Stat 665 (Spring 2011)
Kaizar
Solutions 1
Exercises
40 points total
1. (6 points) Agresti problem 1.8 (use both the score and exact methods – don’t use Wald!)
(a)
ˆ
pi
=
y
n
=
344
1170
= 0
.
294.
(b) The wording of the question is unclear as to the null hypothesis, so I am choosing the
hypotheses:
H
0
:
π
≥
0
.
5
,
H
a
:
π <
0
.
5
•
Score test:
z
=
ˆ
π
−
π
0
radicalBig
π
o
(1

π
0
)
n
=
0
.
294
−
0
.
5
radicalBig
0
.
5(1

0
.
5)
1170
=
−
14
.
09
The pvalue is
P
(
Z
≤ −
14
.
09), where
Z
is N(0,1). Using R, I find this to be
2
.
1
×
10

45
.
I reject the null hypothesis and conclude that a minority would claim to be willing
to accept cuts in their standard of living to protect the environment.
•
Exact test: Using R, I find a pvalue
<
2
.
2
×
10

16
. Again, I reject the null hypothesis
and conclude that a minority would claim to be willing to accept cuts in their standard
of living to protect the environment.
(c)
•
Score interval: From R, I find the interval: (0.26, 0.33)
I am approximately 99% confident that the true proportion of people would claim to
be willing to accept cuts in their standard of living to protect the environment.
•
Exact interval: From R, I find the interval: (0.26, 0.33)
I am at least 99% confident that the true proportion of people would claim to be
willing to accept cuts in their standard of living to protect the environment.
Note that for a true proportion near 0.3, the sample size (1170) appears to be large
enough that the two tests seem to agree well.
2. (8 points) Agresti problem 1.12
•
In this case,
ˆ
pi
= 0. Therefore, the estimated standard error is also 0, which makes the
test statistic z infinite. Obviously this is not a useful test statistic.
•
Since the estimated standard error is zero, the 95% interval is simply the estimate: 0.
That is, according to the Wald procedure, I am 95% confident that the true proportion
of intro stats students who are vegetarian is exactly zero.
•
The score test is:
z
=
ˆ
π
−
π
0
radicalBig
π
o
(1

π
0
)
n
=
0
−
0
.
5
radicalBig
0
.
5(1

0
.
5)
25
=
−
5
Using R, I find the pvalue to be 5
.
7
×
10

7
.
1
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Based on this test, I reject the null hypothesis that half of intro stats students are vege
tarian.
•
Note that a 95% confidence interval is an inversion of a test. That is, the interval is the
collection of values of
π
for which the score test would not reject the null hypothesis.
I would reject the test whenever
z >
1
.
96 and whenever
z <
−
1
.
96.
Notice that as I
increase the hypothesized
π
0
, the value of z decreases (since I’m subtracting a bigger and
bigger value). This is not linear, since the denominator is also changing. I’ve plotted the
relationship between
π
0
and z below, where the dotted lines are at the edges of the rejec
tion region.
0.0
0.4
0.8
60
40
20
0
all possible pi
pi_0
z
0.00
0.10
0.20
3
2
1
0
1
2
interesting range of pi
pi_0[1:50]
z[1:50]
Visually, you can see that the confidence interval should go from the smallest possible
value (zero) to a value at about 0.13. When I actually calculate the value of the score
statistic (z) when
π
0
= 0
.
133, I get:
z
=
ˆ
π
−
π
0
radicalBig
π
o
(1

π
0
)
n
=
0
−
0
.
133
radicalBig
0
.
133(1

0
.
133)
25
=
−
1
.
96
which confirms that 0.133 is the edge of the rejection region.
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 Spring '10
 Staff
 Null hypothesis, Statistical hypothesis testing, Statistical tests, Likelihood function, Likelihoodratio test

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