Understanding sampling distributions is key to the understanding of statistical inference.

A Sampling distribution - is the distribution of all possible values of a statistic, computed from samples of the same size randomly drawn from the same population.

1. From a finite population of size N, randomly draw all possible samples of size n.
2. Compute the statistic of interest (mean, proportion) for each sample.
3. List in one column the different distinct observed values of the statistic, and in another column list the corresponding frequency of occurrence of each distinct observed value of the statistic

Example: Select two values from a population of 5 children, ages 6, 8, 10, 12, 14. There are 25 combinations of the way that the sample of two can be selected.  This is sampling with replacement, so a child can be selected twice.

	Frequency	Combinations
6	1	6,6
7	2	6,8; 8,6
8	3	6,10, 8,8, 10,6
9	4	6,12; 8,10; 10,8; 12,6
10	5	6,14; 8,12; 10,10; 12,8; 14,6
11	4	8,14; 10,12; 12,10; 14,8
12	3	10,14; 12,12; 14,10
13	2	12,14; 14,12
14	1	14,14
Total	25

Calculate the mean of the sampling distribution by summing the values of the mean of the samples, and dividing by the number of samples.

(6+2(7)+3(8)+4(9)+5(10)+4(11)+3(12)+2(13)+14)/25 = 250/25 = 10

The mean of the sampling distribution equals the mean of the population.

Calculate the variance of the sampling distribution by squaring the difference of the mean of each sample from the mean of all sample means and dividing by the number of samples.

((6-10)²+2(7-10)² +3(8-10)² +4(9-10)² +5(10-10)² +4(11-10)² +3(12-10)² +2(13-10)² +(14-10)² )/25 = 100/25 = 4

The variance of the sampling distribution (4) does not equal the variance of the population (8).  Notice that if we divide the variance of the population by the size of the sample, n, we get the variance of the sampling distribution (8/2=4).

Mean:

Variance:

Standard error of the mean:

A sampling distribution from a normally distributed population has the following properties:

1. The distribution of will be normal.
2. The mean, m of the distribution of will equal the mean of the population from which the sample was drawn.
3. The variance s ² of will equal the variance of the population divided by the sample size.

Sampling from Non-normally Distributed Populations

The Central Limit Theorem

Given a population of any nonnormal distribution with a mean m and variance s ^2, the sampling distribution of , computed from samples of size n from this population, will have a mean m and variance s ²/n and will be approximately normally distributed when the sample size is large.

Sampling without replacement - The Finite population Correction

Example: Again select two values from a population of 5 children, ages 6, 8, 10, 12, 14. There are 10 combinations of the way that the sample of two can be selected.  This is sampling without replacement, so no child can be selected twice, and order doesn't matter (6,8=8,6).

	Frequency	Combinations
6	0
7	1	6,8
8	1	6,10
9	2	6,12; 8,10
10	2	6,14; 8,12
11	2	8,14; 10,12
12	1	10,14
13	1	12,14
14	0
Total	10

When sampling without replacement from a finite population, the sampling distribution of will have mean m and variance:



The Standard Normal Transformation of the mean and standard deviation of the sampling distribution:

Transforms the sampling distribution to a standard normal distribution.



Example:  Cranial length in a certain population is normally distributed with a mean of 185.6 mm and a standard deviation of 12.7 mm.  What is the probability that a random sample of 10 from this population will have a mean greater than 190?

Example:  The mean and standard deviation of serum iron values is 120 and 15 micrograms per 100 ml. What is the probability that a random sample of 50 will yield a mean between 115 and 125 mg per 100 ml?

Use the difference of the two sample means as the statistic to estimate the difference in the population means (m ₁-m ₂):

Example:  Population 1 has been exposed to extraterrestrial radiation and a random sample of 15 have a mean intelligence score of 92.  Population 2 was not exposed, and the mean intelligence score for a random sample of 15 is 105.  If the standard deviation of intelligence scores for the 2 populations is 20 and there is no difference between the means of the 2 populations, what is the probability of observing a difference this large or larger?

Distribution of a Sample Proportion

Example:  The percentage of the population that is color blind is 0.08.   If we randomly select 150 from this population, what is the probability that the sample proportion of color blind will be as great as 0.15?

Example: The proportion of illegal drug users in population 1 is .5, while in population 2 the proportion is .33.  What is the probability that random samples of size 100 from each population will yield a difference in the proportion as large as .30?

P(z > 1.89) = 1 - .9706 = .0294