Transcript Part 1 Point Estimators1

STATISTICAL
ANALYSES
IM342
By
Prof. Dr. Ahmed Farouk Abdul Moneim
1
INTRODUCTION
STATISTICS
is concerned with:
1. Collection of Data from Observations, Experiments,…etc
Data Populations are the complete set of data about
Systems attributes.
Data
Samples
are
Subsets
of the Population.
2. Presenting the Collected data in the form of Histograms, Box
Plots, Time series,…….etc
3. Processing these Data in order to render them amenable
to practical applications in:
• Engineering
• Management
and Decision-Making.
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STATISTICAL ANALYSES
are concerned with:
1. Inferences about POPULATIONS
from data SAMPLES collected from Observations and
Experiments
(Induction approach of Science).
‫المنهاج اإلســتقرائى‬
2. TESTING HYPOTHESES About PARAMETRS of Gven
Probability Distributions of Observed Variables
(Mean Life Time of Products), (Comparisons among different Products)
3. DESIGNING EXPERIMENTS and Devicing Methods for
PROCESSING Data Collected from these Experiments
STATISTICAL
ESTIMATIONS
5
THE NEEDS FOR
SAMPLING
Economic and Social Indicators
Material Physical and Chemical Properties
Reliability and Safety Parameters
Ergonomics and Human Factors
6
Economic Indicators
• Per capita income as a measure of prosperity
• Inflation Rate or Price Index Increase Rate
• Unemployment Ratio
Social Indicators
• Illiteracy Rate
• Age Pattern
7
Material Physical and Chemical
Properties
• Ultimate Tensile Strength and Yield Point
Of Metals
• Magnetic Permeability of Metals
• Percentage of Alloying Elements
• Resistance to Corrosion of Different Steels
8
Reliability and Safety
Parameters
Mean Time Between Failures MTBF
Systems Availability
Maintainability of Equipment
Design Factor of Safety
9
Ergonomics and Human
Factors
Anthropometric Measures
Time and Motion Studies
Human Reliability
10
REMEMBER!
Random Variables
Type
Discrete
Continuous
Range
x1, x2 ,…, xn
Continuous from
x1 to xn
Probability
Distribution
Mass Function
f(x)
Probability Density
Function
f(x)
MEAN =
Expected
Value =E(X)
Variance =
Var(x)
E( X ) 
n
x
i
f ( xi )
i 1
Var( X ) 
n
x
i 1
f ( xi ) 
2
i
 xf ( x )dx
x1
Var ( X ) 
xn
x
2
f ( x )dx
x1
- E ( X )
- E ( X )
2
2
 E ( X 2 )  E ( X )
2
Examples
E( X ) 
xn
Uniform, Bernoulli,
Binomial, Geometric,
Negative Binomial,
Poisson
 E ( X 2 )  E ( X )
2
Uniform, Normal,
Exponential, Erlang,
Weibull,…
PROBABILITY DISTRIBUTIONS
N
DISCRETE
E ( X )  (1 / N ) xi var ( X )  (1 / N ) xi2  E ( X )
f ( x)  1 / n
UNIFORM
BINOMIAL
2
i 1
f ( x )  C xN p x (1  p ) N  x E ( X )  Np var ( X )  Np(1  p )
f ( x )  (1  p) x 1 p
GEOMETRIC
E ( X )  1 / p var ( X )  (1  p) / p 2
NEGATIVE BINOMIAL f ( x )  C x r (1  p ) x r p r E ( X )  r / p var( X )  r(1  p) / p 2
r 1
POISSON
f ( x) 
x
x!
e 
E ( X )   var ( X )  
CONTINUOUS
UNIFORM f ( x ) 
NORMAL
1
ba
f (x) =
EXPONENTIAL
In the range from a to b
1
e
s 2p
æ x-m ö2
-ç
÷
è s ø
f ( x )  e  x
E ( X )  (a  b) / 2 var( X )  (b  a ) 2 / 12
With mean µ and variance σ2
With mean 1/λ and variance 1/λ2
POINT
ESTIMATORS
Population
Parameters
Sample
Estimators
Mean

X
Variance
2
S2

and

X
and
S2
Are CONSTANTS BUT Difficult and
2 Even Mostly Impossible to determine
Because of Time and Cost Barriers
Are RANDOM VARIABLES
Why?
They are obtained Easily from SAMPLE DATA
14
X
Is a Random REQUIRED to FIND:
• Its MEAN, VARIANCE and
Variable
• Its PROBABILITY DISTRIBUTION FUNCTION
To be able to do that,
we should consider the following Important Theorem
CENTRAL LIMIT THEOREM
Given
X 1 , X 2 , X 3 ,......, X n
N Random Variables with
Different Arbitrary probability Distribution Functions
Their
Sum
Y  X 1  X 2  X 3  ........ X n
n, Y is a Random Variable Distributed
According to NORMAL DISTRIBUTION
For large values of
15
The Proof of the CENTRAL LIMIT theorem is on an Excel sheet
Find Mean and Variance of
X
1 n
1
X   X i  X 1  X 2  X 3  ...  X n 
n i 1
n
A General Rule
If Y = a X, then
E(Y)= a E(X)
Var(Y) = a2 Var(X)
 
1
E  X 1   E  X 2   E  X 3   ...  E  X n 
n
1
n
E X        ...       
n
n
E X 
 
The n points of Sample are
Taken from the same population
Therefore,
EX 1   EX 2   EX 3   EX n  
 

Population Mean
1
Var X 1   Var X 2   Var X 3   ...  Var X n 
2
n
1 2
n 2  2 Mean of X  
2
2
2
Var X  2       ...    2  
n
n
n Variance of X   2
Var X 
 


n
Standard Deviation of X 

n
16
The Sample Mean
Is used as an
X
ESTIMATOR
of the population Mean

Similarly
The Sample Variance
could be used as
Generally, for each
ESTIMATOR
of the population Variance
POPULATION PARAMETER
There is A corresponding
OR
S2
SAMPLE ESTIMATOR
SAMPLE STATISTIC
2



17
The selection of the most appropriate
SAMPLE STATISTIC
as an ESTIMATOR for a POPULATION PARAMETER
Is governed by the
1.



following Properties
BIAS evaluated as follows:
 

Bias  E   
2.
STANDARD ERROR SE evaluated as follows:

SE  Var 
 
3.
MEAN SQUARE ERROR MSE


MSE  E   

2
 
evaluated as follows:

 Var   Bias 2
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Example
Evaluate the Bias and SE and MSE of the following ESTIMATORS:
X , S 2 and S
For the population parameters:
,  2 , 
 
 
Bias X  E X        0
Then
X
is an Unbiased Estimator of µ
The Standard Error
 
SE  Var X 

n
Find the Mean Square Error of Estimator
 
MSE  Var X  Bias  
2
2
n
0
X
2
n
19
SOLVED EXAMPLES
Example 1
Rubber tires are known to have 20000 km as mean life time and 5000 km as
standard deviation. A Sample of 16 rubber tires are selected randomly, what is
the probability that the SAMPLE MEAN will be in the interval from 17000 to
22000. Find the standard error in using the sample mean as an ESTIMATOR for
the population mean.
P(17000 < X < 22000)
X is normal variable with mean m = 20000 and standard deviation S
5000
s s 5000
SE
=
=
= 1250
S=
=
= 1250
n n 16 16
17000--20000
20000 XX--mm 22000
22000--20000
20000
17000
P(17000
<
X
<
22000)
=
P(
<
<
)
P(17000 < X < 22000) = P(
<
<
)
1250
1250
1250
SS
1250
P(-2.4<<ZZ<<1.6)
1.6)==F(1.6)
F(1.6)--F(-2.4)
F(-2.4)==0.945
0.945-.0082
-.0082==0.937
0.937
P(-2.4
Example 2
Suppose that X is the number of observed “successes” in a sample
of n observations where p is the probability of success on
each observation.
a) Show that p̂ = X/n is an unbiased estimator of p

b) Show that the standard error of p is
p(1  p) / n
X is a Random Variable distributed in BINOMIAL distribution
X is a Binomial Variable then,
E ( X )  np Var ( X )  np (1  p )
X
Estimator pˆ of p : pˆ  , then E ( pˆ )  E ( X ) / n  np / n  p
n
Bias ( pˆ )  E ( pˆ )  p  p  p  0
SE ( pˆ )  Var ( pˆ )  Var ( X / n)  (1 / n 2 )Var ( X )
SE ( pˆ )  (1 / n 2 ) * np (1  p ) 
Estimator
p̂
p (1  p ) / n
is an Unbiased Estimator of
p
Example 3
Let X be a random variable with mean μ and variance б2
Given two independent random samples of sizes n1 and n2, with sample means
2
2
X 1 , X 2 E( X 1 )  E( X 2 )  
Var ( X 1 ) 
Var ( X 2 ) 
n1
n2
Show that X  a X 1  (1  a ) X 2
0  a  1 is an unbiased estimator of µ
If X 1 , X 2 are independent , Find the value of
the standard error of
a
that minimizes
X
Var ( X )  a 2Var ( X 1 )  (1  a ) 2 Var ( X 2 )
a
E ( X )  aE ( X 1 )  (1  a ) E ( X 1 )
 a  (1  a )   
Bias ( X )  E ( X )        0
2
2
n1
 (1  a )
2
2
n2
d
2
2
Var ( X )  2a
 2(1  a )
0
da
n1
n2
a
1 a

0
n1
n2
a
n1
n1  n2
a(
1
1
1

)
n1
n2
n2
Example 4
Prove that Sample Variance
of population Variance
 
 
Bias S 2  E S 2   2

n
1
S2 
Xi  X

n  1 i 1

2
S2
is an unbiased Estimator
2

2
1 n
2

X i  2X i X  X

n  1 i 1

2
2
2
n
n
 1 n 2  2n
 1 b 2
S 
Xi  
X 
X 
Xi  
X


n 1
 n  1 i 1
 n 1
 n  1 i 1
 n 1
2
 
n
2
1
n

2
2 
E S   
E X i  
E X

 n  1 i 1
 n 1
Now Find
 
E Xi
2
and
 
EX
2
Var  X i   E X i2   E  X i   E X i2    2
2
E X i2   Var  X i    2   2   2
then,
   E X
Var X
 
then, E X
2
2
  E X 
2
2
n


2
n
Therefore,
2
 
2
n
 1 n
2 
E S   
E X i  
E X

 n  1 i 1
 n 1
2
2

n
n
n


E S 2  
2 
2 
n 1
n 1
n 1 n
E S 2  
E S
n 1 2

n 1
Then S2 is an Unbiased Estimator of σ2
 
BiasS   E S   
2
2
2
2

n
 
2
 n 1
2
0
25