Education 257 SOLUTIONS HW1 1/23/05
1. Oneway classification
NWK Chap 16, prob 16.12 (Ver4) ; Chap 16 prob 16.9 (Ver5)
a)
step 1 read in data from knee.dat (here done in Minitab)
MTB > read '/usr/class/ed257/HW/knee.dat' c1 c2
24 ROWS READ
ROW C1 C2
1 29 1
2 42 1
3 38 1
4 40 1
. . .
and then get descriptive statistics by group (1,2,3)
MTB > describe c1;
SUBC> by c2.
C2 N MEAN MEDIAN TRMEAN STDEV SEMEAN
C1 1 8 38.00 40.00 38.00 5.48 1.94
2 10 32.00 31.00 31.62 3.46 1.10
3 6 24.00 22.50 24.00 4.43 1.81
C2 MIN MAX Q1 Q3
C1 1 29.00 43.00 32.00 42.00
2 28.00 39.00 29.00 35.00
3 20.00 32.00 20.75 27.50
The group means are 38, 32, and 24 for the below average, average, and
above average groups, respectively. Variances are 30.03, 11.97, and 19.62.
(Note: The group means and SDs are also displayed under the ANOVA table)
b)
here are dotplots below, boxplots or stemandleafs for the 3 groups
would be alternatives. Main point is always look at the data
MTB > dotplot c1;
SUBC> by c2.
C2
1 (below average)
. . . : : .
++++++C1
C2
2 (average)
. : . : . : .
++++++C1
C2
3 (above average)
. . . . . .
++++++C1
20.0 25.0 30.0 35.0 40.0 45.0
These plots illustrate the clustering of the observations in each
group about the group means. The small sample sizes make it difficult
to detect outliers or heteroskedasticity (unequal group variances),
although the observations in the below average group appear to be
somewhat more spread out than are those in the other groups.
c)
run a oneway anova to get the anova table (you could do it by hand
from the descriptive statistics in part a)
here is Minitab output
MTB > oneway c1 c2 resids in c3 fits in c4;
SUBC> tukey.
(Note: the above command tells Minitab to store the residuals in C3
and the fitted values (which are just the group means) in C4. The
words "resids in" and "fits in" are unnecessary; could just write
MTB >oneway c1 c2 c3 c4)
The tukey subcommand above does the calculations for part e
ANALYSIS OF VARIANCE ON C1
SOURCE DF SS MS F p
C2 2 672.0 336.0 16.96 0.000
ERROR 21 416.0 19.8
TOTAL 23 1088.0
INDIVIDUAL 95 PCT CI'S FOR MEAN
BASED ON POOLED STDEV
LEVEL N MEAN STDEV +++
1 8 38.000 5.477 (*)
2 10 32.000 3.464 (*)
3 6 24.000 4.427 (*)
+++
POOLED STDEV = 4.451 24.0 30.0 36.0
The omnibus null hypothesis is
Ho: mu(1)=mu(2)=mu(3)
We test this against the alternative
Ha: not all mu(i) are equal
Test statistic is MSB/MSW = 336/19.8 = 16.96.
Find critical value F(.95,2,21):
MTB > invcdf .95;
SUBC> f 2 21.
0.9500 3.4668
(or use NWK table B.4 (ver4 or ver5)
Since 16.96 > 3.4668, we reject the omnibus null hypothesis and
conclude that there are differences among the three groups.
d) Resids are stored in C3 & fits in C4, from oneway command above.
MTB > plot c3 c4
 *
 *
6.0+
 *
C3  2
 * 2 2
 *
0.0+ *
 * 2
 * *
 2 3

6.0+

 *
 *

++++++C4
25.0 27.5 30.0 32.5 35.0 37.5
We could also plot C3 against C2, or produce aligned dotplots of the
residuals for each group.
Here's one code example to obtain residuals the long way (remember residuals are
just the differences between each observation and the group mean):
MTB > unstack c1 c3c5;
SUBC> subscripts c2.
MTB > let c6=c3mean(c3)
MTB > let c7=c4mean(c4)
MTB > let c8=c5mean(c5)
MTB > stack c6c8 c9
MTB > plot c9 c2.
The plots suggest that the variability of the observations in the
below average group is greater than that for the other groups (the
dotplots and a quick look at the descriptive statistics support this).
Since the sample sizes are a bit unequal, if one wanted to be very careful,
the best analysis here would be to use something like BMDP7D
which we illustrated with the IBS data to use a oneway anova method
that did not require the equal variance assumption.
e. continuing the problem
from tukey subcommand in the main analysis
MTB > oneway c1 c2 resids in c3 fits in c4;
SUBC> tukey.

TUKEY'S multiple comparison procedure
Nominal level = 0.0500
Family error rate = 0.0500
Individual error rate = 0.0199
Critical value = 3.56
Intervals for (mean of column group)  (mean of row group)
1 2
2 0.680
11.320
3 7.943 2.208
20.057 13.792
[check by hand from NWK section 17.5 (ver4 or ver5)
Just presenting the output above, is not really
a complete answer. Much better to saythe confidence interval
for mu(1)  mu(2) has endpoints (.68, 11.32) etc etc
None of these intervals includes zero, so we conclude that each group
means differ from one another.
part (f) Knee problem, design considerations
Note: remember, for Tukey we don't need to worry splitting up the
familywise error rate
width = 5 = 2*q(.95,I,dfw)*sqrt(MSW/n)
= 2*q(.95,3,dfw)*sqrt(19.8/n) using MSW from previous analysis
a little algebra...
2.5 = q(.95,3,dfw)*sqrt(19.8/n)
6.25 = [q(.95,3,dfw)]^2 * 19.8/n
n = 3.168 * [q(.95,3,dfw)]^2
Problem is, we don't know exactly what q is without knowing n, because
q depends on degrees of freedom within. So we should use any prior
information we have to suggest a best guess.
The widths of the intervals in our previous analysis were around 11.
Since we want to cut this width approximately in half, we'll need to
quadruple (approximately) our group sample sizes. So start with n=40
as a best guess, which gives dfw =1203 =117.
Using NWK (ver 4 or 5) Table B.9, q(.95,3,117) = 3.36 (approx.).
Therefore n = 3.168 * (3.36)^2 = 36, which is pretty close to our
original guess. Anything reasonably close to this number is acceptable.

2. Factorial (2way) anova
First order of business is to set up a data file.
typical structure is outcomes in c1, row factor c2,
column factor c3
a. For profile plot get cell means:
MTB > table c2 c3;
SUBC> stats c1.
ROWS: C2 COLUMNS: C3
1 2 3 ALL
1 6 6 6 18
23.167 28.333 12.833 21.444
3.971 4.320 4.792 7.801
2 6 6 6 18
20.500 25.000 32.833 26.111
4.135 2.898 3.430 6.201
ALL 12 12 12 36
21.833 26.667 22.833 23.778
4.108 3.916 11.175 7.337
CELL CONTENTS 
C1:N
MEAN
STD DEV
Profile plot see figures 19.119.5 ver4 or Figs 19.319.7 ver5
I won't try to draw here. Main effects of both Time of Isolation and
Level of Reinforcement are best interpreted keeping in mind the
disordinal interaction indicated in the profile plot. The profile
plot indicates that recall increases steadily with time in isolation
only for verbally reinforced children. For unreinforced children,
recall increases from 20 to 40 minutes of isolation, but decreases
from the 40 minute level (and falls below the 20 minute level) for 60
minutes of isolation.
b. The model for these data is:
y(ijk) = mu + alpha(i) + beta(j) + alphabeta (ij) + epsilon(ijk)
where mu is the grand mean of all observations.
y(ijk) is recall for child k observed within the group
defined by reinforcement level i and isolation level j.
alpha(i) is the effect of reinforcement at level i.
beta(j) is the effect of isolation at level j.
alphabeta (ij) is the effect of the interaction between reinforcement
at level i and isolation at level j .
epsilon(ijk) is a random error
part c.
The ANOVA table below indicates significant main effects and interaction
between the two factors (Time of Isolation and Level of Reinforcement) using
overall error rate .05.
Factor Type Levels Values
C2 fixed 2 1 2
C3 fixed 3 1 2 3
Analysis of Variance for C1
Source DF SS MS F P
C2 1 196.00 196.00 12.42 0.001
C3 2 156.22 78.11 4.95 0.014
C2*C3 2 1058.67 529.33 33.55 0.000
Error 30 473.33 15.78
Total 35 1884.22
To get the critical values for the series of 3 hypothesis tests,
we use inverse cdf function in computer packages (here minitab)
MTB > invcdf .983;
SUBC> f 2 30.
0.9830 4.6817
MTB > invcdf .983;
SUBC> f 1 30.
0.9830 6.3871
This gives us each test at .017 (c.f. alphatot.tab)
We are able to reject the main effects and interaction null hypotheses.

SAS input and output
title 1 'SAS anova HW1 problem 2';
data recall;
input outcome reinforce isolation;
datalines;
26 1 1
23 1 1
28 1 1
19 1 1
18 1 1
25 1 1
30 1 2
25 1 2
27 1 2
36 1 2
28 1 2
24 1 2
6 1 3
11 1 3
17 1 3
10 1 3
14 1 3
19 1 3
15 2 1
24 2 1
25 2 1
16 2 1
22 2 1
21 2 1
24 2 2
29 2 2
23 2 2
26 2 2
27 2 2
21 2 2
31 2 3
29 2 3
35 2 3
38 2 3
34 2 3
30 2 3
;
proc anova data=recall;
class reinforce isolation;
model outcome = reinforceisolation;
run;
output
1 SAS anova HW1 problem 2 1
15:13 Sunday, January 23, 2005
The ANOVA Procedure
Class Level Information
Class Levels Values
reinforce 2 1 2
isolation 3 1 2 3
Number of Observations Read 36
Number of Observations Used 36
1 SAS anova HW1 problem 2 2
15:13 Sunday, January 23, 2005
The ANOVA Procedure
Dependent Variable: outcome
Sum of
Source DF Squares Mean Square F Value Pr > F
Model 5 1410.888889 282.177778 17.88 <.0001
Error 30 473.333333 15.777778
Corrected Total 35 1884.222222
RSquare Coeff Var Root MSE outcome Mean
0.748791 16.70520 3.972125 23.77778
Source DF Anova SS Mean Square F Value Pr > F
reinforce 1 196.000000 196.000000 12.42 0.0014
isolation 2 156.222222 78.111111 4.95 0.0139
reinforce*isolation 2 1058.666667 529.333333 33.55 <.0001

3. Mixed Model NWK Chap 24 (ver4) Chap 25 (ver5)
Model is:
y(ijk) = mu + alpha(i) + beta(j) + alphabeta(ij) + epsilon(ijk)
where mu is the grand mean of all evaluations.
y(ijk) is evaluation k observed within the group defined by
level i of the row factor and level j of the column factor.
alpha(i) is a random effect of patient at level i.
beta(j) is the fixed effect of physician at level j.
alphabeta(ij) is a random effect of the interaction between patient
and physician at level i of patient and level j of physician.
epsilon(ijk) is a random component of error for observation k
at level i of patient and level j of physician.

b.
************So here's the complete data structure
ROW eval MD patient
1 7.2 1 1
2 9.6 1 1
3 8.5 2 1
4 9.6 2 1
5 9.1 3 1
6 8.6 3 1
7 8.2 4 1
8 9.0 4 1
9 7.8 5 1
10 8.0 5 1
11 4.2 1 2
12 3.5 1 2
13 2.9 2 2
14 3.3 2 2
15 1.8 3 2
16 2.4 3 2
17 3.6 4 2
18 4.4 4 2
19 3.7 5 2
20 3.9 5 2
21 9.5 1 3
22 9.3 1 3
23 8.8 2 3
24 9.2 2 3
25 7.6 3 3
26 7.1 3 3
27 7.3 4 3
28 7.0 4 3
29 9.2 5 3
30 8.3 5 3
31 5.4 1 4
32 3.9 1 4
33 6.3 2 4
34 6.0 2 4
35 6.1 3 4
36 5.6 3 4
37 5.0 4 4
38 5.4 4 4
39 6.5 5 4
40 6.9 5 4
So now the mixed model anova (here minitab)
MTB > anova eval = MDpatient;
SUBC> random patient;
SUBC> restricted;
SUBC> ems.
Factor Type Levels Values
MD fixed 5 1 2 3 4 5
patient random 4 1 2 3 4
Analysis of Variance for eval
Source DF SS MS F P
MD 4 3.812 0.953 0.71 0.602
patient 3 180.133 60.044 173.41 0.000
MD*patient 12 16.159 1.347 3.89 0.004
Error 20 6.925 0.346
Total 39 207.028
Source Variance Error Expected Mean Square
component term (using restricted model)
1 MD 3 (4) + 2(3) + 8Q[1]
2 patient 5.9698 4 (4) + 10(2)
3 MD*patient 0.5001 4 (4) + 2(3)
4 Error 0.3463 (4)
you should check always that the computer package produced the
correct test statistics from the anova table
(i.e. choosing the denominators according the the indications
from the E(MS) table and also implicitly that the pvalues
reflect using the correct df from the test statistics ).
Clearly in these data there is
no effect for MD (which could indicate that all doctors are
equally good?), there is a large patient effect and also a
significant interaction (we should look at the profile plot etc
if we were doing more than just using this for practice in
setting up the mixedmodel analysis.)
c. Posthoc comparisons
Now we could the physician means from a table of all 20 cell means
MEANS
MD N eval
1 8 6.5750
2 8 6.8250
3 8 6.0375
4 8 6.2375
5 8 6.7875
Tukey pairwise comparisons for the physician factor (using number
of levels of physician = 5, df(error) use the error term
MSinteraction and df(physician*patient) = 12, and thus q = 4.52
at Type 1 error rate = .05, and number of observations per level
of physician = 8  2 replications times 4 levels of patient).
W = (4.52)*Sqrt[1.347/8] = 1.85. So we can see that all
confidence intervals include 0 (so none of the 10 pairwise
differences are significantly different from 0). We would expect
this result given the anova table above. In real life you
wouldn't bother most likely with a posthoc pairwise comparison
for a main effect this seemingly nonexistent. But for the
exercise it's worth it.....

Another look at part (b) "for old times sake" if the
computing package doesn't automatically produce test
statistics in the mixedmodel specification
Analysis of Variance for eval
Source DF SS MS
patient 3 180.133 60.044
MD 4 3.812 0.953
patient*MD 12 16.159 1.347
Error 20 6.925 0.346
Total 39 207.028
Tests for interaction and main effects;
(t.s. = test statistic).
reference distrib
interaction t.s. 1.347/.346 = 3.89 (F 12,20)
patient t.s. 60.04/.346 =173.5 (F 3,20)
MD t.s. .953/1.347 = .707 (F 4,12)

================================================================
4. Unbalanced designs
NWK Ch 22 (ver4) Ch 23 (ver5, prob 23.9)
Note that this is a 2 way (4 X 3) anova design, with both factors
(subject of course, highest degree earned) fixed and where cell n's
are unequal.
Part a.
A profile plot of the cell means could be drawn with either factor A
or factor B on the xaxis, as below (connect the like symbols between
columns).
Comments about the plots: There appears to be a main effect for
subject of course, with earnings per course lowest for Humanities and
highest for Engineering courses, with Social Sciences and Management
courses at similar intermediate earning levels. This is true
regardless of the highest degree earned by the instructor, suggesting
no important interaction between the factors. There also appears to
be a main effect for highest degree earned, with instructors who have
a PhD earning more than those with either BA or MA degrees. Again
this is true regardless of subject of the course. No important
interaction is visible.
3.80 
 *PhD
E 3.60 
A  *PhD
R 3.40  *PhD
N 
I 3.20 
N 
G 3.00 
S  *MA
2.80  *BA
P  *PhD
E 2.60  *BA*MA
R  *MA*BA
2.40 
C 
O 2.20 
U 
R 2.00  *MA
S 
E 1.80  *BA

1.60 

1 2 3 4
Humanities Engineering
Social Sciences Management
Factor A: Subject of course

3.80 
 *ENG
E 3.60 
A  *SS
R 3.40  *MGT
N 
I 3.20 
N 
G 3.00 
S  *ENG
2.80  *ENG
P  *HUM
E 2.60  *MGT *MGT
R  *SS *SS
2.40 
C 
O 2.20 
U 
R 2.00  *HUM
S 
E 1.80  *HUM

1.60 

1 2 3
BA MA PhD
Factor B: Highest degree earned
Part b.
To test H(o): all alpha(i)s equal 0 vs. H(a): not all alpha(i)s equal
0, the test statistic is MSA/MSE . To find these values,
first the corresponding Degrees of Freedom obscured in the output must
be calculated. We know that a = 4, b = 3 and N(Total) = 45.
Source DF
A (Subject) 3
B (Degree) 2
AB interactions 6
Error 33
Total 44
(You could also find error DF by subtracting other DFs from total DF:
44  3  2  6 = 33.)
(NOTE: For this problem you need not calculate any DFs or Mean
Squares except for Factor A (Subject) and Error.)
The required Mean Squares obscured in the output can be calculated
from the Adjusted Sum of Squares (as you may recall from lecture the
Sequential SS are not
appropriate here as they depend on the order in which the factors are
entered into the GLM procedure; see UNBALANC.LOG).
Source Adj. MS = Adj. SS / DF
A (Subject) 1.4109 4.2326 / 3
B (Degree) 4.1144 8.2287 / 2
AB interaction 0.0074 0.0444 / 6
Error 0.0218 0.7180 / 33
MSA 1.4109
The test statistics is then  =  = 64.72
MSE 0.0218
The critical F value F(0.01;3,33) is approximately 4.46 (see NWK table
B.4 F(.01;3,30) = 4.51
and F(.01;3,40) = 4.31). For purposes here a critical value about 4.5
(or a touch less) is fine here.
Since 64.72 > 4.46 we reject H(o) that all alpha(i)s equal 0.
Part c.
read in data set
MTB > read '/usr/class/ed257/adjprof.dat' into c1c3
45 ROWS READ
ROW C1 C2 C3
1 1.7 1 1
2 1.9 1 1
3 1.8 1 2
4 2.1 1 2
. . .
In order to replicate the GLM analysis:
MTB > glm c1 = c2c3
Factor Levels Values
C2 4 1 2 3 4
C3 3 1 2 3
Analysis of Variance for C1
Source DF Seq SS Adj SS Adj MS F P
C2 3 4.1676 4.2326 1.4109 64.85 0.000
C3 2 8.3825 8.2287 4.1144 189.10 0.000
C2*C3 6 0.0444 0.0444 0.0074 0.34 0.910
Error 33 0.7180 0.7180 0.0218
Total 44 13.3124
Unusual Observations for C1
Obs. C1 Fit Stdev.Fit Residual St.Resid
39 2.30000 2.55000 0.10430 0.25000 2.40R
40 2.80000 2.55000 0.10430 0.25000 2.40R
R denotes an obs. with a large st. resid.
In order to obtain unweighted mean (i.e. ) Miller's approximate solution,
(see also unblanced.log course example)
first carry out a oneway analysis of variance on the cell means. You
can obtain the cell means by issuing the following command.
MTB > table c2 c3;
SUBC> mean c1;
SUBC> count.
ROWS: C2 COLUMNS: C3
1 2 3 ALL
1 1.8000 1.9500 2.7000 2.4250
2 2 8 12
2 2.4500 2.5200 3.4500 2.7846
4 5 4 13
3 2.7500 2.8500 3.7400 3.2364
2 4 5 11
4 2.5500 2.5500 3.4200 3.0333
2 2 5 9
ALL 2.4000 2.5385 3.2364 2.8489
10 13 22 45
CELL CONTENTS 
C1:MEAN
COUNT
Now you want to enter the cell means, the row indicators, and the
column indicators into columns so that you can carry out a oneway
analysis of variance. (An alternative to typing is to use your
editor to cut in the means from above and use set command to
create indices in C2 C3)
MTB > read data into c11c13
DATA> 1.80 1 1
DATA> 1.95 1 2
DATA> 2.70 1 3
DATA> 2.45 2 1
DATA> 2.52 2 2
DATA> 3.45 2 3
DATA> 2.75 3 1
DATA> 2.85 3 2
DATA> 3.74 3 3
DATA> 2.55 4 1
DATA> 2.55 4 2
DATA> 3.42 4 3
DATA> end
12 ROWS READ
Carry out a twoway ANOVA.
MTB > twoway c11c13
ANALYSIS OF VARIANCE C11
SOURCE DF SS MS
C12 3 1.50389 0.50130
C13 2 2.17280 1.08640
ERROR 6 0.01413 0.00236
TOTAL 11 3.69083
In order to obtain the unweighted mean (Miller's) approximate solution,
calculate the harmonic mean for this set of data.
harmonic mean = [(1/12) x (1/2 + 1/2 + 1/8 + 1/4 + 1/5 + 1/4 + 1/2 +
1/4 + 1/5 + 1/2 + 1/2 + 1/5)]^1 = 3.0189 (approx.=3)

or if you like computational overkill, here's an example
Harmonic Mean from Mathematica
< 15/4, HarmonicMean > 160/53, Median > 4}
N[%]
{Mean > 3.75, HarmonicMean > 3.018867924528302,
Median > 4.}

Comparing the GLM and Miller solutions we see:
GLM Solution
SOURCE DF Adj SS Adj MS F
C2 3 4.2326 1.4109 64.85
C3 2 8.2287 4.1144 189.10
Interaction 6 .0444 .0074 .34
Error 33 .7180 .0218
MILLER Solution
SOURCE DF SS MS F
C12 3 3(1.50389)=4.51167 1.50389 68.99
C13 2 3(2.17280)=6.51840 3.25920 149.50
Interaction 6 3(0.01413)=0.04239 .00707 .32
Error 33 .7180 .0218
(Note: The "error" term we obtained from the twoway analysis of variance
on the cell means is actually an estimate of the interaction. Our estimate
of SSE for Miller's solution is obtained by taking the sum of the
squared within cell deviations (i.e. X(ijk)  X(ij bar)).
This procedure gives withincell error estimates identical to GLM.)
As we can see, the GLM solution and Miller's approximate solution yield
similar results. At any reasonable Type I error rate, we can reject the
null hypotheses of no main effects, for both subject matter and degree
earned. Likewise, neither solution offers evidence of an interaction
between subject matter and degree earned. This is pretty good considering
this unbalanced design is a little outside the guidline of nMax/nMin <= 3
for Miller's procedure.

5. Power calcs
NWK ver 4 26.7, ver 5 16.26
What is the power of the test in Problem 1(c)?
Sample sizes were n_1 = 8, n_2 = 10, n_3=6
Test was carried out with Type I error rate alpha set to .01
For the power calculation we are given:
mu_1 = 37 mu_2 = 35 mu_3 = 28
sigma = 4.5
Begin by finding the Weighted Mean following NWK sec 26.4
(also example on p.1055) in Ver4, or section 16.10 pp7178.
Weighted Mean, mu(.) = (8*37 + 10*35 + 6*28)/24 = 33.9167
Next, compute the quantity called phi from NWK p. 1055
phi is a measure of how unequal the mu_i are.
phi = (Sqrt[(8*(37  33.9167)^2 + 10*(35  33.9167)^2 + 6*(28 33.9167)^2)/3]/4.5)
= 2.21418
The degrees of freedom for the omnibus test are 2 and
21
Interpolating Table B.11 to look up the related
value of power
Power = .70 or .71

6. Depression class example, blocking
MTB > Read file DEPRESS.DAT
describe c1c2
Descriptive Statistics
Variable N Mean Median TrMean StDev SEMean
C1 30 6.300 6.000 6.154 2.395 0.437
C2 30 7.567 8.000 7.654 2.596 0.474
Variable Min Max Q1 Q3
C1 3.000 12.000 4.000 8.000
C2 2.000 12.000 5.750 9.250
or "stack" data into 60 rows with outcome group and block as the columns
outcomes and treatmentplacebo indicator and block identifier (1,...30)
Data Display
Row C3 C4 C5
1 6 1 1
2 4 1 2
3 6 1 3
4 7 1 4
5 5 1 5
6 6 1 6
7 8 1 7
8 7 1 8
9 8 1 9
10 3 1 10
11 9 1 11
12 4 1 12
13 8 1 13
14 11 1 14
15 12 1 15
16 6 1 16
17 10 1 17
18 3 1 18
19 5 1 19
20 4 1 20
21 6 1 21
22 7 1 22
23 5 1 23
24 6 1 24
25 3 1 25
26 10 1 26
27 5 1 27
28 4 1 28
29 4 1 29
30 7 1 30
31 4 2 1
32 7 2 2
33 12 2 3
34 10 2 4
35 2 2 5
36 11 2 6
37 9 2 7
38 5 2 8
39 11 2 9
40 8 2 10
41 7 2 11
42 6 2 12
43 8 2 13
44 9 2 14
45 9 2 15
46 8 2 16
47 10 2 17
48 9 2 18
49 8 2 19
50 5 2 20
51 8 2 21
52 7 2 22
53 6 2 23
54 9 2 24
55 3 2 25
56 5 2 26
57 11 2 27
58 7 2 28
59 3 2 29
60 10 2 30
So we have a 2x30 design with 1 replication per cell
Twoway Analysis of Variance
Analysis of Variance for C3
Source DF SS MS
C4 1 24.07 24.07
C5 29 237.73 8.20
Error 29 123.93 4.27
Total 59 385.73
model statement needs to tell anova not
to fit both an interaction AND an error term
For example
anova c3 = c4c5  c4*c5
anova c3 = c4 c5
Analysis of Variance (Balanced Designs)
Factor Type Levels Values
C4 fixed 2 1 2
C5 fixed 30 1 2 3 4 5 6 7 8 9
10 11 12 13 14 15 16 17 18
19 20 21 22 23 24 25 26 27
28 29 30
Analysis of Variance for C3
Source DF SS MS F P
C4 1 24.067 24.067 5.63 0.024
C5 29 237.733 8.198 1.92 0.042
Error 29 123.933 4.274
Total 59 385.733
Analysis of Variance for C3
Source DF SS MS F P
C4 1 24.067 24.067 5.63 0.024
C5 29 237.733 8.198 1.92 0.042
Error 29 123.933 4.274
Total 59 385.733
So if we use Type I error rate .05 to test treatment vs placebo
we find a significant effect
b) calculate how much blocking helps by comparing results
with a randomized design (no matching) or could use
NWK sec 27.7 (ver4) sec 21.9 (ver5).
note ignore blocking
OneWay Analysis of Variance
Analysis of Variance on C3
Source DF SS MS F p
C4 1 24.07 24.07 3.86 0.054
Error 58 361.67 6.24
Total 59 385.73
MSE increases from 4.27 to 6.24 when the
blocking structure is ignored, so relative
efficiency of blocking here was 1.46, about 1.5.
c)
paired ttest (ttest onesample on difference scores for each block)
let c10 = c2 c1
ttest c10
TTest of the Mean
Test of mu = 0.000 vs mu not = 0.000
Variable N Mean StDev SE Mean T PValue
C10 30 1.267 2.924 0.534 2.37 0.024
NOTE the anova test statistic 5.63 = (2.37)^2
these are equivalent.
tinterval c10
Confidence Intervals
Variable N Mean StDev SE Mean 95.0 % C.I.
C10 30 1.267 2.924 0.534 ( 0.175, 2.359)

7. Randomized Block NWK ch 27 (ver4), ch 21 (ver5)
a) Statistical model
Y(ij) = mu + rho(i) + tau(j) + epsilon(ij)
where
Y(ij) is the proficiency of the auditor in block i and treatment j
mu is the grand mean
rho(i) is the main effect for block at level i
tau(j) is the main effect for training method at level j
epsilon(ij) is random error associated with the observation at level i of the
blocking factor and level j of the treatment factor
i=1,...,10
j=1,2,3
Assumptions:
mu is a constant
rho(i) are constants that sum to zero
tau(j) are constants that sum to zero
epsilon(ij) are independent normal (0,sigmasq)
The model generally assumes no block*treatment interaction
b) Profile plot
Here is the treatment*block table:
1 2 3 ALL
1 73.000 81.000 92.000 82.000
2 76.000 78.000 89.000 81.000
3 75.000 76.000 87.000 79.333
4 74.000 77.000 90.000 80.333
5 76.000 71.000 88.000 78.333
6 73.000 75.000 86.000 78.000
7 68.000 72.000 88.000 76.000
8 64.000 74.000 82.000 73.333
9 65.000 73.000 81.000 73.000
10 62.000 69.000 78.000 69.667
ALL 70.600 74.600 86.100 77.100
Plotblocks 110 labeled AJ

 A
90+ BD
 CEG
 F

 A HI
80+
 BD J
 BCE CF
 ADF HI
 EG
70+ J
 G
 I
 H
 J
+++++
1.20 1.60 2.00 2.40 2.80
The plot shows a clear main effect for method; values increase
as j=1,2,3 on horizontal axis increases.
There may be seen some evidence of disordinal interaction between method
and blocks; but it's hard to tell whether it's large enough to be concerned
about. A Tukey 1df for nonaddditivity may be a useful sidecheck here
(but since method is so highly significant, we need not be so worried
about an interaction slightly inflating the error term).
c) ANOVA table and test for main effect
MTB > anova c1=c3 c2
Factor Type Levels Values
method fixed 3 1 2 3
block fixed 10 1 2 3 4 5 6 7 8 9
10
Analysis of Variance for profmeas
Source DF SS MS F P
method 2 1295.00 647.50 103.75 0.000
block 9 433.37 48.15 7.72 0.000
Error 18 112.33 6.24
Total 29 1840.70
Test for main effect for training method
Ho: tau(1)=tau(2)=tau(3)=0 vs. Ha: not all tau(j)=0
Test statistic = MSTR/MSW = 647.50/6.24 = 103.75
Critical value = F(.99,2,18) = 6.0129
Reject Ho; there is a significant effect of training method.
d) Tukey pairwise comparisons for treatment means
The halfwidth of the Tukey interval is (see NWK sec 27.5 ver 4, sec 21.5 ver5)
sqrt(MSW/#blocks)*q(1alpha, #treatments, dfw) = sqrt(6.24/10)*q(.99,3,18)
= sqrt(6.24/10)*4.70 = 3.71
So the interval estimates are
For mu(1)mu(2): 70.674.6 +/ 3.71 = (7.71,.29)
For mu(2)mu(3): 74.686.1 +/ 3.71 = (15.21,7.79)
For mu(1)mu(3): 70.686.1 +/ 3.71 = (19.21,11.79)
None of these intervals contains zero, indicating significant differences
between all pairs of treatment means. Method 3 seems clearly superior
scary as it sounds there may be some reason to go to Chicago.
e) Usefulness of blocking
The easiest way to do this is to run the the oneway ANOVA without the
blocking factor and compare the MSW from this with the MSW above.
MTB > oneway c1 c3
ANALYSIS OF VARIANCE ON profmeas
SOURCE DF SS MS F p
method 2 1295.0 647.5 32.04 0.000
ERROR 27 545.7 20.2
TOTAL 29 1840.7
MSW increases from 6.24 to 20.2 when blocks are not used, so the relative
efficiency of blocking is 20.2/6.24 = 3.24. So we would require at least
3 times as many subjects to achieve the same precision for investigating
differences between the 3 methods with a design that did not use blocking
90 subjects or more.

END 2005 HW1 SOLUTIONS