TUTORIAL 7 SOLUTIONS
#9.11.2 Which of the following hypotheses are simple, and which are composite?
a. X follows a uniform distribution on [0, 1].
b. A die is unbiased.
c. X follows a normal distribution with mean
0 and variance σ 2 > 10.
d. X follows a normal distribution with mean
µ = 0.
Solution
a. This is a simple hypothesis.
b. This is another simple hypothesis.
c. This is a composite hypothesis.
d. This is another composite hypothesis
as the variance is unknown.
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#9.11.6 Consider the coin tossing example of Section 9.1. Suppose that instead of
tossing the coin 10 times, the coin was tossed
until a head came up and the total number
of tosses, X, was recorded.
a. If the prior probabilities are equal, which
outcomes favor H0 and which outcomes
favor H1?
b. Suppose P (H0)/P (H1) = 10. What outcomes favor H0?
c. What is the significance level of a test
that rejects H0 if X ≥ 8?
d. What is the power of this test?
Solution
From Chapter 9.1, we have
H0 : probability of heads is 0.5 versus
H1 : probability of heads is 0.7.
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The following is a Bayesian hypothesis testing approach (in contrast to the NeymanPearson approach).
Let P (H0) be the prior probability that
H0 is true and P (H1) be the prior probability that H1 is true. Clearly
P (H0) + P (H1) = 1.
After observing the number of heads, X, in
10 tosses of a coin, the posterior probability
that H0 is true is given by
P (H0, x)
P (H0|x) =
P (x)
P (x|H0)P (H0)
.
=
P (x)
Here
P (x) = P (x|H0)P (H0) + P (x|H1)P (H1).
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Likewise, the posterior probability that H1
is true is given by
P (H1, x)
P (H1|x) =
P (x)
P (x|H1)P (H1)
=
.
P (x)
This implies that we reject H0 if
P (H0|x) P (H0) P (x|H0)
=
< 1,
P (H1|x) P (H1) P (x|H1)
and accept H0 if
P (H0|x) P (H0) P (x|H0)
=
> 1.
P (H1|x) P (H1) P (x|H1)
Note In the Bayesian hypothesis testing
approach, there is a symmetry between the
null hypothesis H0 and the alternative hypothesis H1.
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Note that X has the geometric distribution with probability of heads equals p where
p = 0.5 under H0 and p = 0.7 under H1.
The pdf of X is
P (X = k) = (1 − p)k−1p, k = 1, 2, . . .
a. If P (H0) = P (H1) = 0.5, then
P (H1|x) P (x|H1)
=
.
P (H0|x) P (x|H0)
We observe that
x
1
2
3
4
P (x|H1)
P (x|H0)
1.4
x
6
5
0.84 0.504 0.302 0.181
7
8
9
10
P (x|H1)
0.109 0.065 0.039 0.024 0.014
P (x|H0)
P (x|H )
Since P (x|H1) > 1 if and only if x ≤ 1, we
0
conclude that X ≤ 1 heads would favor H1
while X ≥ 2 heads would favor H0.
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b. If P (H0)/P (H1) = 10, then we would
reject H0 if and only if
P (H1) P (x|H1)
P (H1|x)
=
P (H0|x)
P (H0) P (x|H0)
P (x|H1)
=
> 1.
10P (x|H0)
This is equivalent to rejecting H0 if and only
if
P (x|H1)
> 10.
P (x|H0)
From the table of ratios of posterior probabilities, we shall always accept H0.
c. The significance level of the test that
rejects H0 when X ≥ 8 is
α = P (X ≥ 8|H0)
∞
X
(0.5)i
=
i=8
= 0.0078.
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d. The power of the test in c. is given by
power = P (RejectH0|H1)
= P (X ≥ 8|H1)
∞
X
(0.3)i
= (0.7)
i=7
= 0.0002187.
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#9.11.10 Suppose that X1, . . . , Xn form
a random sample from a density function,
f (x|θ), for which T is a sufficient statistic
for θ. Show that the likelihood ratio test of
H0 : θ = θ0 versus H1 : θ = θ1
is a function of T . Explain how, if the distribution of T is known under H0, the rejection
region of the test may be chosen so that the
test has level α.
Solution The likelihood ratio test rejects
H0 if
f (x|θ0)
<c
f (x|θ1)
for some constant c. Since T is a sufficient
statistic, by the factorization theorem in Chapter 8.8.1, we have
f (x|θ0) g[T (x), θ0]h(x) g[T (x), θ0]
=
=
.
f (x|θ1) g[T (x), θ1]h(x) g[T (x), θ1]
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This implies that the likelihood ratio test
is a function of T .
Suppose now that the distribution of T is
known under H0 and is, say, T ∼ F .
We further assume that g[T (x), θ0]/g[T (x), θ1]
is a strictly increasing function of T . Then
g[T (x), θ0]
<c
g[T (x), θ1]
if and only if T < c0 for some constant c0.
Then at level α, the critical constant can
be chosen to be c0 = F (1 − α) where F (1 −
α) is the 100αth percentile of F since under
H0 ,
P (T < F (1 − α)) = α.
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#9.11.17 Let X ∼ N (0, σ 2) and consider testing H0 : σ = σ0 versus HA : σ =
σ1, where σ1 > σ0. The values σ0 and σ1
are fixed.
a. What is the likelihood ratio as a function
of x? What values favor H0? What is
the rejection region of a level α test?
b. For a sample, X1, X2, . . . , Xn, distributed
as above, repeat the previous question.
c. Is the test in the previous question uniformly most powerful for testing H0 : σ =
σ0 versus H1 : σ > σ0?
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Solution
a. The likelihood ratio is given by
f (x|σ0)
=
f (x|σ1)
2/(2σ 2 )
1
−x
0
e
2
1/2
(2πσ0 )
2/(2σ 2 )
−x
1
1
e
2
1/2
(2πσ1 )
σ1
x2 1
1
=
exp[ ( 2 − 2 )].
σ0
2 σ1 σ0
Since σ1 > σ0, small values of x2 would
favor H0. I.e., we reject H0 if X 2 > c for
some constant c.
Note that under H0, X 2/σ02 ∼ χ21 distribution.
Consequently at significance level α, we reject H0 if
X 2 > σ02χ21(α)
where χ21(α) is the 100(1 − α)th percentile
of the χ21 distribution.
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b. For a sample, X1, X2, . . . , Xn, distributed as above, the likelihood ratio is
n
Y
f (xi|σ0)
i=1
=
f (xi|σ1)
2/(2σ 2 )
1
−x
0
e i
n
2
Y
1/2
(2πσ0 )
2/(2σ 2 )
−x
1
1
i=1 (2πσ 2)1/2 e i
1 P
n
2
x
1
σ1 n
i=1 i 1
( 2 − 2 )].
= ( ) exp[
σ0
2
σ1 σ0
Thus the likelihood
Pn ratio2 test rejects H0
for large values of i=1 Xi . I.e., reject H0
if
n
X
Xi2 > c
i=1
for some constant c.
Note that under H0,
distribution.
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Pn
2 /σ 2 ∼ χ2
X
n
i=1 i 0
Consequently at significance level α, we reject H0 if
n
X
Xi2 > σ02χ2n(α)
i=1
where χ2n(α) is the 100(1 − α)th percentile
of the χ2n distribution.
c. Yes, the test in the previous question
is uniformly most powerful for testing H0 :
σ = σ0 versus H1 : σ > σ0 because the test
statistic as well as the critical value do not
depend on σ1 > σ0.
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