Question
A probability distribution describing the potential outcomes of a random selection experiment can be used in two ways: Prospectively , it can be used to
A probability distribution describing the potential outcomes of a random selection experiment can be used in two ways:
- Prospectively, it can be used to project the likelihood (probability) of each possible selection experiment outcome.
- Retrospectively, it can be used to assess the likelihood (probability) of having actually obtained a particular outcome. Used in this way, the probability distribution is said to serve as a sampling distribution.
In an association study, we suspect particular values of one property of a set of phenomena tend to co-occur with particular values of a second property, and we assess the association by comparing actual observations to hypothetical observations where the hypothetical observations reflect an assumption (said to be the null hypothesis) that the two properties are not associated. If the actual observations do not match the hypothetical observations exactly, we could use that as evidence that the two properties are not (not associated), and infer they are associated. However, understanding the extent to which properties of phenomena are subject to natural variation, a more prudent approach is to allow for some differences between the actual and hypothetical observations without rejecting the assumption of "no association." This is where we use sampling distributions.
For the particular type of association we are investigating, there will be a sampling distribution describing the probable occurrences of differences between a set of actual observations and the corresponding set of hypothetical observations if the two properties are not associated. Small differences will have a large probability of occurrence, moderate differences will have a moderate probability of occurrence, and large differences will have a small probability of occurrence. These probabilities are said to be p-values. If our actual observations yield a difference that is of sufficiently low probability (less than 0.05) we interpret this as suggesting the differences between our actual observations and our hypothetical observations were not the product of natural variation and we can reject our assumption of "no association" and make the inference that the two properties are associated.
A useful example of a sampling distribution is the binomial distribution. Different binomial distributions can be constructed for different selection experiments as we will discuss.
Consider the flip of a coin. It may have two outcomes: Heads or Tails. We can construct a sampling distribution by considering the number of heads in that single flip of a coin.
X = Number of Heads in a single flip of a coin
Potential Outcomes | Number of Heads | Number of Potential Outcomes | Percentage of Total Number of Outcomes | Probability of Occurrence |
T | 0 | 1 | 1 / 2 | 0.50 |
H | 1 | 1 | 1 / 2 | 0.50 |
Total | 2 | 2 / 2 | 1.00 |
This gives us the following probability distribution:
Number of Heads | Probability of Occurrence |
0 | 0.50 |
1 | 0.50 |
Total | 1.00 |
Now, suppose your friend tells you that she just flipped a coin, and it was Heads. Is this remarkable? No, because with a probability of 0.50, it was not improbable.
Now, consider a series of two flips of a coin and the resulting number of Heads:
X = Number of Heads in a single flip of a coin
Potential Outcomes | Number of Heads | Number of Potential Outcomes |
T T | 0 | 1 |
T H | 1 | 1 |
H T | 1 | 1 |
H H | 2 | 1 |
Total | 4 |
Consolidating the results:
Number of Heads | Number of Potential Outcomes | Percentage of Total Number of Potential Outcomes |
0 | 1 | 1 / 4 = 0.25 |
1 | 2 | 2 / 4 = 0.50 |
2 | 1 | 1 / 4 = 0.25 |
Total | 4 | 4 / 4 = 1.00 |
This gives us the following probability distribution:
Number of Heads | Probability of Occurrence |
0 | 0.25 |
1 | 0.50 |
2 | 0.25 |
Total | 1.00 |
Now, suppose your friend tells you that she just flipped a coin twice and got two Heads. Is this remarkable? No, because with a probability of 0.25, it was not improbable.
Now, consider a series of three flips of a coin and the resulting number of Heads:
X = Number of Heads in a single flip of a coin
Potential Outcomes | Number of Heads | Number of Potential Outcomes |
T T T | 0 | 1 |
T T H | 1 | 1 |
T H T | 1 | 1 |
H T T | 1 | 1 |
H H T | 2 | 1 |
H T H | 2 | 1 |
T H H | 2 | 1 |
H H H | 3 | 1 |
Total | 8 |
Consolidating the results:
Number of Heads | Number of Potential Outcomes | Percentage of Total Number of Potential Outcomes |
0 | 1 | 1 / 8 = 0.125 |
1 | 3 | 3 / 8 = 0.375 |
2 | 3 | 3 / 8 = 0.375 |
3 | 1 | 1 / 8 = 0.125 |
Total | 8 | 8 / 8 = 1.00 |
This gives us the following probability distribution:
Number of Heads | Probability of Occurrence |
0 | 0.125 |
1 | 0.375 |
2 | 0.375 |
3 | 0.125 |
Total | 1.00 |
Now, suppose your friend tells you that she just flipped a coin three times and got three Heads. Is this remarkable? No, because with a probability of 0.125, it was not improbable.
Now, consider a series of four flips of a coin and the resulting number of Heads. Complete the following tables, using as many rows as necessary.
X = Number of Heads in a single flip of a coin
Potential Outcomes | Number of Heads | Number of Potential Outcomes |
T T T T | 0 | 1 |
Total |
Consolidating the results:
Number of Heads | Number of Potential Outcomes | Percentage of Total Number of Potential Outcomes |
0 | ||
Total |
This gives us the following probability distribution:
Number of Heads | Probability of Occurrence |
0 | |
Total | 1.00 |
Now, suppose your friend tells you that she just flipped a coin four times and got four Heads. Is this remarkable? Is the probability of that occurrence less than 0.05?
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