Statistics 252 Winter 2006 Midterm #3 Solutions
1. (a) By denition, the signicance level is the probability of a Type I error; that is, the proba2
bility under H0 that H0 is rejected. Hence, since 4S2 2 ,
4
4S 2
4 1.945
>
1
1
= PH0 (reject H0 ) = P (S 2
Stat 252.01 Winter 2007
Assignment #6 Solutions
1. Textbook
(9.73) If Y1 , . . . , Yn are i.i.d. exponential(), then each Yi has common density
1
fY (y|) = ey/ ,
y > 0.
Therefore, the likelihood function is
n
n
fY (yi |) =
L() =
i=1
i=1
1
1 yi /
e
= n exp
Stat 252 Winter 2007
Assignment #5 Solutions
1. Textbook
(8.40) (a) By denition, if 0 < y < , then
y
FY (y) =
y
fY (u) du =
0
2( u)
(2u u2 )
du =
2
2
y
=
0
(2y y 2 )
2y y 2
=
2.
2
That is,
FY (y) =
0,
2y
y 0,
y2
,
2
0 < y < 1,
y .
1,
(b) If U = Y /, then
Stat 252 Winter 2007
Assignment #4 Solutions
2. Textbook
(8.30) If = Y , then E() = E(Y ) = so that Y is an unbiased estimator of . Since the
is
standard error of
= Var(Y ) =
,
n
a natural guess for the estimated standard error is
=
.
n
(8.36) (a) If
Stat 252 Winter 2007
Assignment #2 Solutions
1. If the population mean is , then E(Yi ) = , i = 1, . . . , n. Hence,
E Y =E
1
n
n
Yi
=
i=1
1
n
n
E(Yi ) =
i=1
n
= .
n
(That is, Y is an unbiased estimator of .) In order to show that S 2 is an unbiased estim
Stat 252.01 Winter 2007
Assignment #8 Solutions
1. As always, = PH0 (reject H0 ) and = PHA (accept H0 ). Since X has an Exponential()
distribution so that f (x|) = 1 ex/ , x > 0, and since our rejection region is cfw_X < c, we
nd that
c
= PH0 (reject H0
Stat 252.01 Winter 2007
Assignment #10 Solutions
1. (a) To nd the method of moments estimator of we must solve the equation E(Y ) = Y for
. Since E(Y ) = , we conclude
MOM = Y .
1. (c) The likelihood function is
n
fY (yi |) =
L() =
3
n
2n
i=1
n
exp
yi
i
Stat 252.01 Winter 2007
Assignment #7 Solutions
Important Remark: The factorizations of L into L = g h are not unique. Many answers are
possible.
Important Remark: Any one-to-one function of a sucient statistic for is also sucient for .
(9.30) If Y1 , . .
Stat 252.01 Winter 2007
Assignment #9 Solutions
1. We nd that f0 (y) = ey for y > 0 and fA (y) = 2e2y for y > 0. Therefore, the likelihood
ratio is
ey
1
f0 (y)
= 2y = ey
(y) =
fA (y)
2e
2
for y > 0, and so the rejection region is
1 Y
e <c
2
RR = cfw_(Y )