In Battalio, Kagel, and Jiranyakul, 1990, participants had the choice between the risky gamble (payoffs in \$)
$A: \begin{cases} 20 & \text{ with probability } 0.7\\ 0 & \text{ with probability } 0.3\\ \end{cases}$
and \$14 for sure
$B: \begin{cases} 14& \text{ with probability } 1.0\\ \end{cases}$.
There were two samples. Respondents in the first sample answered a hypothetical question. 16 out of 29 participants chose $B$. The respondents' choices in the second sample had real consequences. Each respondent played the preferred lottery. 24 out of 31 participants chose $B$.
If we assume that the proportion of people choosing B is the same for real and hypothetical questions, how likely are the observed frequencies? Battalio, Kagel, and Jiranyakul (1990) reported a test statistic of 3.34 and a p-value of 0.07 for this item.
Let's look at the data in the form of a contingency table:
| B | A | |
|---|---|---|
| Real | 24 | 7 |
| Hypothetical | 16 | 13 |
import scipy
import matplotlib
import numpy as np
from matplotlib import pyplot as plt
from scipy import stats
print('numpy version: ' + np.__version__)
print('scipy version: ' + scipy.__version__)
print('matplotlib version: ' + matplotlib.__version__)
numpy version: 1.19.2 scipy version: 1.6.2 matplotlib version: 3.5.0
As a numpy array:
obs = np.array([[24, 7], [16, 13]])
obs
array([[24, 7],
[16, 13]])
We can extend the data with the row and column sums (calculating "marginal sums").
| B | A | Sum | |
|---|---|---|---|
| Real | 24 | 7 | 31 |
| Hypothetical | 16 | 13 | 29 |
| Sum | 40 | 20 | 60 |
If our null hypothesis of independence is true, our best estimator of the fraction of people choosing $B$ is $40/60=2/3$ (and $1/3$ for $A$). Then, the expected frequencies given the sizes of the two groups are:
| B | A | Sum | |
|---|---|---|---|
| Real | 20.67 | 10.33 | 31 |
| Hypothetical | 19.33 | 9.67 | 29 |
| Sum | 40 | 20 | 60 |
scipy agrees.
stats.contingency.expected_freq(obs)
array([[20.66666667, 10.33333333],
[19.33333333, 9.66666667]])
Similar to the one-way chi square test, the test statistic is defined as the sum of the differences between observed and expected frequencies, $\sum \frac{(observed - expected)^2}{expected}$, sometimes written as $\chi^2 = \sum_i \frac{(E_i-O_i)^2}{E_i}$. The more the observed frequencies deviate from the expected ones, the higher the test statistic. In our example:
(24-20.67) ** 2 / 20.67 + (16-19.33) ** 2 / 19.33 + (7-10.33) ** 2 / 10.33 + (13-9.67) ** 2 / 9.67
3.3303336453568035
This is the test statistic reported in Battalio, Kagel, and Jiranyakul (1990). (There is small approximation error.) scipy's documentation suggests to call
chi2, p, dof, ex = stats.chi2_contingency(obs, correction=False)
to obtain the test statistic
chi2
3.337041156840934
the p-value
p
0.0677363129657488
and the expected frequencies under the null hypothesis
ex
array([[20.66666667, 10.33333333],
[19.33333333, 9.66666667]])
Note: In R, this test is one line of code: "prop.test(x=c(24,16), n=c(31,29), correct=F)". Try it online.
Bottom Line: Using R, you can add a statistical test to your thesis just by running a single line of code.
Optional: You may set up a small simulation to estimate the p-value, see Simple Tests for Theses I: One-Way Chi-Square Test.
Literature: