Download Exam 2

Exam 2
STA 4322
Mathematical Statistics II
Fall 2000
Name:
KEY
SSN:
Instructions: Unless otherwise indicated, you must show your work to receive credit on
this exam. Circle your final answers. On problems which say SET UP ONLY, you should
write the answer in its final form, having filled in all of the numbers required to complete
the calculation, but you need not finish the numerical calculation.
1. Let Y1 , . . . , Yn denote a random sample from a population with density
f (y) = θy θ−1 ,
0 ≤ y ≤ 1,
where θ > 0. Find the method-of-moments estimator of θ.
E(Y ) =
Z
1
y θy
0
θ−1
dy = θ
Z
0
1
y θ dy = θ
1
1
θ
y θ+1 =
.
θ+1
θ+1
y=0
θ
Y
= Y =⇒ θ = Y (θ + 1) = Y θ + Y =⇒ θ(1 − Y ) = Y =⇒ θ =
θ+1
1−Y
θ̂MME =
Y
1−Y
1
(12 pts)
STA 4322 Exam 2
2
2. Suppose that (Y1 , . . . , Yn ) denote a random sample from the exponential distribution with
density
1
f (y) = e−y/θ ,
θ
y > 0.
Note that E(Yi ) = θ and Var(Yi ) = θ2 .
Show that
Y −θ
√
Y/ n
converges in distribution to a standard normal random variable. Give reasons for any claims
you make, including the names (or abbreviations) of any “named theorems” that you use.
(12 pts)
Since
µ = E(Yi ) = θ
σ 2 = V (Yi ) = θ2 < ∞,
and
the central limit theorem (CLT) applies to give
√
√
n Y −µ
n Y −θ d
=
−→ N (0, 1).
σ
θ
(∗)
Since
E(Yi ) = θ
and
V (Yi ) = θ2 < ∞,
the weak law of large numbers (WLLN) implies that
P
Y → θ =⇒
Y P θ
→ = 1.
θ
θ
Now by Slutsky’s theorem, (∗) and (†) together imply that
√
n Y −θ
Y −θ d
θ
√ −→ N (0, 1).
=
Y
Y/ n
θ
(†)
STA 4322 Exam 2
3
3. Let Y1 , . . . , Yn be a random sample of n observations from a population with density function
f (y) =
1 2 −y/β
y e
,
2β 3
y ≥ 0.
Note that this is a gamma density with α = 3.
(a) Find a sufficient statistic for β.
(10 pts)
!
( n
)
n n
Y
Y
X
1 2 −yi /β
L=
f (yi ) =
y e
= 2−n β −3n
yi2 exp −
yi /β
3 i
2β
i=1
i=1
i=1
i=1
!
n
n
X
Y
where u =
yi .
= β −3n e−u/β · 2−n
yi2
| {z }
i=1
i=1
g(u,β)
{z
}
|
n
Y
h(y1 ,...,yn )
Thus by the factorization criterion,
U=
n
X
n
Yi
is sufficient for β, as is
Y =
i=1
1X
Yi .
n i=1
(b) Find the MVUE of θ = β 2 . Hint: Recall that the a gamma distribution has mean αβ
and variance αβ 2 .
(10 pts)
We must find a function of a (complete) sufficient statistic that is unbiased for β 2 . Note
that Y = U/n is sufficient. Also,
E(Y ) = µ = αβ = 3β
and
V (Y ) =
σ2
αβ 2
3β 2
=
=
,
n
n
n
so that
i2 3β 2
1 2 3(3n + 1) 2
2
E(Y ) = V (Y ) + E(Y ) =
+ (3β) = 3 3 +
β =
β
n
n
n
2
h
Thus
n
2
E
Y
3(3n + 1)
2
= β2
=⇒
θ̂MVUE =
nY
3(3n + 1)
STA 4322 Exam 2
4
4. Suppose that Y1 , . . . , Yn denote a random sample from a geometric distribution with success
probability θ, where 0 < θ < 1. Then the Yi ’s have common probability function
p(y) = θ(1 − θ)y−1 ,
y = 1, 2, . . . .
Find the maximum likelihood estimator of θ.
L=
n
Y
i=1
n
(12 pts)
n
Y
Pn
p(yi ) =
θ(1 − θ)yi −1 = θn (1 − θ) i=1 (yi −1)
i=1
Pn
i=1 yi −n
= θ (1 − θ)
P
where u = ni=1 yi .
= θn (1 − θ)u−n
ln(L) = n ln(θ) + (u − n) ln(1 − θ)
n u−n
d
ln(L) = −
dθ
θ
1−θ
d
n
u−n
ln(L) = 0 =⇒
=
=⇒ n(1 − θ) = (u − n)θ
dθ
θ
1−θ
=⇒ n − nθ = uθ − nθ =⇒ n = uθ =⇒ θ =
Thus the MLE is
θ̂MLE =
n
n
= Pn
=
U
i=1 Yi
1
n
1
Pn
i=1
Yi
=
1
.
Y
n
u
STA 4322 Exam 2
5
5. A state regulatory agency suspects that the mean length of stay at a particular hospital is
greater than 6 days. A random sample of 100 admissions to the hospital yields a sample
mean length of stay of 6.8 days with a standard deviation of 4 days.
(a) Define the parameter of interest in terms of the real life setting of the problem. (2 pts)
µ = mean length of stay for all patients at the hospital
(b) Give the null and alternative hypotheses of interest to the state agency.
H0 : µ = 6
(3 pts)
Ha : µ > 6
(c) What is the rejection region of the test if α = .01.
(2 pts)
{Z ≥ z.01 } = {Z ≥ 2.33}
(d) Calculate an appropriate test statistic.
z=
(4 pts)
y − µ0
6.8 − 6
.8
√ = √
=
= 2.0
.4
s/ n
4/ 100
(e) Does the test reject or fail to reject the null hypothesis?
(2 pts)
2.0 6≥ 2.33 =⇒ Fail to reject H0
(f) Restate the conclusion in part (e) in terms appropriate to the problem.
(2 pts)
There is insufficient evidence to conclude at the α = .01 level of significance that the
mean length of stay is greater than six days.
STA 4322 Exam 2
6
(g) Find the p-value of the test.
(4 pts)
p-value = P (Z ∗ ≥ 2.0) = .0228 (from normal table)
(h) What is the power of the test with rejection region given in part (c) against the alternative that the mean length of stay is 7.6 days? (Assume that σ ≈ 4.)
(5 pts)
Y −6
√
≥ 2.33µ = 7.6
1−β =P
4/ 100
Y −6
√
≥ 2.33 ⇐⇒ Y ≥ 6 + (2.33 × .4)
4/ 100
Y − 7.6
6 + (2.33 × .4) − 7.6
−1.6
⇐⇒
≥
=
+ 2.33 = −4 + 2.33 = −1.77
.4
.4
.4
1 − β = P Z ≥ −1.77 = 1 − .0384
| {z } = .9616
∗
table
(i) What sample size is required to achieve a power of 1 − β = .90 against the alternative
that the mean length of stay is 7.6 days? (Assume σ ≈ 4.) (SET UP ONLY) (5 pts)
√
σ
σ
µ0 + zα √ = µa − zβ √ =⇒ µa − µ0 = (zα + zβ )σ/ n
n
n
2
√
(zα + zβ )σ
(zα + zβ )σ
=⇒ n =
=⇒ n =
µ a − µ0
µa − µ0
zα = z.01 = 2.33
zβ = z.10 = 1.28
2
2
(zα + zβ )σ
(2.33 + 1.28)(4)
=
= 9.0252 = 81.45 ≈ 82.
n=
µa − µ0
7.6 − 6.0
STA 4322 Exam 2
6. Define each of the following.
7
(5 pts apiece)
(a) Convergence in probability (give a precise mathematical definition of the statement “Yn
converges in probability to c as n → ∞”):
Yn converges in probability to c as n → ∞ if
P Yn − c ≥ → 0 as n → ∞, for all > 0.
Equivalently,
P Yn − c < → 1 as n → ∞, for all > 0.
(b) Sufficient statistic:
A statistic U is sufficient for a parameter θ if the conditional distribution of the data
given U does not depend on θ.
(c) Significance level (α-level):
The significance level α of a test of hypothesis is the probability of type I error, that is,
the probability of rejecting the null hypothesis when in fact the null is true.