1. SOLVED EXAMPLES

2. Example 1
Rubber tires are known to have 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 to Find the standard error in using the sample mean to ESTIMATE the population mean.

3. 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.
Show that = X/n is an unbiased estimator of p
Show that the standard error of is

4. 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
Show that is an unbiased estimator of µ
If are independent , Find the value of that minimizes
the standard error of

5. Now Find
and

6. Therefore,
Then S2 is Unbiased Estimator of σ2

7. Z is Standard Normal variable
To proceed further to find SE(S2), Bias (S) and MSE(S),
a new important distribution should be considered.
Standard Normal
σ2=1 Z
It is the CHI-SQUARE distribution
Consider the variable ζ1 = Z2 where, n
Z is Standard Normal variable 2 ζ
ζ1 is a Chi-square variable with one degree of freedom
has mean=1 and variance = 2 1
Chi-square
ζ2 is a Chi-square variable with two degrees of freedom
has mean =2 and variance=4
ζn is a Chi-square variable with n degrees of freedom
has mean =n and variance = 2n

8. is a Chi-square variable with (n-1) degrees of freedom ζn-1 .
Mean = n and Variance = 2 (n-1)
Then

9. Find the Standard Error SE of Estimator

10. BIAS and MEAN SQUARE ERROR OF ESTIMATOR S
Chi-Square distribution
The Probability density function of ζn-1 is given by:
Then
S is a Biased Estimator of σ

12. NONCONVENTIONAL PARAMETERS
ESTIMATION OF
NONCONVENTIONAL PARAMETERS
Conventional Parameters are mainly:
Mean, Variance, Standard Deviation, Quartiles
Consider the following Population Probability Density Functions
Parameters Ө,β and η in the above distribution are Nonconventional
There are TWO Methods for Deriving ESTIMATORS
For these PARAMETERS
1- Method of MOMENTS
2- Method of MAXIMUM LIKELIHOOD FUNCTION

13. METHOD OF MOMENTS

14. Consider the following Probability Density Function
Required to find an expression to Evaluate an Estimator of the Parameter θ
By the method of Moments

15. MAXIMUM LIKELIHOOD
FUNCTION

16. Given a SAMPLE of size N: x1, x2, x3,…, xN
Taken from the population with known distribution f(x|Ө)
Likelihood Function is the Joint Probability Function Of
All points of the SAMPLE
The idea of the method of Maximum Likelihood is to determine
The expression of the parameter Ө that makes L(Ө) MAXIMUM
This could be done by Differentiating L(Ө) w.r.t Ө and equating to zero

17. A population is defined by the following probability density function
Example 1
A population is defined by the following probability density function
Find the Maximum Likelihood Estimator of Ө
Now, the Likelihood Function of the population is given by:
Taking the logarithm of L(Ө)
Differentiate w.r.to Ө and equate to zero, we find
Consider the first moment of the f(x)
Therefore

18. Example 2
A population is defined by the BINOMIAL distribution as follows:
Find the Maximum Likelihood Estimator of p
The Likelihood Function of the population is given by:
Taking the logarithm of
the likelihood function
Differentiate w.r.to
and equate to zero

19. A population is defined by the following probability density function
Example 3
A population is defined by the following probability density function
Find the Maximum Likelihood Estimator of Ө
Now, the Likelihood Function of the population is given by:
Taking the logarithm of L(Ө)
Differentiate w.r.to Ө and equate to zero, we find
Therefore, the Maximum Likelihood Estimator MLE
of the parameter Ө
can be readily found as follows:

20. Example 4
A population is defined by the following Normal Distribution
Find the Maximum Likelihood Estimators of µ, σ
Now, the Likelihood Function of the population is given by:
Taking the logarithm of L(µ, σ)
Differentiate partially w.r.to µ and σ and equate to zero, we find