Test 2, St 711, Dickey
1. (4 pts.) I have seven evaluators, available for a 3 day period, and 7 employees to
evaluate. Evaluation takes a day so each evaluator will evaluate 3 employees, rating each
on a 1 to 100 scale. Name a design that would be good for this, assuming there may be
random day effects, random evaluator effects, and fixed employee effects.
2. (24 points) Lagoons (large ponds) on hog farms are used to dispose of waste and
hence contain some toxic substances. There are hundreds of these farms in NC. In my
lab, I want to compare 4 chemicals for purifying the lagoon water using water from actual
lagoons. I expect much lagoon to lagoon variation, but not much within a lagoon. I want
my results to hold for the whole state and I’ll need 20 replicates of each treatment (80
runs in all). Explain how I should collect water samples, how I should run the
experiment, and which factors I would consider fixed and which random.
Lay out the analysis of variance table with sources and degrees of freedom.
For your design, are the confidence intervals for either of the following affected by
whether or not the lagoon effects are considered random?
(a) The mean for a single chemical
– confidence interval affected? (yes no)
(a) The difference in means between 2 chemicals
– confidence interval affected? (yes no)
(3) A Graeco-Latin square has lower case letters (a,b,c,d) corresponding to levels of
weight and capital letters (A,B,C,D) corresponding to levels of height in an experiment in
which 4 subjects (the rows) lifted the different weights to the different heights. Columns
correspond to the order of lifts. The response Y is stress on the spine of the lifter. Here
are the data with stress numbers Y labeled by weight and height levels.:
Subject
1
2
3
4
Lift1
32
56
84
80
(a,A)
(c,B)
(d,C)
(b,D)
Lift2
44
64
92
96
(b,B)
(d,A)
(c,D)
(a,C)
Lift3
76
20
32
84
(c,C)
(a,D)
(b,A)
(d,B)
Lift4
68
64
52
100
(d,D)
(b,C)
(a,B)
(c,A)
(A) (18 pts.) List all subjects, if any, who lifted weight b to height C __________ and
compute the mean stress ________ for weight b averaged over the other factors.
(B) I used this code to analyze the data:
proc glm; class row column height weight;
model Y = row column height weight/solution; run;
and got this analysis of variance table and partial (Type III) sums of squares list:
DF
Sum of
Squares
___
___
___
Source
row
column
height
weight
Source
Model
Error
Corrected Total
Mean Square
F Value
Pr > F
8764.000000
___________
8887.000000
730.333333
41.000000
17.81
0.0183
DF
Type III SS
Mean Square
F Value
Pr > F
3
3
__
3
3683.000000
1059.000000
___________
___________
1227.666667
353.000000
___________
********
29.94
8.61
_____
*****
0.0098
0.0552
0.0420
0.0151
(a) (36 pts.) Fill in the 9 blanks in the output above. The means for heights A, B,
C, and D were 57, 59, 80, and 65 and the overall mean 65.25.
(b) (8 pts.) List all of the Type I (sequential) sums of squares, if any, that can be
computed from the given information and your computations so far. Give the numbers:
Type I SS for: row ________ column ________ height _________ weight _________.
(c) (10 pts.) The difference in mean stress for height C versus B is 80-59=21.
Assuming random subject and lift time effects, fixed height and weight effects, give the
standard error __________of this difference.
ANSWERS
1. A Latin Square would be great if evaluators had 7 days, but with only 3 days A
Youden Square would be the next best thing with 3 rows (days) and 7 column
(evaluators). Note that the rows here have a meaning so a balanced incomplete block
design (where the entries in each column are randomized) would NOT capture the day
effects.
2. Collect large water samples for each of 20 randomly selected lagoons. Split each into
4 parts and assign the 4 treatments randomly to the 4 parts. Lagoons are random,
chemicals fixed. This of course is a randomized complete block design.
ANOVA
Source
df
Blocks
Trt
Error
19  (lagoons)
2 (chemicals)
38
Yes, No
3. (A) Subject 2 (4th lift). (80+44+32+64)/4 = 55. The point here, of course, is that this
simple mean works because of the DESIGN, that is, the simple mean has one effect from
every row and one for every column and so is automatically “averaged over the other
effects”.
(B) (a) Height df is 3 so model df is 3+3+3+3=12. Error df is then total df (15) minus
12 so 15-12=3 is error df and error SSq is 3(41)=123. Two missing sums of squares add
up to 8764-3683-1059=4022 so we need only compute one of them directly. Either one
is OK, but since height means are already computed, that one would be faster. Taking
differences from the overall mean we get -8.25, -6.25, 14.75. and -0.25. We thens quare
these, add them together and multiply by 4 (4 observations in each mean) getting 1299 as
our height SS then 4022-1299 = 2723 is SS(weight). The mean square 1299/3 = 433 and
F=433/41 = 10.56 follow.
(b) Type I and Type III are the same for all orthogonal designs like this. The
sums of squares do not depend on order of entry into the model so they are all the same
as what they would have been if fitted last (Type I = Type III). This of course is not
always the case. Since you know Type III, just fill those in here.
(c) This again is an easier computation because of the balance inherent in Latin
squares. The row and column effects difference out of this expression and we are left
with the difference of two averages of error terms (e’s), each an average 4 e’s so the
variance is 2(41)/4 = 20.5 and thus the standard error is 4.5277.