This exam is open book, open notes; you may use a calculator. Show your work to receive full credit. Use the back if more space is needed. Each problem or part is worth 10 points. There are several tables at the back of the exam.
1. (Data from Kuehl.) An experiment was conducted to study the effect of nitrogen on lettuce growth. Twenty plots were divided at random into five groups of four, with each group getting a different amount of fertilizer. The response is number of heads of lettuce per plot. The data and some summary calculations follow.
Cmd> anova("heads=n") \# n is a factor for the nitrogen groups
Model used is heads=n
DF SS MS
CONSTANT 1 4.067e+05 4.067e+05
n 4 4994.8 1248.7
ERROR1 15 3338 222.53
Cmd> tabs(heads,n)
component: mean
(1) 112 145.5 149 157.5 149
component: var
(1) 445.33 395.67 38.667 51 182
component: count
(1) 4 4 4 4 4
(a) Test the null hypothesis that all the treatment means are the same.
(b) Test the null hypothesis that the mean number of heads in the plots with nitrogen added is the same as the mean number of heads in the plot with no nitrogen added.
(c) Use the Tukey HSD procedure at the 0.05 level to make all paired comparisons. Summarize with an underline diagram.
(d) Assuming for the moment that all assumptions are met, what other aspects of these data would you examine and how would you go about it?
(e) Here is a plot of residuals versus predicted values and the
output of boxcoxvec. Is any transformation needed? If a transformation
is needed, which one and why; if no transformation is needed, why not?
Cmd> boxcoxvec("n",heads,powers:run(0,3,.25))
component: power
(1) 0 0.25 0.5 0.75 1
(6) 1.25 1.5 1.75 2 2.25
(11) 2.5 2.75 3
component: SS
(1) 3805 3647.3 3519 3416.9 3338
(6) 3280 3240.8 3218.8 3212.6 3221.2
(11) 3243.5 3279 3327.1
2. (Data from Nelson, 1988.) A chemical reaction uses both a base and an alcohol. A completely
randomized experiment was run using the six factor/level combinations
of 2 bases and 3 alcohols, with each factor/level combination used for
four runs (units). Some summary information follows:
Cmd> anova("y=base*alcohol")
Model used is y=base*alcohol
DF SS MS
CONSTANT 1 1.9601e+05 1.9601e+05
base 1 6.5104 6.5104
alcohol 2 5.3958 2.6979
base.alcohol 2 22.566 11.283
ERROR1 18 36.758 2.0421
Cmd> tabs(y,base,alcohol)\$mean
(1,1) 90.825 89.8 88.925
(2,1) 89.125 92.275 91.275
(a) Find 
.
(b) Interpret the results of this experiment (you may assume that all assumptions were met).
3. Here are the Yates' effects from a 
factorial with one
replication. What conclusions would you draw?
Cmd> yts<-yates(y);yts (1) 0.13688 -0.018125 -0.026875 0.15488 -0.026375 (6) 0.032625 0.003375 0.23638 -0.021375 0.035125 (11) -0.005625 0.031625 -0.028625 -0.022125 0.001625 Cmd> chplot(rankits:rankits(yts),yts)
Cmd> chplot(halfnorm:halfnorm(yts),absyts:abs(yts))
4. Suppose in Question 2 that the first data value was missing,
making the problem unbalanced. Here are some summaries.
Cmd> anova("ymiss=alcohol*base")
Model used is ymiss=alcohol*base
WARNING: summaries are sequential
DF SS MS
CONSTANT 1 1.8767e+05 1.8767e+05
alcohol 2 6.5014 3.2507
base 1 8.5925 8.5925
alcohol.base 2 18.778 9.3891
ERROR1 17 36.457 2.1445
Cmd> anova("ymiss=base*alcohol")
Model used is ymiss=base*alcohol
WARNING: summaries are sequential
DF SS MS
CONSTANT 1 1.8767e+05 1.8767e+05
base 1 7.9032 7.9032
alcohol 2 7.1907 3.5954
base.alcohol 2 18.778 9.3891
ERROR1 17 36.457 2.1445
Find the F-statistics for testing the main effects of base and alcohol.
5. Suppose that I have two treatments, that my sample sizes are
very large, and
that my noncentrality parameter 
is 8 (note, this is not 
as shown in Montgomery).
In a second experiment, I have eight treatments, my samples sizes
are very large, and my noncentrality parameter 
is again 8. In which experiment do I have greater power, the
experiment with two treatments or the experiment with 8 treatments? Why?