Sampling Distributions
Suppose that we draw all possible samples of size n from a given population. Suppose further that we compute a statistic (e.g., a mean, proportion, standard deviation) for each sample. The probability distribution of this statistic is called a sampling distribution.
Variability of a Sampling Distribution
The variability of a sampling distribution is measured by its variance or its standard deviation. The variability of a sampling distribution depends on three factors: * N: The number of observations in the population. * n: The number of observations in the sample. * The way that the random sample is chosen.
If the population size is much larger than the sample size, then the sampling distribution has roughly the same sampling error, whether we sample with or without replacement. On the other hand, if the sample represents a significant fraction (say, 1/10) of the population size, the sampling error will be noticeably smaller, when we sample without replacement.

Sampling distribution:- A sampling distribution is a distribution of all of the possible values of a statistic (say sample mean) for a given size sample selected from a population.
Sample Distribution of the Mean is an Unbiased Estimate of the Population Mean
If all possible samples of a certain size, n, are selected from a population, the mean of these sample means (the grand mean) would be equal to the population mean. Types of sampling distribution:-
There are two types of sampling distribution: 1) With replacement 2) Without replacement

1) With replacement
Suppose we have a bowl of 100 unique numbers from 0 to 99. We want to select a random sample of numbers from the bowl. After we pick a number from the bowl, we can put the number aside or we can put it back into the bowl. If we put the number back in the bowl, it may be selected more than once; if we put it aside, it can selected only one time.
When a population element can be selected more than one time, we are sampling with replacement
2) Without replacement
Suppose we have a bowl of 100 unique numbers from 0 to 99. We want to select a random sample of numbers from the bowl. After we pick a number from the bowl, we can put the number aside or we can put it back into the bowl. If we put the number back in the bowl, it may be selected more than once; if we put it aside, it can selected only one time.
When a population element can be selected only one time, we are sampling without replacement.

THE TOPIC WHICH WE HAVE SELECTED IS WITHOUT REPLACEMENT SAMPLING DISTRIBUTION
Example:
Draw all possible samples of size 2 without replacement from a population consisting of 3, 6, 9, 12, 15. Form the sample distribution of sample means and verify the results.
Solution:
We have population values 3, 6, 9, 12, 15, population size N=5 and sample size n=2. Thus, the number of possible samples which can be drawn without replacement is
= =10

Sample No. | Sample Values | Sample Mean
(X) | Sample No. | Sample Values | Sample Mean
(X) | 1 | 3, 6 | 4.5 | 6 | 6, 12 | 9.0 | 2 | 3, 9 | 6.0 | 7 | 6, 15 | 10.5 | 3 | 3, 12 | 7.5 | 8 | 9, 12 | 10.5 | 4 | 3, 15 | 9.0 | 9 | 9, 15 | 12.0 | 5 | 6, 9 | 7.5 | 10 | 12, 15 | 13.5 |
The sampling distribution of the sample mean X and its mean and standard deviation are: X | f | f(X) | Xf(X) | X2f(X) | 4.5 | 1 | 1/10 | 4.5/10 | 20.25/10 | 6.0 | 1 | 1/10 | 6.0/10 | 36.00/10 | 7.5 | 2 | 2/10 | 15.0/10 | 112.50/10 | 9.0 | 2 | 2/10 | 18.0/10 | 162.00/10 | 10.5 | 2 | 2/10 | 21.0/10 | 220.50/10 | 12.0 | 1 | 1/10 | 12.0/10 | 144.00/10 | 13.5 | 1 | 1/10 | 13.5/10 | 182.25/10 | Total | 10 | 1 | 90/10 | 877.5/10 |
E(X)=∑Xf(X)=90/10=9
Var(X¯)=∑X2f(X)−[∑Xf(X)]2=887.510−(9010)2=87.75−81=6.75

The mean and variance of the population are: X | 3 | 6 | 9 | 12 | 15 | ∑X=45 | X2 | 9 | 36 | 81 | 144 | 225 | ∑X2=495 | μ=∑X/N=455=9 and σ2=∑X2/N− (∑X/N)2 =4955−(455)2=99−81=18
Verification:
(i) E(X)=μ=9 (ii) Var(X)=σ2 /n(N−n/N−1)=18/2(5−2/5−1)=6.75
Example:
If random samples of size three and drawn without replacement from the population consisting of four numbers 4, 5, 5, 7. Find sample mean X for each sample and make sampling distribution of X. Calculate the mean and standard deviation of this sampling distribution. Compare your calculations with population parameters.
Solution:
We have population values 4, 5, 5, 7, population size N=4 and sample size n=3. Thus, the number of possible samples which can be drawn without replacement is = =4 Sample No. | Sample Values | Sample Mean (X) | 1 | 4, 5, 5 | 14/3 | 2 | 4, 5, 7 | 16/3 | 3 | 4, 5, 7 | 16/3 | 4 | 5, 5, 7 | 17/3 |
The sampling distribution of the sample mean X and its mean and standard deviation are: X | f | f(X) | Xf(X) | X2f(X) | 14/3 | 1 | 1/4 | 14/12 | 196/36 | 16/3 | 2 | 2/4 | 32/12 | 512/36 | 17/3 | 1 | 1/4 | 17/12 | 289/36 | Total | 4 | 1 | 63/12 | 997/36 | μX=∑Xf(X)=6312=5.25

σX=∑X2f(X)−[∑Xf(X)]2
=99736− (6312)2
=0.3632
The mean and standard deviation of the population are: X | 4 | 5 | 5 | 7 | ∑X=21 | X2 | 16 | 25 | 25 | 49 | ∑X2=115 | μ=∑X/N=21/4=5.25 and σ2=∑X2/N−(∑X/N)2
=115/4−(21/4)2
=1.0897 σ/n/(N−n/N−1) =1.08973(4−3/4−1)
=0.3632