Al-Azhar University – Gaza MATHEMATICAL STATISTICS

MATHEMATICAL STATISTICS Gaza –Azhar University -Al

Dr. Abuzaid. A. H. 1

Al- Azhar University - Gaza

Faculty of Science

Department of Mathematics

Course Synopsis

Bachelor. Degree Program:

MATH 4322- Mathematical Statistics ( 3 credits) Course:

English Medium of Instruction:

According to the timetable. Lecture Day/Time:

Assistant. Prof. Dr. Ali Hassan Abu Zaid Lecturer:

[email protected] Mobile / E-mail:

MATH 3316, Probability Theory. Pre-requisite:

Sampling distributions - Estimation theory - Applications to statistical

estimation -Testing hypothesis - Applications to statistical hypotheses.

Synopsis:

1. I. Miller, M. Miller, John E. Freund's Mathematical Statistics (6th

Edition), Prentice Hall, 2003. (Textbook.)

2. R. E. Walpole, R. H. Myers, S. L. Myers, Probability and

Statistics for Engineers and Scientists, Prentice Hall College,

1997.

3. Any Introductory Mathematical Statistics book.

Main Reference:

Lecture notes. Teaching Material:

Lectures and discussions. Teaching Method:

(Week 9) 35% Mid Term Test Evaluation Mode:

15% Quizzes and home work

50% Final Exam

mailto:[email protected]



Al- Azhar University - Gaza

Faculty of Science

Department of Mathematics

MATH 4322- Mathematical Statistics

Teaching Schedule

Week LECTURE TOPIC / COURSE ASSIGNMENT

1 Introduction to mathematical statistics.

2 Introduction to sampling distributions

3 Sampling distribution of the mean, Law of large numbers,

4 . Central limit theorem, Sampling distribution of sample variance.

5 Sampling distribution for the variance and variance–ratio.

6 Introduction to estimation, biasness, efficiency and consistency of estimators

7 Method of estimation (Moments).

8 Method of estimation (maximum likelihood).

9 Mid Term Test.

10 Confidence interval for the mean, and differences between means.

11 Confidence interval for proportions and differences between proportions.

12 Confidence interval for variances the ratio of two variances

13 Logical argument of hypothesis, Errors in hypothesis testing, power of the test.

14 Neyman-pearson lemma, Tests concerning means and differences between means.

15 Tests concerning means and differences between variances.

16 Revision

Final Exam.



MATHEMATICAL STATISTICS

Common Probability Distributions

Name pdf Mean Variance mgf

Bernoulli

elsewhere

xp

xp

pxf

0

01

1

),(

10 p

p )1( pp tpeq

)1( pq

Binomial xnxqp

x

npnxf

),,(

nx ,...,1,0

np npq ntpeq

Geometric pqpxf x 1),(

10 p , ,...1,0x p

1

2p

q

t

t

qe

pe

1

Poisson

!),(

x

exf

x

,...1,0x , np

))1(exp( te

Uniform

abbaxf

1),,(

bxa 2

ba

12

2ab

)( abt

ee tatb

Normal

2

2

2

)(exp

2

1),,(

xxf

,x , 0

2

2exp

22tt

Gamma

x

exxf

1

)(

1),,(

0,,0 x

2 )1( t

Exponential

elsewhere0

01

),( xexf

x

0

2 1̀)1( t

Chi - square

elsewhere0

022

1

),(22

2

2xex

xfx

2 2)21( t



CHAPTER 8

SAMPLING DISTRIBUTIONS

Objective: To study the probability distributions of various sample statistics.

8.1 Introduction:

There are two types of statistics:

• Descriptive statistics: deal with the enumeration, organization, and graphical representation of

data.

• Inferential statistics: concerned with reaching conclusions from incomplete information, that is,

generalizing from the specific sample

Population: A large set or collection of items that have something in common.

Parameter: every calculated value for the population, which is a constant value for the same

population.

Sample: is a set of observable random variables, nXXX ,...,, 21 , where n is the sample size.

Most of inferential procedures deal with random samples.

The random variables, nXXX ,...,, 21 are identically distributed if every iX has the same

probability distribution.

Random Sample of size n is a set of n independent and identically distributed (i.i.d.) observable

random variables, nXXX ,...,, 21 .

Statistic: is a function T of observable random variables, nXXX ,...,, 21 that does not depend on

any unknown parameters.

Example:

Sample mean,

n

i

ixn

x1

1, Sample variance

n

ii xx

ns

1

22 )(1

1

Different samples from same population give different values of certain statistics. Thus, a

sample statistic is a random variable. Hence, the statistic has a probability distribution.

The probability distribution of a sample statistic is called the Sampling Distribution.

Example:

Let ={1,2,3,4,5}. Consider all possible samples consisting of 3 numbers randomly chosen

without replacement. Obtain the distribution of the sample mean.

Solution: There are 103

5

samples of size 3.

(1,2,3), (1,2,4), (1,2,5),(1,3,4), (1,3,5), (1,4,5), (2,3,4), (2,3,5), (2,4,5), (3,4,5)

The sampling distribution of x is given below

x 6/3 7/3 8/3 9/3 10/3 11/3 12/3

)(xp 0.1 0.1 0.2 0.2 0.2 0.1 0.1



8.2 Sampling Distribution of the Mean: Infinite Population (with replacement).

Theorem: Let nXXX ,...,, 21 be a random sample of size n from a population with mean and

variance 2 . Then )(xE and .)var(2

nx

Proof:

We denote xxE )( and .)var( 2xx

n

x

is called (standard error) of the mean. It is the standard deviation of all possible

sample means.

The variance of the sample mean is n

2 which is smaller than the population variance 2 , for

.2n

Theorem: (Law of Large numbers)

Let 0 be any positive number, then 2

2

1

nxP .

Proof:



Theorem: (Central Limit Theorem)

If nXXX ,...,, 21 is a random sample from an infinite population with mean and variance 2 ,

and the moment generating function )(tM x then the n

xZ

n /lim

.

Proof:



Results of Central Limit Theorem for Several Populations.

CLT does justify approximating the distribution of x with a normal distribution having mean

and the variance n

2 when n is large.

In practice, this approximation is used when .30n

Example: A random sample of size n = 64 is taken from a normal population with = 51.4 and

= 6.8. What is the probability that the mean of sample will

(a) exceed 52.9 (b) fall between 50.5 and 52.3 (c) be less than 50.6.

Solution:

(a)

(b)

(c)



Example:

A particular brand of drink has an average of 12 ounces per can. As a result of randomness, there

will be small variations in how much liquid each bottle really contains. It has been observed that the

amount of liquid in these bottles is normally distributed with σ = 0.8 ounce. A sample of 10 bottles

of this brand of soda is randomly selected from a large lot of bottles, and the amount of liquid, in

ounces, is measured in each. Find the probability that the sample mean will be within 0.5 ounce of

12 ounces.

Solution:

Theorem: If x is the mean of a random sample of size n from the normal distribution, with

mean, and variance, 2 , then the sampling distribution of x is also normally distributed

with mean, and variance, n

2.

Proof:



8.3 The Distribution of the Mean: Finite Population (without replacement).

If nXXX ,...,, 21 are random variables from a finite population with size n, then nxxx ,...,, 21

constitute a random sample. The number of such possible ordered samples is

)1)....(1()!(

!

nNNN

nN

NPn

N

The joint probability distribution is given by:

!

)!(

)1)....(1(

1),...,( 1

N

nN

nNNNxxf n

.

Regardless of the order, the number of all un ordered possible sample is

)!(!

!

nNn

N

n

NCn

N

Thus, the joint probability distribution is

)1)....(1(

!

!

)!(!1),...,( 1

nNNN

n

N

nNn

n

Nxxf n

Definition: The joint marginal probability distribution of any two of the random variables, say rX

and sX is given by

)1(

1),(

NNxxg

sr

Theorem: If rX and sX are the thr and ths random variables of a random sample size n drawn

from finite population },...,,{ 21 Nccc then

1),cov(

2

NXX

sr

Proof:



Theorem: If x is the mean of a random sample of size n from a finite population of size N with

mean and variance 2 . Then )(xE and .1

)var(2

N

nN

nx

Proof:

1N

nN is called the correction factor, where 1

1lim

N

nN

N

Exercises:

Q.8.1 If nXXX ,...,, 21 is a random sample from infinite population. Show that

0),cov( XXX r , for r =1, 2, …,n.



Q.8.3 If two samples 1

X and 2

X with sizes 1

n and 2

n are generated from two normal

populations ),( 2

11N and ),( 2

22N , respectively. Show that

2

2

2

1

2

1

2121,~

nnNXX .

.

Q. 8.13 If a random sample of size n is selected from the finite population which consists of the

integers 1,2, …., N, show that:

(a) the mean of X is ;2

1N

(b) The variance of X is ;12

))(1(

n

nNN



(c) The mean and variance of XnY are 2

)1()(

NnYE and ;

12

))(1()var(

nNNnY

Q 8.18 How many different samples of size n=3 can be drawn from a finite population of size (a)

N=12

Q 8.19 What is the probability of each possible sample if (a) a random sample of size n = 4 is to be

drawn from a finite population of size N=12.

Q. 8.33 The Actual proportion of families in Gaza who own rather than rent their home is 0.7. If 84

families are interviewed at random and their answers to the question whether they own their

home are recorded. What is the probability that the sample proportion ̂ will fall between

0.64 and 0.76?



Tutorial: Q.8.2, Q.8.4, Q8.5, Q.8.14 →Q8.33

8.4 Sampling Distribution of Sample Variance

Chi-square Distribution:

A random variable X has the chi-square distribution ( 2 ) with degree of freedom if its

probability density is given by:

elsewhere

xexxfx

0

022

1),(

22

2

2

The mean is and variance is 2 . The moment generating function is 2)21()( ttM X .

Chi-square distribution plays an important role in problem of sampling from normal populations.



Example: If =7 in chi-square distribution, find 2 , where 95.0)( 2 xP .

Solution:

Theorem: If X has the standard normal distribution, then 2X has the chi-square distribution with

degree of freedom =1.

Proof:



Theorem: If nXXX ,...,, 21 are independent random variables having the standard normal

distributions, then

n

iiXY

1

2

has the chi-square distribution with = n degrees of freedom.

Proof:

Theorem: If nXXX ,...,, 21 are independent random variables having chi-square distributions with

n ,...,, 21 degrees of freedom, then

n

ii

XY1

has the chi-square distribution with n ...21 degrees of freedom.

Proof:

Theorem: (Sampling distribution of the sample variance)

If X and 2S are the mean and the variance of a random sample of size n from a normal population

with the mean and the standard deviation , then

1. X and 2S are independent.

2. The random variable 2

2)1(

Sn has a chi-square distribution with n-1 degrees of freedom.



Proof:

1. proof is omitted.

2.

Example: If 1021 ,...,, xxx is a random sample from the normal distribution )9,(N . Find

)17( 2 SP .

Solution:



Example: The claim that the variance of a normal population is 2 =4 is to be rejected if the

variance of a random sample of size 9 exceeds 7.7535. What is the probability that this claim will

be rejected even though 2 = 4 ?

Solution:

8.5 Sampling Distribution of X With Unknown Variance 2

The Student t-Distribution:

In practice, the population standard deviation is unknown. Thus, it is necessary to replace with

an estimate, which is usually the sample standard deviation s. Thus, the suggested transformation of

the Central Limit Theorem will be ns

X

/

and it leads to new probability distribution.

Theorem: If Y and Z are independent random variables, Y has a chi-square distribution with

degrees of freedom, and Z has the standard normal distribution, then the distribution of /Y

ZT

is given by 2

1

2

1

2

2

1

)(

ttf for t

and it is called the t distribution with degrees of freedom.

T- distribution was originally developed by W. S. Gosset in 1908. Because his employers, the

Guinness brewery, would not permit him to publish this important work in his own name, he used

the pseudonym “Student.” Thus, the distribution is known as the Student t-distribution.

Example: If X has t- distribution with =1. Find )078.3( tP

Solution:



Theorem: If X and 2S are the mean and the variance of a random sample of size n from a normal

population with the mean and the standard deviation , then

nS

XT

/

has the t distribution with n- 1 degrees of freedom.

Proof:

The following Figure illustrates the behavior of the t-distributions for n = 2, 10, 20, and 30.

It is clear that as n becomes larger and larger, it is almost impossible to distinguish the graphs.

It can be shown that the t-distribution tends to a standard normal distribution as the degrees

of freedom (equivalently, the sample size n) tend to infinity.

In fact, the standard normal distribution provides a good approximation to the t-distribution

for sample sizes of 30 or more. We will use this approximation in the statistical inference problems

for n ≥ 30.



Example: A manufacturer of fuses claims that with 20% overload, the fuses will blow in less than

10 minutes on the average. To test this claim, a random sample of 20 of these fuses was subjected to

a 20% overload, and the times it took them to blow had the mean of 10.4 minutes and a sample

standard deviation of 1.6 minutes.

It can be assumed that the data constitute a random sample from a normal population. Do they tend

to support or refute the manufacturer’s claim?

Solution:

8.6 Sampling Distribution for the Variance–Ratio

The F-distribution was developed by Fisher to study the behavior of two variances from random

samples taken from two independent normal populations.

we may be interested to know whether the population variances are equal or not?

Theorem: If U and V are independent random variables having chi-square distributions with 1

and 2

degrees of freedom, then

2

1

/

/

V

UF

is a random variable having an F distribution that has a density given by

211

1

2

1

2

11

2

2

2

1

21

21

1.

22

2)(

fffg

for 0f and 0)( fg elsewhere.

A graph of )(xf for various values of n is given in the following page:



Theorem: If 2

1S and 2

2S are the variances of independent random samples of size

1n , and

2n from

normal populations with the variances 2

1 and 2

2 , then

2

2

2

1

2

1

2

2

2

2

2

2

2

1

2

1

/

/

S

S

S

SF

is a random variable having an F distribution with (1

n - 1), and (2

n - 1) degrees of freedom.

Proof:

Example: If 1

S and 2

S are the standard deviations of independent random samples of size 1

n = 61,

and 2

n = 31 from normal populations with 2

1 =12 and 2

2 =18, find

16.1

2

2

2

1

S

SP .

Solution



Exercises:

Q. 8.37 Show that for random samples of size n from a normal population with the variance 2 ,

the sampling distribution of the sample variance has the mean 2 and the variance 1

2 4

n

.

Q 8.60 The claim that the variance of a normal population is 2 = 25 is to be rejected if the

variance of a random sample of size 16 exceeds 54.668 or is less than 12.102. What is the

probability that this claim will be rejected even though 2 = 25?



Q. 8.38: Show that if n

XXX ,...,,21

are independent random variables having the Chi-square

distribution with 1 and nn

XXXY ...21

, then the limiting distribution of n

n

Y

Z

n

2

1

as

n is the standard normal distribution.

Solution:



The values of the standard normal distribution; )0( zZP

Al-Azhar University – Gaza MATHEMATICAL STATISTICS

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