Download Solutions for Homework 4

U.C. Berkeley — Stat 135 : Concepts of Statistics
Professor: Noureddine El Karoui
Homework 4
Due March 1 2007
Solutions for Homework 4
Problems
Chapter 4: 17, 18
Chapter 9: 3, 4, 5, 9, 10, 11, 20
Problem 4.17
Referring to Example C in Section 3.7, we see that the density function of the kth order statistic is given by
n!
xk−1 (1 − x)n−k .
(k − 1)!(n − k)!
So we have X(k) ∼ Beta(k, n − k + 1)
So the integrations have already been done for us:
k
n+1
k(n − k + 1)
Var(X(k) ) =
(n + 1)2 (n + 2)
E(X(k) ) =
Problem 4.18
Assume that X1 , ..., Xn are iid from a uniform distribution on (0,1).
Assume that Z1 , ..., Zn are iid from a uniform distribution on (a,b).
Then Zi = a + (b − a)Xi in distribution.
So Z(i) = a + (b − a)X(i) in distribution. Therefore
E(Z(n) − Z(1) ) = EZ(n) − EZ(1) = (b − a)(EX(n) − EX(1) ) = (b − a)(EX(n) − EX(1) ) = (b − a)(
n−1
)
n+1
Where the last step follows from the previous problem.
Problem 9.3
α is the probability that the test rejects the null hypothesis when it is true. Under H0 , X is Bin(100, .5),
which is approximately N (50, 25). Thus,
α = Pr0 (|X − 50| > 10) = Pr0 (|X − 50| > 10) = Pr0 (|X − 50|/5 > 2) ≈ 2Φ(−2) ≈ .046
1
Power is the probability that the test rejects the null hypothesis when it is false (1 minus the probability β
of a Type II error). Under Ha , X is Bin(100, p), which is approximately N (100p, 100p(1 − p)). Thus, the
power is
1 − β = PrA (|X − 50| > 10)
= 1 − PrA (−10 ≤ X − 50 ≤ 10)
= 1 − PrA (40 ≤ X ≤ 60)
curve( 1 - ( pnorm(60, mean=100*x, sd=sqrt(100*x*(1-x))) pnorm(40, mean=100*x, sd=sqrt(100*x*(1-x))) ),
xlab=’p’, ylab=’Power’ )
0.6
0.2
0.4
Power
0.8
1.0
We are lazy and let R standardize X for us.
0.0
0.2
0.4
0.6
0.8
1.0
p
Problem 9.4
The likelihood ratio is the likelihood of xi under H0 divided by the likelihood under H0 , i.e. Λ = Pr0 (X =
xi )/PrA (X = xi ). Its distributions are given in the table below.
X
x4
x2
x1
x3
H0
.2
.3
.2
.3
HA
.4
.4
.1
.1
Λ
1/2
3/4
2
3
Following the Neyman-Pearson paradigm, our test will reject H0 in favor of HA for small values of Λ. To
construct a level α test we need find a critical value c such that the probability of a Type I error
Pr0 (Λ ≤ c) = α
2
For α = .2 we see that the critical value c should be c = 1/2 because
Pr0 (Λ ≤ 1/2) = .2
Similarly, for α = .5 the critical value is c = 3/4.
Problem 9.5
a
False. The null hypothesis is either true or false. The significance level is the probability that the test
incorrectly rejects the null hypothesis when it is true.
b
False. Decreasing the significance level of a test makes it less likely to reject the null hypothesis. The power
would be expected to decrease.
c
False. See (a).
d
False. That probability that the null hypothesis is falsely rejected is the significance level.
e
False. A type I error occurs when the test statistics falls in the rejection region of the test AND the null
hypothesis is true.
f
False. A type I error is more serious than a type II error.
g
False. The power of a test is the probability that the null hypothesis is rejected when it is false. This
probability is determined by the alternative distribution.
h
True. The likelihood ratio depends on the outcome of an experiment.
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Problem 9.9
Suppose X1 , ..., X25 ∼ N (µ, 100) are iid.
H0 : µ = µ0 = 0
H1 : µ = µA = 1.5
Since µA > µ0 we know from Example A on page 333 that the Likelihood Ratio is small when X̄ is big. This
implies that we will reject our null hypothesis when X̄ > c for some constant c.
Under the null, X̄ ∼ N (µ, 100
25 ) So X̄ = 2Z in distribution where Z is standard normal.
Pr0 (Reject Null) = Pr0 (X̄ > c) = Pr(Z >
c
c
) = 1 − Φ( )
2
2
Now define cα by α = 1 − Φ( c2α ), so we have an alpha level test.
Let β = PrA (Accept Null) = PrA (X̄ < c)
Under the alternative, Xi ∼ N (1.5, 100) so X̄ ∼ N (1.5, 4). This implies that X̄ = 1.5 + 2Z in distribution
where Z is standard normal. Therefore,
β = Pr(1.5 + 2Z < c) = Pr(Z <
c − 1.5
)
2
P OW ER(cα ) = 1 − βα = 1 − Φ(
cα − 1.5
)
2
Now looking at the normal table and plugging we see c.1 = 2.56 has power .2981 and c.01 = 4.66 has power
.0571.
Problem 9.10
By the factorization theorem, we can write f (x1 , ..., xn |θ) = g(t, θ) ∗ l(x1 , ..., xn ).
So the likelihood ratio can be written as LR(x1 , ..., xn ) =
g(T,θ0 )
g(T,θ1 )
Let h(T ) be the function on the RHS above. Then the likelihood ratio test would reject the null for h(T ) < c
where c is chosen according to Pr0 (h(T ) < c) = α.
If we know the distribution of T under the null, we may be able to calculate the distribution of h(T ) under
the null as well. However, if h is a complicated function, it may be difficult. Instead, one might simulate M
random variables with distribution T, calculate h(T) for each of them, then look at the proportion less than
c for various values of c to pick the right significance level. If you made M very large, you could make this
very accurate no matter how complicated the function h.
Once you know the value of c, you look for the values of T such that h(T) is less than c. This is your rejection
region for T.
Problem 9.11
As in Example 9.4A, the rejection region for an α level test of this type is given by
α
|X̄| ≥ 2z( )
2
4
Under the alternative with parameter µ, X̄ ∼ N (µ, 4), so X̄ = µ + 2Z in distribution where Z is a standard
normal. Therefore,
α α µ
α
µ
α
β(µ) = Prmu |X̄| ≤ 2z( ) = Pr |µ + 2Z| ≤ 2z( ) = Pr − − z( ) ≤ Z ≤ − + z( )
2
2
2
2
2
2
We are asked to plot the power (1 − βµ ) for significance levels α = .1, .05. Looking at our normal table, we
.05
see that z( .1
2 ) = 1.65 and z( 2 ) = 1.96
After that we are asked to increase our sample size. In that case, X̄ = µ + Z in distribution.
Here is the code and the graphic:
a = seq(-10,10,.05)
par(mfrow=c(2,1))
plot( a, 1 - pnorm( -a/2 + 1.96, 0 , 1 ) + pnorm( -a/2 - 1.96 , 0 , 1), main="Sample size 25", type
lines( a, 1 - pnorm( -a/2 + 1.65, 0 , 1 ) + pnorm( -a/2 - 1.65 , 0 , 1), lty=2 )
legend(x=0,y=.6, legend = c(".05 level", ".10 level"), lty=c(1,2))
plot( a, 1 - pnorm( -a + 2*1.96, 0 , 1 ) + pnorm( -a - 2*1.96 , 0 , 1), main="Sample size 100", type
lines( a, 1 - pnorm( -a + 2*1.65, 0 , 1 ) + pnorm( -a - 2*1.65 , 0 , 1), lty=2 )
legend(x=0,y=.6, legend = c(".05 level", ".10 level"), lty=c(1,2))
0.8
0.6
0.4
.05 level
.10 level
0.2
1 − pnorm(−a/2 + 1.96, 0, 1) + pnorm(−a/2 − 1.96, 0, 1)
1.0
Sample size 25
−10
−5
0
5
10
5
10
a
0.8
0.6
0.2
0.4
.05 level
.10 level
0.0
1 − pnorm(−a + 2 * 1.96, 0, 1) + pnorm(−a − 2 * 1.96, 0, 1)
1.0
Sample size 100
−10
−5
0
a
5
Problem 9.20
If x and c are positive, Then
1
2x
< c iff
1
2c
< x. So, to find c for a level α test, we note that
Pr0 (Likelihood Ratio < c) = Pr0 (
Some algebra implies that the cα =
1
2(1−α) .
1
1
1
< c) = Pr0 ( < x) = 1 −
2x
2c
2c
For our case, c = .9.
Now, the power of this test is
Z
.9
2xdx = 1 − .81 = .19
1 − Pr1 (Don’t Reject) = 1 − Pr1 (X < .9) = 1 −
0
The Neyman-Person lemma tells us that no other test can do better.
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