STAT 409
Fall 2015
A. Stepanov
Homework #9
( due Friday, November 20, by 4:00 p.m. )
Please include your name ( with your last name underlined ), your NetID,
and your discussion section number at the top of the first page.
No credit will be given without supporting work.
1 – 2.
Let X have a Binomial distribution with the number of trials n = 15 and with
probability of “success” p. We wish to test H 0 : p = 0.30 vs. H 1 : p ≠ 0.30.
ˆ
Recall that p =
1.
a)
X
n
Find the values of
is the maximum likelihood estimator of p.
Λ(x) =
L( p 0 = 0.30 ; x )
ˆ
L( p; x )
for x = 0, 1, 2, … , 14, 15.
If you are using a computer, include the printout.
b)
2.
Suppose we observe X = 7. Find the p-value of this test.
c)
Likelihood Ratio Test:
Reject H 0 if Λ ( x ) ≤ c.
Let c = 0.15. Find
(i)
the significance level,
(ii)
power when p = 0.20,
(iii)
power when p = 0.40
for the corresponding rejection region.
3.
Let X have a Binomial distribution with the number of trials n = 15 and with
probability of “success” p. We wish to test H 0 : p = 0.30 vs. H 1 : p > 0.30.
a)
Find the best Rejection Rule with the significance level α closest to 0.05.
b)
Find the power of the test from part (a) at p = 0.40 and at p = 0.50.
c)
Suppose we observe X = 7. Find the p-value of this test.
4 – 6.
Let λ > 0 and let X 1 , X 2 , … , X n be independent random variables, each with
the probability density function
f (x; λ ) =
λ
x λ +1
x ≥ 1.
,
n
Recall
Y=
∑ ln X i
is a sufficient statistic for λ ,
∑ ln X i
has a Gamma distribution with α = n and θ = 1 λ ,
i =1
n
Y=
/
i =1
n
ˆ
the maximum likelihood estimator of λ is λ =
n
=
∑ ln X i
n
Y
.
i =1
4.
We wish to test H 0 : λ = 3 vs. H 1 : λ ≠ 3.
a)
Likelihood Ratio Test:
Reject H 0 if Λ ( y ) ≤ c.
Show that Λ ( y ) ≤ c is equivalent to Y
b)
Let k = e
–3
.
5.
c)
n –3Y
e
e
≤ k.
Then Y = 1 is one ( obvious ) solution of Y
Suppose n = 7.
Y
n –3Y
n –3Y
e
=e
–3
.
Find ( to the fourth decimal place ) the value of d such that
≤ e– 3
⇔
Y ≤ 1 or Y ≥ d
Find
(i)
the significance level,
(ii)
power when λ = 2,
(iii)
power when λ = 4
for the rejection region in part (b).
6.
We wish to test H 0 : λ = 3 vs. H 1 : λ < 3.
a)
Suppose n = 7. Find the uniformly most powerful rejection region with a 5%
level of significance. Round the critical value for Y to the fourth decimal place.
b)
Find the power of the test in part (a) when λ = 2 and when λ = 1.
7 – 8.
7.
Crosses of mice will produce either gray, brown, or albino offspring. Mendel’s
model predicts that the probability of a gray offspring is 9/16 , the probability
of a brown offspring is 3/16 , and the probability of an albino offspring is 4/16 .
An experiment to assess the validity of Mendel’s theory produces the following data:
36 gray offspring; 21 brown offspring; and 23 albino offspring. Test
H 0 : p 1 = 9/16 , p 2 = 3/16 , p 3 = 4/16
a)
8.
at α = 0.05,
b)
vs
H 1 : not H 0
at α = 0.10.
Suppose the experiment in Problem 7 is repeated, but with twice as many observations.
Suppose also that we happened to get the same proportions, namely, 72 gray offspring;
42 brown offspring; and 46 albino offspring. Repeat Problem 7 in this case, using
α = 0.01.
EXCEL functions you may find helpful:
= BINOMDIST( x , n , p , 0 )
gives
P( X = x )
= BINOMDIST( x , n , p , 1 )
gives
P( X ≤ x )
= CHIINV ( α , v )
gives
2
χ α (v ) for χ 2 distribution with v degrees
of freedom, x s.t. P ( χ 2 (v ) > x ) = α.
= CHIDIST ( x , v )
gives
χ 2 distribution
with v degrees of freedom, P ( χ 2 (v ) > x ).
the upper tail probability for
= POISSON( x , λ , 0 )
gives
P( X = x )
= POISSON( x , λ , 1 )
gives
P( X ≤ x )
