Sampling Distribution of
Proportion:
Where x is the number of successes (values with a specified characteristic) in a sample of size n.
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Infinite Population with
Replacement:
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or alternatively
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Finite Population without
Replacement:
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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 post-graduates in the population.
(ii) Select all possible samples of two members from the population without replacement, and compute the proportion of post-graduate members in each sample.
(iii) Compute the mean (μp) and the SD (σp) 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) = NCn = 7C2 = 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 |
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|
|
|
4,5 |
4,6 |
4,7 |
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|
|
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5,6 |
5,7 |
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|
|
|
|
|
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 |
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|
|
|
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0.5 |
0.5 |
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|
|
|
|
|
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 |
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Total |
|
21 |
1 |
|
p |
P(p) |
p.P(p) |
|
|
|
p2.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 |
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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 n1 and n2 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 ‘t-value’ from ‘t-table’.
Properties of
t-distribution:

Sampling Distribution of
Variances:
Population Variance:
or alternatively
Mean of sampling distribution of S2
(
):
Example:
A population consists of the following
numbers: 1,3,5,7. Find the
population variance (σ2) 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) = Nn = 42 = 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 S2:
|
S2 |
Tally
Marks |
f |
f.S2 |
|
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 σ12
and σ22 are replaced by the estimators S12
and S22 the distribution of
can be standardised provided that
the samples are large (n1 and n2 > 30).
Weighted
Average of S12 and S22:
Where (n1 + n2 – 2) is the degree of freedom.
Where Sp is pooled SD.
Where the degree of freedom ν is as follows: