Sampling Distribution of
Proportion:
Where x is the number of successes (values with a specified characteristic) in a sample of size n.
Infinite Population with
Replacement:

or alternatively

Finite Population without
Replacement:

Example:
A coordination team consists of seven members. The education of each member as follows: (G = Graduate, PG = Post Graduate)
Members 
1 
2 
3 
4 
5 
6 
7 
Education 
G 
PG 
PG 
PG 
PG 
G 
G 
(i) Determine the proportion of postgraduates in the population.
(ii) Select all possible samples of two members from the population without replacement, and compute the proportion of postgraduate members in each sample.
(iii) Compute the mean (μ_{p}) and the SD (σ_{p}) of the sample proportion computed in (ii).
Solution:
(i) Proportion of PG in the population:
N = 7
No. of PG = 4
π = 4/7 = 0.57
(ii) No. of possible samples (without replacement) = ^{N}C_{n} = ^{7}C_{2} = 21 samples.
1,2 
1,3 
1,4 
1,5 
1,6 
1,7 

2,3 
2,4 
2,5 
2,6 
2,7 


3,4 
3,5 
3,6 
3,7 



4,5 
4,6 
4,7 




5,6 
5,7 





6,7 
The corresponding sampling proportions are:
0.5 
0.5 
0.5 
0.5 
0 
0 

1 
1 
1 
0.5 
0.5 


1 
1 
0.5 
0.5 



1 
0.5 
0.5 




0.5 
0.5 





0 
Sampling Distribution of Proportion
p 
Tally
Marks 
f 
P(p) 
0 
 
3 
3/21 = 1/7 = 0.143 
0.5 
 
12 
12/21 = 4/7 = 0.571 
1 
 
6 
6/21 = 2/7 = 0.286 
Total 

21 
1 
p 
P(p) 
p.P(p) 



p^{2}.P(p) 
0 
0.143 
0 
–0.5715 
0.32661 
0.04671 
0 
0.5 
0.571 
0.2855 
–0.0715 
0.00511 
0.00292 
0.14275 
1 
0.286 
0.286 
0.4285 
0.18361 
0.05251 
0.286 
Total 

0.5715 


0.10214 
0.42875 
(iii) Mean ( ) and SD ( ) of sample proportion distribution:
or alternatively
The results are verified as below:
Shape of the Sampling Distribution
of Proportion p:
The central limit theorem also holds for the random variable p, which states that:
(i) The sampling distribution of proportion p approaches a normal distribution with mean and SD (with replacement)
(ii) If the random sampling is without replacement and the sampling fraction , the f.p.c. must be used as below in the formula of SD:
(iii) When n ≥ 50 and both n.π and n(1 – π) are greater than 5, the sampling distribution can be considered ‘normal’.
(iv) When the distribution of p is normal, the following statistic will be standard normal variable:
Sampling Distribution of
Difference between Two Proportions:
as n_{1} and n_{2} increase.
Moreover:
will be standard normal variable.
and the estimated standard error as below:
Sampling Distribution of t:
Therefore, the standard error is equal to :
In the above equation the (n – 1) is called ‘Degree of Freedom’ or simply d.f., through which we can obtain ‘tvalue’ from ‘ttable’.
Properties of
tdistribution:
Sampling Distribution of
Variances:
Population Variance:
or alternatively
Mean of sampling distribution of S^{2}
(
):
Example:
A population consists of the following numbers: 1,3,5,7. Find the population variance (σ^{2}) and the mean of sampling distribution of variances ( ), if all samples are drawn with replacement of size 2 from the population.
Solution:
No. of possible samples (with replacement) = N^{n} = 4^{2} = 16 samples
Samples:
1,1 
1,3 
1,5 
1,7 
3,1 
3,3 
3,5 
3,7 
5,1 
5,3 
5,5 
5,7 
7,1 
7,3 
7,5 
7,7 
Means of samples:
1 
2 
3 
4 
2 
3 
4 
5 
3 
4 
5 
6 
4 
5 
6 
7 
Variances of samples:
0 
1 
4 
9 
1 
0 
1 
4 
4 
1 
0 
1 
9 
4 
1 
0 
Sampling Distribution of S^{2}:
S^{2} 
Tally
Marks 
f 
f.S^{2}^{ } 
0 
 
4 
0 
1 
 
6 
6 
4 
 
4 
16 
9 
 
2 
18 
Total 

16 
40 
Pooled Estimate of Variance:
Thus, the π will be equal to:
and it will be a standard normal variable.
When the σ_{1}^{2}
and σ_{2}^{2} are replaced by the estimators S_{1}^{2}
and S_{2}^{2} the distribution of
can be standardised provided that
the samples are large (n_{1} and n_{2} > 30).
Weighted
Average of S_{1}^{2} and S_{2}^{2}:
Where (n_{1} + n_{2} – 2) is the degree of freedom.
Where S_{p} is pooled SD.
Where the degree of freedom ν is as follows: