1
$\begingroup$

I have a data set that contains binomial samples within two conditions and several blocks, with a random, small number of samples for each condition-block combination. For example:

╔═════════╦═══════════════════╦══════════════════╗
║         ║ Condition 1       ║ Condition 2      ║
╠═════════╬═══════════════════╬══════════════════╣
║ Block 1 ║ 3 of 11 successes ║ 4 of 6 successes ║
║ Block 2 ║ 0 of 12 successes ║ 3 of 5 successes ║
║ Block 3 ║ 9 of 9 successes  ║ 2 of 7 successes ║
║ ...     ║                   ║                  ║
║ Block 40║ 1 of 4 successes  ║ 6 of 6 successes ║
╚═════════╩═══════════════════╩══════════════════╝

I'd like to test whether condition has a significant effect on success rate, how should I approach this analysis?

From what I've found the analysis I need feels similar to Cochran's Q Test, but doesn't quite fit since Cochran's Q Test only permits a single value (success/failure) per condition-block combination. If I had larger sample sizes for each entry in the table I'd convert them to percentages and use the Friedman Test to test for differences, but I'm concerned that the small sample size invalidates this approach. As a result, my data set seems to fall into a no man's land between tests, so I'm not sure how to analyze it.

$\endgroup$
2
  • $\begingroup$ Are the different blocks assumed to be independent? $\endgroup$
    – Wolfgang
    Oct 20, 2015 at 18:09
  • $\begingroup$ @Wolfgang I believe I can safely assume independence across blocks. $\endgroup$
    – Adam Bosen
    Oct 20, 2015 at 18:21

1 Answer 1

3
$\begingroup$

Since these are binomial data, I would analyze them as such, using appropriate (mixed-effects) logistic regression models. Your dataset will have a block id variable, a condition dummy variable (0 for condition 1, 1 for condition 2), and the number of successes and failures within each condition within each block. For example, for the data you posted (using R):

dat <- data.frame(block = c(1,1,2,2,3,3,40,40),
                  condition = c(0,1,0,1,0,1,0,1),
                  successes = c(3,4,0,3,9,2,1,6),
                  failures = c(8,2,12,2,0,5,3,0))
dat

This looks like this:

  block condition successes failures
1     1         0         3        8
2     1         1         4        2
3     2         0         0       12
4     2         1         3        2
5     3         0         9        0
6     3         1         2        5
7    40         0         1        3
8    40         1         6        0

Then we can fit a logistic regression model with a block factor and the condition dummy:

res1 <- glm(cbind(successes, failures) ~ factor(block) + condition, data=dat, family=binomial)
summary(res1)

This yields (cutting out some stuff here):

Coefficients:
                Estimate Std. Error z value Pr(>|z|)
(Intercept)      -0.7006     0.5542  -1.264    0.206
factor(block)2   -1.1749     0.8209  -1.431    0.152
factor(block)3    1.1215     0.7479   1.500    0.134
factor(block)40   1.0269     0.8722   1.177    0.239
condition         0.9361     0.6029   1.552    0.121

The idea here is that the overall success rate may differ across blocks, so we allow for such differences by including the block factor. The coefficient for the condition dummy is the estimated log odds ratio for condition 2 versus 1 within blocks. If the success rate is unaffected by condition, then this implies a log odds ratio of 0, so the p-value given here is what you are looking for (i.e., it tests the null hypothesis that the true log odds ratio is 0).

The model above assumes that the true log odds ratio is constant across blocks. That may not be the case. We can model such differences by adding a random effect for the condition dummy:

library(lme4)
res2 <- glmer(cbind(successes, failures) ~ factor(block) + condition + (condition - 1 | block), data=dat, family=binomial)
summary(res2)

This yields (again cutting out some things):

Random effects:
 Groups Name      Variance Std.Dev.
 block  condition 15.38    3.921   
Number of obs: 8, groups:  block, 4

Fixed effects:
                Estimate Std. Error z value Pr(>|z|)  
(Intercept)      -0.9907     0.6722  -1.474   0.1405  
factor(block)2   -3.3084     2.4908  -1.328   0.1841  
factor(block)3    4.0845     1.8769   2.176   0.0295 *
factor(block)40   0.2032     1.2622   0.161   0.8721  
condition         2.1526     2.4080   0.894   0.3713  
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Now we test if the average log odds ratio is significantly different from 0 (note that with only 4 blocks of data, this is stretching things a bit, but with 40 blocks, you should be fine).

You can test if the second model is actually a significant improvement over the first with:

anova(res2, res1)

This yields:

     Df    AIC    BIC  logLik deviance Chisq Chi Df Pr(>Chisq)    
res1  5 49.418 49.815 -19.709   39.418                            
res2  6 38.118 38.595 -13.059   26.118  13.3      1  0.0002655 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

In essence, this is testing the null hypothesis that the variance of the random condition effect is 0. Again, with so little data, this is just an illustration, but the results above suggest that there is significant variability in the condition effect. Assuming that you find something similar in the full dataset, you may need to qualify your conclusions a bit more (i.e., you can say something about whether condition affects the outcome on average).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.