Paired Binary Data

Imagine now that our two samples of Bernoulli populations aren’t independent, but paired in some way.

$Y_i,\dots ,Y_n\sim \text{Bernoulli}\left(p_Y\right)$
$X_i,\dots ,X_n\sim \text{Bernoulli}\left(p_X\right)$

but $(Y_i,X_i)$ are paired.

Examples:

• $n$ subjects with a disease, and $n$ without a disease are sampled then matched (based on demographic factors), response is presence of some risk factor
• Sibling (or twin) studies: $n$ pairs of related people where one falls in one group, and the other falls in the other group, observe some binary response on every person.
• Binary before and after measurements on the same person

Paired Binary Data

Gather sample of $n=40$ voters.

Before debate: Will you vote for candidate A?
After debate: Will you vote for candidate A?

Just a $2\times 2$ table?

How to analyse?

Option 1: Treat like paired two sample data and do a paired t-test

Option 2: McNemar’s test

Paired t-test

Null hypothesis: $H_0:p_{\text{before}}=p_{\text{after}}$

Look at (per voter) differences:

Paired t-test calculations

$$\overline{D}=\frac{1}{n}\left((-1\times 9)+(0\times 20)+(1\times 11)\right)=\frac{b-c}{n}=\frac{2}{40}=0.05$$

$$\begin{array}{cc}{s}_D^2& =\frac{1}{n-1}\left(9(-1-\overline{D}{)}^2+20(0-\overline{D}{)}^2+11(1-\overline{D}{)}^2\right)\\ & =\frac{1}{n-1}\left(c+b-\frac{(b-c{)}^2}{n}\right)\\ & =\frac{1}{40-1}\left(9+11-\frac{(11-9{)}^2}{n}\right)\\ & =0.51\end{array}$$

Paired t-test calculations

## 
##  One Sample t-test
## 
## data:  df$diff
## t = 0.4427, df = 39, p-value = 0.6604
## alternative hypothesis: true mean is not equal to 0
## 95 percent confidence interval:
##  -0.1784514  0.2784514
## sample estimates:
## mean of x 
##      0.05

McNemar’s test

Null hypothesis: $H_0:p_{\text{before}}=p_{\text{after}}$

Conditions on the number of discordant pairs, $b+c$.

Under Null hypothesis, we expect the number of discordant pairs (e.g. people who change their minds during debate) should be equally split between $b$ and $c$.

McNemar’s test

Conditional on $b+c$, $$b\sim \text{Binomial}(b+c,0.5)$$

Do, one sample Z-test for proportions, leads to $$Z=\frac{b-c}{\sqrt{b+c}}\dot{\sim }N(0,1)\text{under null hypothesis}$$ (sometimes people square this statistic, and compare to ${\chi }_1^2$)

Example: McNemar’s

$$Z=\frac{b-c}{\sqrt{b+c}}=\frac{11-9}{\sqrt{11+9}}=0.45$$

Compare to N(0,1)

Final points

• McNemar’s test is equivalent to the paired t-test, in the sense that the two test statistics are monotone transformations of each other.

• For large sample sizes, the two test statistics get closer and closer to the same value: asymptotically equivalent.