Stat 252.01 Winter 2005
Assignment #9 Solutions
(11.56) For the quadratic model Y = 0 + 1 x + 2 x2 + given in the problem, we know that
the least squares estimators i can be found by solving the matrix equation
= (X X)1 X Y
where
1
1
1
X = 1
1
1
1
y1
y
Stat 252.01 Winter 2005
Assignment #7 Solutions
1. (a) As proved in class, the least squares estimate 0 is given by
0 = y 1 x.
Thus,
n
n
n
(yi 0 1 xi ) =
(yi yi ) =
i=1
i=1
n
yi n0 1
i=1
xi = ny n0 n1 x
i=1
= ny n(y 1 x) n1 x
= 0.
(b) To prove that the le
Stat 252.01 Winter 2005
Assignment #2 Solutions
1. (a) Since Y1 U(0, ), we have
0 y ,
otherwise.
1/,
0,
fY (y) =
(b) If f (t) denotes the density of = min(Y1 , . . . , Yn ), then f (t) = F (t), where
F (t) = P ( t) = 1 P ( > t) = 1 P (Y1 > t, . . . , Yn >
Stat 252.01 Winter 2005
Assignment #5 Solutions
1. (a) If = P (Type II error), then we dene the power of a test to be
power = 1 = 1 PHA (accept H0 ) = PHA (reject H0 ).
Now, with the alternative given, we reject H0 if
Z=
or, equivalently, if
Y 0
> 1.65,
Stat 252.01 Winter 2005
Assignment #4 Solutions
(10.2) (a) It might be helpful to have clear null and alternative hypotheses in mind. Therefore,
let H0 be the drug dosage level induces sleep in 80% of people suering from insomnia and let
HA be the drug do
Stat 252.01 Winter 2005
Assignment #3 Solutions
(8.9) (a) Let = Var(Y ), and = n(Y /n)(1 Y /n). To prove is unbiased, we must show
= . Since
that E()
E() = E(n(Y /n)(1 Y /n) = E(Y )
1
E(Y 2 ),
n
and since Y is Binomial(n, p) so that E(Y ) = np, E(Y 2 )
Stat 252.01 Winter 2005
Assignment #8 Solutions
(11.17) Using standard properties of covariances, and the formul derived in class for 0 , 1 ,
we compute
Cov(0 , 1 ) = Cov(Y 1 x, 1 )
= Cov(Y , 1 ) x Cov(1 , 1 )
1
n
= Cov
=
i
1
Cov
nSxx
1
=
nSxx
=
Yi
Statistics 252 Practice Midterm Solutions Winter 2005
1. (a) Since
log f (y|) = 2 log() 2 y,
we nd
2
2
log f (y|) = 2 2y.
2
Thus,
I() = E
2
log f (Y |)
2
2
4
+ 2E(Y ) = 2
2
=
since E(Y ) = 2 . (This is because Y Exp(2 ).)
(b) Let Z = 2 Y so that
2 z
P (Z
Statistics 252 Practice Midterm #2 Solutions Winter 2005
1. (a) By denition, the signicance level is the probability of a Type I error; that is, the
probability under H0 that H0 is rejected. Hence, since Y N (, 4/n),
Y 0
3.92/ n 0
>
= PH0 (reject H0 ) =
Statistics 252 Winter 2005 Midterm #2 Solutions
1. (a) By denition, the signicance level is the probability of a Type I error; that is, the
probability under H0 that H0 is rejected. Hence, since Y N (, 9/n),
0.05 = PH0 (reject H0 ) = P (Y > c| = 0) = P
Y
Stat 252.01 Winter 2005
Assignment #10 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 , .
Statistics 252Mathematical Statistics
Winter 2005 (200510)
Final Exam Solutions
Instructor: Michael Kozdron
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. (b) By denition