STAT 557
1. (a)
(b)
(c)
(d)
(e)
Solutions to Assignment 1
Fall 2002
Adults in the State of Iowa with telephones
An adult from the population
a binary (yes, no) response
nominal
Since the sample size (n = 825) is fairly large and the point estimate of p is not too extreme (np > 5 and
n(1 p) > 5), the approximation would be reasonable if the random digit dialing approximately produces
a simple random sample from the population of adults with telephones in Iowa. This should be case if
they talk to only one adult on each call. Then, with p = 0:4630, an approximate 95 % condence interval
is (0:429; 0:497): If they talk to every adult present in the household on each call, however, it would be
closer to a one-stage cluster sample than a simple random sample and the number of \yes" responses may
not have a binomial distribution. In that case, using the binomial variance formula in the construct of a
condence interval would not be appropriate.
2. (a) Patients in the national registry of records for individuals who have been diagnosed as suering from a
certain liver disease
(b) a patient from the population
(c) disease severity measured on a ve point scale
(d) ordinal
(e) Since the sample size (n = 825) is fairly large and the point estimate of p is not too extreme (np > 5 and
n(1 p) > 5), the approximation would be reasonable. With p = 0:171, the approximate 95 % condence
interval is (0:124; 0:218):
3. (a) Iowa State students
(b) A student and his or her parents
(c) There are three potentially correlated binary response variables corresponding to right (or left) handedness
for a student and his or her two parents. For each student/parent trio you could think of this as a
multinomial response variable with trio would belong to exactly on the 8 categories, if information on
both parents is available.
(d) nominal
(e) The Pearson chi-square test can be used to compare proportions obtained from independent binomial
random variables. Since the binomial variables in this problem are correlated, the use of the Pearson
chi-square test statistic is not reasonable.
4. (a) Schools in Iowa with sixth grade classes
(b) This is a two stage sampling scheme where schools are sampled from the population of schools in Iowa
with sixth grades students. Then one class is randomly selected from the sixth grade classes in each of the
selected schools. An entire sixth grade class is the unit of response. Since students in a class are inuenced
by the same teacher and they largely share the same educational background, two students from the same
class will tend to respond in a more similar way than two students from dierent schools. To use sixth
grade students as the units of response, correlations among responses from classmates would have to be
taken into account. This could be diÆcult because it would require a model they would allow for dierent
types or strengths of relationships between dierent pairs of classmates.
(c) For each of the 25 animals there are two response variables (each with ve categories) corresponding to
attitudes toward the animal before and after the visit to the wildlife center.
(d) ordinal
1
p
(e) The original data was used to create a binary response for each student (did their attitude toward snakes
become more positive?) As mentioned in part (a), students' response may not be independent and the
total number of positive responses not be the binomial. Thus, p(1 p) may underestimate the variance
of the sample proportion and p (1:96) p(1 p) may produce intervals that are too short to provide
95% condence.
p
5. Let 1 and 2 be the survival rates of the standard treatment and the new treatment, respectively. Then the
testing problem is "H0 : 1 = 2 : HA : 1 < 2 ". The alternative is one-sided because the objective of the
study is to demonstrate that the new treatment is more eective. Here, = 0:05, = 0:20, z = 1:64485,
z = 0:84162, 1 = 0:55, 2 = 0:75.
s
p = (1 + 2 )=2 = 0:65
2p(1 p)
= 1:02273
r =
1 (1 1 ) + 2 (1 2 )
(z + z r)2 (1 (1 1 ) + 2 (1 2 ))
= 69:27
n =
(1 2 )2
Then, 70 subjects are needed in each treatment.
6. (a) Note that p = 4=933 = 0:004287 and np = 4 < 5. Thus a normal approximation would not be reasonable.
The exact condence interval is computed as (0.001169, 0.010940) using the formulas on pages 73-76 of
the lecture notes.
(b) Similarly as in part (a), the normal approximation would not be reasonable. The exact condence interval
is (0.0000585, 0.0017427).
7. (a) The table of counts would have a multinomial distribution if simple random sampling with replacement
was used. Since the population is large, a multinomial distribution would provide a good approximation to
the distribution of the counts if simple random sampling without replacement was used. The log-likelihood
function is
`(; Y) = log(n!)
X2 X2
i=1 j =1
log(Yij !) +
X2 X2
i=1 j =1
Yij log(ij )
(b) Under the independence model ij = i+ +j . Substituting this into the log-likelihood function shown
above, we have
`(; Y) = log(n!)
X2 X2
i=1 j =1
log(Yij !) +
X2
i=1
Yi+ log(i+ ) +
X2
j =1
Y+j log(+j )
Maximizing this log-likelihood with respect to the constraints 1+ + 2+ = 1 and +1 + +2 = 1. The
formula for the maximum likelihood estimates for the expected counts is
Yi+ Y+j
m
^B
;
ij = Y
++
and
i = 1; 2andj = 1; 2
(m
^B
^B
^B
^B
11 ; m
12 ; m
21 ; m
22 ) = (799:5; 220:5; 295:2; 81:5):
(c) Under the general alternative, the mle's of the expected counts will be same as the observations; i.e.,
(m
^ C11 ; m
^ C12 ; m
^ C21 ; m
^ C22 ) = (784; 236; 311; 66):
(d)
G2 = 2
X2 X2 Y
i=1 j =1
ij log
Yij
= 5:321 on 1 d.f. with p-value=.021
m
^B
ij
2
X2 = 2
X2 X2 Y
( ij
i=1 j =1
2
m
^B
ij )
= 5:15 on 1 d.f. with p-value=.023
m
^B
ij
The data suggest that opinions on gun registration are not held independently of opinions on the death
penalty. In particular, people who oppose the death penalty are more likely to favor gun registration than
people who favor the death penalty.
(e) The log-likelihood function is
`(; Y) = log(n!)
X2 X2
log(Yij !) +
= log(n!)
X2 X2
log(Yij !) + Y11 log(2 ) + (Y12 + Y21 ) log((1
= log(n!)
X2
i=1 j =i
i=1 j =i
j =i
X2 X2 Y
i=1 j =i
ij log(ij )
)) + 2Y22 log((1 )2 )
log(Yij !) + (2Y11 + Y12 + Y21 ) log() + (Y12 + Y21 + 2Y22 ) log((1
))
(f) Solve the likelihood equation
0=
@`(; Y) 2Y11 + Y12 + Y21
=
@
Y12 + Y21 + 2Y22
1 The maximum likelihood estimate is
Y +Y
2Y + Y + Y
^ = 11 12 21 = 1+ +2 = 0:757
2Y++
2Y++
(g) Compute m.l.e.'s for expected counts:
2
m
^A
11 = Y++ ^ = 800:5055
m
^A
^A
12 = m
21 = Y++ ^(1 ^) = 256:9944
2
m
^A
22 = Y++ (1 ^) = 82:5055
(h) The deviance statistic is
G2 = 2
X2 X2 Y
^A
ij log(Yij =m
ij ) = 16:2823
i=1 j =i
with 2 d.f. and p-value = .0003. It is not surprising that the data do not support this model because the
independence model was rejected in a previous part of this problem and this model is a restricted form of
the independence model.
(i) An analysis of deviance table
Comparison
Model A vs. Model B
Model B vs. Model C
Model A vs. Model C
d.f.
1
1
2
Deviance
10.962
5.321
16.283
p-value
.0009
.021
.0003
Note that the deviance statistic for \Model A vs. Model B" is given by
G2 = 2
X2 X2 Y
i=1 j =i
^B
^A
ij log(m
ij =m
ij ) = 10:962
Although Model B is a signicant improvement over Model A, neither Model A nor Model B is appropriate
in this case.
3
> #=========================================================================#
> #
Problem 5
#
> #=========================================================================#
> pi1 <- .55
> pi2 <- .75
>
> alpha <- .05
> beta <- .20
> z.alpha <- qnorm( 1 - alpha )
> z.beta <- qnorm( 1 - beta )
>
> p <- (pi1+pi2)/2
> r <- sqrt( 2*p*(1-p)/(pi1*(1-pi1)+pi2*(1-pi2)) )
> n <- (z.beta+z.alpha*r)^2*(pi1*(1-pi1)+pi2*(1-pi2))/(pi1-pi2)^2
> c(p,r,n)
[1] 0.65000 1.02273 69.27248
>
> #=========================================================================#
> #
Problem 6
#
> #=========================================================================#
>
#The following function uses F-distribution to construct a
>
#100*a% "exact" confidence interval for the probability of
>
#success in a binomial distribution;
>
> binci.f<-function(x, n, a){
+
# x = observed number of successes; +
# n = number of trials;
+
# a = level of confidence (e.g. 0.95);
+
p <- x/n
+
a2 <- 1-((1-a)/2)
+
if (x > 0)
+
f1 <- qf(a2, 2*(n-x+1), 2*x)
+
else
+
f1 <- 1
+
if (n > x)
+
f2 <- qf(a2, 2*(x+1), 2*(n-x))
+
else
+
f2 <- 1
+
plower <- x / (x + (n-x+1)*f1)
+
pupper <- (x+1)* f2 / ((n-x)+(x+1)*f2)
+
cbind( plower=plower, pupper=pupper )
+ }
> binci.f(x=4, n=933, a=0.95)
plower
pupper
[1,] 0.001169328 0.01094038
> binci.f(x=2, n=4143, a=0.95)
plower
pupper
[1,] 0.00005846764 0.001742731
>
>
> #=========================================================================#
> #
Problem 7
#
> #=========================================================================#
> #-----------------------------------------------------------------------(b)
> Y <- matrix( c( 784, 236, 311, 66), 2, 2, byrow=T )
> dimnames(Y) <- list( c("GR.Favor","GR.Oppose"), c("DP.Favor","DP.Oppose") )
> Y
DP.Favor DP.Oppose
GR.Favor
784
236
GR.Oppose
311
66
>
> row.total <- Y["GR.Favor",] + Y["GR.Oppose",]
> col.total <- Y[,"DP.Favor"] + Y[,"DP.Oppose"]
> total <- sum(Y)
>
#------------------------------------#
>
# Expected counts; Independence Model#
>
#------------------------------------#
> mB.hat <- (col.total %o% row.total) / total
> mB.hat
DP.Favor DP.Oppose
GR.Favor 799.4989 220.50107
GR.Oppose 295.5011 81.49893
> #------------------------------------------------------------------------(d)
>
#------------------------------------#
>
# Pearson X^2 Test: ModelB vs ModelC #
>
#------------------------------------#
> chisq.test(Y, correct=F)
Pearson's chi-square test without Yates' continuity correction
data: Y X-square = 5.1503, df = 1, p-value = 0.0232
>
# See (i) for deviance G^2 test
4
#
> #------------------------------------------------------------------------(f)
>
#------------------------------------#
>
# MLE of theta
#
>
#------------------------------------#
> theta <- (2 * Y[1,1] + Y[1,2] + Y[2,1]) / ( 2 * sum(Y) )
> theta
[1] 0.7569792
> #------------------------------------------------------------------------(g)
>
#------------------------------------#
>
# Expected counts by Model A
#
>
#------------------------------------#
> piA.hat <- matrix( c(theta^2, theta*(1-theta),
+
theta*(1-theta), (1-theta)^2 ),
2, 2 )
> mA.hat <- total * piA.hat
> mA.hat
[,1]
[,2]
[1,] 800.5055 256.99445
[2,] 256.9945 82.50555
> #------------------------------------------------------------------------(i)
> table2 <- matrix(0, 3, 3)
> dimnames(table2) <- list( c( "A vs B", "B vs C", "A vs C"),
+
c("df", "deviance", "pvalue") )
>
#------------------------------------#
>
#
MODEL A vs MODEL B
#
>
#------------------------------------#
> table2["A vs B", "deviance"] <- 2 * sum( Y * log( mB.hat/mA.hat ) )
> table2["A vs B", "df"]
<- 1
> table2["A vs B", "pvalue"] <- 1 - pchisq(table2["A vs B", "deviance"],
+
table2["A vs B", "df"] )
>
#------------------------------------#
>
#
MODEL B vs MODEL C
#
>
#------------------------------------#
> table2["B vs C", "deviance"] <- 2 * sum( Y * log( Y / mB.hat ) )
> table2["B vs C", "df"]
<- 1
> table2["B vs C", "pvalue"] <- 1 - pchisq(table2["B vs C", "deviance"],
+
table2["B vs C", "df"] )
>
#------------------------------------#
>
#
MODEL A vs MODEL C
#
>
#------------------------------------#
> table2["A vs C", "deviance"] <- 2 * sum( Y * log( Y /mA.hat ) )
> table2["A vs C", "df"]
<- 2
> table2["A vs C", "pvalue"] <- 1 - pchisq(table2["A vs C", "deviance"],
+
table2["A vs C", "df"] )
>
#------------------------------------#
>
# Analysis of Deviance Table
#
>
#------------------------------------#
> table2
df deviance
pvalue
A vs B 1 10.96130 0.0009303443
B vs C 1 5.32065 0.0210741513
A vs C 2 16.28195 0.0002913527
5