.1363118
2.9889270
Pr > |t|
0.7009
0.0003
0.0004
LSMEAN
Number
1
2
3
Least Squares Means for Effect trt
t for H0: LSMean(i)=LSMean(j) / Pr > |t|
Dependent Variable: y
i/j
1
2
3
10.2
1
3.877496
0.0263
4.739498
0.0118
2
-3.8775
0.0263
4.587099
0.0135
3
-4.7395
0.0118
-4.5871
0.0135
Suppose we are interested in determining if there are regional effects on the salaries of teachers. We
obtain from DASL a dataset containing average teacher salaries for each state. The dataset also has
another variable that is known to be related to teacher salary and that is average expenditure per student.
Using the output below (and the following questions as guideposts), analyze the data to determine if
there are regional differences in salary. Note: The entire dataset and description may be found at
http://lib.stat.cmu.edu/DASL/Stories/EducationalSpending.html.
a. Consider the assignment of regions (see below). Comment.
b. Consider the analysis of covariance (see page 3-4). Perform the analysis and draw appropriate
conclusions. (Side question: Why are the Type I / III SS different?)
c. If we had just done an ANOVA and ignored the covariate (see page 5), would we have drawn any
incorrect conclusions? Consider the one-way ANOVA what conclusions would you draw?
d. Is there any additional output that you wish you had to examine?
proc freq; tables region; run;
Cumulative
Cumulative
Region
Frequency
Percent
Frequency
Percent
ENC
5
9.80
5
9.80
ESC
4
7.84
9
17.65
MA
3
5.88
12
23.53
MN
8
15.69
20
39.22
NE
6
11.76
26
50.98
PA
5
9.80
31
60.78
SA
9
17.65
40
78.43
WNC
7
13.73
47
92.16
WSC
4
7.84
51
100.00
ADDITIONAL REGIONAL INFORMATION
ENC = Indiana, Michigan, Ohio, Illinois, Wisconsin,
ESC = Kentucky, Alabama, Tennessee, Mississippi
MA = New York, Pennsylvania, New Jersey
MN = Wyoming, New Mexico, Utah, Montana, Idaho, Colorado, Arizona, Nevada
NE = Maine, Vermont, Rhode Island, New Hampshire, Massachusettes, Connecticut
PA = Washington, Oregon, California, Alaska, Hawaii
SA = Maryland, Virginia, North Carolina, Georgia, Deleware, District of Columbia,
West Virginia, South Carolina, Florida
WNC = Minnesota, Missouri, South Dakota, Kansas, Iowa, North Dakota, Nebraska
WSC = Arkansas, Oklahoma, Louisiana, Texas
Source
Model
Error
Total
R-Square
0.765853
Source
Spend
Region
DF
9
41
50
SumofSquares
679.7466365
207.8215988
887.5682353
Coeff Var
9.191611
DF
1
8
Mean Square
75.5274041
5.0688195
Root MSE
2.251404
Type I SS
606.4901533
73.2564832
F Value
14.90
Pr > F
<.0001
Pay Square
606.4901533
9.1570604
F Mean
24.49412
Mean Value
119.65
1.81
Pr > F
<.0001
0.1037
Source
DF
Type III SS
Mean Square
F Value
Spend
1
382.2488496
382.2488496
75.41
Region
8
73.2564832
9.1570604
1.81
The GLM Procedure
Least Squares Means
Adjustment for Multiple Comparisons: Tukey-Kramer
Pr > F
<.0001
0.1037
Region
Pay LSMEAN
Standard
Error
Pr > |t|
LSMEAN
Number
ENC
26.6220761
1.0069028
<.0001
1
ESC
MA
MN
NE
PA
SA
WNC
WSC
24.5768606
23.2072148
25.1035843
22.7431288
26.3157257
24.2792555
23.5503181
23.9822325
1.1986906
1.4091828
0.8025782
0.9279276
1.0771737
0.7504688
0.8573429
1.1572955
<.0001
<.0001
<.0001
<.0001
<.0001
<.0001
<.0001
<.0001
2
3
4
5
6
7
8
9
Least Squares Means for Effect Region
t for H0: LSMean(i)=LSMean(j) / Pr > |t|
Dependent Variable: Pay
i/j
1
3
4
5
6
7
8
9
1.308531
0.9230
1
2
1.968314
0.5729
1.179985
0.9563
2.831038
0.1374
0.20742
1.0000
1.865574
0.6399
2.32408
0.3521
1.722737
0.7295
0.696142
0.9986
-0.37278
1.0000
1.182952
0.9557
-1.0189
0.9818
0.210387
1.0000
0.710787
0.9984
0.37201
1.0000
-1.1453
0.9632
0.282011
1.0000
-1.88155
0.6296
-0.67163
0.9989
-0.20379
1.0000
-0.40748
1.0000
1.90748
0.6127
-0.88334
0.9927
0.750146
0.9976
1.333031
0.9152
0.807512
0.9960
-2.57577
0.2281
-1.28729
0.9294
-0.63366
0.9993
-0.82263
0.9955
1.551596
0.8242
1.967604
0.5733
1.418739
0.8841
0.639696
0.9992
0.215306
1.0000
2
-1.30853
0.9230
3
-1.96831
0.5729
-0.69614
0.9986
4
-1.17998
0.9563
0.37278
1.0000
1.145301
0.9632
5
-2.83104
0.1374
-1.18295
0.9557
-0.28201
1.0000
-1.90748
0.6127
6
-0.20742
1.0000
1.018901
0.9818
1.881553
0.6296
0.883341
0.9927
2.575774
0.2281
7
-1.86557
0.6399
-0.21039
1.0000
0.671629
0.9989
-0.75015
0.9976
1.287291
0.9294
-1.5516
0.8242
8
-2.32408
0.3521
-0.71079
0.9984
0.203789
1.0000
-1.33303
0.9152
0.633655
0.9993
-1.9676
0.5733
-0.6397
0.9992
9
-1.72274
0.7295
-0.37201
1.0000
0.40748
1.0000
-0.80751
0.9960
0.822631
0.9955
-1.41874
0.8841
-0.21531
1.0000
-0.30403
1.0000
0.304026
1.0000
ONE-way ANOVA
Sum of
Squares
297.4977869
590.0704484
887.5682353
Source
Region
Error
Corrected Total
DF
8
42
50
R-Square
0.335183
Root MSE
3.748239
Coeff Var
15.30261
Mean Square
37.1872234
14.0492964
Pr > F
0.0190
Pay Mean
24.49412
Tukey's Studentized Range (HSD) Test for Pay
Alpha
0.05
Error Degrees of Freedom
42
Error Mean Square
14.0493
Critical Value of Studentized Range 4.62238
Minimum Significant Difference
7.7029
Harmonic Mean of Cell Sizes
5.059112
NOTE: Cell sizes are not equal.
GRP
B
B
B
B
B
B
B
B
Means with the same letter are not significantly different.
10.3
F Value
2.65
Mean
29.640
27.933
26.540
24.289
24.213
23.850
22.643
21.650
21.000
A
A
A
A
A
A
A
N
5
3
5
9
8
6
7
4
4
Region
PA
MA
ENC
SA
MN
NE
WNC
WSC
ESC
This problem is adapted from Devore, Probability and Statistics for Engineering. We want to test the
power consumption for five different brands of a dehumidifier. Because power consumption will be
different at different humidity levels, it was determined to control the humidity levels for this
experiment and block on this variable. Thus each brand was tested at each of four different humidity
levels (ranging from moderate to heavy).
a. Pages 5-6 of the output show the analysis for the humidifier data. Using the output provided,
determine the results of this analysis.
b. Suppose that the interaction plot looked like the one shown at the bottom of page 6. How would
your analysis change?
ANALYSIS FOR HUMIDIFIER DATA
proc sort data=h; by brand humidity;
symbol1 v=dot i=join;
proc gplot data=ex1807;
plot power*humidity = brand;
run;
proc glm;
class humidity brand;
model power = humidity brand;
lsmeans brand / adjust=tukey pdiff tdiff;
output out=diag p=pred r=resid;
run;
p o we r
1100
1000
900
800
700
600
1
2
3
4
h u mi d i t y
br and
1
2
3
4
5
Source
humidity
brand
Error
Corrected Total
Sum of
Squares
53231.0000
116217.7500
1671.0000
171119.7500
DF
4
3
12
19
R-Square
0.990235
Coeff Var
1.362242
Mean Square
13307.7500
38739.2500
139.2500
Root MSE
11.80042
F Value
95.57
278.20
power Mean
866.2500
The GLM Procedure
Least Squares Means
Adjustment for Multiple Comparisons: Tukey
brand
1
2
3
4
5
LSMEAN
Number
1
2
3
4
5
power LSMEAN
797.500000
843.500000
839.000000
914.000000
937.250000
Least Squares Means for Effect brand
t for H0: LSMean(i)=LSMean(j) / Pr > |t|
Dependent Variable: power
i/j
1
1
2
3
4
5
5.512838
0.0010
4.973539
0.0024
13.96186
<.0001
16.74824
<.0001
2
3
4
5
-5.51284
0.0010
-4.97354
0.0024
0.539299
0.9813
-13.9619
<.0001
-8.44902
<.0001
-8.98832
<.0001
-16.7482
<.0001
-11.2354
<.0001
-11.7747
<.0001
-2.78638
0.0978
-0.5393
0.9813
8.449023
<.0001
11.2354
<.0001
8.988323
<.0001
11.7747
<.0001
2.78638
0.0978
Hypothetical Interaction Plot for LAST question.
p o we r
1100
1000
900
800
700
600
1
2
3
4
h u mi d i t y
br and
1
2
3
4
5
Pr > F
<.0001
<.0001