SIE 430/530 Engineering Statistics
Homework 1
Solution
I. Textbook Problems（Casella and Berger 2nd Ed）:
1.4 For events A and B , find formulas for the probabilities of the following events in terms
of the quantities P(A), P (B) , and P(A n B) .
(a)
(b)
(c)
(d)
either A or B or both
either A or B but not both
at least one of A or B
at most one of A or B
Answer:
(a) either A or B or both
Solution:
A or B or both = Au B, so by Thm 1.2.9b
P(A or B or both)
= P(A u B) = P(A) + P(B) - P(A n B)
(b) either A or B but not both
Solution:
A or B but not both = (AU B) n (An B)C, so using Thm 1.2.9a and b:
P(A or Bbut not both) = P ((Au B) n (An Bt)
= p (Au B) - p ((Au B) n (An B))
= P (A) + P (B) - 2P (An B).
( c) at least one of A or B
Solution:
At least one of A and B = A U B
P (AU B)
1
= P(A) + P(B) - P(A n B)
(d) at most one of A or B
At most one of A or B = only A or only B or neither= (An Bt, so by Thm l.2.8c
P (At most A or B) = P ((An Bt) = 1 - P (An B)
Note that it is incorrect to say "At most A or B" is equivalent to part b, because "at most" includes
the event that an element is in neither A or B while part b does not. An alternative approach takes
the set in part band adds the event that an element is in neither:
1.21 A closet contains n pairs of shoes. If 2r shoes are chosen at random {2r < n), what is
the probability that there will be no matching pair in the sample?
Answer:
The probability is
(f1;
2
2
)
r .
There are (;~) ways of choosing 2r shoes from a total of 2n shoes.
Thus there are (;~) equally likely sample points. The numerator is the number of sample points
for which there will be no matching pair. There are (;~) ways of choosing 2r different shoes
styles. There are two ways of choosing within a given shoe style (left shoe or right shoe), which
gives 22r ways of arranging each one of the (~) arrays. The product of this is the numerator
G~)22r_
1.26 A fair die is cast until a 6 appears. What is the probability that it must be cast more
than five times?
Answer:
For a fair 6-sided die, the probability of rolling a 6 is
¼, and the probability of not rolling a 6 is
~- Consider
the event that it takes x rolls to get a 6: this means your first x - 1 rolls were not a six and on your final
roll you got a 6. Each roll of the die is indepedent, so you can simply multiply the probability associated
with each individual roll to get the total probability:
P(x rolls to get a 6) =
2
(65)x-l 6
1
Let X represent the number of rolls it takes to get a 6, then you are interested in the probability X > 5:
P(X > 5)
=1-
P(X '.S 5)
5
= 1- LP(X = i)
i=l
(5)i-11
6
6
1 (5)- 1
=1--L
5
=1
-L
i=l
4
6 j=O
=1- ~1-
6
(
i/ ()
6 1- 6
=l- (1 -Gt)
=Gt
II. Additional Problems:
C
C
C
1. Show that P( A  B  C )  1  P( A )  P( B )  P (C ) . (Hint: using Bonferroni Inequality)
Answer:
From Bonferroni’s inequality, we know that,
n
P(n~ 1 Ai) 2: LP(Ai) - (n -1)
i=l
Therefore
( ⋂ ⋂ ) ≥ ( ) + ( ) + ( ) − (3 − 1)
, ( ⋂ ⋂ ) ≥1− (
)+1− (
, ( ⋂ ⋂ )≥ 1− (
)− (
3
)+1− (
)− (
)
)−2
SIE 430/530 Engineering Statistics
Homework 2
Solution
I. Textbook Problems（Casella and Berger 2nd Ed）:
1.33 Suppose that 5% of men and .25% of women a.re color-blind. A person is chosen a.t
random and that person is color-blind. What is the probability that the person is
n:ia.le? (Assume males a.nd females to be in equal numbers.)
Answer:
Let C be the event a person is colorblind , M be the event a person is male, and F be the event a person is
female . From the problem statement:
P(C IM) = 0.05
P(CIF) = 0.0025
P(M)
= 0.5
P(F)
= 0.5
The probability of interest is the probability a person chosen at random is male, given that we know they
are colorblind. The sets M and F are a partition of the sample space, so you can use Bayes Rule to find
this:
P(MIC) P(CIM)P(M)
- P(CIF)P(F) + P(CIM)P(M)
0.05 X 0.5
0.0025 X 0.5 + 0.05 X 0.5
0.025
0.02635
250
2625
~ 0.9524.
1
1.41 As in Example 1.3.6, consider telegraph signals "dot" and "dash" sent in the proportion
3:4, where erratic transmissions ca.use a dot to become a dash with probability ¾ and
a dash to become a dot with probability ½.
(a) If a dash is received, what is the probability that a dash has been sent?
Answer:
(a) If a dash is received, what is the probability that a dash has been sent?
Solut ions:
. d
P (dash received I dash sent )P (dash sent)
Pd
( as h sent I d as h receive ) = ~ - - - - ~ - - - ~ ~ - - ~ - - - - - - ~ - - ~ ~ - ~
P (dash received I dash sent) P (dash sent)+ P(dash received I dot sent )P (dot sent )
-
~Xt
-~xt+¼x ¥
32
41
II. Additional Problems:
1. Suppose that Y has density function
if y > 1
otherwise
a. Show that ( ) is a pdf
b. Find the cdf
Answer:
a. Here
( ) ≥ 0 for all
Therefore,
b.
( )=∫
Therefore,
and ∫
( )
=∫
=
( ) is a pdf
( )
=∫
( )=1−
=
;1 ≤
=1−
<∞
2
=1
2. Suppose that Y has density function
(1 − ),
0 ≤ ≤ 1,
0,
elsewhere.
( )=
a. Find the value of k that makes ( ) a probability density function.
b. Find (.4 ≤ ≤ 1)
c. Find (.4 ≤ | ≤ 0.8)
Answer:
I
f
a). ky(l - y )dy = [.!_ ky 2
2
0
-
.!_ ky 3 ] ~ = 1 => k = 6
3
I
b) . p(0.4::; Y::; 1) =
f 6y (l - y )dy = 0.648
0.4
08
c).P(Y:s;0.8) =
2
r.
04
2
P(Y::; 0.4) = f . 6y (l - y )dy = 6[L l~.4 -
Jo
3
6y(l - y )dy = 6[L J~·s _ 2'.__ l~-s ] = 0.896
Jo
2
3
2
3
L
3
l~.4 ] = 0.352
P (Y::; 0.4 Jy::; 0.8) = P(Y::; 0.4, y::; 0.8) = P(Y::; 0.4) = 0.393
P(Y::; 0.8)
P(Y::; 0.8)
3
SIE 430/530 Engineering Statistics
Homework 3
Solution
I. Textbook Problems（Casella and Berger 2nd Ed）:
, ~.24 Compute EX and Va.r X for ea.ch of the following pro ba.bility distributions.
(a.) Jx(x)
(b) Jx(x)
= axa-l, 0 < x < 1, a> 0
= ¼, x = 1, 2, ... , n, n > 0 an integer
Solution:
a. EX
a+1 1l
= f 01 xaxa- 1 dx = fo1 axadx = a:+l
O=
EX 2
= Jor1 x 2 axa- 1 dx = Jor1 axa+ 1 dx =
VarX
=
a:2 -
( a:l )
2
=
(a+2)(a+1) 2
a:l .
a+2 11
a+2 O
Qd2__
=
___Q_.
a+2
•
= °"n
~ = _l °" n
X = _l n(n+l) _ n+l .
L...- x=l n
n L...- x=l
n
2
2
EX2 = °"n
x = _l °" n
X2 = .l n(n+1)(2n+l) = (n+1)(2n+l)
L...-i=l n
n L...-i=l
n
6
6
_
VarX = (n+1)(2n+l) _ (n+1) 2 = 2n 2 +3n+l _ n 2 +2n+l = n 2 1
6
2
6
4
12 .
b EX
·
2
II. Additional Problems:
(i)
The density function of X is given by
f(x)
If E[X]
=
=
l
a+ bx2
0
0S
X
'.S 1
otherwise
i, find a, b.
Solution:
Since f is a density it integrates to 1 and so a+ b/3 = 1. In addition 3/5 = E[X] =
f01 x(a + bx2 )dx = a/2 + b/4. Hence, a= 315 and b = 6/5.
1
(ii)
The joint density of X and Y is given by
f (x,y)
=
1
12 x(2
5
-
X -
0 <
J)
0
X
< 1, 0 < J < 1
.
otherwise
Compute the conditional density of X, given that Y
Solution:
For O < x < I, 0 < y < I, we have
fe 1y(xly)
=
f(x,y)
fy(y)
f(x,y)
f~oof(x,y) dx
x(2 -
f01 x(2 -
X -
J)
x - y) dx
x(2- x- y)
i -y/2
6x(2 -
X -
J)
4-3y
2
= y, where O < y
< l.
(iii)
The joint probability density function of X and Y is given by
f ( x, y ) 
a.
b.
c.
d.
6 2 xy
( x  ),0  x  1, 0  y  2
7
2
Verify that this is indeed a joint density function.
Compute the density function of X.
Find P(X>Y)
Find P(Y>0.5|X<0.5)
Solution:
(a) Verify that this is indeed a joint density fun ction.
Proof. To how f( x, y) is a joint PDF, we show that the double integral is equal to 0.
[ 2 fl
6(
lo lo 7
x2
xy)dx dy
[ 2 ~ ( x3 + x 2y)] 1dy
+T
10
1
[ 2~
lo
1
3
4
(! + ! _ )
3
0
dy
4
6(1- y +~2)] 2
-
7
3
8
0
- IT]
(b) Compute the density function of X.
The density function of X is the same thing as the marginal density function for X .
\Ve integrate out y. \Ve can simply use the limits of the domain of x as the limits of
integration because we will still get a function in t erms of x .
fx( x ) =
2
lo[ 76 ( x 2 + 2x y ) dy
2
xy2 ) 1o
76 ( x2y +4
I~(2x + x) I
2
3
(c) Find P(X > Y).
To find the probability that X > Y we integrate the joint distribution. \Ve have to
be careful about the limits of integration. First , we know that O < x < l and that
0 < y < 2. Since X > Y , the value of y can only vary from its lower bound of O up to
its upper bound which is x . X has no restriction so we just integrate over the domain
for X: from O to 1.
P(X
> Y)
1151
~
4
( I) Find P(l
> ½IX < ½).
·in I3n · . · Rul h w w w ul l u ·unll · u · it w ul l pr lu
th f. 11 win,... xpr
1 · Jt · th in
rrnl
int rnl f h j in .
limi f r int "ratio,...
r
P (l > ¼n
- \' < ¼
p (X < ½)
5
SIE 430/530 Engineering Statistics
Homework 4
Solution
I. Textbook Problems（Casella and Berger 2nd Ed）:
2,23 Let X have the pdf
.
1
f(x) = 2(1 +x),
(a) Find the pdf of Y = X 2 .
-1
< x < 1.
(b) Find E Y and Var Y.
Solution:
2.23 a. Use Theorem 2.1.8 with Ao = {0} , A 1 = (-1, 0) and A 2 = (0, 1). Then g 1 (x) = x 2 on A 1
and g 2 (x) = x 2 on A 2 . Then
1 y - 1/ 2,
fy(y) -- 2
b. EY
= f01 yfy(y)dy = ½
EY2
=
O < y < l.
fo1 y 2 f y(y)dy = ½
1 (1)2
4
VarY = 53 = 45·
3.2 A manufacturer receives a. lot of 100 parts from a vendor. The lot will be unacceptable if
more than five of the parts a.re defective. The manufacturer is going to select randomly
K parts from the lot for inspection and the lot will be accepted if no defective parts
a.re found in the sample.
(a) How large does K have to be to ensure that the probability that the manufacturer
accepts an unacceptable lot is less than .10?
(b) Suppose the manufacturer decides to accept the lot if there is at most one defective
in the sample. How large does K have to be to ensure that the probability that
the manufacturer accepts an unacceptable lot is less than .10?
Solution:
Let X = number of defective parts in the sample. Then X ~ hypergeometric(N
where M = number of defectives in the lot and K = sample size.
= 100, M, K)
a. If there are 6 or more defectives in the lot, then the probability that the lot is accepted
(X = 0) is at most
P(X
= 0 I M = 00 N = 6 K) = (g)(i;t) = (lOO - K) ..... (lOO - K - 5)
l
'
'
cio)
100 . .. . . 95
By trial and error we find P(X = 0) = .10056 for K = 31 and P(X = 0) = .09182 for
K = 32 . So the sample size must be at least 32.
b. Now P(accept lot) = P(X = 0 or 1), and , for 6 or more defectives, the probability is at
most
P(X
= 0 or
(g) (i;t)
m(l~1)
11M = 100, N = 6,K) = cio) + cio)
By trial and error we find P(X = 0 or 1) = .10220 for K
for K = 51. So the sample size must be at least 51.
= 50 and P(X = 0 or 1) = .09331
3.8 Two movie theaters compete for the business of 1,000 customers. Assume that each
customer chooses between the movie theaters independently and with "indifference."
Let N denote the number of seats in each theater.
(a) Using a binomial model, find an expression for N that will guarantee that the
probability of turning away a customer (because of a full house) is less than 1%.
Solution:
.01 where X ~ binomial(lO00, 1/2). Since the 1000 customers choose
1/2. We thus require
(1000) (-1) 1- -1) < .01
P(X > N) = ~
We want P(X > N)
randomly, we take p
<
=
lOOO
~
x=N+l
which implies that
(-12 )
1000
X (
X
lOOO
~
~
2
1000-X
2
(1000) < .01.
X
x=N+l
This last inequality can be used to solve for N, that is, N is the smallest integer that satisfies
I: (1000) < .01.
1000
x = N+l
The solution is N
= 537.
X
II. Additional Problems:
(1) Derive the variance of the Poisson Distribution
Solution:
Expected value and variance of Poisson random variables. We said that A is the expected
value of a Poisson(>.) random variable, but did not prove it. We did not (yet) say what
the variance was. For t he expected value, we calculate for X t hat is a Poisson (>.) random
variable:
since the x = 0 term is itself 0
divided on top and bottom by x
x i:- 1
- >.e-" ~ - - -
-
;:i (x -
(
>,l
),0
factor out C" and >. too
1)!
>,2
)
= >.e- " '0. + '1. + '2. + ...
),X
= >.e- " I : x =O
x!
= >.e- "e"
= >,
So in summary E(X ) = >.. For Var(X ) = E (X 2 ) - (E (X )) 2 = E (( X )(X - 1) + X ) (E(X)) 2 = E ((X) (X - 1)) + E(X) - (E (X )) 2 = E ((X )(X - 1)) + >, - >. 2. Now we calculate
e-,\ ),X
E (( X )(X - 1)) = L (x)(x - 1)- 1x=O
X.
c" ).X
= L (x)(x - 1)- 1x= 2
x.
because x = 0 and x = l terms are themselves 0
divide out by x and x - l
factor out e- " and >. 2
(I had extra
e-" in the video on this line)
In summary, Var(X ) = >, 2 + >, - >, 2 = >..
So both the expected value and the variance of X are equal to >..
(2) Suppose that Y ~ exponential( ) the pdf of Y is ( ) =
/
,
> 0.
Let U = √ . Find the pdf of U
Solution:
First, we note that the transformation g(Y) = -JV is a continuous strictly
increasing function ofy over Ry = {y: y > O}, and, thus, g(Y) is one-to-one. Next, we need
to find the domain of U. This is easy since y > 0 implies u = .Jy > 0 as well.
Thus, Ru
= {u: u > O}. Now, we find the inverse transformation:
g(y)
= u = fy <=> y = g- 1 (u) = u 2
(by inverse transformation)
and its derivative:
d
d
- g - 1 (u) = - (u 2 ) = 2u.
du
du
Thus, for u > 0,
Summarizing,
u>O
otherwise.
This is a Weibull distribution. The Weibull family of distributions is common in life
science (survival analysis), engineering and actuarial science applications. □
SIE 430/530 Engineering Statistics
Homework 5
Solution
I. Textbook Problems（Casella and Berger 2nd Ed）:
2.30 Find the moment generating function corresponding to
(b) / ( x)
=~
C
1
0 < x < c.
(d) P (x = x) =(r + xx- 1 )pr (l- p)z,
x ;;;; 01
,
1
••• 1
0< p <1 I r >0 a.n m
· t eger .
Answer:
b · E(e tX ) =
fi e ...lE.etxdx
2
2 ·
= -(ctetc
- etc+ 1)
0 c
c t
2
2
2
•
PrL:.o (r+:- 1 ) ((1-p)etr. Now use the fact
(1- p)etr = 1 for (1-p)et < 1, since this is just the
d. E (etX) = I.::.oetx(r+:- 1 )pr(1- p)x =
I_::_0 (r+:- 1)
p)ir (1 sum of this pmf, to get E(etX) = ( I-(l~p)e' ) r, t < that
((1 -
log(l - p).
jl.20 Let the random variable X have the pdf
_ -2 e/( X ) -
z 2 /2
$
'
0<
X
<
00.
(a) Find the mean and variance of X. (This distribution is sometimes called a folded
normaL)
(b) If X has the folded normal distribution, find the transformation g(X) = Y and
values of a: and /3 so that Y
gamma(a:, /3).
~
Answer:
(a)
The mean of X is, by integration by parts,
Also, by integration by parts,
EX 2
=
100
=
_2_
x 2f (x)dx
/oo
x2e-x,/2dx
-12rrlo
2 100 e -x2 /2d x
= -2- [-xe-x2 /2] oo + -~
o
~o
= (-0+ 0) + 1
= 1
Therefore, the variance of X is
Var.X
= EX2 -(EX)2
=
=
1-(-2
)
-/2rr
1- -
2
2
'Tr
(b)
The given pdf is written as
fx(x)=(
2
1
r (½) X (½)
2
Assuming the transformation g (X) = Y = X,.
fy(y)
,
112
)exp(-x2 ) x l(O<x<oo)
then
= fx(g- 1 (y)) I:Yg- 1 (y)I x I (0 < y < oo)
1
(
r (½) X (½) 112
)
(r (½) xl(½)1/2)
exp (-2y)
y½-1exp
Therefore, when t he transformation g (X) = Y = X,.
with parameters o = ft = ½-
2
,
ly- 1x I (0 < y < oo)
1 2
/
(-1;2) xI
(0 < Y < oo)
a random variable Y has a gamma distribution
Alternatively,
b. Let Y
= g(X) = X 2 .
Thus X
= g- 1 (Y) =
I
v'Y,where X, Y ~ 0.
d
Jy(y) = fx(g-1(y)) -g-1(y)
= -2e-y/2. - 1 = -1y-1f2e-y/2 = r(l/2, 2)
dy
v'2ii
2..fy v'2-i
I
where a= 1/2, /3
= 2.
3.28 Show that each of the following families is an exponential family.
(d) Poisson family
Answer:
The probability mass function (i.e. , the density respect to counting measure) of a Poisson
random variable is given as follows:
Rewriting this expression we obtain:
1
p(xl>.) = 1 exp{xlog>.->.} .
X.
Thus the Poisson distribution is an exponential family distribution, with:
log>.
'T/
T(x)
= X
= >. = e11
1
= x.
A(77)
,·
h(x )
Moreover, we can obviously invert the relationship between 77 and >.:
>.
= eTJ.
II. Additional Problems:
1. Show that if X i follows a binomial distribution with ni trials and probability of success
n
pi  p, i  1,2,..., n and X i are independent then
X
i
follows a binomial distribution.
i 1
(Hint: Use the moment generating function of Bernoulli random variable)
Answer:
The probability of success p is the same for all i.
Proof. '\!ve know that the moment generating function for a Bernoulli random variable
~ is Myi(t) = 1 - p + pet. vVe also know that a binomial random variable is just a
sum of Bernoulli random variables, Y = Y1 + ... + Yni . Then by the properties of
moment-generating functions,
My(t)
= MY1+ ... +Yn , (t) = Myl (t) X • • • X
'
Myn ,(t)
'
So analogously,
All of the
Xis
are independent, then similarly,
n
M.._.n
L..,i=l X '·
i=l
(1 - p + pifl ... (1 - p + petrn
(1 - p + pifl +...+nn
(1 - p + pi)Lf=l n i
Therefore, L~=l Xi is binomial with parameters N = L~=l ni and p.
n
M.._.n
L..,i=l X·•
i=l
(1 - p + pif 1 ••• (1 - p + petrn
(1 - p + pifl +... +nn
(1 - p + pi)Lf=l ni
Therefore, L~=l Xi is binomial with parameters N = L~=l ni and p.
SIE 430/530 Engineering Statistics
Homework 6
Solution
I. Textbook Problems（Casella and Berger 2nd Ed）:
4.24 Let X and Y be independent random variables with X ~ gamma( r, 1) and Y ~
gamma(s, 1). Show that Z 1 = X + Y and Z2 = X/(X + Y) are independent, and find
the distribution of each.
Answer:
Let z1
= x + y,
z2
=
x~y'
then x
JJJ =
= z1z2, y = z1(l 8x
8x
8z1
8z2
8z1
8z2
8y
z2) and
8y
The set {x > 0,y > 0} is mapped onto the set {z1 > 0,0
< z2 < l}.
fz 1 ,z2 (z 1 ,z2 ) can be factored into two densities. Therefore Z 1 and Z2 are independent and
Z 1 ~ gamma(r + s, 1), Z 2 ~ beta(r, s).
4.27 Let X ~ n(µ, o 2 ) and let Y ~ n('y, o 2 ). Suppose X and Y are independent. Define
U = X + Y and V = X - Y. Show that U and V are independent normal random
variables. Find the distribution of each of them.
Answer:
The joint pelf of X and Y is
fx,y(x, y)
1
( (x-µ)2+(y-,)2)
2c,2
'
= 21rc,2 exp -
Consider the transformation U = g1 (X, Y) = X
-00
+ Y where u =
<
X
< 00
1
-00
<y<
00.
g1 (x, y) = x + y and
V = g2(X,Y) = X -Ywherev = g2(x,y) = x-y.
Define A= {(x, y): fx ,y(x, y) > O} and B = {(u, v) : u = g1 (x, y) & v = g2(x, y) for some (x, y) EA}.
Then A= {(x,y): -oo < x < oo, -oo < y < oo} = !R2 and
B
= {(u, v) : - oo < u < oo, -oo < v < oo} = !R2 •
Here, u = x + y and v = x - y are uniquely solved with respect to x and y, i.e., the unique solution is
x = ~ and y = u 2v for any u and v. It implies that the transformation U = g1 (X, Y) and V = g2 (X, Y)
is a one-to-one transformation from A onto B.
Therefore, the joint pdf of (U, V) is
fu,v(u,v)
= fx ,Y(u;v,u;v)IJI
21r1c,2exp [-2! 2
( ..• J =
(
(u;v -µr + (u;v
I £:,1,g: i: I= I 1½
~
&v.&v
2
½
_l
I=
2
-,r)] 1-~I
X
-~)
2
2
2
1 2 exp [ -2c,12 ((u+v)
41rc,
- 2 - 2µ (u+v)
- 2 +µ 2 + (u-v)
- 2 - 2, (u-v)
- 2 +,2)]
2
2
1
2 2
- 1-exp
[4m,- 2
2c,-2 (2 (~)
2 + 2 (~)
2 - µ(u+ v) - 1 (u-v) + µ +, )]
_l_exp [--l-(u2 -u(µ+,)+ (µ+,)2 + v2 -v(µ-,)+ (µ-,)2)]
41rc, 2
2c, 2 2
2
2
2
=
_l_exp [--1 ((u-(µ+,))2+(v-(µ-,))2)]
41ro-2
4o-2
~~./2c,) exp [ 2 (~c,) 2 (u-(µ+,)) 2] x ~~./20-) exp [ 2 ()-20') 2 (v- (µ-,)) 2]
Now, the marginal pdf of U is
fu(u)
[ : fu,v (u, v)dv
1
e-
•(,Aa)' (u-(µ+-Y))' X
~(./20-)
1
e~ (-/20-)
•(~a)' (u-(µ+"f))'
roo
l
e- •(,Aa)' (v-(µ---,))' dv
J_oo ~(-/20-)
Similary, the marginal pdf of V is
fv(v)
=
1_:
fu,v (u,v)du
1
e-
2(,hv)' (v -(µ--y))2 X
./27r (v'2cr)
ro
1
e-
2(,hv)' (1L-(µ.-n))2 du
L oo ./27r (v'2cr)
1
- ~ ( v - (µ - -y)) 2
- - - - e 2(~v)
./27r (v'2cr)
Accordingly, fu,v(u ,v) factors into fu(u) and fv(v). It implies that U and V are independent by
Definition 4.2.5.
5.13 Let X1, ... ,Xn be iid n:(µ,o- 2 ). Find a functi~n of 82 , the sample variance, say g(S 2 ),
that satisfies Eg(S2 ) o-. (Hint: Try g(S2 ) = c./82, where c is a constant.)
=
Answer:
E
(cv182)
c ~E
V~
~
(✓ S 2 (n-1)
)
2
a
r=y'qr (n;-1) 2(n-1) /2 q( n-1 )- 1e- q/2
1
cy ~ lo
JS
2 (n - 1)/a2 is the square root of a
Since
be another x2 pdf and get
E
( v182)
s C
✓ o-2 .
C
2
x2
r(n/2)2n/ 2
random variable. Now adj ust the integrand to
{ 00
1
n - I r( (n - I) /2)2((n- 1)/2 l o r( n /2)2n/2 q
=l
Soc=
r( n;-l ) ✓n - 1
v12r(¥-)
.
gives E(cS) = a.
dq,
(n- 1)/2 -
since
x;.
pdf
~ - q/2
2e
dq.
II. Additional Problems:
1. Let be , , … a random sample of size n from the population ( ) with mean (finite)
and variance (finite). Please show that
( )=
/ .
a.
( )=
where
is the sample variance.
b.
Answer:
(a)
A random sample x1, x2 , ... , Xn from a distribut ion f( x) is a set of independently and identically variables with Xi~ f (x) for all i. Their joint p.d.f is
n
f (x1 , x2 , . .. , Xn)
=
f (x 1)f( x2) · · · f (x2)
=
IT f( xi) i=l
The variance of a sum of independent variables is t he sum of their variances, since covariances
are zero. Therefore
_ = V (L -n
Xi ) = - 1
V (x)
n2
L V (xi) =
2
a.
-na 2 = n2
n
(b)
E(s 2 ) = E { ~ L(xi - x) 2 }
=E [~L{(xi -µ)+(µ- x)} 2 ]
= E [~
=
L {(xi -
µ) 2 + 2(xi - µ)(µ - x) + (µ - x) 2 }]
V( x ) - 2E{(x - µ) 2 }
+ E{(x -
µ) 2 }
=
V( x ) - V( x ).
Here, we have used the result that
E{
¾I::( xi -
µ)(µ -
x)}
= -E{(µ -
x)2 }
= -V(x ).
It follows that
E(s 2) = V(x) - V( x ) = a 2 - a 2 = a 2 (n - l).
n
n
SIE 430/530 Engineering Statistics
Homework 7
Solution
I. Textbook Problems（Casella and Berger 2nd Ed）:
5.17 Let X be a random variable with an Fp,q distribution.
(c) Show that 1/X has an Fq,p distribution.
Answer:
c. Write X =*then½=*~ Fq,p, since U ~
x;, V
~
x~ and U and V
are independent.
6.22 Let X 1 , ... , Xn be a random sample from a population with pdf
f(xl0)
= 0x 8 -
1,
0 < x < 1,
0 > 0.
(a) Is EX; sufficient for 0?
(b) Show that ∏
is a complete statistic for .
Answer:
a. The sample density is Ili 0xf- 1 = 0n(ITi xi) 0 - 1 , so Ili Xi is sufficient for 0, not I:i Xi.
b. Because Ili f(xij0) = 0ne(B-l) log(ITi xi ), log (Ili Xi) is complete and sufficient by Theorem
6.2.25. Because Ili Xi is a one-to-one function of log (Ili Xi), Ili Xi is also a complete
sufficient statistic.
II. Additional Problems:
1. Let = ( , , … , ) be a random sample, where is a Poisson random variable with mean .
+ ⋯+
is a sufficient statistic for .
Show that the sample sum given by =
Answer:
Pxn1r (X nl t ) -- lP'(Xn -_
X
nlT(Xn) -_ t ) -_ P(Xn
=Xn-and
T =
---t)
.
P(T = t)
But
Hence,
p (xn = xn) = nn
i=l
Thus,
xi!
P(Xn = xn)
P(T = t)
which does not depend on ,\.
,\
e -.X;_xi
=
e-n.X ;_ I: x;
IT (xi !)
=
e -n.X;_t
IT (xi !)
t!
So T = I:i Xi is a sufficient statistic for
2. Let = ( , , … , ) be a random sample, where
mean where is unknown. Find the MLE for .
is an exponential random variable with
Answer:
Solution. For
X
E
[O, 00
r, let
S
=
L(>-.;x) =
X1
+ · · · + Xn-
Observe that
l
IJ >.e-xi/>._
n
i=l
Thus
f'(>.;x) = s/>. 2 -n/>..
Solving£'(>.; x) = 0 gives >. = s/n. Elementary calculus justifies that
this is indeed a global maximum. Thus the mle for >. is given by the
sample mean.
3. Let , , … ,
be gamma random variables with parameters
density function is:
(i)
(ii)
Find the Method of Moment estimators of
Given that = 2, find the MLE estimator of .
and , so that the probability
.
Answer:
(i)
The first theoretical moment about the origin is:
E(X;) =
ae
And the second theoretical moment about the mean is:
Var(Xj)
= E(Xi - µ )2= o.02
Again, since we have two parameters for which we are trying to derive method of moments
estimators, we need two equations. Equating the first theoretical moment about the origin with the
corresponding sample moment, we get:
1
E(X) = a0 = -
n
n
I: X i =
-
X
i= l
And, equating the second theoretical moment about the mean with the corresponding sample
moment, we get:
Var(X)
1
=
n
a0 2 = - I::(Xi - .X) 2
n i= l
Now, we just have to solve for the two parameters a and 0. Let's start by solving for a in the first
equation (E(X)). Doing so, we get:
x
a=-
0
x
Now, substituting a= -
0
into the second equation (Var(X)), we get:
a0 2
= (
x-
0
) 02
=
X0
=
_!_
n
f, (Xi i= l
X) 2
Now, solving for 0 in that last equation, and putting on its hat, we get that the method of moment
estimator for 0 is:
A
0MM
1
n
nX
i=l
- 2
= -_ I:(Xi -X)
And, substituting that value of 0 back into the equation we have for a, and putting on its hat, we
get that the method of moment estimator for a is:
A
O'.MM
=
-
A
x _ __
X = ___n_
-
0 MM
n
(1/ nX) I: (Xi - X) 2
I:(Xi - X) 2
i= l
i= l
(ii)
The likelihood function for a gamma distribution is the following:
L(a, 0lx) = J11(xla, 0) =
[ 11
TI
Taking the log, we obtain:
e(a, 0)
=
f
( ; 0 ( x;)a-i exp(- X;)
a • l1a i=1
i=1 a
-n -log(f(a)) - na -log(0) + (a - 1)
11
1 11
E
log(x;) - 0 ;~ x;
1
Taking the derivative of the log likelihood with respect to 0 and setting it equal to o:
ae(a, 0)
na
1 11
a0 = - - + - 2 I: x; = o
0
0 i=l
Solving for 0, we obtain the following MLE for 0:
11
L X;
i=l
0la= A
a· n
Now,
=
1 -X11
a
|( = 2) = x
SIE 430/530 Engineering Statistics
Homework 8
Solution
I. Textbook Problems（Casella and Berger 2nd Ed）:
7.12 Let X 1 , ... , Xn be a random sample from a population with pmf
x
= 0 or 1,
0
<
0<
-
(a) Find the method of moments estimator and MLE of 0.
(b) Find the mean squared errors of each of the estimators.
(c) Which estimator is preferred? Justify your choice.
Answer:
X;
~ iid Bernoulli(0) , 0 :S 0 :S 1/2 .
a. method of moments:
1"'
-
EX = 0 = - L.., Xi = X
n .
MLE: In Example 7.2.7, we showed that L(0lx) is increasing for 0 ::; x and is decreasing
for 0 2:'. x. Remember that O :S 0 :S 1/ 2 in this exercise. Therefore, when X :S 1/ 2, X is
the MLE of 0, because X is the overall maximum of L(0lx). When X > 1/ 2, L(0lx) is an
increasing function of 0 on (0, 1/ 2] and obtains its maximum at the upper bound of 0 which
is 1/ 2. So the MLE is 0 = min { X, 1/ 2}.
b. The MSE of 0 is MSE(0) = Var0 + bias(0) 2 = (0(1 - 0) / n)
simple formula for MSE(0), but an expression is
MSE(0)
0(1 - 0) /n. There is no
= E(0 - 0)2 = t(0 - 0)2 (n) 0Y(l - 0r-y
y=O
=
where Y =
+ 02 =
[n/2)
L ( ~ -0)
y=O
2 (
y
n ) 0Y(l-0)n-y +
y
I:i Xi ~ binomial(n, 0)
Ln
(
y=[n/2)+1
~-0 )
2 (
n ) 0Y(l-0)n-y,
y
and (n/2] = n/2, if n is even, and (n/ 2] = (n - 1) / 2, if
n is odd.
c. Using the notation used in (b), we have
MSE(0)
= E(X -
0)2
=
t (~ -
0)2 ( n )0Y(1 -0r-y.
y=O
y
Therefore,
MSE( 0) - MSE( 0)
=
t (~
y=(n/2)+1
+~ -
20) (~ -D ( ;)0Y(1- 0r-y.
The facts that y/n > 1/ 2 in the sum and 0 :S 1/ 2 imply that every term in the sum is positive.
Therefore MSE(0) < MSE(0) for every 0 in 0 < 0 S 1/2. (Note: MSE(0) = MSE(0) = 0 at
0 = 0.)
!_
2
'1.24 Let X1, ... ,Xn be iid Poisson(>.), and let>. have a gamma(o,,B) distribution, the
conjugate family for the Poisson.
(a) Find the posterior distribution of>..
(b) Calculate the posterior mean and variance.
Answer:
For n observations, Y
= I:i Xi~ Poisson(n>.).
a. The marginal pmf of Y is
m(y)
=
r)O (n>.)Ye-n>.
h
y!
nY
y!I'(a)~OI.
AOl.-le->.//3 d>.
1
r(a)~OI.
r=>.(y+o,_)-Ie-
Jo
/3/ ( n~+l)
d>.
Thus,
_ f(yl>.)1r(>.) _
1r(>.ly) -
( )
my
-
). (Y+ 01.) -1 e- /3/(n~+ l)
y+o,_
(-/3- )
I'( Y+a ) n/3+1
~
(
gamma y + a,
-~-)
a
n/J+l
b.
E(>.ly)
Var(>.ly)
(y+a)n:+1
~2
(y + a) (n~+l)2.
~
1
n~+l y + n~+l (a~)-
.
II. Additional Problems:
(i) Calculate the maximum likelihood estimate of the intensity parameter of the Poisson
distribution,
f ( y | ) 
e   y
,  0, y  0,1, 2, 3, 4,...
y!
For the data: [7,4,3,4,7,6,9,11,21,3]
Answer:
n
n
L( )  ln  f (k i |  )   ln(
i 1
i 1
n
n
e   ki
)
ki !
 n  ( k i ) ln( )   ln(k i !)
i 1
i 1
Set1storderderivativeof equaltozero
ˆ 

1 n
 ki ;doublecheckingsec ondderivative
n i 1
 2L n
   2 k i  0
 2 i 1
1 n
 ˆ   k i istheMLEestimateforPoisson
n i 1
Giventhesamplevalues
Theˆ 
1
                 
10
SIE 430/530 Engineering Statistics
Homework 9
Solution
(i) Let , , … ,
be i.i.d from the Poisson distribution P( ) with an unknown parameter
.
> 0. Please prove that the UMVUE of is T/n, T = ∑
Answer:
Then T(X) = I:f= 1 Xi is sufficient and complete for 0 > 0 and has the Poisson distribution
P(n0).
Since E(T) = n0, the UMVUE of 0 is T /n.
(ii) Show that for X 1 , X 2 ,..., X n ~ N ( ,  2 ) ,  2 is known. The LRT of H 0 :    0 , versus
H1 :    0 has the rejection region of
( X  0 )
n
 /✓
 c ,where c is a positive number.
Answer:
To build LRT for H 0 :   0 vs. H1 :   0 , the max value of
likelihood function under  0 and unrestricted MLE estimator for normal distribution need to be derived.
we know that x is the unrestricted MLE, and the restricted MLE is 0 if x  0 , x if x  0 ;so the LRT is :
  ( x i 0 )2 / 2 2

2  n /2
i
2
2
 (2 ) e
= e  n (x 0 ) /2 if x   0

  (x i  x )2 / 2 2
 (x) = 
2 n/2
i
 (2 ) e
1
if x  0
so for test which rejects null hypothesis when  (x)  c,  0  c  1)
we have e n (x 0 )
2
/2 2
 c which is equal as (x   0 ) / ( / ✓
n)  c'
(iii) Derive the LRT test of H 0 : p = p0 vs H1 :p ¹ p0 for random variable X1, …, Xn that follow
Bernoulli distribution with parameter p. (Assume the threshold value c is given.)
Answer:
n
assume we have x1 ,..., x n as bernoulli r.v. and denote Sn =  x i
i =1
Sn
The likelihood function is p (1  p)
n Sn
, with p the probability of x being 1.
and therefore the unrestricted MLE is calculated by max pSn (1  p) n Sn .
S
ˆ n
= MLE estimator is p=
n
Sn
n Sn
p (1  p0 )
=  (x)= 0
, and we reject H 0 if  (x)  c, which can be further written as
S
S
( n )Sn (1  n ) n Sn
n
n
S
S
Sn log(p0 )+(n-Sn )log(1-p0 )  c',where c'=log[c  ( n )Sn (1  n ) n Sn ]
n
n
SIE 430/530 Engineering Statistics
Homework 10
Solution
I. Textbook Problems（Casella and Berger 2nd Ed）:
8.18 Let X 1 , .• ,,Xn be a random sample from a n(0,a2 ) population, a 2 known. An LRT
of Ho: 0 = 80 versus H1 : () i- 0o is a test that rejects Ho if IX - 0ol/(a / ..jn) > c.
(a) Find an expression, in terms of standard normal probabilities, for the power function of this test.
(b) The experimenter desires a Type I Error probability of .05 and a maximum Type
II Error probability of .25 at 0 = 0o + a. Find values of n and c that will achieve
this.
Answer:
(a)
P. (
0
(3(0)
IX-0ol )
a/jn >c
1 - P0 ( -
1- P.0 (
where Z
1 _ P. (IX-0ol <
e a/vn -
~ ::; X -0o ::;
~)
- c(J/.Jii, +00 -0 < X-0
c(J/.Jii,+0
0 -0)
-<---(J/vn
- (Jlvn (Jlvn
------
l-P
(-c+ (J1v1n
_0_o-_0 < Z < c+ _0_o-_0)
- (J1v1n
1 + q?
( -C
~ n(O, 1)
c)
+ : 1~)
and
q?
-
q? ( C
+ : 1~)
,
is the standard normal cdf.
(b)
The size is .05
type II error) is
= {3( 00 ) = 1 + <I>(-c) - <I>(c) which implies c = 1.96. The power (1 -
.75 ::; /3( 0o +a-)
=
1 + <I>(-c - ../n) - <I>( c - ../n)
=
1 + <I>( -1.96-../n) -<I>(l.96 - ../n).
::::;Q
<I>(-.675)
~
.25 implies 1.96 - y'n = -.675 implies n
=
6.943
~
7.
8.20 Let X be a random variable whose prnf under Ho and H1 is given by
X
f(xlHo)
f(xlH1)
1
.01
2
.01
3
.01
4
.01
5
.01
.06
.05
.04
.03
.02
6
.01
.01
7
.94
.79
Use the Neyman-Pearson Lemma to find the most powerful test for Ho versus H1 with
size a = .04. Compute the probability of Type II Error for this test.
Answer:
By the Neyman-Pearson Lemma, the UMP test rejects for large values of J(xlH1 )/ J(xlHo).
Computing this ratio we obtain
X
1 2 3 4 5 6
6
5
4
3
2
1
7
.84
The ratio is decreasing in x. So rejecting for large values of J(xlH 1 )/ f(xlHo) corresponds to
rejecting for small values of x. To get a size a test, we need to choose c so that P(X :S
clHo) = a. The value c = 4 gives the UMP size a = .04 test. The Type II error probability is
P(X = 5, 6, 7IH1) = .82.
8.38 Let X1, ... , Xn be iid n(8, o- 2 ), where 0o is a specified value of 8 and o- 2 is unknown.
We are interested in testing
Ho: 0 = Oo
versus
(a) Show that the test that rejects Ho when
IX-Ool > t n - 1 , o / 2 ~
is a test of size o.
(b) Show that the test in part (a) can be derived as an LRT.
Answer:
a.
Size
P0 0
{
IX -
0o
I> tn-1,a/2 ../S27n}
1 - P0 0 {-tn-l,a/2../S2/n :S
1- Pe0 {-tn-1 a/2 <
'
1- (1- a)
-
X - 0o :S tn-l,a/2../S2/n}
~
s In <- tn-1 'a/2 }
2
(
~
s In "'tn-1 under Ho)
2
= a.
b. The unrestricted MLEs are 0 = X and 8- 2 = Li(Xi - .X) 2/n. The restricted MLEs are
0o = 0o and 8-5 = Li(Xi - 0o) 2 /n. So the LRT statistic is
>.(x)
=
2
(21r8-o)-n/ exp{-n8-~/(W~)}
l
Li (xi-x)2 n/2
[ Li (xi-0o) 2
For a constant c, the LRT is
reject Ho if
l
(21r8-)-n/2exp{-n8-2 /(28-2)}
L·i (x i.- x)2
[ Li (xi-x) 2 + n(x-0O)2
[
l
Li (xi-x)2
n/2
2
Li (xi-x) + n(x-00)2
- - - - ~1- - - - ~ < c2/n
1 + n(x-00)2 I Li (xi - x)2
After some algebra we can write the test as
reject Ho if
Ix - 0o I> [ (c- 2/n -
112
1) (n - 1) : ]
We now choose the constant c to achieve size a, and we
reject if
Ix -
0o
I> tn-l ,a/2.Js2ln-
SIE 430/530 Engineering Statistics
Homework 11
Solution
I. Textbook Problems（Casella and Berger 2nd Ed）:
8.6 Suppose that we have two independent random samples: X 1, ..
and Y1, ... , Yrn are exponential(µ).
. ,
Xn are exponential( 0),
(a) Find the LRT of Ho: 0 =µversus H 1 : 0 -=j=. µ.
(b) Show that the test in part (a) can be based on the statistic
T=
EXi
.
EXi + EY;
(c) Find the distribution of T when Ho is true.
Answer:
8.6 a.
,\(x,y)
sup 80 L(0lx, y)
sup 8 L(0lx, y)
sup 0 ,µ
IT'i=l
.' !ex;/0 IT'!" le-Y; f µ
0
J=l µ
SUPe
Differentiation will show that in the numerator
denominator
,\(x, y)
0 = x andµ= fi.
err.1+n exp { - ( L~=l Xi + L7=1 Yi)/ 0}
Bo = (Li Xi + Li Yi)/ (n + m), while in the
Therefore,
( L: x,)nexp { - ( L: x, ) L i Xi} (L:Y;) rn exp { - ( L:Y;) Li Yi}
And the LRT is to reject Ho if >.(x, y) ::; c.
b.
A= (n+mf+rn (
L i Xi
)n (
LiYi
)rn
nnmrn
Li Xi + Li Yi
Li Xi + Li Yi
= (n+mf+rnTn(l-T)rn.
nnmrn
Therefore ,\ is a function of T. ,\ is a unimodal function of T which is maximized when
T = rn~n. Rejection for ,\ :S c is equivalent to rejection for T :::; a or T 2: b, where a and b
are constants that satisfy an(l - a)rn = bn(l - b)rn.
c. When Ho is true, Li X i ~ gamma(n, 0) and Li Yj ~ gamma(m, 0) and they are independent. So by an extension of Exercise 4.19b, T ~ beta(n, m) .
II. Additional Problems:
1. Boys of a certain age are known to have a mean weight of = 85 pounds. A complaint is
made that the boys living in a municipal children's home are underfed. As one bit of
evidence, = 25 boys (of the same age) are weighed and found to have a mean weight
of ̅ = 80.94 pounds. It is known that the population standard deviation is 11.6 pounds
(the unrealistic part of this example!). Based on the available data, what should be
concluded concerning the complaint?
Answer:
Solution. The null hypothesis is Ho:µ= 85, and the alternative hypothesis is HA:µ< 85. In general,
we know that if the weights are normally distributed, then:
X-µ
Z=--
ulyii,
follows the standard normal N(O, I) distribution. It is actually a bit irrelevant here whether or not the
weights are normally distributed, because the same size n = 25 is large enough for the Central Limit
Theorem to apply. In that case, we know that Z, as defined above, follows at least approximately the
standard normal distribution. At any rate, it seems reasonable to use the test statistic:
X-µo
Z=---
ulyii,
for testing the null hypothesis
Ho:µ= µo
against any of the possible alternative hypotheses HA : µ ¥=, µo, HA : µ < µo, and HA : µ > µo.
For the example in hand, the value of the test statistic is:
Z = 80.94 - 85 = -1.75
11.6/\1"25
The critical region approach tells us to reject the null hypothesis at the a= 0.05 level if Z < -1.645.
Therefore, we reject the null hypothesis because Z = -1. 75 < -1.645, and therefore falls in the
rejection region:
•·
I
1 ••
As always, we draw the same conclusion by using the P-value approach. Recall that the P-value
approach tells us to reject the null hypothesis at the a= 0.05 level if the P-value::: a= 0.05. In this
case, the P-value is P(Z < -1.75) = 0.0401:
- I
:,5
2. It is assumed that the mean systolic blood pressure is = 120 mm Hg. In the Honolulu
Heart Study, a sample of = 100 people had an average systolic blood pressure of 130.1
mm Hg with a standard deviation of 21.21 mm Hg. Is the group significantly different
(with respect to systolic blood pressure!) from the regular population?
Answer:
Solution. The null hypothesis is H 0 : µ = 120, and
because there is no specific direction implied, the
alternative hypothesis is HA:µ -=ft 120. In general, we know that if the data are normally distributed,
then:
X-µ
T=-Sl..jn
follows a !-distribution with n-1 degrees of freedom. Therefore, it seems reasonable to use the test
statistic:
X-µo
T=--Sl..jn
for testing the null hypothesis Ho : µ = µ 0 against any of the possible alternative hypotheses
HA : µ '# µo , HA : µ < µo, and HA : µ > µo. For the example in hand, the value of the test statistic
is:
_ 130.1 - 120 _
62
t- - - - - - 4 .7
21.21/ v'fOO
The critical region approach tells us to reject the null hypothesis at the a = 0.05 level if t
~ to.025,99 =
1.9842 or if t::; to.025,99 = -1.9842. Therefore, we reject the null hypothesis because t = 4. 762 >
1.9842, and therefore falls in the rejection region:
Again, as always, we draw the same conclusion by using the P-value approach. The P-value
approach tells us to reject the null hypothesis at the a= 0.05 level if the P-value::; a= 0.05. In this
case, the P-value is 2 x P(T99 >4.762) < 2 x P(T99 >1.9842) = 2(0.025) = 0.05:
4.1LL
As expected, we reject the null hypothesis because P-value ~ 0.01 <a= 0.05.
By the way, the decision to reject the null hypothesis is consistent with the one you would make
using a 95% confidence interval. Using the data, a 95% confidence interval for the meanµ is:
x ± to.02s,99 ( ; ) = 130.1 ± 1.9842 (
~)
which simplifies to 130.1 ± 4.21. That is, we can be 95% confident that the mean systolic blood
pressure of the Honolulu population is between 125.89 and 134.31 mm Hg. How can a population
living in a climate with consistently sunny 80 degree days have elevated blood pressure?!
Anyway, the critical region approach for the a = 0.05 hypothesis test tells us to reject the null
hypothesis thatµ= 120:
ift
=X -
µo > 1.9842
sl,Jn, -
or if t
=x-
µo < -1.9842
sf ,Jn, -
which is equivalent to rejecting:
if
x - µo ~
1.9842 ( ;
or
if
)
which is equivalent to rejecting:
if
or if
which, upon inserting the data for this particular example, is equivalent to rejecting:
if
µ0
::;
125.89
or if
µo
~
134.31
which just happen to be(!) the endpoints of the 95% confidence interval for the mean. Indeed, the
results are consistent!
3. A manufacturer claims that the thickness of the spearmint gum it produces is 7.5 one
hundredths of an inch. A quality control specialist regularly checks this claim. On one
production run, he took a random sample of = 10 pieces of gum and measured their
thickness. He obtained:
7.65
7.60
7.65
7.70
7.55
7.55
7.40
7.40
7.50
7.50
The quality control specialist's hypotheses are:
:
= 7.5
:
≠ 7.5
Please calculate the P-value for the given hypothesis test.
Answer:
Here
= 10,
̅ = 7.55,
= 0.10274,
= 7.5
Since the population variance in unknown, we will use T statistic for the hypothesis test
We know that,
=
After putting the values of ̅ , , ,
P-Value = 2 × (
we have
̅−
⁄√
= 1.54
> 1.54) = 2 × 0.0790 = 0.158
Since the P value > 0.05, there is not enough evidence to reject null hypothesis
4. Let X1 , X 2 ,..., X n be iid observations from a N (, 2 ) distribution with known variance.
:
The LRT of
=
rejects when
( ̅
)
⁄√
≥
⁄
where
is the level of significance.
Find the confidence interval of ̅
Answer:
N(0 ,1)
z
100(1-a)o/o CI forµ, 0<a<l
(e.g. a=0.05
P( -Za/2
95% C.1.)
s Z s Z a/2 ) = 1- a
P( -Za/ 2 S
-
⇒
X-µ
a/ ✓n S Z a/
CY
2)
= 1- a
-
CY
P(-X - Z a/ 2 · --,Jn <
- -µ <- -X + Z a / 2 · --,Jn )
-
= 1- a
CY
-
CY
:. the 100(1-a)o/o C.I. forµ is [X -Za;2 • ✓
n'X +Za;2 • ✓
n]
SIE430/530 Engineering Statistics
Sample Exam I
NOTES:
1. One page formula sheet is allowed. Calculators are allowed.
2. Write your answer on this exam paper. DO NOT use your own paper. If you need
more space, write on the back side of this exam paper.
3. The test is worth 100 points. The assigned points for each question are given beside
it.
4. There are 11 pages in this exam paper.
5. Sign the Honor Code below:
I have neither given nor received aid on this examination, nor have I concealed
any violation of the Honor Code
Name(print):
____________________________
Student ID:
____________________________
Signature, Date: ___________________________
===============================================================
This sample exam is intended to give you an idea of
possible format of Exam I only. There is no indication
of coverage.
1
Question 1: (24 points)
Given P(A) = 0.5, P(B) = 0.3 and P( A ∩ B) = 0.1 , find:
a) P(A|B). (8 points)
b) P(A|B). (8 points)
c) P(A|A∩B)? (8 points)
2
Question 2: (18 points)
A random variable X follows a distribution with probability density function
f X ( x) = kx(1 − x), 0 < x < 1.
(i) Determine the value of k. (6 points)
(ii) Find the probability P(0.4 ≤ x < 1). (6 points)
(iii) Find the value of a such that the cumulative density function F(a) = 0.6. (6 points)
3
(Additional page for Question 2)
4
Question 3: (22 points)
Suppose X follows a distribution with the density function:
f X ( x | λ ) = λ e − λ x , x > 0, λ > 0.
(*)
a) Find its moment generating function. (6 points)
b) Find E(X) and Var(X). (8 points)
c) Let X and Y be independent random variables that follow density functions f X ( x |1)
and fY ( y |1) , respectively. Please note that the functions are both defined in the form
of (*). Show that Z1=X + Y and Z2= X/(X+Y) are independent. (8 points)
5
(Additional page for Question 4)
6
Question 4: (21 points)
Three shafts are made and assembled in a linkage. The length of each shaft, in
centimeters, is distributed as follows:
Shaft 1: l1~N(75,0.09)
Shaft 2: l2~N(60,0.16)
Shaft 3: l3~N(25,0.25)
The length of the linkage is calculated by l = l1+ l2+ l3. Assume the shafts’ length are
independent to each other:
a) What is the distribution of sample mean l?(9 points)
b) How can we define the boundaries such that P(l ≤b1 or l ≥ b2 ) = 0.00270? (12
points)
7
(Additional page for Question 4)
8
Question 5: (15 Points, 3 points for each question) Each of the following statements
can be TRUE or FALSE. Mark in the “(
)” with a “T” is you think it is TRUE, and
with an “F” if you think it is FALSE. (No need to show calculation procedures or
explanations).
1) (
2) (
) If X and Y are independent random variables, then FX,Y(x,y) = FX(x)FY(y).
) From the definition of probability, we know that
• 0 ≤ FX(x) ≤ 1, and
• 0 ≤ fX(x) ≤ 1.
3) (
) For every two non-zero probability events A and B, P(A|B) + P(Ac|B) = 1.
4) (
) Let X and Y be independent and identically distributed random variables and let
U = max(X, Y) and V = min(X, Y). Suppose X has mean 0, Then E(UV) = 0.
) Let X 1,…, Xn be random sample of size n collected from a normal population
X −μ
N(μ,σ2), with sample mean X . Then
~ N (0,1) .
n
σ /✓
5) (
9
Appendix
r
Appendix II
cp (z) =
Cumulative Standard Nonna[ Distribution
- l
-~./2ir
e_,,'12
__ du
~
0
z
0.0
0. 1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
10
I.I
1.2
1.3
1.4
1.5
1.6
l.7
18
J.9
2.0
2. 1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
3.0
3.1
3.2
3.3
3.4
3.5
3.6
3.7
3.8
3.9
0.00
0.01
0.02
0.03
0.04
0.50000
0.53983
0.57926
0.61791
0.65542
0.69146
0.72575
0.75803
0.78814
0.81594
0.84134
0.86433
0.88493
0.90320
0.91924
0.933 19
0.94520
0.95543
0.96407
0.97 128
0.97725
0.98214
0.98610
0.98928
0.99 180
0.99379
0.99534
0.99653
0.99744
0.99813
0.99865
0.99903
0.99931
0.99952
0.99966
0.99977
0.99984
0.99989
0.99993
0.99995
0.50399
0.54379
0.58317
0.62172
0.65910
0.69497
0.72907
0.76115
0.79103
0.8 1859
0.84375
0.86650
0.88686
0.90490
0.92073
0.93448
0.94630
0.95637
0.96485
0.97193
0.97778
0.98257
0.98645
0.98956
0.99202
0.99396
0.99547
0.99664
0.99752
0.99819
0.99869
0.99906
0.99934
0.99953
0.99968
0.99978
0.99985
0.99990
0.99993
0.99995
0.50798
0.54776
0.58706
0.62551
0.62276
0.69847
0.73237
0.76424
0.79389
0.82 121
0.846 13
0.86864
0.88877
0.90658
0.92219
0.93574
0.94738
0.95728
0.96562
0.97257
0.9783 1
0.98300
0.98679
0.98983
0.99224
0.994 13
0.99560
0.99674
0.99760
0.99825
0.99874
0.99910
0.99936
0.99955
0.99969
0.99978
0.99985
0.99990
0.99993
0.99996
0.51197
0.55172
0.59095
0.62930
0.66640
0 70194
0.73565
0.76730
0.79673
0.82381
0.84849
0.87076
0.89065
0.90824
0.92364
0.93699
0.94845
0.95818
0.96637
0.97320
0.97882
0.98341
0.98713
0.99010
0.99245
0.99430
0.99573
0.99683
0.99767
0.99831
0.99878
0.999] 3
0.99938
0.99957
0.99970
0.99979
0.99986
0.99990
0.99994
0..99996
0.51595
0.55567
0.59483
0.63307
0.67003
0.70540
0.73891
0.77035
0.79954
0.82639
0.85083
0.87285
0.89251
0.90988
0.92506
0.93822
0.94950
0.95907
0.967 11
0.97381
0.97932
0 98382
0.98745
0.99036
0.99266
0.99446
0.99585
0.99693
0.99774
0.99836
0.99882
0.99916
0.99940
0.99958
0.99971
0.99980
0.99986
0.99991
0.99994
0.99996
z
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
l.l
1.2
1.3
J.4
1.5
l.6
J.7
1.8
1.9
2.0
2. 1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
3.0
3.1
3.2
3.3
3.4
3.5
3.6
3.7
3.8
3.9
10
Appendix 11
<l>(z)
-
~
0.0
0. 1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
J. I
1.2
1.3
l.4
1.5
1.6
I. 7
1.8
1.9
2.0
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
3.0
3.1
3.2
3.3
3.4
3.5
3.6
3.7
3.8
3.9
=
(Continued)
f ffir
~
-
I
0.05
0.5 1994
0.55962
0.59871
0.63683
0.67364
0.70884
0.74215
0.77337
0.80234
0.82894
0.853 14
0.87493
0.89435
0.91149
0.92647
0.93943
0.95053
0.95994
0.96784
0.97441
0.97982
0.98422
0.98778
0.99061
0.99286
0.9946]
0.99598
0.99702
0.99781
0.99841
0.99886
0.999 18
0.99942
0.99960
0.99972
0.9998 1
0.99987
0.99991
0.99994
0.99996
. du
e- u·n
0.06
0.52392
0.56356
().60257
0.64058
0.67724
0.71226
0.74537
0.77637
0 80510
0.83147
0.85543
0.87697
0.89616
0.9 1308
0.92785
0.94062
0.95154
0.96080
0.96856
0.97500
0.98030
0.9846 1
0.98809
0.99086
0.99305
0.99477
0.99609
0.99711
0.99788
0.99846
0.99889
0.9992 1
0.99944
0.99961
0.99973
0.99981
0.99987
0.99992
0.99994
0.99996
0.07
0.08
0.09
z
0.52790
0.56749
0.60642
0.6443]
0.68082
0.71566
0.74857
0.77935
0.80785
0.83397
0.85769
0.87900
0.89796
0.9 1465
0.92922
0 94179
0.95254
0.96164
0.96926
0.97558
0.98077
0.98500
0.98840
0.99 111
0.99324
0.99492
0.99621
0.99720
0.99795
0.99851
0.99893
0.99924
0.99946
0.99962
0.99974
0.99982
0.99988
0.99992
0.99995
0.99996
0 53188
0.57142
0.61026
0.64803
0.68438
0.7 1904
0.75175
0.78230
0.8 1057
0.83646
0.85993
0 88100
0.89973
0.91621
0.93056
0.94295
0.95352
0.96246
0.96995
0.97615
0.98 124
0.98537
0.98870
0.99134
0.99343
0.99506
0.99632
0.99728
0.9980 1
0.99856
0.99897
0.99926
0.99948
0.99964
0.99975
0.99983
0.99988
0.99992
0.99995
0.99997
0.53586
0.57534
0.61409
0.65173
0.68793
0.72240
0.75490
0.78523
0.81327
0.83891
0.86214
0.88297
0.90 147
0.91773
0.93 189
0.94408
0.95448
0.96327
0.97062
0.97670
0.98 169
0.98574
0.98899
0.99158
0.99361
0.99520
0.99643
0.99736
0.99807
0.99861
0.99900
0.99929
0.99950
0.99965
0.99976
0.99983
0.99989
0.99992
0.99995
0.99997
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1. 1
1.2
1.3
1.4
l.5
1.6
1.7
1.8
1.9
2.0
2.1
2 .2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
3.0
3.1
3.2
3.3
3.4
3.5
3.6
3.7
3.8
3.9
11
Question 1: (24 points)
GivenP(A) = 0.5, P(B) = 0.3 and P(AnB) = 0.1, find:
a) P(AIB). (8 points)
b) Ef.41B). (8 points)
c)
1( B/A)
P(AIAnB)? (8 points)
_l_
0. '
o.:, -
~
'Pc A) An If>) :: 7 [ A fl{ A() 8)] _ 1'< An B)
""
f(A(li,)
w~
,1
-
<j>cA0B)"'
7rA/AuB)
_ 1[An (AuB)]
?( A UB)
t>CA')
7>cA )-t -P( B)-'Pt AAB)
o.i;
0.C--t d- ~ -O. 1 ::
r
7
2
I
Question 2: (18 points)
A random variable X follows a distribution with probability density function
fx(x)
= kx(l-x),
0 < x < l.
(i) Determine the value of k. ( 6 points)
(ii) Find the probability P(0.4::; x < 1). (6 points)
(iii) Find the value of a such that the cumulative density functio F(a) = 0.6. (6 points)
;) f
~0
X ( ,C) :,__ /i_71 CI- ?I)
(0<
?{ <
I)
... "- 'lo
w Ck (-x
J /"I. ==- I
(/-?I)
I 2.
}j !o=~k-~r==f
' ,
I
k [ 21\-3.X
~ /< =- 6
~> ?-ro.4~-x~/)
1
--
( I
} d-4
61C (1- 'X) /l'>(
( ~ ?< l. - 2 ){} ) /
iii)
,:4
fco.)= o.6
r
9
6X{l-~)/..X =0.6
i'X~ ~
(;-t,'-- ~)[ u = (). t
~ tlJ..- 2t\ , =-
o. 6
= o. G
}
{ll1-(~-2t{)
-------ti)
Question 3: (22 points)
Suppose X follo ws a distribution with the density function:
f x(x /J) =Je-,1,x , x> 0, l >0.
a) Find its moment generating function. (6 points)
b) Find E(X) and Var(X). (8 points)
c) L et X and Yb e independent random variables that follow density functions f x (x / 1)
and fr CY 11), respectively. Show that Z 1=X + Yand Z2= Xl(X+Y) are independent. (8
I
DD
point ')
() )/iA;t>=f:{_ e,-frX) =-
')(
-A f
1\-"t
(~-ii)·
C
-
~~
1
f
{
=- -
(
A,-6-
"f-X Pw-J.. Vt\A' X
X, Y( ~ '~ )=
x+Xy
i:.,_ "' Hy
2,
,~
l
VJ
Ft<,\]
~ d.-0/:'~l)
~ P.. , b p ~
4 10,rw at- ~-tJ
C)
~
\ )1L
I
?
A
/
T
- (A-t):L ~ 'i-..-t)L iJ=-o ~ Y\
X'- -(FX)'l_
E y- =- lECe-b'f-)
'31 v' 6
. ttlt
t) '1,
f ~~)( d.M-tri'~
'
e- (A'--0)7< ix
\)
J.Ece-tJx) _ ot ( -x=r
tkt
A
-/\ _
O
- 7?f
VM X =
/Vufu ·
=
0
- )-
bi c. x =
JX
C>o
O
==
e:"'
e,-rx . A
rfJ
¾,
2)- /
2
- (A-rJ1 r--<J ~= 7f
~M
~vri•~
.
~,¼- ~
X
i,v,'ll
wurl/4
61-
Ae-k X • Ae-,\ i:l :: 1\ 1_ e A
=
(Additional page for Question 4)
()(-t ~)
::;;,
{
x= ~, EL
) °' ;r, (/-Z,..)
5
l)
l
1,
=
~1
11d~
-~x
az1.
_a
aY
d'2 I
~,
"ii-
1-il..
~2~
;::
-z,
- ~ 2 , Z.1- - i ,( 1- Zl. ") r
~'
_\". e-}- I Xtlj) .
'P~ ,·n
/J ',("' 0 ,
,j,. 0
X;::E , i!L , '{::: ~,((...
- )t. e -,A [z,i.L -ti ,( /-.El.-)]
- ,\L. e---A ~, . z,
~1)
.~ ,
A;:_ I
O
.-.
2' ,
27<
z~
~
,·~
<-zl..<=/
Question 4: (21 points)
Three shafts are made and assembled in a linkage. The length of each shaft, in
centimeters, is distributed as follows:
Shaft 1: li~N(15,0.09)
Shaft 2: /i~N(60,0.16)
Shaft 3: h~N(25,0.25)
The length of the linkage is calculated by / = /i+ /2+ /J. Assume the shafts' length are
independent to each other:
a) What is the distribution of sample mean n(4 points)
b) How can we define the boundaries such that P(l ~1 or l ~ b2 ) = 0.00270? (7
points)
c) Assume that under abnormal production condition, the mean length of shaft 1
changes to 80 with the variance unchanged, and distribution parameters of the
other two shafts are not changed. Suppose a sample of size n (n =25) linkage is
randomly selected from the abnormal production condition, what is the
distribution of the new sample mean, X ? (4 points)
d) Suppose the two boundaries, b1 and b2, are still used, find P(b 1 $.X $.b2)? (6
points)
')
~=
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-4-
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= 0.5
h)
b). - 160
m
Question 5: (15 Points, 3 points for each question) Each of the following statements
can be TRUE or FALSE. Mark in the "(
)" with a "T" is you think it is TRUE, and
with an "F" if you think it is FALSE. (NQ need to show calculation procedures or
explanations).
1) (
2)(
"'f' )If X and Y are independent random variables, then Fx,r(x,y) = Fx(x)Fy(y).
f) From the definition of probability, we know that
•
0:SFx(x):Sl,and
• o:sJx<..x) :s t. X
3) (
T) For every two non-zero probability events A pnd B, P(AIB) + P(AclB) = 1.
M~ B
-fJw,
~,wli,-, -;w,, ~ ~
~ ...
4) ( \ ) Let X and Ybe iJldependent and identically distrib9ted random variables ancl let
U= max(X, Y) and V= min(X, Y). Su~~oseXhas mean 0, Then E(UV) = 0.
5) (
T ) Let X
V~ \/:. ")(""- T ~ Er U "- V )-=-F-( X., '-() -:: Q-c_l '(
1, ••• ,
Xn
be random sam_ple of size n collected from a normal population
N(µ,cl), with sample mean
X. Then X
J.
al n
J
/}UM,14'.
~ N(0,1).
8
,A-
/!'I"
cJ
A
C
Midterm Exam Solution 2018
SIE 430/530
October 2018
1
Solution 1
P ( A ∪B) = P (A) + P (B) − P (A ∩ B)
⇒ 0 · 6 = 0 · 4 + 0 · 5 − P (A ∩ B)
⇒ P (A ∩ B) = 0 · 3
(a)P (B|A) =
P (A∩B)
P (A)
=
0.3
0.4
=
3
4
(b)P (A|B) =
P (A∩B)
P (B)
=
0·3
0·5
=
3
5
(c)P (A|A ∩ B) =
2
(i)
P (A∩(A∩B))
P (A∩B)
=
P (A∩B)
P (A∩B)
=1
Solution 2
R0
−1
k(1 + x)dx +
h
⇒k x+
i0
x2
2 −1
R1
h 0
k(1 − x)dx = 1
i1
x2
h2 0
+k x−
h
⇒ k 0 − (−1) −
⇒ 21 k + 12 k = 1
⇒k=1
1
2
i
=1
1
2
i
+k 1− −0 =1
h
ix
x2
2 −1
x2
2
+
x
1
−1 (1 + x)dx + 0 (1 − x)dx = 2 + x −


0
x < −1



 x + x2 + 1 −1 ≤ x < 0
2
2
F (x) =
x2
1

x
−
+
0≤x≤1


2
2

 1
x>1
x
2
(ii)F (x) =
F (x) =
Rx
−1 (1
+ x)dx = x +
R0
F (0.5) = (0.5) −
h
R
(0.5)2
2
+
=x+
1
2
=
7
8
(iii)F or x = m(median)
when −1 ≤ x < 0
2
m + m2 + 21 = 12
2
⇒ m + m2 =0
⇒ m 1 + m2 = 0
m = 0 or m = −2
1
1
i2
2 x
0
=
1
2
+x−
x2
2
when 0 ≤ x ≤ 1
2
m − m2 + 12 = 12
2
⇒ m − m2 = 0
⇒ m 1 − m2 = 0
m = 0 or m = 2
Therefore m = 0
3
Solution 3
(a) M x (t) =
=
λr
(λ−t)r
here
R ∞ tx λr
e
0
Γ(r)
R ∞ (λ−t)r r−1 −(λ−t)x
x e
0
Γ(r)
(λ−t)r r−1 −(λ−t)x
x e
Γ(r)
(b)E(x) =
E (x2 ) =
dM x(t)
dt
t=0
d2 M x(t)
dt
t=0
f (z1 , z2 ) =
=
(1)r
Γr
(
⇒
z1
−z1
λ
λ−t
r
Γ(r)
=
xr−1 e−(λ−t)x dx
1
1−t/λ
r
is a Gamma pdf
with α = r, β =
1
λ−t
r(r+1)
λ2
V ar(x) = E (x2 ) − (E(x))2 =
z1 = x + y
(c)
x
z2 = x+y
"
z2
J = det
1 − z2
|J| = z1
N ote z1 > 0,
=
0
r
λ
=
=
R ∞ λr
xr−1 e−λx dx =
r
λ2
x = z1 z2
y = z1 − z1 z2
#
= −z1
0 < z2 < 1
s
(z1 z2 )r−1 e−z1 z2 (1)
[z1 (1 − z2 )]s−1 e−z1 (1−z2 ) · z1
Γs
1
z r+s−1 e−z1 Γ(r+s)
z r−1
Γ(r+s) 1
ΓrΓs 2
(1 − z2 )s−1
where z1 > 0, 0 < z2 < 1
Since f (z1 , z2 ) = h (z1 ) · g (z2 ) ,
1
z r+s−1 e−z1
Γ(r+s) 1
Γ(r+s) r−1
z
ΓrΓs 2
z1
and z2
is Gamma(r + s, 1)
(1 − z2 )s−1
is Beta(r, s)
(d)z1 ∼ Gamma(r + s, 1)
E (z1 ) = r + s V ar (z1 ) = r + s
2
are independent
4
Solution 4
(a) We know that, for normally distributed Samples with mean = µ and variance = σ 2
2
sample mean ( X) also distributed normally with mean = µ and variance = σn
Here, µ = 40; σ 2 = 16; n = 16
Therefore, X ∼ N (40, 1)
(b)P (x ≤ b1 , or x ≥ b2 ) = 0.0027 ⇒ P (b1 < x < b2 ) = 0.9973
P
b1√
−40
1
<z<
b2√
−40
1
P − b2 −40
<z<
1
b2 −40
1
= 0.9973; where Z ∼ N (0, 1)
= 0.9973; where b1 = 80 − b2
φ(b2 − 40) − φ (− (b2 − 40)) = 0.9973. ⇒ 2φ (b2 − 40) = 0.9973 + 1
⇒ φ (b2 − 40) = 0.99865
From z Table
5
(a) F
b2 − 40 = 3 ⇒ b2 = 43;Therefore b1 = 37
Solution 5
(b) T
(c) F
(d) T
(e) F
3
SIE430/530 Engineering Statistics
Sample Exam II
NOTES:
1. Two page formula sheet is allowed. Calculators are allowed.
2. Write your answer on this exam paper. DO NOT use your own paper. If you need
more space, write on the back side of this exam paper.
3. The test is worth 100 points. The assigned points for each question are given beside
it.
4. There are 10 pages in this exam paper.
5. Sign the Honor Code below:
I have neither given nor received aid on this examination, nor have I concealed
any violation of the Honor Code
Name(print):
____________________________
Student ID:
____________________________
Signature, Date: ___________________________
===============================================================
This sample exam is intended to give you an idea of
possible format of Exam I only. There is no indication
of coverage.
1
Question 1: (20 points)
An experiment was conducted to investigate the filling capability of packaging
equipment at a winery in Newberg, Oregon. A sample of 10 bottles of Pinot Gris was
randomly selected and the fill volume (in ml) measured. Assume that fill volume has
normal distribution. The mean fill volume of this sample is 749 ml. From the prior
experience, the filling process variance is known as 1 ml.
a) Do the sample data support the claim that the mean fill volume is less than 750?
Use   0.05 . (8 points)
b) Find a one-side 95% upper confidence bound on mean fill volume. (6 points)
c) What is the p-value of this hypothesis test? (6 points)
2
Question 2: (20 points)
Let X1, X2, …, Xn be a random sample from a N(,) distribution, i.e., the pdf of X is
given by
f ( x | , 2 ) 
1
2
2
e x   
2
/ 2 2
, -  x   .
Assume that >0 is known.
a) Using the moment-generating function of X, find the distribution of T  i 1 X i .
n
(8 points)
b) Show that T is sufficient for . (6 points)
c) It is easy to show that T is also complete for . Find the UMVUE of . (6 points)
3
(Additional page for Question 2)
4
Question 3: (24 points)
Let X1, X2, …, Xn be iid observations from a Exponential() distribution, i.e.,
1
f ( x |  )  - e x /  , 0  x  ,   0 .

a) Find the Maximum Likelihood Estimator (MLE) for . (14 points)
n
b) Show that Y  i 1 X i has a Gamma(n, ) distribution. (4 points)
c) Calculate the mean-square-error of the MLE. (6 points)
5
(Additional page for Question 3)
6
Question 4: (16 points)
Let X1, X2, …, Xn be iid observations from a Exponential() distribution,
a) Formulate the Likelihood Ratio Test for testing H0: = vs. H1: ≠. (10 points)
b) Find the equations needed to determine the sample size n and the critical value c
such that the size of the test when = is 0.05 and the power for detecting =2 is
0.80. (If you are not confident with the likelihood ratio statistic you derived in a),
please use T( X )) (6 points)
7
(Additional page for Question 4)
8
Question 5: (20 points) (Textbook problem 8.6)
Let X1, X2, …, Xn be iid observations from a Exponential(), and Y1, Y2, …, Yn be iid
observations from a Exponential(),
a) Find the LRT of H0: = vs. H1: ≠. (10 points)
b) Show that the test in part a) can be based on the statistic (4 points)
 Xi
T
 X i   Yi
c) Find the distribution of T when H0 is true. (6 points)
9
(Additional page for Question 5)
10
SIE430/530 Engineering Statistics
Sample Exam II Solution
Question 1: Solution
n = 10; x = 749; σ = 1
H 0 : μ = 750
(a)
H 1 : μ < 750
Test statistic z 0 =
x − μ0
749 − 750
= −3.16
n
1 10
Reject H 0 if z 0 < − zα = − Z 0.05 = −1.64
Conclusion: Reject H 0 : μ = 750 at α=0.05. There is enough evidence that mean is smaller than 750 at the
σ
=
significance level of 0.05.
(b) 95% upper C.I.:
μ < x + Zα σ
n
= 749 + 1.64 × 1
(c) p = Φ ( Z 0 ) = Φ ( −3.16) = 0.00079
10
= 749.51
So~
(J._2.
; , F~
r::.rt
r
XA/ Nr~,
~~)
Acm == e .,,wt -t tr'-t1/.l-' J
t. := ~" 'x/
'xI ,·,,·I".
t'\.,-
,M,. rt> :::
,,
ft1 ~,+--+ x~ t-t-J-= .f, M~· ttJ
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Q,
:s
ctJ,n-r' I )
x
. . '!. ==
1iw,e-~'dub)
ttu M LE
-vl
¼ i·u·w
1,4111
AQcw--- , we vv,e M~F
MX (~) "'
/-
of
~ /\ .
~
-/;i,
,~k
--tA,,.
1""1,~
slwvv rlJ-.
-Ix,
'>di
r\
=
,TI Mx,· rt)
= (- I -
It.;.. ~ -t,l.l Mfff
C)
ex )
f
E{
=
*)
Xt )'1
cf
=
6 ANn IY'll. c 'Yl , A )
ii- E y = *.
)1 .\
::::-
:.
~"'
0-
)
1/M DJ =
~
[
y~a~ (YI,\~
}\.
UNBtASED -es-6ivY\ltw·
y)
'
Vad fl =- It" VMr '()
,. IA sE d )= VM ( \ ) + [ B,AA ( ;
- A/it1
I
~ ~ · Y\ · A = n,
)f
J_
~L