Sampling Distributions: Foundation of Statistical Analysis

Unlock the power of sampling distributions to make accurate inferences about populations. Learn essential concepts like the Central Limit Theorem and confidence intervals for robust statistical reasoning.

Sampling distributions

A sampling distribution is a probability distribution of a statistic obtained through a large number of samples drawn from a specific population. The sampling distribution of a given population is the distribution of frequencies of a range of different outcomes that could possibly occur for a statistic of a population.

Sample proportion will estimate population proportion
Sample mean will estimate population mean

$\quad N = population \;size$
$\quad n = sample \;size$
$\quad \mu = population \;mean$
$\quad \overline{x} = sample \;mean$
$\quad p = population \;proportion$
$\quad \hat{p} = sample\; proportion$

Sampling distribution model

Example 1

Understanding Proportions and Sample Size
There are 7 supercars competing on a racetrack. The cars are labelled {1, 2, 3, 4, 5, 6, 7}.
Find:

From our population of cars: car 1, car 2, car 3, car 4, car 5, car 6 and car 7, there are only 4 odd numbered cars out of the seven cars in the population; therefore, the population proportion of odd numbered cars is just 4 out of 7, which is written as a ratio as follows:

$\large p =\frac{4}{7}$

If the sample that we will take out of the population of 7 cars mentioned above will contain 3 cars, and with replacement, we have 7 different possible cars as outcomes for each of these 3 cars; the amount of different possible samples that could be obtained is calculated by multiplying the 7 possible outcomes for each car that will be picked:

different samples with replacement $=777=7^{3} =343$
And so, there are a total of 343 possible different samples of the 7-car population when it is allowed to repeat cars within the sample (with replacement).

Without replacement? Since we cannot replace outcomes that we have already had in one of the spaces of the sample, then the possible amount of samples that can be obtained in this case follows the calculation for a combination:

$nC_{r} =$ number of selections of $r$ items taken from a set of $n$ distinct items
$nC_{r} =$ $\large \frac{n!}{(n-r)!\,r!}$
Therefore, in this case we have that:

$7C_{3}=$ $\large \frac{7!}{(7-3)!\cdot 3!} = \frac{7\times6\times5\times4\times3\times2\times1}{(4\times3\times2\times1)(3\times2\times1)} = \frac{7\times6\times5}{3\times2\times1} = \frac{210}{6}$ $= 35$
We will talk about combinations extensively in later lessons, for now, you can think of it as the method to obtain the possible amount of different samples that can be obtained from a population when there is no replacement (which happens to be the most logical scenario in many cases), just as it has been defined with this problem.

Example 2

Alice, Bob, Cole, Daisy, and Eve are all the students in a 10th grade class. Alice, Cole and Eve brought a lunch to class while the rest of the class did not.

We know the population proportion is written as the ratio of the sample size to the population size, therefore:

$p = \frac{n}{N}=\frac{3}{5} = 0.6$
Notice that this is because three people from the class brought lunch, out of a total of 5. So our total population was 5 people, and the sample who brought lunch was 3.

This is the same as looking for all the possible combinations of two students in the classroom, therefore, on this case we must use the method to find a number of samples without replacement because we cannot have a person to be repeated itself. The calculation follows the equation for a combination as shown above, and so:

$nC_{r} = 5C_{2} =$ $\large \frac{5!}{(5-2)!\cdot 2!} = \frac{5\times4\times3\times2\times1}{(3\times2\times1)(2\times1)} = \frac{5\times4}{2\times1} = \frac{20}{2}$ $= 10$
And so, there are 10 possible samples of two people at a time coming from the 10th grade class forming the next list:

{Alice, Bob}
{Alice, Cole}
{Alice, Daisy}
{Alice, Eve}
{Bob, Cole}
{Bob, Daisy}
{Bob, Eve}
{Cole, Daisy}
{Cole, Eve}
{Daisy, Eve}
If we think of this scenario as one following a completely random selection, then it means that each student and therefore each pair sample, will have the same probability to occur.
With that in mind, since there are a total of 10 possible different pair samples, then each of those pairs has a 1/10 of probability of occurring (or being selected).

For this case we will look at the sample proportion for bringing lunch to class of each pair, in other words, from each of the possible pairs in the classroom, what proportion of each pair brought lunch?: If both the students brought lunch, then it means their sample proportion is 1 (or 2/2); if only one out of the two students in a pair brought lunch, then the sample proportion of this pair is 0.5 (or ½); if none of the students from a pair brought lunch, then this pairs sample proportion is zero (or 0/2). Since Alice, Cole and Eve brought lunch and Bob and Daisy did not, then the sample proportions result as follows (third column):

Combination	P(x) = probability	$\hat{p}$ = sample proportion
{Alice, Bob}	1/10	0.5
{Alice, Cole}	1/10	1
{Alice, Daisy}	1/10	0.5
{Alice, Eve}	1/10	1
{Bob, Cole}	1/10	0.5
{Bob, Daisy}	1/10	0
{Bob, Eve}	1/10	0.5
{Cole, Daisy}	1/10	0.5
{Cole, Eve}	1/10	1
{Daisy, Eve}	1/10	0.5

Averaging the sample proportions from the figure above we obtain:

$\mu_{\hat{p}} =$ $\large \frac{0.5+1+0.5+1+0.5+0+0.5+0.5+1+0.5}{10} =$ $6/10 = 0.6$
As you can see, the average of the sample proportions for the people who brought lunch to school is the same as the proportion of the population that brought lunch. Therefore, in here we reaffirm what we mentioned at the beginning of this lesson: Sample proportion will estimate population proportion.

Example 3

Uncle Sammy grows prize winning big zucchinis. The weights of his four largest zucchinis are given in the table below:



We obtain the average easily as we have learned before:

Average weight = $\large \frac{7+9+11+5}{4}=\frac{32}{4}$ $=8$
And so the average weight of uncle Sammys zucchinis is 8lbs.

Once again, we need to find the possible combinations of two zucchinis which represent all the possible different samples of size 2 from the zucchinis in the total population:

$nC_{r} = 4C_{2} =$ $\large \frac{4!}{(4-2)!\, 2!} = \frac{4\times3\times2\times1}{(2\times1)(2\times1)} = \frac{4\times3}{2\times1} = \frac{12}{2}$ $=6$
And so, there are 6 different possible samples of two zucchinis from the total of uncle Sammys ones and they conform the next list:

{A, B}
{A, C}
{A, D}
{B, C}
{B, D}
{C, D}
Since they would be completely randomly selected, they all have the same chance at being picked. Therefore, they all have a chance of getting picked since there are a total of 6 possible outcomes all with the same probability of being picked.

Let us make a table with the combinations, their probability of being picked and then their average weight in each of its columns. Notice that the average weight for each sample is obtained with the simple formula for average (adding all of the items from a list, and dividing by the number of items in the list).
Remember, the average weight of each sample is what we call the sample mean.

Combination	P(x) = probability	$\overline{x}$ = Avg. weight
{A, B}	1/6	$\overline{x} = \frac{7+9}{2} =8 lbs$
{A, C}	1/6	$\overline{x} = \frac{7+11}{2} = 9 lbs$
{A, D}	1/6	$\overline{x} = \frac{7+5}{2} = 6 lbs$
{B, C}	1/6	$\overline{x} = \frac{9+11}{2} = 10 lbs$
{B, D}	1/6	$\overline{x} = \frac{9+5}{2} = 7 lbs$
{C, D}	1/6	$\overline{x} = \frac{11+5}{2} = 8 lbs$

Simply calculating the average of all the sample means shown in the table above:

$\mu_{\hat{x}} =$ $\large \frac{8+9+6+10+7+8}{6} =$ $48/6 = 8$
And so, we reaffirm what was mentioned at the beginning of this lesson: Sample mean will estimate population mean.

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To finalize this lesson here are a few recommendations to you: A lot of terminology is covered on this lesson for the sampling distribution of the sample mean, along with a simple example. Then, this page on the sample proportion provides a review on the formulas we have used during this lesson. Both links can be useful to you to reaffirm what was learnt today and to aid in your independent studies.

This is it for our lesson of today, see you in the next one!