Virtual University of Pakistan Lecture No. 31 of the course on Statistics and Probability by Miss Saleha Naghmi Habibullah
IN THE LAST LECTURE, YOU LEARNT •Normal Distribution. •Mathematical Definition •Important Properties •The Standard Normal Distribution •Direct Use of the Area Table •Inverse Use of the Area Table •Normal Approximation to the Binomial Distribution
TOPICS FOR TODAY
• Sampling Distribution of X • Mean and Standard Deviation of the Sampling Distribution of X • Central Limit Theorem
In today’s lecture, we begin the third and last part of this course, i.e. INFERENTIAL STATISTICS --- that branch of Statistics which enables us to to draw conclusions or inferences about various phenomena on the basis of real data collected on sample basis.
In this regard, the first point to be noted is that statistical inference can be divided into two main branches --- estimation, and hypothesis-testing. Estimation itself can be further divided into two branches --point estimation, and interval estimation.
Statistical Inference
Estimation
Point Estimation
Interval Estimation
Hypothesis Testing
The second important point is that the concept of sampling
distributions forms the basis for both estimation hypothesis-testing,
and
SAMPLING DISTRIBUTION The probability distribution of any statistic (such as the mean, the standard deviation, the proportion of successes in a sample, etc.) is known as its sampling distribution.
In this regard, the first point to be noted is that there are two ways of sampling --sampling with replacement, and sampling without replacement.
In case of a finite population containing N elements, the total number of possible samples of size n that can be drawn from this population with replacement n is N .
On the other hand:
In case of a finite population containing N elements, the total number of possible samples of size n that can be drawn from this population without N replacement is . n
We begin with the sampling distribution of 鵃出:
We illustrate the concept of the sampling distribution of X with the help of the following example:
EXAMPLE Let us examine the case of an annual Ministry of Transport test to which all cars, irrespective of age, have to be submitted. The test looks for faulty breaks, steering, lights and suspension, and it is discovered after the first year that approximately the same number of cars have 0, 1, 2, 3, or 4 faults.
The above situation is equivalent to the following:
Let X denote the number of faults in a car. Then X can take the values 0, 1, 2, 3, and 4, and the probability of each of these X values is 1/5.
Hence, we have the following probability distribution:
No. of Faulty Items (X) 0 1 2 3 4 Total
Probability f(x) 1/5 1/5 1/5 1/5 1/5 1
In order to compute the mean and standard deviation of this probability distribution, we carry out the following computations:
No. of Faulty Items (x) 0 1 2 3 4 Total
Probability f(x)
x f(x)
x f(x)
1/5 1/5 1/5 1/5 1/5 1
0 1/5 2/5 3/5 4/5 10/5=2
0 1/5 4/5 9/5 16/5 30/5=6
2
MEAN AND VARIANCE OF THE POPULATION DISTRIBUTION:
µ = E( X ) = ∑ xf ( x ) = 2 2 2 2 σ = Var ( X ) = E( X ) − [ E ( X ) ] 2 2 = ∑ x f ( x) − [∑ x f ( x)] 2
= 6−2 = 6−4 = 2
∴ σ = 2 = 1.414
Practically speaking, only a sample of the cars will be tested at any one occasion, and, as such, we are interested in considering the results that would be obtained if a sample of vehicles is tested.
Let us consider the situation when only two cars are tested after being selected at the roadside by a mobile testing station. The following table gives all the possible situations:
NO. OF FAULTY ITEMS Second Car First Car 0 1 2 3 4
0
1
2
3
4
(0,0) (1,0) (2,0) (3,0) (4,0)
(0,1) (1,1) (2,1) (3,1) (4,1)
(0,2) (1,2) (2,2) (3,2) (4,2)
(0,3) (1,3) (2,3) (3,3) (4,3)
(0,4) (1,4) (2,4) (3,4) (4,4)
The above situation is equivalent to drawing all possible samples of size 2 from this probability distribution (i.e. the population) REPLACEMENT.
WITH
From the above list of 25 samples, we can work out all the possible sample means. These are indicated in the following table:
SAMPLE MEANS Second Car First Car 0 1 2. 3 4
0
1
2
3
4
0.0 0.5 1.0 1.5 2.0
0.5 1.0 1.5 2.0 2.5
1.0 1.5 2.0 2.5 3.0
1.5 2.0 2.5 3.0 3.5
2.0 2.5 3.0 3.5 4.0
It is immediately evident that some of these possible sample means occur several times. In view of this, it would seem reasonable and sensible to construct a frequency distribution from the sample means. This is given in the following table:
Sample Mean x
0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Total
No. of Samples f 1 2 3 4 5 4 3 2 1 25
If we divide each of the above frequencies by the total frequency 25, we obtain the probabilities of the various values of鵃出. (This is so because every one of the 25 possible situations is equally likely to occur, and hence the probabilities of the various possible values of鵃出 can be computed using the classical definition of probability i.e. m/n --- number of favourable outcomes divided by total number of possible outcomes.)
Hence, we obtain the following probability distribution:
Sample Mean x
0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Total
No. of Samples f 1 2 3 4 5 4 3 2 1 25
Probability P(鵃出 =鵃台) 1/25 2/25 3/25 4/25 5/25 4/25 3/25 2/25 1/25 25/25=1
The above is referred to as the
SAMPLING of the mean.
DISTRIBUTION
The visual picture of the sampling distribution is as follows:
Sampling Distribution of鵃出 for n = 2 P( x ) 5/25 4/25 3/25 2/25 1/25 0
0.0 0.5
1.0
1.5
2.0
2.5 3.0
3.5 4.0
X
Next, we wish to compute the mean and standard deviation distribution.
of
this
As we are already aware, for the probability distribution of a random variable X, the mean is given by µ = E(X) = ∑x f(x) and the variance is given by σ = Var(X) = E(X ) - [E(X)] 2
2
2
The point to be noted is that, in case of the sampling distribution of 鵃出, our random variable is not X but鵃出. Hence, the mean and variance of our sampling distribution are given by:
MEAN AND VARIANCE OF THE SAMPLING DISTRIBUTION OF X:
µ x = E ( X ) = ∑ xf ( x ) σ
2
x
= Var ( X ) = E ( X ) − [ E ( X ) ] 2
= ∑ x f ( x) − [∑ x f ( x)] 2
2
2
The square root of the variance is the standard deviation, and the standard deviation of a sampling distribution is termed as its standard error.
In order to find the mean and standard error of the sampling distribution of 鵃出 in this example, we carry out the following computations:
Sample Mean x
0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Total
Probability f(x)=P(X =x) 1/25 2/25 3/25 4/25 5/25 4/25 3/25 2/25 1/25 25/25=1
x f(x) 0 1/25 3/25 6/25 10/25 10/25 9/25 7/25 4/25 50/25=2
2
(x) f(x) 0 1/50 6/50 18/50 40/50 50/50 54/50 49/50 32/50 250/50=5
Hence, in this example, we have:
µ x = E ( X ) = ∑ xf ( x ) = 50 / 25 = 2
AND
σ
2
x
= Var ( X ) = E ( X ) − [ E ( X ) ] 2
= ∑ x f ( x) − [∑ x f ( x)] 2
2
= 5− 2 = 5− 4 =1 Hence
σx = σ
2
x
= 1 =1
2
2
These computations lead to the following two very important properties of the sampling distribution of 鵃出:
Property No.1 In the case of sampling with replacement as well as in the case of sampling without replacement, we have:
Âľx = Âľ
In this example:
µ=2 and µx = 2 Hence µx = µ
Property No.2 In case of sampling with replacement:
Ďƒ Ďƒx = n
In this example: σ= 2 σ 2 ∴ = =1 n 2 and σ x = 1 σ Hence σ x = n
IMPORTANT NOTE:
In case of sampling without replacement from a finite population: σ N−n σx = n N −1 The factor N − n
N −1
is known as the finite population correction (fpc).
The point to be noted is that, if the sample size n is much smaller than N − n the population size N, then N −1
is approximately equal to 1, and, as such, the fpc is not required. Hence, in sampling from a finite population, we apply the fpc only if the sample size is greater than 5% of the population size.
Next,
shape
we
consider
of the distribution of 鵃出.
the
sampling
As indicated by the line chart, the above sampling distribution is absolutely symmetric and triangular.
But let us consider what will happen to the shape of the sampling distribution with if the sample size is increased.
If in the car tests instead of taking samples of 2 we had taken all possible samples of size 3, our sampling distribution would contain 3 5 = 125 sample means, and it would be in the following form:
SAMPLING DISTRIBUTION FOR SAMPLES OF SIZE 3 x
0.00 0.33 0.67 1.00 1.33 1.67 2.00 2.33 2.67 3.00 3.33 3.67 4.00
No. of Samples 1 3 6 10 15 18 19 18 15 10 6 3 1 125
f(яге x) 1/125 3/125 6/125 10/125 15/125 18/125 19/125 18/125 15/125 10/125 6/125 3/125 1/125 1
The graph of this distribution is as follows:
Sampling Distribution of鵃出 for n = 3 P( x ) 20/125 16/125 12/125 8/125
3.00 3.33 3.67 4.00
0
0.00 0.33 0.67 1.00 1.33 1.67 2.00 2.33 2.67
4/125
X
If in the car tests instead of taking samples of 2 we had taken all possible samples of size 4, our sampling distributions would contain 4 5 = 625 sample means, and it would be in the following form:
SAMPLING DISTRIBUTION FOR SAMPLES OF SIZE 4 x No. of Samples f(яге x) 0.00 1 1/625 0.25 4 4/625 0.50 10 10/625 0.75 20 20/625 1.00 35 35/625 1.25 52 52/625 1.50 68 68/625 1.75 80 80/625 2.00 85 85/625 2.25 80 80/625 2.50 68 68/625 2.75 52 52/625 3.00 35 35/625 3.25 20 20/625 3.50 10 10/625 3.75 4 4/625 4.00 1 1/625 625 1
The graph of this distribution is as follows:
Sampling Distribution of鵃出 for n = 4 P( x ) 100/625 80/625 60/625 40/625
0
0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00 2.25 2.50 2.75 3.00 3.25 3.50 3.75 4.00
20/625
X
As in the case of the sampling distribution of 鵃出 based on samples of size 2, x each of these two distributions has a mean of 2 defective items.
It is clear from the above figures that as larger samples are taken, the shape of the sampling distribution undergoes discernible changes. In all three cases the line charts are symmetrical, but as the sample size increases, the overall configuration changes from a triangular distribution to a bellshaped distribution.
When relatively large samples are taken, this bell-shaped distribution assumes the form of a ‘normal’ distribution (also called the ‘Gaussian’ distribution), and this happens irrespective of the form of the parent population. (For example, in the problem currently under consideration, the population of defective items in a car is rectangular.)
This leads us to the following fundamentally important theorem:
CENTRAL LIMIT THEOREM: The theorem states that: “If a variable X from a population has mean µ and finite variance σ 2, then the sampling distribution of the sample meanX approaches a normal distribution with mean µ and variance σ 2/n as the sample size n approaches infinity.”
As n → ∞, the sampling distribution ofX approaches normality:
X
µx = µ
σx =
σ n
Due to the Central Limit Theorem, the normal distribution has found a central place in the theory of statistical inference. (Since, in many situations, the sample is large enough for our sampling distribution to be approximately normal, therefore we can utilize the mathematical properties of the normal distribution to draw inferences about the variable of interest.)
The rule of thumb in this regard is that if the sample
size, n, is greater than or equal to 30, then we can assume that the sampling distribution of 鵃出 is approximately normally distributed.
On the other hand:
If the POPULATION sampled is normally distributed , then the sampling distribution of X will also be normal regardless of sample size. In other words, X will be
normally distributed with 2 mean µ and variance σ /n.
IN TODAY’S LECTURE, YOU LEARNT
• Sampling Distribution of X • Mean and Standard Deviation of the Sampling Distribution of X • Central Limit Theorem
IN THE NEXT LECTURE, YOU WILL LEARN
• Sampling Distribution of pˆ • Sampling Distribution of X1 − X2