Download HW #7 Solutions

Math 4650/5820 - Spring 2010
HW #7 Solutions
Problem #2
a. This is a simple hypothesis because the uniform distribution on [0,1]
is completely specified.
b. This is a simple hypothesis because the unbiasedness of a die completely specifies the probability distribution of a die.
c. This is a composite hypothesis because µ = 0 and σ 2 > 10 do not
completely specify the normal distribution.
d. This is a composite hypothesis because µ = 0 does not completely
specify the normal distribution.
Problem #5
a) False. The significance level of a test is the probability of rejecting H0
when it is true, which is not equal to the probability that the null hypothesis
is true.
b) False. If the significance level of a test is decreased, the probability of
a type II error would in general increase, thus the power would be expected
to decrease.
c) False. If a test is rejected at the significance level α, then the probability
of making a type I error is α, which is not the probability that the null
hypothesis is true.
d) False. The probability that the null hypothesis is falsely rejected is in
general not equal to the power of the test (most often, it is smaller than the
power of the test).
e) False. A type I error occurs when the null hypothesis is true, but we
reject it. This is different from the statement that the test statistic falls in
the rejection region of the test.
f) False. A type I error is in general more serious than Type II error.
g) False. The power of a test is determined by the alternative distribution
of the test statistic.
h) True. the likelihood ratio is a random variable.
Problem #9
To find the rejection region, we use the likelihood ratio test. It follows
from some algebra that
f0 (X)
<C
f1 (X)
if and only if
X̄ > C.
Therefore, when α = 0.10, we obtain
P (X̄ > C|H0 ) = P (Z >
C
√ ) = 0.10
σ/ n
Plugging n = 25 and σ = 10 into the above expression and using Table 2 of
the Appendix, we have
√
C
C 25
),
0.10 = P (Z > √ ) = 1 − Φ(
σ/ n
10
and
√
C 25
= 1.28
10
Therefore, C = 2.56 and the rejection region is given by {(X1 , · · · , X2 5) :
X̄ > 2.56}. The power of the test is given by
2.56 − 1.5
X̄ − 1.5
√ >
√
|Ha )
σ/ n
σ/ n
= P (Z > 0.53) = 1 − 0.7019 = 0.2981.
P ower = P (X̄ > 2.56|Ha ) = P (
By the same argument, when α = 0.01, we obtain the rejection region
{(X1 , · · · , X2 5) : X̄ > 4.66}, and the power= P (Z > 1.58) = 1 − 0.9429 =
0.0571.
Problem #11
This is a two-sided alternative. The generalized likelihood ratio test re√ > z(α/2) (as given in Example A of section 9.4 of
jects H0 if and only if σ/|X̄|
n
our text). For α = 0.1, we have z(α/2) = 1.645. Plugging n = 25 and σ = 10
in the above test, we see that our test rejects H0 if and only if |X̄| > 3.29.
The power of the test is
P ower = P (|X̄| > 3.29|Ha) = 1 − P (|X̄| ≤ 3.29|Ha)
3.29 − µ
−3.29 − µ
√
√ )
≤Z≤
= 1 − P(
σ/ n
σ/ n
−3.29 − µ
3.29 − µ
= 1 − P(
≤Z≤
)
2
2
−3.29 − µ
3.29 − µ
) − Φ(
)]
= 1 − [Φ(
2
2
In the case of n = 100, we obtain
P ower = P (|X̄| > 1.645|Ha ) = 1 − P (|X̄| ≤ 1.645|Ha )
1.645 − µ
−1.645 − µ
√
√ )
≤Z≤
= 1 − P(
σ/ n
σ/ n
= 1 − P (−1.645 − µ ≤ Z ≤ 1.645 − µ)
= 1 − [Φ(1.645 − µ) − Φ(−1.645 − µ)].
Similarly for α = 0.05, we have z(0.05/2) = 1.96, and
P ower = P (|X̄| > 3.92|Ha ) = 1 − P (|X̄| ≤ 3.92|Ha )
−3.92 − µ
3.92 − µ
) − Φ(
)]
= 1 − [Φ(
2
2
For n = 100, we have
P ower = P (|X̄| > 1.96|Ha ) = 1 − P (|X̄| ≤ 1.96|Ha )
= 1 − [Φ(1.96 − µ) − Φ(−1.96 − µ)]
From the graphs of the power function, we see that the powers for n = 100
(dotted line) are larger than the powers for n = 25 for both cases of α = 0.1
and α = 0.05, whereas for fixed sample sizes, the powers for α = 0.05 are
smaller than the powers for α = 0.1.
0.7
0.6
Power, (α =0.1)
0.5
0.4
0.3
0.2
0.1
0
−2
−1.5
−1
−0.5
0
µ
0.5
1
1.5
2
−1.5
−1
−0.5
0
µ
0.5
1
1.5
2
0.7
0.6
Power, (α =0.05)
0.5
0.4
0.3
0.2
0.1
0
−2