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Distributions of sampling
statistics
Chapter 6
Sample mean & sample variance
Sample vs. population


A population is a large collection of items
that have measurable values associated
with the experimental study
A proper sampling technique is adopted to
select items, that is so called a sample,
from the large collection in order to draw
some conclusions about the population.

From selecting the small one to prospecting
the large whole
Definition of sample




If X1,X2,X3,…,Xn are i.i.d. variables with the
distribution F, then they constitute a sample from
the distribution F.
The population distribution F is usually not specified
completely. And sometimes it is supposed that F is
specified up to a set of unknown parameters.
The parametric inference problem emerges.
A statistics is a random variable whose value is
determined by the sample data and used to
inference the supposed parameter.
The sample mean
X 1  X 2  ... X n
X 
n
_





E[X]=E[(X1+X2+…Xn)/n]=(1/n)(E[X1]+E[X2]+…E[Xn])
=μ
Var(X)=Var{(X1+X2+…Xn)/n}
=(1/n2)(nσ2 )=σ2/n
c.f. population mean & variance: μ,σ2
If the sample size n increases, then the sample
variance of X will decrease. See fig. 6.1
The central limit theorem
Let X1,X2, …, Xn be a sequence of i.i.d. random variables
each having mean μ and variance σ2
•The sum of a large number of independent random
variables has a distribution that is approximately normal
X 1  X 2  ... X n is approximat ely normal with mean n and variance n 2
X 1  X 2  ... X n  n
is approximat ely a standard normal random variable
 n
See example 6.3b and p.206 the binomial trials
Approximate distribution of
the sample mean
_
n
X   X i / n,
i 1
_
X   n 

 Z (0,1)
/ n
•See example 6.3d, 6.3e
How large a sample is needed?


If the underlining population distribution
is
_
normal, then the sample mean X will also be
normal regardless of the sample size.
A general thumb is that one can be confident
of the normal approximation whenever the
sample size n is at least 30.
The sample variance
n
sample variance S 2 
(X
i 1
_
i
 X )2
sample standard deviation S 
 ( n  1) S
n
(X

2
i 1
_
i
,
n 1
 X)
2

S2 ,
n
X
i 1
_
i
_
 n
2
 ( n  1) E[ S ]  E  X i   nE[ X
 i 1


 E[ X
i 1
2
i
_
2
] nE[ X
]
n
 {Var ( X
i 1
_
 n
2
 n
 n
2
 n 2  n(
2
 E[ S 2 ]  
,
2
2
n
2
nX
2
]
i
_
 n{Var ( X )  E[ X ]2 }
2
2
_
)  E[ X i ] }  nE[ X
2
/ n)  n 2  ( n  1)
, an unbiased estimator
2
,
2
]
_
Joint distribution of X and S
n
 ( X i  )
i 1

n
2

2
_
2
(
X

X
)
 i
i 1

2

 Xi   
 Xi  X

  

 

i 1 
i 1 

n
_
2
n
A chi-square distribution
with n degree of freedom
2
_

n( X   ) 2

2




  


2
,

n( X  ) 



_
2
A chi-square distribution
with 1 degree of freedom
A chi-square distribution
with n-1 degree of freedom
Implications
If X1,X2,X3,…,Xn, is a sample from a normal
population having mean μ and
variance σ2, then

_
2
X 
X ~ N ( ,
),
~ Z (0,1)
n
/ n
2
2
S
( n  1)
~
x
( n  1),
2
_

_
n
_
( X  )
( X  )

~ t n 1
S
S
n
Sampling from a finite
population
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A binomial random variable


2
E[X]=np,
σ
=np(1-p)
_
If X is the proportion of the sample that has a
special characteristic and equal to X/n, then
_
E[ X ]  E[ X ] / n  p, S   / n 
p(1  p) / n
By approximation:
_
( X  )
Z 

S
 p)
n
~ Z (0,1)
p (1  p ) / n
(X
Homework #5

Problem 8,10,,15,23,28