Binomial Test
Research Scenario
Question
Pam knows that children seated in a room with four doors—one to their front, one to their back and one on each side—are equally likely to pick any door when asked which door they would choose when leaving the room. As part of a larger research program, Pam wants to demonstrate that children labeled as “oppositional” (oppositionally defiant) are more likely to select the door behind them. To determine whether or not her hypothesis is correct, she conducts a study with 80 oppositional children and asks them to pick one of four doors to leave the room.
Instrument and Scoring
During the research study, Pam captures the door selections for each of the 80 children and codes her data such that children who select the door behind them are scored with a “1” and children who select any of the other three doors (the door in front of them and either door to their side) are scored a “0”. Given that Pam is interested in children who select the door behind them, and since it (the door behind) is one of four possible options available to each child, the test proportion for door behind is 0.25 since it is one out of four possible doors, (1/4 = 0.25).
Null Hypothesis
Pam is interested in understanding if “oppositional” children select the door behind them at a significantly higher rate than the expected 25%. With that in mind, the null hypothesis for this study, would is that oppositional children do not select the door behind them at a significantly higher rate than other, non-oppositional children. Since this research question is interested in understanding whether oppositional children show a higher rate in rear door selection, it is a directional-type hypothesis in which a one-tailed p-value is used.
Assumptions
The cases represent a random sample from the population, and the scores on the test variable are independent of each other.
Check
There is not much that can be done to check this assumption, rather this is something that best addressed through implementing proper research/sampling practices–simply put, if this assumption is found to be violated in the data analysis phase there is little the researcher can do to address the issue.
Binomial Test Overview
The Binomial Test compares the observed frequencies of the two categories of a dichotomous variable to the frequencies that are expected under a binomial distribution with a specified probability parameter.
Binomial Test Overall Equation
\[P(k) = \frac{n!}{(n - k)!\cdot k!} \cdot p^k \cdot (1 - p)^{n - k}\]
Simplified Equation
\[P(\text{success}) = \frac{n_\text{trials}!}{(n_\text{trials} - k_\text{successes})!\cdot k_\text{successes}!} \cdot \text{probability of success}^{k_\text{successes}}\cdot (1 - \text{probability of success})^{\text{trials} - \text{successes}}\]
Data Needed to Complete the Binomial Test
To conduct a binomial test Pam needs three pieces of information.
- the number of trials (sample size) (n) = 80
- the number of successes; in this case the number of times an oppositional child select the door behind (k) = 25
- the hypothesized test proportion (p) = 0.25
Step 1: Input All of the Data
\[P(25) = \frac{80!}{(80 - 25)!\cdot 25!} \cdot 0.25^{25}\cdot (1 - 0.25)^{80 - 25}\]
If the numbers above were computed as is, they would be very large, so for the time being, they will not be shown here, and instead the simplified answer is presented below.
Step 2: Compute the Probability of Selecting Door Behind 25 Times
\[ P(25) \approx 0.043 \]
However, the analysis isn’t yet completed. As outlined in the research question, Pam is interested in understanding if oppositional children choose the door behind them greater than 25% of the time. Reviewing the equation above, it should become clear (if not already) that Pam currently only has the probability of oppositional defiant children choosing the door behind them 25 times out of 80.
To answer the question fully, probability scores must be computed for all possible number of times the door behind them is selected equal and greater than 25 times and up to the maximum number of trials, 80. Resulting probability scores should then be summed together to determine the associated p-value.
Rather than repeat the calculation from above 55 more times, a table is presented below which illustrates the computed probabilities for each door selection outcome group up to the maximum.
Step 3: Review Probability Output
| No. Of Times Selecting Door Behind | Probability | One-Tailed p-Value |
|---|---|---|
| 25 | 0.043 | 0.124 |
| 26 | 0.031 | 0.080 |
| 27 | 0.020 | 0.050 |
| 28 | 0.013 | 0.029 |
| 29 | 0.008 | 0.017 |
| 30 | 0.004 | 0.009 |
| 31 | 0.002 | 0.005 |
| 32 | 0.001 | 0.002 |
| 33 | 0.001 | 0.001 |
| 34 | 0.000 | 0.000 |
| 35 | 0.000 | 0.000 |
| 80 | 0.000 | 0.000 |
| Note: | ||
| This table does not include all possible outcomes the dataset. Selections greater than 25 with probabilities of zero, except for 80, were suppressed. |
Binomial Test Conclusion
Reviewing the table above, it would appear that while oppositional children selected the door behind them at a higher rate than expected, ~0.31 versus 0.25, the difference was not statistically significant since the one-tailed exact p-value was ~0.12; well above 0.05. However, had 28 children selected the door behind them, the difference would have been statistically significant. This is illustrated in the table above as well as the chart of the binomial distribution scores presented below.
A work by Alex Aguilar
aaguilar@thechicagoschool.edu