Download PSTAT 120B Probability and Statistics - Week 2

PSTAT 120B Probability and Statistics - Week 2
Fang-I Chu
University of California, Santa Barbara
October 11, 2012
Fang-I Chu
PSTAT 120B Probability and Statistics
couple notes about hw1
about #3(6.14):
Fang-I Chu
PSTAT 120B Probability and Statistics
couple notes about hw1
about #3(6.14):
uses transformation method. We can begin with CDF to do
the transformation, or set the Jacobian and use the
transformation formula.
Fang-I Chu
PSTAT 120B Probability and Statistics
couple notes about hw1
about #3(6.14):
uses transformation method. We can begin with CDF to do
the transformation, or set the Jacobian and use the
transformation formula.
carefully compute integral.
Fang-I Chu
PSTAT 120B Probability and Statistics
couple notes about hw1
about #3(6.14):
uses transformation method. We can begin with CDF to do
the transformation, or set the Jacobian and use the
transformation formula.
carefully compute integral.
this type of problem is very IMPORTANT. Same type of
problem came up again in hw2 #1.
Fang-I Chu
PSTAT 120B Probability and Statistics
couple notes about hw1
about #3(6.14):
uses transformation method. We can begin with CDF to do
the transformation, or set the Jacobian and use the
transformation formula.
carefully compute integral.
this type of problem is very IMPORTANT. Same type of
problem came up again in hw2 #1.
about #2(3.148) and #6(6.40)
Fang-I Chu
PSTAT 120B Probability and Statistics
couple notes about hw1
about #3(6.14):
uses transformation method. We can begin with CDF to do
the transformation, or set the Jacobian and use the
transformation formula.
carefully compute integral.
this type of problem is very IMPORTANT. Same type of
problem came up again in hw2 #1.
about #2(3.148) and #6(6.40)
Answer or compute in the way what question asked:
Fang-I Chu
PSTAT 120B Probability and Statistics
couple notes about hw1
about #3(6.14):
uses transformation method. We can begin with CDF to do
the transformation, or set the Jacobian and use the
transformation formula.
carefully compute integral.
this type of problem is very IMPORTANT. Same type of
problem came up again in hw2 #1.
about #2(3.148) and #6(6.40)
Answer or compute in the way what question asked:
1. #2(3.148) ask you to derive V (Y ) using ln m(t).
Fang-I Chu
PSTAT 120B Probability and Statistics
couple notes about hw1
about #3(6.14):
uses transformation method. We can begin with CDF to do
the transformation, or set the Jacobian and use the
transformation formula.
carefully compute integral.
this type of problem is very IMPORTANT. Same type of
problem came up again in hw2 #1.
about #2(3.148) and #6(6.40)
Answer or compute in the way what question asked:
1. #2(3.148) ask you to derive V (Y ) using ln m(t).
2. #6(6.40) ask you to find the density function of
U = Y12 + Y22 , so if your answer leaves as MGF of U, that’s
incomplete answer.
If you turned in your hw1, but didn’t get it back from section
today, please talk to me after section.
Fang-I Chu
PSTAT 120B Probability and Statistics
Topics for review
The Bivariate Transformation Method
Hint for #1
Hint for #2
Sample distributions
Hint for #3 (Exercise 7.12)
Hint for #4 (Exercise 7.20)
similar problem (Exercise 7.43) for #5 (Exercise 7.42)
Hint for #6 (Exercise 7.54)
Fang-I Chu
PSTAT 120B Probability and Statistics
The Bivariate Transformation Method
Fang-I Chu
PSTAT 120B Probability and Statistics
The Bivariate Transformation Method
Assumption: Y1 and Y2 are random variables with joint
density function fY1 ,Y2 (y1 , y2 ) and let
u1 = h1 (y1 , y2 ), u2 = h2 (y1 , y2 ) is a one-to-one transformation
from (y1 , y2 ) to (u1 , u2 ) with inverse
y1 = h1−1 (u1 , u2 ), y2 = h2−1 (u1 , u2 ).
Fang-I Chu
PSTAT 120B Probability and Statistics
The Bivariate Transformation Method
Assumption: Y1 and Y2 are random variables with joint
density function fY1 ,Y2 (y1 , y2 ) and let
u1 = h1 (y1 , y2 ), u2 = h2 (y1 , y2 ) is a one-to-one transformation
from (y1 , y2 ) to (u1 , u2 ) with inverse
y1 = h1−1 (u1 , u2 ), y2 = h2−1 (u1 , u2 ).
Content:
Fang-I Chu
PSTAT 120B Probability and Statistics
The Bivariate Transformation Method
Assumption: Y1 and Y2 are random variables with joint
density function fY1 ,Y2 (y1 , y2 ) and let
u1 = h1 (y1 , y2 ), u2 = h2 (y1 , y2 ) is a one-to-one transformation
from (y1 , y2 ) to (u1 , u2 ) with inverse
y1 = h1−1 (u1 , u2 ), y2 = h2−1 (u1 , u2 ).
Content:
Jacobian is |J| = ∂h1−1
∂u1
∂h2−1
∂u1
Fang-I Chu
∂h1−1
∂u2
∂h2−1
∂u2
PSTAT 120B Probability and Statistics
The Bivariate Transformation Method
Assumption: Y1 and Y2 are random variables with joint
density function fY1 ,Y2 (y1 , y2 ) and let
u1 = h1 (y1 , y2 ), u2 = h2 (y1 , y2 ) is a one-to-one transformation
from (y1 , y2 ) to (u1 , u2 ) with inverse
y1 = h1−1 (u1 , u2 ), y2 = h2−1 (u1 , u2 ).
Content:
∂h1−1
∂h1−1
∂u1
∂u2
∂h2−1
∂h2−1
∂u1
∂u2
of U1 and U2
Jacobian is |J| = The joint density
is
fU1 ,U2 (u1 , u2 ) = fY1 ,Y2 (h1−1 (u1 , u2 ), h2−1 (u1 , u2 ))|J|
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #1
#1
Suppose that Y1 and Y2 are independent Gamma random variables
with the same beta parameter value β but possibly different alpha
1
parameter values α1 and α2 . Define S = Y1 + Y2 and R = Y1Y+Y
.
2
(i) Derive the joint PDF of (R, S).
(ii) Hence show that S has a Gamma distribution- specify the
parameters- and that R has a Beta distribution.
(iii) Are R and S independent? Why?
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #1(i)
1(i)
Derive the joint PDF of (R, S).
Hint for(i):
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #1(i)
1(i)
Derive the joint PDF of (R, S).
Hint for(i):
Known:
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #1(i)
1(i)
Derive the joint PDF of (R, S).
Hint for(i):
Known:
Y1 ∼ Γ(α1 , β) and Y2 ∼ Γ(α2 , β).
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #1(i)
1(i)
Derive the joint PDF of (R, S).
Hint for(i):
Known:
Y1 ∼ Γ(α1 , β) and Y2 ∼ Γ(α2 , β).
Y1 and Y2 are independent.
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #1(i)
1(i)
Derive the joint PDF of (R, S).
Hint for(i):
Known:
Y1 ∼ Γ(α1 , β) and Y2 ∼ Γ(α2 , β).
Y1 and Y2 are independent.
Facts:
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #1(i)
1(i)
Derive the joint PDF of (R, S).
Hint for(i):
Known:
Y1 ∼ Γ(α1 , β) and Y2 ∼ Γ(α2 , β).
Y1 and Y2 are independent.
Facts:
Y1 and Y2 are independent, so we have
fY1 ,Y2 (y1 , y2 ) = fY1 (y1 )fY2 (y2 )
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #1(i)
1(i)
Derive the joint PDF of (R, S).
Hint for(i):
Known:
Y1 ∼ Γ(α1 , β) and Y2 ∼ Γ(α2 , β).
Y1 and Y2 are independent.
Facts:
Y1 and Y2 are independent, so we have
fY1 ,Y2 (y1 , y2 ) = fY1 (y1 )fY2 (y2 )
1
S = Y1 + Y2 , R = Y1Y+Y
, implying Y1 = RS and
2
Y2 = (1 − R)S
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #1(i)
1(i)
Derive the joint PDF of (R, S).
Hint for(i):
Known:
Y1 ∼ Γ(α1 , β) and Y2 ∼ Γ(α2 , β).
Y1 and Y2 are independent.
Facts:
Y1 and Y2 are independent, so we have
fY1 ,Y2 (y1 , y2 ) = fY1 (y1 )fY2 (y2 )
1
S = Y1 + Y2 , R = Y1Y+Y
, implying Y1 = RS and
2
Y2 = (1 − R)S
0 < r < 1, 0 < s < ∞
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #1(i)
1(i)
Derive the joint PDF of (R, S).
Hint for(i):
Known:
Y1 ∼ Γ(α1 , β) and Y2 ∼ Γ(α2 , β).
Y1 and Y2 are independent.
Facts:
Y1 and Y2 are independent, so we have
fY1 ,Y2 (y1 , y2 ) = fY1 (y1 )fY2 (y2 )
1
S = Y1 + Y2 , R = Y1Y+Y
, implying Y1 = RS and
2
Y2 = (1 − R)S
0 < r < 1, 0 < s <
∞
dy1 dy1 s
r dr
ds = |J| = dy
dy2 2
−s 1 − r = s
dr
ds
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #1(i)
1(i)
Derive the joint PDF of (R, S).
Hint for(i):
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #1(i)
1(i)
Derive the joint PDF of (R, S).
Hint for(i):
Way to obtain joint pdf of (R, S).
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #1(i)
1(i)
Derive the joint PDF of (R, S).
Hint for(i):
Way to obtain joint pdf of (R, S).
Apply bivariate transformation method, you will obtain the
answer!
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #1(ii)
1
(ii)Hence show that S has a Gamma distribution- specify the
parameters- and that R has a Beta distribution.
(iii) Are R and S independent? Why?
Hint for (ii):
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #1(ii)
1
(ii)Hence show that S has a Gamma distribution- specify the
parameters- and that R has a Beta distribution.
(iii) Are R and S independent? Why?
Hint for (ii):
Known:
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #1(ii)
1
(ii)Hence show that S has a Gamma distribution- specify the
parameters- and that R has a Beta distribution.
(iii) Are R and S independent? Why?
Hint for (ii):
Known:
obtained joint pdf of (R, S):
1
s
α1 +α2 −1
s
exp
−
β α1 +α2 Γ(α1 + α2 )
β
Γ(α1 + α2 ) α1 −1
×
r
(1 − r )α2 −1
Γ(α1 )Γ(α2 )
fR,S (r , s) =
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #1(ii)
1
(ii)Hence show that S has a Gamma distribution- specify the
parameters- and that R has a Beta distribution.
(iii) Are R and S independent? Why?
Hint for (ii):
Known:
obtained joint pdf of (R, S):
1
s
α1 +α2 −1
s
exp
−
β α1 +α2 Γ(α1 + α2 )
β
Γ(α1 + α2 ) α1 −1
×
r
(1 − r )α2 −1
Γ(α1 )Γ(α2 )
fR,S (r , s) =
Note the joint pdf of (R, S) is obtained in (a), and you will
NEED to show complete computation how you get your joint
pdf in your hw for 1(a) part.
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #1(ii)
1
(ii)Hence show that S has a Gamma distribution- specify the
parameters- and that R has a Beta distribution.
(iii) Are R and S independent? Why?
Hint for (ii):
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #1(ii)
1
(ii)Hence show that S has a Gamma distribution- specify the
parameters- and that R has a Beta distribution.
(iii) Are R and S independent? Why?
Hint for (ii):
Observe and think:
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #1(ii)
1
(ii)Hence show that S has a Gamma distribution- specify the
parameters- and that R has a Beta distribution.
(iii) Are R and S independent? Why?
Hint for (ii):
Observe and think:
Do you recognize the joint pdf of (R, S) is composed by a
Gamma pdf and a Beta distribution?
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #1(ii)
1
(ii)Hence show that S has a Gamma distribution- specify the
parameters- and that R has a Beta distribution.
(iii) Are R and S independent? Why?
Hint for (ii):
Observe and think:
Do you recognize the joint pdf of (R, S) is composed by a
Gamma pdf and a Beta distribution?
Specify the parameters for Gamma distribution and for Beta
distribution!
Hint for (iii):
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #1(ii)
1
(ii)Hence show that S has a Gamma distribution- specify the
parameters- and that R has a Beta distribution.
(iii) Are R and S independent? Why?
Hint for (ii):
Observe and think:
Do you recognize the joint pdf of (R, S) is composed by a
Gamma pdf and a Beta distribution?
Specify the parameters for Gamma distribution and for Beta
distribution!
Hint for (iii):
Does the range of R and S independent of each other?
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #1(ii)
1
(ii)Hence show that S has a Gamma distribution- specify the
parameters- and that R has a Beta distribution.
(iii) Are R and S independent? Why?
Hint for (ii):
Observe and think:
Do you recognize the joint pdf of (R, S) is composed by a
Gamma pdf and a Beta distribution?
Specify the parameters for Gamma distribution and for Beta
distribution!
Hint for (iii):
Does the range of R and S independent of each other?
What does Theorem 5.5 state?
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #2
#2
The random variables X and Y are independent Exponential both
X
with mean 1. Define R = X +Y
. Obtain the CDF of R. What is
this distribution?
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #2
#2
The random variables X and Y are independent Exponential both
X
with mean 1. Define R = X +Y
. Obtain the CDF of R. What is
this distribution?
Known: X ∼ exp(1) and Y ∼ exp(1)
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #2
#2
The random variables X and Y are independent Exponential both
X
with mean 1. Define R = X +Y
. Obtain the CDF of R. What is
this distribution?
Known: X ∼ exp(1) and Y ∼ exp(1)
X
≤ r)
X +Y
= P(X ≤ r (X + Y ))
P(R ≤ r ) = P(
= P((1 − r )X ≤ rY )
1−r
= P(Y ≥
X)
r
Z Z
∞
∞
e −y dye −x dx
=
0
Fang-I Chu
1−r
r
X
PSTAT 120B Probability and Statistics
Hint for #2
#2
The random variables X and Y are independent Exponential both
X
with mean 1. Define R = X +Y
. Obtain the CDF of R. What is
this distribution?
Known: X ∼ exp(1) and Y ∼ exp(1)
X
≤ r)
X +Y
= P(X ≤ r (X + Y ))
P(R ≤ r ) = P(
= P((1 − r )X ≤ rY )
1−r
= P(Y ≥
X)
r
Z Z
∞
∞
e −y dye −x dx
=
0
Fang-I Chu
1−r
r
X
PSTAT 120B Probability and Statistics
Some important distributions
Fang-I Chu
PSTAT 120B Probability and Statistics
Some important distributions
Let Y1 , Y2 , . . . , Yn be a random sample of size n from a
normal P
distribution with mean µ and variance σ 2 . Then
1
Ȳ = n ni=1 Yi is normally distributed with mean µȲ = µ and
2
variance σȲ2 = σn .
Fang-I Chu
PSTAT 120B Probability and Statistics
Some important distributions
Let Y1 , Y2 , . . . , Yn be a random sample of size n from a
normal P
distribution with mean µ and variance σ 2 . Then
1
Ȳ = n ni=1 Yi is normally distributed with mean µȲ = µ and
2
variance σȲ2 = σn .
Then Zi = Yiσ−µ has standard normal distribution. i.e.
Z ∼ N (0, 1).
Fang-I Chu
PSTAT 120B Probability and Statistics
Some important distributions
Let Y1 , Y2 , . . . , Yn be a random sample of size n from a
normal P
distribution with mean µ and variance σ 2 . Then
1
Ȳ = n ni=1 Yi is normally distributed with mean µȲ = µ and
2
variance σȲ2 = σn .
Then Zi = Yiσ−µ has standard normal distribution. i.e.
Z ∼ N (0, 1).
Pn
Pn Yi −µ
has a χ2 distribution with n degrees of
i=1 Zi =
i=1 σ
freedom.
Fang-I Chu
PSTAT 120B Probability and Statistics
Some important distributions
Let Y1 , Y2 , . . . , Yn be a random sample of size n from a
normal P
distribution with mean µ and variance σ 2 . Then
1
Ȳ = n ni=1 Yi is normally distributed with mean µȲ = µ and
2
variance σȲ2 = σn .
Then Zi = Yiσ−µ has standard normal distribution. i.e.
Z ∼ N (0, 1).
Pn
Pn Yi −µ
has a χ2 distribution with n degrees of
i=1 Zi =
i=1 σ
freedom.
(n−1)S 2
σ2
has a χ2 distribution with (n − 1) degrees of freedom.
1 Pn
2
Recall S 2 = n−1
i=1 (Yi − Ȳ )
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #3(Exercise 7.12)
7.12
Suppose the forester in Exercise 7.11 would like the sample mean
to be within 1 square inch of the population mean, with probability
0.90. How many trees must be measure in order to ensure this
degree of accuracy?
7.11
A forester studying the effects of fertilization on certain pine
forests in the Southeast is interested in estimating the average
basal area of pine trees. In studying basal areas of similar trees for
many years, he has discovered that these measurements (in square
inches) are normally distributed with standard deviation
approximately 4 square inches. If the forester samples n = 9 trees,
find the probability that the sample mean will be within 2 square
inches of the population mean.
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #3(Exercise 7.12)
Think:
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #3(Exercise 7.12)
Think:
Do you understand the problem?
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #3(Exercise 7.12)
Think:
Do you understand the problem?
What does this question really ask us?
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #3(Exercise 7.12)
Think:
Do you understand the problem?
What does this question really ask us?
Are you able to write out our goal in mathematical
expression?
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #3(Exercise 7.12)
Hint for #4
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #3(Exercise 7.12)
Hint for #4
Known: σ = 4 and Z =
X̄ −µ
σ
√
n
Fang-I Chu
(page 354)
PSTAT 120B Probability and Statistics
Hint for #3(Exercise 7.12)
Hint for #4
Known: σ = 4 and Z =
X̄ −µ
σ
√
n
(page 354)
Goal: how large must n be to make P(|X̄ − µ| ≤ 1) ≥ 0.9 (
to ensure the degree of accuracy means we need to have at
least probability 0.90, which implying our probability need to
be greater or equal to 0.9)
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #3(Exercise 7.12)
Hint for #4
Known: σ = 4 and Z =
X̄ −µ
σ
√
n
(page 354)
Goal: how large must n be to make P(|X̄ − µ| ≤ 1) ≥ 0.9 (
to ensure the degree of accuracy means we need to have at
least probability 0.90, which implying our probability need to
be greater or equal to 0.9)
Fact: P(|Z | ≤ 1.645) = 0.9( table of page 848)
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #3(Exercise 7.12)
Hint for #4
Known: σ = 4 and Z =
X̄ −µ
σ
√
n
(page 354)
Goal: how large must n be to make P(|X̄ − µ| ≤ 1) ≥ 0.9 (
to ensure the degree of accuracy means we need to have at
least probability 0.90, which implying our probability need to
be greater or equal to 0.9)
Fact: P(|Z | ≤ 1.645) = 0.9( table of page 848)
Step to find the value of n
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #3(Exercise 7.12)
Hint for #4
Known: σ = 4 and Z =
X̄ −µ
σ
√
n
(page 354)
Goal: how large must n be to make P(|X̄ − µ| ≤ 1) ≥ 0.9 (
to ensure the degree of accuracy means we need to have at
least probability 0.90, which implying our probability need to
be greater or equal to 0.9)
Fact: P(|Z | ≤ 1.645) = 0.9( table of page 848)
Step to find the value of n
Find n to satisfy P( |X̄√−µ|
≤
σ
n
Fang-I Chu
√
n
σ )
= 0.9
PSTAT 120B Probability and Statistics
Hint for #3(Exercise 7.12)
Hint for #4
Known: σ = 4 and Z =
X̄ −µ
σ
√
n
(page 354)
Goal: how large must n be to make P(|X̄ − µ| ≤ 1) ≥ 0.9 (
to ensure the degree of accuracy means we need to have at
least probability 0.90, which implying our probability need to
be greater or equal to 0.9)
Fact: P(|Z | ≤ 1.645) = 0.9( table of page 848)
Step to find the value of n
Find n to satisfy P( |X̄√−µ|
≤
σ
n
√
n
σ )
since P(|Z | ≤ 1.645) = 0.9 and
that
√
n
σ
= 0.9
X̄ −µ
σ
√
n
= Z , We need to find n
= 1.645
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #3(Exercise 7.12)
Hint for #4
Known: σ = 4 and Z =
X̄ −µ
σ
√
n
(page 354)
Goal: how large must n be to make P(|X̄ − µ| ≤ 1) ≥ 0.9 (
to ensure the degree of accuracy means we need to have at
least probability 0.90, which implying our probability need to
be greater or equal to 0.9)
Fact: P(|Z | ≤ 1.645) = 0.9( table of page 848)
Step to find the value of n
√
Find n to satisfy P( |X̄√−µ|
≤
σ
n
σ )
n
since P(|Z | ≤ 1.645) = 0.9 and
√
n
σ
= 1.645
that
since σ = 4, we solve n in
Fang-I Chu
= 0.9
X̄ −µ
σ
√
n
= Z , We need to find n
√
n
4
= 1.645
PSTAT 120B Probability and Statistics
Hint for #4 (Exercise 7.20)
7.20(a)
If U has a χ2 distribution with v df, find E (U) and V (U).
Hint for (a):
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #4 (Exercise 7.20)
7.20(a)
If U has a χ2 distribution with v df, find E (U) and V (U).
Hint for (a):
MGF for U ∼ χ2 (v ): MU (t) =
Fang-I Chu
1
1−2t
v
2
PSTAT 120B Probability and Statistics
Hint for #4 (Exercise 7.20)
7.20(a)
If U has a χ2 distribution with v df, find E (U) and V (U).
Hint for (a):
MGF for U ∼ χ2 (v ): MU (t) =
1
1−2t
v
2
Take derivatives with respect to t, then set t = 0, obtain
E (U) and V (U)
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #4 (Exercise 7.20)
7.20(a)
If U has a χ2 distribution with v df, find E (U) and V (U).
Hint for (a):
MGF for U ∼ χ2 (v ): MU (t) =
1
1−2t
v
2
Take derivatives with respect to t, then set t = 0, obtain
E (U) and V (U)
similar to hw1, #2.
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #4 (Exercise 7.20)
7.20(b)
(b). Using the results of Theorem 7.3, find E (S 2 ) and V (S 2 ) when
Y1 , Y2 , . . . , Yn is a random sample from a normal distribution with
mean µ and variance σ 2
Hint for (b):
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #4 (Exercise 7.20)
7.20(b)
(b). Using the results of Theorem 7.3, find E (S 2 ) and V (S 2 ) when
Y1 , Y2 , . . . , Yn is a random sample from a normal distribution with
mean µ and variance σ 2
Hint for (b):
2
1 Pn
2
results of Theorem 7.3: (n−1)S
=
2
i=1 (Yi − Ȳ ) has a
σ
σ2
χ2 distribution with (n − 1)df. Note Ȳ and S 2 are
independent random variables.
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #4 (Exercise 7.20)
7.20(b)
(b). Using the results of Theorem 7.3, find E (S 2 ) and V (S 2 ) when
Y1 , Y2 , . . . , Yn is a random sample from a normal distribution with
mean µ and variance σ 2
Hint for (b):
2
1 Pn
2
results of Theorem 7.3: (n−1)S
=
2
i=1 (Yi − Ȳ ) has a
σ
σ2
χ2 distribution with (n − 1)df. Note Ȳ and S 2 are
independent random variables.
S2 ∼
σ2 2
n−1 χn−1 .
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #4 (Exercise 7.20)
7.20(b)
(b). Using the results of Theorem 7.3, find E (S 2 ) and V (S 2 ) when
Y1 , Y2 , . . . , Yn is a random sample from a normal distribution with
mean µ and variance σ 2
Hint for (b):
2
1 Pn
2
results of Theorem 7.3: (n−1)S
=
2
i=1 (Yi − Ȳ ) has a
σ
σ2
χ2 distribution with (n − 1)df. Note Ȳ and S 2 are
independent random variables.
σ2 2
n−1 χn−1 .
σ2
E (S 2 ) = n−1
E (χ2n−1 ).
S2 ∼
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #4 (Exercise 7.20)
7.20(b)
(b). Using the results of Theorem 7.3, find E (S 2 ) and V (S 2 ) when
Y1 , Y2 , . . . , Yn is a random sample from a normal distribution with
mean µ and variance σ 2
Hint for (b):
2
1 Pn
2
results of Theorem 7.3: (n−1)S
=
2
i=1 (Yi − Ȳ ) has a
σ
σ2
χ2 distribution with (n − 1)df. Note Ȳ and S 2 are
independent random variables.
σ2 2
n−1 χn−1 .
σ2
E (S 2 ) = n−1
E (χ2n−1 ).
Obtain E (χ2n−1 ) using result
S2 ∼
Fang-I Chu
from (a).
PSTAT 120B Probability and Statistics
Hint for #4 (Exercise 7.20)
7.20(b)
(b). Using the results of Theorem 7.3, find E (S 2 ) and V (S 2 ) when
Y1 , Y2 , . . . , Yn is a random sample from a normal distribution with
mean µ and variance σ 2
Hint for (b):
2
1 Pn
2
results of Theorem 7.3: (n−1)S
=
2
i=1 (Yi − Ȳ ) has a
σ
σ2
χ2 distribution with (n − 1)df. Note Ȳ and S 2 are
independent random variables.
σ2 2
n−1 χn−1 .
σ2
E (S 2 ) = n−1
E (χ2n−1 ).
Obtain E (χ2n−1 ) using result
σ2
2
Var(S 2 ) = (n−1)
2 Var(χn−1 ).
S2 ∼
Fang-I Chu
from (a).
PSTAT 120B Probability and Statistics
Hint for #4 (Exercise 7.20)
7.20(b)
(b). Using the results of Theorem 7.3, find E (S 2 ) and V (S 2 ) when
Y1 , Y2 , . . . , Yn is a random sample from a normal distribution with
mean µ and variance σ 2
Hint for (b):
2
1 Pn
2
results of Theorem 7.3: (n−1)S
=
2
i=1 (Yi − Ȳ ) has a
σ
σ2
χ2 distribution with (n − 1)df. Note Ȳ and S 2 are
independent random variables.
σ2 2
n−1 χn−1 .
σ2
E (S 2 ) = n−1
E (χ2n−1 ).
Obtain E (χ2n−1 ) using result
σ2
2
Var(S 2 ) = (n−1)
2 Var(χn−1 ).
S2 ∼
from (a).
Obtain Var(χ2n−1 ) using result from (a).
Fang-I Chu
PSTAT 120B Probability and Statistics
Similar problem (Exercise 7.43) for #5 (Exercise 7.42)
7.43
An anthropologist wishes to estimate the average height of men for
a certain race of people. If the population standard deviation is
assumed to be 2.5 inches and if she randomly samples 100 men,
find the probability that the difference between the sample mean
and the true population mean will not exceed 0.5 inch.
Solution:
Fang-I Chu
PSTAT 120B Probability and Statistics
Similar problem (Exercise 7.43) for #5 (Exercise 7.42)
7.43
An anthropologist wishes to estimate the average height of men for
a certain race of people. If the population standard deviation is
assumed to be 2.5 inches and if she randomly samples 100 men,
find the probability that the difference between the sample mean
and the true population mean will not exceed 0.5 inch.
Solution:
Known:
Fang-I Chu
PSTAT 120B Probability and Statistics
Similar problem (Exercise 7.43) for #5 (Exercise 7.42)
7.43
An anthropologist wishes to estimate the average height of men for
a certain race of people. If the population standard deviation is
assumed to be 2.5 inches and if she randomly samples 100 men,
find the probability that the difference between the sample mean
and the true population mean will not exceed 0.5 inch.
Solution:
Known: σ = 2.5, n = 100
Fang-I Chu
PSTAT 120B Probability and Statistics
Similar problem (Exercise 7.43) for #5 (Exercise 7.42)
7.43
An anthropologist wishes to estimate the average height of men for
a certain race of people. If the population standard deviation is
assumed to be 2.5 inches and if she randomly samples 100 men,
find the probability that the difference between the sample mean
and the true population mean will not exceed 0.5 inch.
Solution:
Known: σ = 2.5, n = 100
Goal: find the probability that the difference between sample
mean and the true population mean will not exceed 0.5 inch.
(Are you able to write out the goal in mathematical
expression?):
Fang-I Chu
PSTAT 120B Probability and Statistics
Similar problem (Exercise 7.43) for #5 (Exercise 7.42)
7.43
An anthropologist wishes to estimate the average height of men for
a certain race of people. If the population standard deviation is
assumed to be 2.5 inches and if she randomly samples 100 men,
find the probability that the difference between the sample mean
and the true population mean will not exceed 0.5 inch.
Solution:
Known: σ = 2.5, n = 100
Goal: find the probability that the difference between sample
mean and the true population mean will not exceed 0.5 inch.
(Are you able to write out the goal in mathematical
expression?):Find P(|Ȳ − µ| ≤ 0.5)
Fang-I Chu
PSTAT 120B Probability and Statistics
Similar problem (Exercise 7.43) for #5 (Exercise 7.42)
Solution:
Fang-I Chu
PSTAT 120B Probability and Statistics
Similar problem (Exercise 7.43) for #5 (Exercise 7.42)
Solution:
Step to approach: By the Central Limit Theorem (what is
this? review it! ),
Fang-I Chu
PSTAT 120B Probability and Statistics
Similar problem (Exercise 7.43) for #5 (Exercise 7.42)
Solution:
Step to approach: By the Central Limit Theorem (what is
this? review it! ),
P(|Ȳ − µ| ≤ 0.5) = P(−0.5 ≤ Ȳ − µ ≤ 0.5)
= P(
= P(
−0.5
σ
√
n
−0.5
2.5
10
≤
Ȳ − µ
σ
√
n
≤Z ≤
≤
−0.5
2.5
10
0.5
σ
√
n
)
= P(−2 ≤ Z ≤ 2)
= 0.9544
Fang-I Chu
PSTAT 120B Probability and Statistics
)
Hint for #6 (Exercise 7.54)
7.54
Many bulk products - such as iron, ore, coal, and raw sugar - are
sampled for quality by a method that requires many small samples
to be taken periodically as the material is moving along a conveyor
belt. The small samples are then combined and mixed to form one
composite sample. Let Yi denote the volume of ith small sample
from a particular lot and suppose that Y1 , Y2 , . . . , Yn constitute a
random sample, with each Yi value having mean µ (in cubic
inches) and variance σ 2 . The average volume µ of the samples can
be set by adjusting the size of the sampling device. Suppose that
the variance σ 2 of the volumes of the samples is known to be
approximately 4. The total volume of the composite sample must
exceed 200 cubic inches with probability approximately 0.95 when
n = 50 small samples are selected. Determine a setting for µ that
will allow the sampling requirement to be satisfied.
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #6 (Exercise 7.54)
Hint for #6:
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #6 (Exercise 7.54)
Hint for #6:
Known:
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #6 (Exercise 7.54)
Hint for #6:
Known: σ 2 = 4, n = 50
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #6 (Exercise 7.54)
Hint for #6:
Known: σ 2 = 4, n = 50
Goal: the probability that the composite sample must exceed
200 cubic inches is approximately 0.95.
Find µ satisfy such sampling requirement. (Are you able to
write out the goal in mathematical expression?):
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #6 (Exercise 7.54)
Hint for #6:
Known: σ 2 = 4, n = 50
Goal: the probability that the composite sample must exceed
200 cubic inches is approximately 0.95.
Find µ satisfy such sampling requirement. (Are you able to
write out the goal in mathematical expression?):
P(Ȳ > 200) ≈ 0.95
Fang-I Chu
PSTAT 120B Probability and Statistics
Hint for #6 (Exercise 7.54)
Hint for #6:
Known: σ 2 = 4, n = 50
Goal: the probability that the composite sample must exceed
200 cubic inches is approximately 0.95.
Find µ satisfy such sampling requirement. (Are you able to
write out the goal in mathematical expression?):
P(Ȳ > 200) ≈ 0.95
Step to approach: Apply the Central Limit Theorem as in
Exercise 7.43.
Fang-I Chu
PSTAT 120B Probability and Statistics