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Cochran's Q test
data: y
Q = 0, df = 2, p-value = 132 Cochrans Q Test
32.1 Cochran’s Q Test
Cochran’s Q Test is a non-parametric statistical test used to determine whether there are significant differences in the frequencies of a binary outcome across three or more related groups or conditions. It is an extension of the McNemar test for scenarios involving more than two related groups and is commonly used for repeated measures where the response variable is dichotomous. This test is useful for analyzing data from studies where the same subjects are under different conditions, such as different time points or different treatments.
32.1.1 Understanding Cochran’s Q Test:
1. Null and Alternative Hypotheses:
- Null Hypothesis (H0): The null hypothesis states that the proportions of the binary outcome are the same across all groups or conditions.
- Alternative Hypothesis (H1): The alternative hypothesis suggests that there is a significant difference in the proportions of the binary outcome across at least one of the conditions.
2. Test Statistic:
- The Cochran’s Q test statistic is based on the number of times each subject has the characteristic of interest across all conditions and the total number of characteristics observed for all subjects across all conditions.
- The test statistic follows a chi-squared distribution with (k - 1) degrees of freedom under the null hypothesis, where (k) is the number of related groups or conditions.
3. Calculation of Test Statistic:
- Let (n) be the total number of subjects, and (k) be the number of conditions. The Cochran’s Q test statistic is calculated by comparing the variance of the total scores across conditions with the variance expected by chance.
- The formula for Cochran’s Q test statistic is: \[ Q = \frac{(k-1)(k\sum{T_j^2} - (\sum{T_j})^2)}{k\sum{t_i} - \sum{T_j^2}} \] where (T_j) is the total number of times the characteristic appears in the (j)-th condition and (t_i) is the total number of times the characteristic appears for the (i)-th subject.
4. Interpretation of Results:
- If the calculated (Q) value is greater than the critical value from the chi-squared distribution with (k - 1) degrees of freedom at the chosen significance level (commonly (= 0.05)), then the null hypothesis is rejected, indicating significant differences across conditions.
32.1.2 Applications of Cochran’s Q Test:
A. Medical Research:
- In clinical trials, the Cochran’s Q test is used to assess the consistency of treatment effects observed at different time points or under different conditions within the same group of patients.
B. Psychology:
- Psychologists may apply the Cochran’s Q test to evaluate the consistency of binary responses (like success or failure) across repeated measures or different experimental conditions.
C. Quality Control:
- In industrial settings, the Cochran’s Q test can be used to compare the pass/fail rates of products or processes across different shifts or batches.
32.1.3 Considerations:
- Cochran’s Q test assumes that the observations are independent within subjects but not between subjects.
- The test is only applicable to binary (dichotomous) outcomes.
- The Cochran’s Q test may lose power if the sample size is small, and alternative methods should be considered in such cases.
In summary, Cochran’s Q test offers a robust method for analyzing differences in binary outcomes across more than two related groups or conditions. It is especially valuable for repeated measures design where the same subjects are exposed to different conditions, allowing researchers to investigate the consistency of an effect or response across those conditions.
32.1.4 Example Problem:
A software company wants to test the reliability of three versions of a software application (Version A, Version B, and Version C) under the same conditions. They have 5 testers that each test all three versions for reliability. The outcome is binary: Pass (if the software version works reliably during the test) or Fail (if it does not). The results are as follows:
| Tester | Version A (Pass=1, Fail=0) | Version B (Pass=1, Fail=0) | Version C (Pass=1, Fail=0) |
|---|---|---|---|
| 1 | 1 | 0 | 1 |
| 2 | 1 | 1 | 1 |
| 3 | 0 | 0 | 0 |
| 4 | 1 | 1 | 1 |
| 5 | 0 | 1 | 0 |
The company wants to know if there is a significant difference in reliability between the three software versions.
32.1.5 Calculation of Cochran’s Q Test:
-
Calculate the totals for each version (sum across rows):
- \(T_A = 1 + 1 + 0 + 1 + 0 = 3\)
- \(T_B = 0 + 1 + 0 + 1 + 1 = 3\)
- \(T_C = 1 + 1 + 0 + 1 + 0 = 3\)
-
Calculate the totals for each tester (sum across columns):
- \(t_1 = 1 + 0 + 1 = 2\)
- \(t_2 = 1 + 1 + 1 = 3\)
- \(t_3 = 0 + 0 + 0 = 0\)
- \(t_4 = 1 + 1 + 1 = 3\)
- \(t_5 = 0 + 1 + 0 = 1\)
-
Compute the sums needed for the Q statistic:
- \(\sum{T_j^2} = T_A^2 + T_B^2 + T_C^2 = 3^2 + 3^2 + 3^2 = 27\)
- \((\sum{T_j})^2 = (T_A + T_B + T_C)^2 = (3 + 3 + 3)^2 = 81\)
- \(\sum{t_i} = t_1 + t_2 + t_3 + t_4 + t_5 = 2 + 3 + 0 + 3 + 1 = 9\)
- \(k\sum{t_i} = 3 * 9 = 27\)
- \(k = 3\) (number of software versions)
- Calculate the Q statistic: \[ Q = \frac{(k-1)(k\sum{T_j^2} - (\sum{T_j})^2)}{k\sum{t_i} - \sum{T_j^2}} \] \[ Q = \frac{(3-1)(3*27 - 81)}{3*9 - 27} \] \[ Q = \frac{2(81 - 81)}{27 - 27} \] This gives us an indeterminate form, which suggests there is no variability between the versions as they all have the same total of passes.
Given this result, we can infer that there is no statistically significant difference in reliability between the three software versions, according to the Cochran’s Q test.
In this case, since all versions had the same number of passes and the calculated Q is zero, there is no need to consult a chi-square distribution table as the Q value will not exceed the critical value for any conventional level of alpha.
32.1.6 R Code
First, make sure that you have the necessary package installed for conducting the Cochran’s Q test:
Then, use the following R code:
This will calculate Cochran’s Q test for the given matrix where each row represents a tester and each column represents a version of the software.
32.1.7 Python Code
In Python, you can use the pingouin library, which provides a function to perform Cochran’s Q test. First, ensure pingouin is installed using pip:
!pip3 install pingouinThen, use the following Python code:
Code
import pingouin as pg
import pandas as pd
# Creating a DataFrame for the data
# Each row represents a tester's responses to the different software versions
data = {
'Tester': ['Tester1', 'Tester2', 'Tester3', 'Tester4', 'Tester5'],
'Version_A': [1, 1, 0, 1, 0],
'Version_B': [0, 1, 0, 1, 1],
'Version_C': [1, 1, 0, 1, 0]
}
df = pd.DataFrame(data).set_index('Tester')
# Perform Cochran's Q test
result = pg.cochran(df)
print(result) Source dof Q p-unc
cochran Within 2 0.0 1.0This code block creates a pandas DataFrame for the test results, where rows are indexed by tester, and each column represents their response to a particular software version. The pg.cochran() function is then used to perform Cochran’s Q test on the DataFrame.
32.2 Post-hoc Tests for Cochran’s Q Test:
After performing Cochran’s Q test, if the result is significant, it implies that there are differences in the binary outcomes across the related groups. However, Cochran’s Q test does not specify which groups differ from each other. To identify the specific groups between which these differences occur, you would perform post-hoc pairwise comparisons.
For Cochran’s Q test, one common approach for post-hoc analysis is to use pairwise comparisons with a Bonferroni correction to adjust for multiple testing.
Let’s consider a hypothetical example involving Cochran’s Q test and its subsequent post-hoc analysis:
32.2.1 Example Problem:
A researcher is conducting a study on the efficacy of four different diets on weight loss. They have 10 participants who try each diet for a month, and the outcome is binary: Success (if the participant loses a significant amount of weight) or Failure (if they do not).
The study’s results are summarized as follows:
- Diet A: 6 successes, 4 failures
- Diet B: 7 successes, 3 failures
- Diet C: 3 successes, 7 failures
- Diet D: 5 successes, 5 failures
The researcher conducts a Cochran’s Q test and finds a significant difference in the success rates of the diets but now needs to conduct post-hoc tests to determine where the differences lie.
32.2.2 R Code for Post-hoc Test:
Code
install.packages("rcompanion")Code
library(rcompanion)
# Create data in long format for post-hoc analysis
long_data <- data.frame(
Participant = rep(1:10, times = 4),
Diet = factor(rep(c("A", "B", "C", "D"), each = 10)),
Success = c(1,1,1,0,1,1,1,0,0,0, 1,1,1,1,1,1,0,1,0,0, 0,1,0,0,0,0,0,1,0,1, 1,0,1,0,1,0,1,0,1,1)
)
# Create contingency tables for each pair of diets and perform pairwise comparisons
# Here's how you might create a contingency table for diets A and B:
dietA_dietB <- with(long_data, table(DietA = Success[Diet == "A"], DietB = Success[Diet == "B"]))
# Now perform a McNemar test on this contingency table
mcnemar_test_AB <- mcnemar.test(dietA_dietB)
# Print the result
print(mcnemar_test_AB)
McNemar's Chi-squared test with continuity correction
data: dietA_dietB
McNemar's chi-squared = 0, df = 1, p-value = 1Code
# Here's how you might create a contingency table for diets A and C:
dietA_dietC <- with(long_data, table(DietA = Success[Diet == "A"], DietC = Success[Diet == "C"]))
# Now perform a McNemar test on this contingency table
mcnemar_test_AC <- mcnemar.test(dietA_dietC)
# Print the result
print(mcnemar_test_AC)
McNemar's Chi-squared test with continuity correction
data: dietA_dietC
McNemar's chi-squared = 0.57143, df = 1, p-value = 0.4497Code
# Here's how you might create a contingency table for diets A and D:
dietA_dietD <- with(long_data, table(DietA = Success[Diet == "C"], DietD = Success[Diet == "D"]))
# Now perform a McNemar test on this contingency table
mcnemar_test_AD <- mcnemar.test(dietA_dietD)
# Print the result
print(mcnemar_test_AD)
McNemar's Chi-squared test with continuity correction
data: dietA_dietD
McNemar's chi-squared = 0.57143, df = 1, p-value = 0.449732.2.3 Python Code for Post-hoc Test:
Code
from statsmodels.stats.contingency_tables import mcnemar
import scipy.stats as stats
import pandas as pd
# Define the function for pairwise comparisons with Bonferroni correction
def pairwise_mcnemar(dataframe, alpha=0.05):
columns = dataframe.columns
p_values = []
for i in range(len(columns)):
for j in range(i+1, len(columns)):
table = pd.crosstab(dataframe[columns[i]], dataframe[columns[j]])
result = mcnemar(table, exact=False, correction=True)
p_values.append((columns[i], columns[j], result.pvalue))
# Apply Bonferroni correction
corrected_alpha = alpha / len(p_values)
print("Using a corrected alpha of {:.5f} for {} comparisons".format(corrected_alpha, len(p_values)))
# Print results
for comparison in p_values:
print("Comparison between {} and {}: p-value = {:.5f}".format(comparison[0], comparison[1], comparison[2]))
if comparison[2] < corrected_alpha:
print("The difference between {} and {} is statistically significant.\n".format(comparison[0], comparison[1]))
else:
print("No significant difference between {} and {}.\n".format(comparison[0], comparison[1]))
# Example DataFrame for our problem
df = pd.DataFrame({
'Diet_A': [1,1,1,0,1,1,1,0,0,0],
'Diet_B': [1,1,1,1,1,1,0,1,0,0],
'Diet_C': [0,1,0,0,0,0,0,1,0,1],
'Diet_D': [1,0,1,0,1,0,1,0,1,1]
})
pairwise_mcnemar(df)Using a corrected alpha of 0.00833 for 6 comparisons
Comparison between Diet_A and Diet_B: p-value = 1.00000
No significant difference between Diet_A and Diet_B.
Comparison between Diet_A and Diet_C: p-value = 0.44969
No significant difference between Diet_A and Diet_C.
Comparison between Diet_A and Diet_D: p-value = 0.61708
No significant difference between Diet_A and Diet_D.
Comparison between Diet_B and Diet_C: p-value = 0.22067
No significant difference between Diet_B and Diet_C.
Comparison between Diet_B and Diet_D: p-value = 1.00000
No significant difference between Diet_B and Diet_D.
Comparison between Diet_C and Diet_D: p-value = 0.44969
No significant difference between Diet_C and Diet_D.In this example, we first conduct pairwise comparisons for each pair of diets using McNemar’s test. Then, we apply the Bonferroni correction to adjust the significance level for the number of comparisons to control the family-wise error rate. By comparing each diet to the others, we can conclude which specific diets differ in terms of their effectiveness for weight loss. The output of these post-hoc tests will help the researcher to understand whether the significant result from Cochran’s Q test is due to a particular diet being more effective or less effective compared to the others.