4.2 Sampling Distributions Associated with
Normal Populations
Ulrich Hoensch
Monday, March 8, 2010
Linear Combinations of Normal Distributions
Theorem 4.2.1
Let X1 , X2 , . . . , Xn be independent random variables with
Xi ∼ P
N(µi , σi2 ). Let Y be a linear combination of the Xi ’s, that is:
Y = ni=1 ai Xi with real constants
Pn ai . Then, Y has a normal
distribution
with
mean
µ
=
Y
i=1 ai µi and variance
P
σY2 = ni=1 ai2 σi2 .
Corollary 4.2.2
Let X1 , X2 , . . . , Xn be a random sample of size n from a normally
distributed population with mean µ and variance σ 2 . Then,
X
σ2
X ∼ N µ,
, and
X ∼ N nµ, nσ 2 .
n
Remark. This corollary asserts that if the population has a normal
distribution, the statements of the Central Limit Theorem are true
for any sample size, not only for “n large.”
Chi-Square Distributions
Theorem 4.2.3
Let X1 , X2 , . . . , Xk be independent
Pkrandom variables with
2
Xi ∼ χ (ni ). Then the sum V = i=1 Xi has the distribution
V ∼ χ2 (n1 + n2 + . . . + nk ).
Theorem 4.2.4
Let X1 and X2 be independent random variables, and suppose
X1 ∼ χ2 (n1 ) and Y = X1 + X2 ∼ χ2 (n) with n > n1 , Then,
X2 = Y − X1 ∼ χ2 (n − n1 ).
Remarks.
I
Recall that for X ∼ χ2 (n), µ = E (X ) = n,
σ 2 = Var (X ) = 2n, and MX (t) = (1 − 2t)−n/2 .
I
X ∼ χ2 (n) is a special case of a gamma distribution with
α = n/2 and β = 2.
Chi-Square Distributions
Theorem 4.2.5
If X has a gamma distribution with parameters α > 0 and β > 0,
then
2X
∼ χ2 (2α).
Y =
β
Theorem 4.2.6
If X ∼ N(0, 1), then X 2 ∼ χ2 (1).
Theorem 4.2.7
Let X1 , X2 , . . . , Xn be a random sample of size n from a normally
distributed population with mean µ and variance σ 2 . Then,
Zi = (Xi − µ)/σ are independent N(0, 1)-distributed random
variables, and
n
n X
X
Xi − µ 2
2
∼ χ2 (n).
Zi =
σ
i=1
i=1
Distribution of the Sample Variance
Theorem 4.2.8
Let X1 , X2 , . . . , Xn be a random sample of size n from a normally
distributed population
with mean µ and variance σ 2 . Let
P
S 2 = (1/(n − 1)) ni=1 (X − X )2 be the sample variance. Then,
(n − 1)S 2
∼ χ2 (n − 1).
σ2
(b) X and S 2 are independent random variables.
(a)
Remarks.
I
Corollary 4.2.2 and Theorem 4.2.8 give us information about
the distribution of the sample mean and the sample variance
of a random sample taken from a normally distributed
population.
I
We can use χ2 cdf (a, b, n) to find the probability that a
chi-square distributed random variable with n degrees of
freedom is between a and b.
Example
Suppose X1 , X2 , . . . , X10 is a random sample taken from an
N(12, 25)-distributed population.
I
Find the range for the middle 90% of the distribution of the
sample mean X .
Example
I
Find the probability that the sample variance S 2 is between 20
and 30.
Example
I
Find the range for the middle 90% of the distribution of the
sample variance.
Student t-Distribution
Definition 4.2.2
2
Suppose Y and Z are independent
p random variables, Y ∼ χ (n)
and Z ∼ N(0, 1). Then, T = Z / Y /n has a (Student)
t-distribution with n degrees of freedom. We write T ∼ t(n) in
this situation.
Remarks.
I
The PDF of T ∼ t(n) is symmetric about the line t = 0 and
equal to
Γ((n + 1)/2)
f (t) = √
πnΓ(n/2)
−(n+1)/2
t2
1+
.
n
I
If T ∼ t(n), then E (T ) = 0 and Var (T ) = n/(n − 2).
I
For n ≥ 30, we have that a t(n)-distribution can be
well-approximated by a N(0, 1)-distribution.
Student t-Distribution
n=2
n=5
n=10
n=30
-4
-2
0
The PDF of T ∼ t(n) for n = 2, 5, 10, 30.
2
4
Student t-Distribution
Theorem 4.2.9
Suppose X is the mean and S 2 is the variance of a random sample
of size n taken from a normally distributed population with mean µ
and variance σ 2 . Then the random variable
T =
X −µ
√ ∼ t(n − 1).
S/ n
Proof: By Corollary 4.2.2, X ∼ N(µ, σ 2 /n), so
√
Z = (X − µ)/(σ/ n) ∼ N(0, 1). By Theorem 4.2.8,
Y = (n − 1)S 2 /σ 2 ∼ χ2 (n − 1). Since X and S 2 are independent
by Theorem
p 4.2.8, so are Z and Y . Thus, by Definition 4.2.2,
T = Z / Y /(n − 1) ∼ t(n − 1). But now
X −µ
√
σ/ n
X −µ
√
σ/ n
X −µ
√ .
T =p
=q
=
=
2
(n−1)S
S/σ
S/ n
Y /(n − 1)
2
Z
σ (n−1)
Example
A random sample of 20 soda can yields an average of 235 ml, with
a sample standard deviation of 10 ml. Assuming that the volume
of soda in the cans is normally distributed and that the population
mean is 240 ml, find the probability that the sample mean is equal
to or less than the observed value.
F -Distribution
Definition 4.2.3
Suppose U and V are independent random variables, U ∼ χ2 (n1 )
and V ∼ χ2 (n2 ). Then, F = (U/n1 )/(V /n2 ) has a (Fisher)
F -distribution with n1 numerator and n2 denominator degrees
of freedom. We write F ∼ F (n1 , n2 ) in this situation.
Remarks.
I
The PDF of F ∼ F (n1 , n2 ) is zero if x ≤ 0 and for x > 0 it is
equal to
Γ((n1 + n2 )/2)
f (x) =
Γ(n1 /2)Γ(n2 /2)
I
n1
n2
n1 /2
x
(n1 /2)−1
n1
1+ x
n2
−(n1 +n2 )/2
The F -distribution is skewed to the right, and if Fα (n1 , n2 ) is
chosen so that P(F > Fα (n1 , n2 )) = α, then
Fα (n1 , n2 )F1−α (n2 , n1 ) = 1.
.
F -Distribution
0
1
2
3
4
5
The PDF of F ∼ F (10, 5). From Table AIV.5, we find that
F0.1 (10, 5) ≈ 3.30. The shaded area equals 10%. (You can verify
that Fcdf (3.30, 1E 99, 10, 5) ≈ 0.1.)
F -Distribution
Theorem 4.2.10
Suppose a random sample of size n1 is drawn from a normally
distributed population with variance σ12 , and (independently)
another random sample of size n2 is drawn from a normally
distributed population with variance σ22 . If S12 is the variance of the
first sample, and S22 is the variance of the second sample, then
F =
S12 /σ12
∼ F (n1 − 1, n2 − 1).
S22 /σ22
Proof: By Theorem 4.2.8, U = (n1 − 1)S12 /σ12 ∼ χ2 (n1 − 1) and
V = (n2 − 1)S22 /σ22 ∼ χ2 (n2 − 1). Note that U and V are
independent. By Definition 4.2.3,
F = (U/(n1 − 1))/(V /(n2 − 1)) ∼ F (n1 − 1, n2 − 1). But now
F =
S 2 /σ 2
U/(n1 − 1)
= 12 12 .
V /(n2 − 1)
S2 /σ2
Example
Two random samples are taken from two different normally
distributed populations. It is assumed that these populations have
the same variance σ 2 = σ12 = σ22 . Suppose the size of the first
sample is n1 = 20 and the size of the second sample is n2 = 30.
I
Find the probability that the ratio of the sample variances
S12 /S22 is between 0.8 and 1.2.
Example
I
Find two values a and b so that P(a ≤ S12 /S22 ≤ b) = 0.90.
Homework Problems for Section 4.2 (Points)
p.204-206: 4.2.1 (3), 4.2.7 (2), 4.2.8 (2), 4.2.10 (3), 4.2.11 (2),
4.2.19 (3).
Homework problems are due at the beginning of the class on
Wednesday, March 17.