STAT 511
Final Exam Solutions
Spring 1999
Note that there may be than one way to approach a problem. Sometimes several approaches are
considered below, although students were expected to give only one answer. The solutions given
below sometimes include comments or details that were not expected in answers provided by
students. Not every possible solution is considered below.
1. a. (6 points)
β0 is the upper bound or asymptote for the mean responsiveness as the dosage
level becomes large.
β1
is the dosage level required to achieve 50% of the upper bound to the mean
responsiveness. It is a scaling parameter in this sense.
b. (10 points) Standard errors are obtained from the large sample normal distribution for the least
∧
∧
−1 

squares estimator β , i.e., β ~ N β, σ 2 D T D  , where D is an nxp matrix in which


(
 ∂E(Yi )
the i-th row is 
 ∂β 0
∂E(Yi )
∂β1
∂E(Yi ) 
.
∂β 2 
)
The formulas for these derivatives can
∧
be found in the S-Plus code provided on page 2 of the exam. The covariance matrix for β
is estimated as
∧  ∧ ∧
σ 2 D T D


−1
∧
∧
∧
, where D is D evaluated at β and
σ 2 is the residual
2
mean square, i.e.,




∧


∧
∧
1 19 
β0
2

σ =
∑  Yj - β 0 ∧  .
19 - 3 j=1
∧ β2 


1 +  x j β 1 


 

c. (4 points) Deviance is used as the label for various quantities in S-Plus. In this case, we are
dealing with least squares estimation and the deviance is the sum of squared residuals:
1
2




∧


19
∧
β0


deviance = ∑ Yj - β 0 ∧  .

j=1
∧ β 2 


1 +  x j β 1 


 

d. (10 points) Find the dosage level, call it X0, at which the mean response is E(Y0) = 80. Using
the least squares estimate of the model, the estimated dosage is
∧


 Y0 
X 0 = β 1 ∧

 β 0 - Y0 
∧
∧
1 β2



 1
 Y0  
= β 1 exp  ∧ log ∧
  = 5.986 .
β 2
 β 0 - Y0  

∧
Use the delta method to obtain a standard error based on the large sample normal
∧
approximation to the distribution of X 0 . Compute
 ∂X
G =  0
 ∂β 0
∂X 0
∂β1
∂X 0 

∂β 2 

-X 0
= 
 β 2 (β 0 − Y0 )
 Y0  
-X 0
log

2
 β 0 − Y0  
β2
X0
β1
∧
∧
∧
and let G denote G evaluated at β = β . Then, a standard error for X 0 is
S∧
X0
=
∧
∧∧
∧  −1 ∧
σ G  D D


2
T
GT .
Some students estimated the dosage at which E(Y0) = 80β 0 . Full credit was given for this
solution, although the details are not shown here.
2
(e) (8 points) A lack-of-fit test can be performed in the following way. Note that two
responses were obtained at six of the dosage levels. Let Yjk denote the k-th response
obtained for the j-th dosage level. Here j = 1,2,....,13 and k = 1, n j with
n1 = n 2 = n 3 = n10 = n11 = n12 = n13 = 1 and n 4 = n 5 = n 6 = n 7 = n 6 = n 7 = 2.
Following the way Venables & Ripley use approximate F-tests in the least squares analysis
of non-linear models, a lack-of-fit test is obtained by rejecting the proposed model
model A:
Yjk = β 0 -
(
β0
1 + X j β1
)
β2
+ ε jk
in favor of the general alternative (that specifies a separate mean response for each dosage
level)
model B:
Yjk = µ j + ε jk
if
F =
(deviance model A - deviance model B ) / (16 − 6)
> F(10,6), α .
(deviance mod el B ) / (6)
Note that the least squares estimates of the means under model B are simply the sample
means at the individual dosage levels and
9 2 
_ 2
deviance model B = ∑ ∑  Yjk - Y j•  .

j=4 k =1 
Some students mentioned residual plots, but they would provide essentially the same
information provided by the plot on page 3 of the exam. Other students proposed likelihood
ratio tests, but there is no likelihood function because the distribution of the random errors
was not specified. Furthermore, the Pearson chi-square test, that we used for binomial and
Poisson counts, would not necessarily have an approximate chi-square distribution in this
situation, regardless of the sample size.
2. (10 points) Almost everyone received full credit for this problem. The basic steps in computing
a bootstrapped estimate of a standard error are:
1. Use simple random sampling with replacement to select a bootstrap sample of
n=100 values from the original sample.
2. Compute the 20% trimmed mean for the bootstrap sample.
3. Repeated steps 1 and 2 a large number of times (B=200 should be sufficient,
but the computations are quickly done so I would use B=5000)
4. Compute the standard deviation of the B values for the trimmed mean obtained
from the bootstrap samples.
3
3. (a.) (6 points) For (Y1, X1) , (Y2 , X2 ) ,...., (Yn , X n ) , where Yj is a random response and
X j is a vector of fixed explanatory variables, the basic features of a generalized linear
model are:
•
•
Y1 ,...., Yn are independent
The distribution of Yj is a member of the exponential family of distributions
•
There is a smooth link function h( ) such that h(E( Yj |X j )) = X jT β .
(b) (8 points) Use Yj to denote the number of male turtles that emerge from the n j = 20 eggs
incubated at the j-th temperature. Assuming that Yj ~ Bin(n j , π j ) j=1,2,...,6 , and
the observations are independent, the test statistic tests the null hypothesis
H o : log(1 - log(1 - π j )) = β 0 + β1X j
j = 1,2,...,6
against the alternative
Ha : log(1 - log(1 - π j )) = α j ,
j = 1,2,...,6
The alternative places no restrictions on the probabilities other than 0 < π j < 1, for
j=1,2,...,6. This is a lack-of-fit test. For large sample sizes, this test statistic approximately
has a central chi-square distribution with (6-2) = 4 degrees of freedom when the null
hypothesis is true.
(c) (8 points) Like many situations you may encounter as a practicing statistician, the statement of
the objectives in this question is somewhat vague. There are several approaches one could
take. First you must decide how you will use the 120 eggs from the Illinois turtles. Most
students avoided this issue, but those who addressed it decided to replicate the experiment
that was performed on the New Mexico eggs. In that case, 20 of the Illinois eggs are
incubated at each of the six temperatures used for the New Mexico eggs. We will assume
that this experiment was done in remainder of this discussion.
One approach was to compare male hatch rates for the New Mexico and Illinois eggs at the
six temperatures without presupposing any particular relationship with temperature.
Let YNM,j and YI,j denote the number of males that hatch from the n j = 20 New Mexico
eggs and the n j = 20 Illinois eggs, respectively, incubated at the j-th temperature. Under
the null hypothesis of no difference in locations, the mle for the probability that a male
turtle hatches from an egg is p j = (YNM, j + YI, j ) / (2n j ) . Under the general
4
alternative, the mle’s for the probabilities of a male turtle emerging from an egg incubated
at the j-th temperature are simply the observed proportions, p NM,j = (YNM, j / n j ) and
p I,j = ( YI,j / n j ) , respectively. Then, a large sample chi-squared test can be obtained
from the log of the ratio of the likelihoods (for independent binomial counts) multiplied by
-2, i.e.,
6
(
G 2 = 2 ∑ YNM, j log( p NM, j / p j ) + (n j - YNM,j )log((1 − p NM, j ) / (1 − p j ))
j=1
6
(
+ 2 ∑ YI,j log( p I, j / p j ) + (n j - YI, j )log((1 − p I, j ) / (1 − p j ))
j=1
)
)
or the Pearson statistic
X
2
(YNM,j - n jp j )2
= ∑
n jp j
j=1
6
(n j - YNM, j - n j (1 - p j ))2
+ ∑
n j (1 - p j )
j=1
6
(YI,j - n jp j )2
+ ∑
n jp j
j=1
6
(n j - YI,j - n j (1 - p j ))2
+ ∑
n j (1 - p j )
j=1
6
Reject the null hypothesis if the test statistic is larger than the upper α percentile of a central
chi-square distribution with 12-6=6 degrees of freedom.
Another suggestion was to perform six separate chi-square tests, one at each temperature
level, to test if the probabilities of a male are the same for New Mexico and Illinois eggs at
each of the six temperatures used in this study. Either a deviance statistic (G 2 ) or a
Pearson statistic could be used. Some students used a bootstrap procedure to do this. Some
referred to the use of the conservative Bonferroni procedure to control the experimentwise
Type I error level for the set of six tests.
Most students took a more parametric approach. One idea was to fit the complimentary loglog model suggested in this problem to the data for the New Mexico eggs and independently
fit the model to the data for the Illinois eggs, yielding two sets of parameter estimates. Use
the test from part (b) to check if the complimentary log-log model is adequate for both
populations of eggs. If the model is not rejected for either location, test the hypothesis that
the two intercepts are equal and test the null hypothesis that the two slopes are equal. Some
students indicated they would use a Bonferroni procedure to control the Type I error level.
Some students proposed a bootstrap procedure for making confidence intervals for the
difference in the two intercepts and the difference in the two slopes.
5
Another approach is to compare the fit of a nested set of models. (The methods we
previously discussed are also included in this general strategy.) Here, the complimentary
log-log model proposed in the question is fit to the combined data from the two experiments
(call this model A). Then, the parameters in the complimentary log-log model are allowed
to be different for the two locations (call this model B). Large sample chi-squared tests are
provided by the deviance test
6
(
G 2 = 2 ∑ YNM, j log( p NM ( B), j / p A, j ) + (n j - YNM,j )log((1 − p NM( B), j ) / (1 − p A, j ))
j=1
6
(
+ 2 ∑ YI,j log( p I( B), j / p A, j ) + (n j - YI,j )log((1 − p I( B), j ) / (1 − p A, j ))
j=1
)
)
or the Pearson statistic
(n jp NM(B),j - n jp A, j )2
n j p A, j
j=1
6
X2 = ∑
(n j (1 - p NM(B),j ) - n j (1 - p A, j ))2
n j (1 - p A, j )
j=1
6
+ ∑
(n jp I(B),j - n jp A, j )2
n j p A, j
j=1
6
+ ∑
(n j (1 - p I(B), j ) - n j (1 - p A, j ))2
n j (1 - p A, j )
j=1
6
+ ∑
Here, p NM( B), j and p I ( B), j denote the mle’s for the probability that a male emerges from
an egg incubated at the j-th temperature, under model B, for eggs from the New Mexico
and Illinois locations, respectively. Furthermore, p A, j denotes the mle for the probability
that a male emerges form an egg incubated at the j-th temperatue under model A. When
the null hypothesis is true (model A is correct) and the sample sizes are large enough, both
of these test statistics have approximate central chi-square distributions with 4-2 = 2
degrees of freedom.
Credit was given for any of these approaches as long as it was well described and
implemented in a reasonable way. Other suggestions were made by students that are not
reviewed here. Although this is not part of the solution, you should think about the
potential advantages and disadvantages of the procedures outlined above. Would they
necessarily yield the same inferences? When would one approach be preferred to another?
6
4. (a) (8 points) Here there are three rabbits (random effects) and two treatments (fixed effects).
The proposed mode can be expressed as a mixed model:
Y11 
1 0 
Y 
0 1
 12 
 
Y21 
1 0  µ1 

 =   
Y22 
0 1 µ 2 
Y31 
1 0 


 
Y32 
0 1
where
Var(u) = σ 2η I3× 3
1
1

0
+ 
0
0

0
and
0
0
  η1 
0  
η
0   2 
η 
1  3 

1
0
0
1
1
0
0
+
ε11 
ε 
 12 
ε 21 
  = Xβ + Zu + ε
ε 22 
ε 31 
 
ε 32 
Var(ε ) = σ 2ε I 6 × 6 .
(b) (8 points) Since X has full column rank, β is estimable and the unique blue for β is the
_ 
∧
−
Y•1
1
generalized least squares estimator β GLS = X T Σ −1X X T Σ −1Y =  _ 
Y 
 •2 
(
where
)
Σ = σ 2η Z T Z + σ 2ε I 6 × 6 is a block diagonal matrix where each 2x2 block is
σ 2ε + σ 2η

 σ2
η



2
2
σε + σ η 
σ 2η
Given the distributions of the random effects listed in the statement of the problem,
∧

β GLS ~ N β,

(XT Σ −1X)
−1 
.

Note that the inverse of X T Σ −1X exists in this case and it is the unique generalized
inverse. In fact,
(X
T −1
Σ X
)
−1
2
 2
1 σ ε + σ η
=
3  σ2
η

7


2
2
σε + σ η 
σ 2η
(I expected that the generalized least squares estimator would be given for the blue.
Note that in this special case, the generalized least squares estimator is equivalent to
the ordinary least squares estimator. You can easily show this. Hence, the ordinary least
squares estimator is also an acceptable answer proving you correctly specify its covariance
matrix.)
(c) (8 points) An appropriate set of error contrasts would be M( I − PX )Y , where M is a kxn
matrix with row rank equal to k=n-rank(X). In this case, k=6-2=4, and the elements of
_

Y
Y
 11 _ •1
Y12 - Y
_ •2
Y - Y • 1
(I − PX )Y =  21 _
Y22 - Y
_ •2
Y31 - Y•1
_
Y - Y
•2
 32









satisfy two linear constraints. The basic idea of REML is to obtain an estimator of the
variance components with less bias than the mle. In the simple examples we looked at in
class, the mle produced estimators for variance components with negative bias because it
divided a sum of squared deviations by the sample size n instead of the appropriate degrees
of freedom. REML makes an adjustment for degrees of freedom by effectively reducing the
number of observations from the original sample size n to n-rank(X), the number of error
contrasts.
(d.) (6 points) I anticipated (and awarded full credit) for a general statements about the generalized
least squares estimator. In particular, that it may no longer be a linear estimator, or an
unbiased estimator, or a best linear unbiased estimator. For large samples its distribution
would be close to the normal distribution described in part (b).
In this special case, however,
_ 
Y •1
β GLS = X T Σ −1X X T Σ −1Y = X T X X T Y =  _ 
Y 
 •2 
is not a function of the variance components and inserting REML estimates of variance
components into Σ has no effect. Consequently, the generalized least squares estimator
remains a best linear unbiased estimator with the exact finite sample normal distribution
described in part (b). (I did not expect anyone to realize this during the exam.)
∧
(
)
(
−1
8
)
−1