# What is the difference between McNemar's test and the chi-squared test, and how do you know when to use each?

I have tried reading up on different sources, but I am still not clear what test would be the appropriate in my case. There are three different questions I am asking about my dataset:

1. The subjects are tested for infections from X at different times. I want to know if the proportions of positive for X after is related to the proportion of positive for X before:

             After
|no  |yes|
Before|No  |1157|35 |
|Yes |220 |13 |

results of chi-squared test:
Chi^2 =  4.183     d.f. =  1     p =  0.04082

results of McNemar's test:
Chi^2 =  134.2     d.f. =  1     p =  4.901e-31


From my understanding, as the data are repeated measures, I must use McNemar's test, which tests if the proportion of positive for X has changed.

But my questions seems to need the chi-squared test - testing if the proportion of positive for X after is related to the proportion of positive for X before.

I am not even sure if I understand the difference between McNemar's test and the chi-squared correctly. What would be the right test if my question were, "Is the proportion of subjects infected with X after different from before?"

2. A similar case, but where instead of before and after, I measure two different infections at one point in time:

        Y
|no  |yes|
X|No  |1157|35 |
|Yes |220 |13 |


Which test would be right here if the question is "Does higher proportions of one infections relate to higher proportions of Y"?

3. If my question was, "Is infection Y at time t2 related to infection X at time t1?", which test would be appropriate?

              Y at t2
|no  |yes|
X at t1|No  |1157|35 |
|Yes |220 |13 |


I was using McNemar's test in all these cases, but I have my doubts if that is the right test to answer my questions. I am using R. Could I use a binomial glm instead? Would that be analogous to the chi-squared test?

• Did you try to read stats.stackexchange.com/questions/tagged/mcnemar-test topics here on Mcnemar test? – ttnphns Nov 18 '13 at 14:58
• What do you mean by "relation between two probabilities"? – Michael M Nov 18 '13 at 15:01
• @ttnphns I did go through them, but was unable to reformulate it to my question. After more thinking, it seems I can answer two questions based on the Q1:Chi-sq would tell me if the proportion of +ve X after is related to the proportion of +ve X before while Mcnemar's would tell me if there has been a change in the proportions. Am I right? – Anto Nov 18 '13 at 16:01
• You cannot use a standard $\chi^2$-test of independence here because each person is represented by two values causing non-random samples. – Michael M Nov 18 '13 at 18:35
• Thanks @MichaelMayer. I was using mcnemar's until I saw this. Where the Mcnemar's is explained, he says what the doing a Chi-sq would answer in the same case. I am quite perplexed. The way what each test tell us is framed on this page, I must go for the Chi-sq but since they are measurements on the same subject, I must choose McNemar's! – Anto Nov 20 '13 at 14:40

It is very unfortunate that McNemar's test is so difficult for people to understand. I even notice that at the top of its Wikipedia page it states that the explanation on the page is difficult for people to understand. The typical short explanation for McNemar's test is either that it is: 'a within-subjects chi-squared test', or that it is 'a test of the marginal homogeneity of a contingency table'. I find neither of these to be very helpful. First, it is not clear what is meant by 'within-subjects chi-squared', because you are always measuring your subjects twice (once on each variable) and trying to determine the relationship between those variables. In addition, 'marginal homogeneity' is barely intelligible (I know what this means and I have a hard time moving from the words to the meaning). (Tragically, even this answer may be confusing. If it is, it may help to read my second attempt below.)

Let's see if we can work through a process of reasoning about your top example to see if we can understand whether (and if so, why) McNemar's test is appropriate. You have put:

This is a contingency table, so it connotes a chi-squared analysis. Moreover, you want to understand the relationship between ${\rm Before}$ and ${\rm After}$, and the chi-squared test checks for a relationship between the variables, so at first glance it seems like the chi-squared test must be the analysis that answers your question.

However, it is worth pointing out that we can also present these data like so:

When you look at the data this way, you might think you could do a regular old $t$-test. But a $t$-test isn't quite right. There are two issues: First, because each row lists data measured from the same subject, we wouldn't want to do a between-subjects $t$-test, we would want to do a within-subjects $t$-test. Second, since these data are distributed as a binomial, the variance is a function of the mean. This means that there is no additional uncertainty to worry about once the sample mean has been estimated (i.e., you don't have to subsequently estimate the variance), so you don't have to refer to the $t$ distribution, you can use the $z$ distribution. (For more on this, it may help to read my answer here: The $z$-test vs. the $\chi^2$ test.) Thus, we would need a within-subjects $z$-test. That is, we need a within-subjects test of equality of proportions.

We have seen that there are two different ways of thinking about and analyzing these data (prompted by two different ways of looking at the data). So we need to decide which way we should use. The chi-squared test assesses whether ${\rm Before}$ and ${\rm After}$ are independent. That is, are people who were sick beforehand more likely to be sick afterwards than people who have never been sick. It is extremely difficult to see how that wouldn't be the case given that these measurements are assessed on the same subjects. If you did get a non-significant result (as you almost do) that would simply be a type II error. Instead of whether ${\rm Before}$ and ${\rm After}$ are independent, you almost certainly want to know if the treatment works (a question chi-squared does not answer). This is very similar to any number of treatment vs. control studies where you want to see if the means are equal, except that in this case your measurements are yes/no and they are within-subjects. Consider a more typical $t$-test situation with blood pressure measured before and after some treatment. Those whose bp was above your sample average beforehand will almost certainly tend to be among the higher bps afterwards, but you don't want to know about the consistency of the rankings, you want to know if the treatment led to a change in mean bp. Your situation here is directly analogous. Specifically, you want to run a within-subjects $z$-test of equality of proportions. That is what McNemar's test is.

So, having realized that we want to conduct McNemar's test, how does it work? Running a between-subjects $z$-test is easy, but how do we run a within-subjects version? The key to understanding how to do a within-subjects test of proportions is to examine the contingency table, which decomposes the proportions:
\begin{array}{rrrrrr} & &{\rm After} & & & \\ & &{\rm No} &{\rm Yes} & &{\rm total} \\ {\rm Before}&{\rm No} &1157 &35 & &1192 \\ &{\rm Yes} &220 &13 & &233 \\ & & & & & \\ &{\rm total} &1377 &48 & &1425 \\ \end{array} Obviously the ${\rm Before}$ proportions are the row totals divided by the overall total, and the ${\rm After}$ proportions are the column totals divided by overall total. When we look at the contingency table we can see that those are, for example:
$$\text{Before proportion yes} = \frac{220 + 13}{1425},\quad\quad \text{After proportion yes} = \frac{35 + 13}{1425}$$ What is interesting to note here is that $13$ observations were yes both before and after. They end up as part of both proportions, but as a result of being in both calculations they add no distinct information about the change in the proportion of yeses. Moreover they are counted twice, which is invalid. Likewise, the overall total ends up in both calculations and adds no distinct information. By decomposing the proportions we are able to recognize that the only distinct information about the before and after proportions of yeses exists in the $220$ and $35$, so those are the numbers we need to analyze. This was McNemar's insight. In addition, he realized that under the null, this is a binomial test of $220/(220 + 35)$ against a null proportion of $.5$. (There is an equivalent formulation that is distributed as a chi-squared, which is what R outputs.)

There is another discussion of McNemar's test, with extensions to contingency tables larger than 2x2, here.

Here is an R demo with your data:

mat = as.table(rbind(c(1157, 35),
c( 220, 13) ))
colnames(mat) <- rownames(mat) <- c("No", "Yes")
names(dimnames(mat)) = c("Before", "After")
mat
margin.table(mat, 1)
margin.table(mat, 2)
sum(mat)

mcnemar.test(mat, correct=FALSE)
#  McNemar's Chi-squared test
#
# data:  mat
# McNemar's chi-squared = 134.2157, df = 1, p-value < 2.2e-16
binom.test(c(220, 35), p=0.5)
#  Exact binomial test
#
# data:  c(220, 35)
# number of successes = 220, number of trials = 255, p-value < 2.2e-16
# alternative hypothesis: true probability of success is not equal to 0.5
# 95 percent confidence interval:
#  0.8143138 0.9024996
# sample estimates:
# probability of success
#              0.8627451


If we didn't take the within-subjects nature of your data into account, we would have a slightly less powerful test of the equality of proportions:

prop.test(rbind(margin.table(mat, 1), margin.table(mat, 2)), correct=FALSE)
#  2-sample test for equality of proportions without continuity
#  correction
#
# data:  rbind(margin.table(mat, 1), margin.table(mat, 2))
# X-squared = 135.1195, df = 1, p-value < 2.2e-16
# alternative hypothesis: two.sided
# 95 percent confidence interval:
#  0.1084598 0.1511894
# sample estimates:
#    prop 1    prop 2
# 0.9663158 0.8364912


That is, X-squared = 133.6627 instead of chi-squared = 134.2157. In this case, these differ very little, because you have a lot of data and only $13$ cases are overlapping as discussed above. (Another, and more important, problem here is that this counts your data twice, i.e., $N = 2850$, instead of $N = 1425$.)

1. The correct analysis is McNemar's test (as discussed extensively above).
2. This version is trickier, and the phrasing "does higher proportions of one infections relate to higher proportions of Y" is ambiguous. There are two possible questions:

• It is perfectly reasonable to want to know if the patients who get one of the infections tend to get the other, in which case you would use the chi-squared test of independence. This question is asking whether susceptibility to the two different infections is independent (perhaps because they are contracted via different physiological pathways) or not (perhaps they are contracted due to a generally weakened immune system).
• It is also perfectly reasonable to what to know if the same proportion of patients tend to get both infections, in which case you would use McNemar's test. The question here is about whether the infections are equally virulent.
3. Since this is once again the same infection, of course they will be related. I gather that this version is not before and after a treatment, but just at some later point in time. Thus, you are asking if the background infection rates are changing organically, which is again a perfectly reasonable question. At any rate, the correct analysis is McNemar's test.
Edit: It would seem I misinterpreted your third question, perhaps due to a typo. I now interpret it as two different infections at two separate timepoints. Under this interpretation, the chi-squared test would be appropriate.
• @Alexis As far as I can make out, you and gung seem to be talking past each other. Even the so-called "unpaired" or "independent samples" t-test, or the "one-way" or "independent samples ANOVA", actually requires paired data in gung's sense: for each subject, you must record both a categorical group membership variable and a continuous outcome variable. (If the group membership variable has two levels, we usually use the unpaired t-test; for 3+ levels you need one-way ANOVA). – Silverfish Mar 12 '15 at 9:46
• When explaining which test to use, I show both ways of looking at it - if you have observations of a continuous variable, one for each subject, and the subjects come from 2 (or 3+) groups and you're interested in differences between groups, then use the independent-samples t-test (or one-way ANOVA). Then confirm your choice by looking at your data table: do you have, for each subject, two pieces of information: category for group membership and the continuous variable. We can even turn things around and say the t-test is a kind of test of association between binary and continuous variable. – Silverfish Mar 12 '15 at 9:50
• Paired t-test (or correlated samples ANOVA) is used if, for each subject, you have two (or 3+) continuous readings, taken under different conditions, and you want to test for differences between conditions. This is "paired" in a different sense. But in this question, we have two categorical variables recorded for each subject. Looking at the data table, the recorded values of those categorical variables must come in pairs. But this doesn't mean that the study design itself is paired. This is confusing (as gung notes). But if you know your study design, this can resolve it (as alexis notes) – Silverfish Mar 12 '15 at 10:05
• @Silverfish If you have two observations (of the same nominal variable) made on each subject, in what sense is that not a paired design? – Alexis Mar 12 '15 at 15:26
• @Alexis It's that "of the same variable" which is key - and potentially confusing. You might know it represents the same variable, albeit under different conditions or at different times, but depending on the way we lay the data table out, they may appear to be recorded as different variables (eg a separate "before" and "after" variable). – Silverfish Mar 12 '15 at 16:59

Well, it seems I've made a hash of this. Let me try to explain this again, in a different way and we'll see if it might help clear things up.

The traditional way to explain McNemar's test vs. the chi-squared test is to ask if the data are "paired" and to recommend McNemar's test if the data are paired and the chi-squared test if the data are "unpaired". I have found that this leads to a lot of confusion (this thread being an example!). In place of this, I have found that it is most helpful to focus on the question you are trying to ask, and to use the test that matches your question. To make this more concrete, let's look at a made-up scenario:

You walk around a statistics conference and for each statistician you meet, you record whether they are from the US or the UK. You also record whether they have high blood pressure or normal blood pressure.

Here are the data:

mat = as.table(rbind(c(195,   5),
c(  5, 195) ))
colnames(mat)        = c("US", "UK")
rownames(mat)        = c("Hi", "Normal")
names(dimnames(mat)) = c("BP", "Nationality")
mat
#         Nationality
# BP        US  UK
#   Hi     195   5
#   Normal   5 195


At this point, it is important to figure out what question we want to ask of our data. There are three different questions we could ask here:

1. We might want to know if the categorical variables BP and Nationality are associated or independent;
2. We might wonder if high blood pressure is more common amongst US statisticians than it is amongst UK statisticians;
3. Finally, we might wonder if the proportion of statisticians with high blood pressure is equal to the proportion of US statisticians that we talked to. This refers to the marginal proportions of the table. These are not printed by default in R, but we can get them thusly (notice that, in this case, they are exactly the same):

margin.table(mat, 1)/sum(mat)
# BP
#    Hi Normal
#   0.5    0.5
margin.table(mat, 2)/sum(mat)
# Nationality
#  US  UK
# 0.5 0.5


As I said, the traditional approach, discussed in many textbooks, is to determine which test to use based on whether the data are "paired" or not. But this is very confusing, is this contingency table "paired"? If we compare the proportion with high blood pressure between US and UK statisticians, you are comparing two proportions (albeit of the same variable) measured on different sets of people. On the other hand, if you want to compare the proportion with high blood pressure to the proportion US, you are comparing two proportions (albeit of different variables) measured on the same set of people. These data are both "paired" and "unpaired" at the same time (albeit with respect to different aspects of the data). This leads to confusion. To try to avoid this confusion, I argue that you should think in terms of which question you are asking. Specifically, if you want to know:

1. If the variables are independent: use the chi-squared test.
2. If the proportion with high blood pressure differs by nationality: use the z-test for difference of proportions.
3. If the marginal proportions are the same: use McNemar's test.

Someone might disagree with me here, arguing that because the contingency table is not "paired", McNemar's test cannot be used to test the equality of the marginal proportions and that the chi-squared test should be used instead. Since this is the point of contention, let's try both to see if the results make sense:

chisq.test(mat)
#  Pearson's Chi-squared test with Yates' continuity correction
#
# data:  mat
# X-squared = 357.21, df = 1, p-value < 2.2e-16
mcnemar.test(mat)
#  McNemar's Chi-squared test
#
# data:  mat
# McNemar's chi-squared = 0, df = 1, p-value = 1


The chi-squared test yields a p-value of approximately 0. That is, it says that the probability of getting data as far or further from equal marginal proportions, if the marginal proportions actually were equal is essentially 0. But the marginal proportions are exactly the same, $50\%=50\%$, as we saw above! The results of the chi-squared test just don't make any sense in light of the data. On the other hand, McNemar's test yields a p-value of 1. That is, it says that you will have a 100% chance of finding marginal proportions this close to equality or further from equality, if the true marginal proportions are equal. Since the observed marginal proportions cannot be closer to equal than they are, this result makes sense.

Let's try another example:

mat2 = as.table(rbind(c(195, 195),
c(  5,   5) ))
colnames(mat2)        = c("US", "UK")
rownames(mat2)        = c("Hi", "Normal")
names(dimnames(mat2)) = c("BP", "Nationality")
mat2
#         Nationality
# BP        US  UK
#   Hi     195 195
#   Normal   5   5
margin.table(mat2, 1)/sum(mat2)
# BP
#     Hi Normal
#  0.975  0.025
margin.table(mat2, 2)/sum(mat2)
# Nationality
#  US  UK
# 0.5 0.5


In this case, the marginal proportions are very different, $97.5\%\gg 50\%$. Let's try the two tests again to see how their results compare to the observed large difference in marginal proportions:

chisq.test(mat2)
#  Pearson's Chi-squared test
#
# data:  mat2
# X-squared = 0, df = 1, p-value = 1
mcnemar.test(mat2)
#  McNemar's Chi-squared test with continuity correction
#
# data:  mat2
# McNemar's chi-squared = 178.605, df = 1, p-value < 2.2e-16


This time, the chi-squared test gives a p-value of 1, meaning that the marginal proportions are as equal as they can be. But we saw that the marginal proportions are very obviously not equal, so this result doesn't make any sense in light of our data. On the other hand, McNemar's test yields a p-value of approximately 0. In other words, it is extremely unlikely to get data with marginal proportions as far from equality as these, if they truly are equal in the population. Since our observed marginal proportions are far from equal, this result makes sense.

The fact that the chi-squared test yields results that make no sense given our data suggests there is something wrong with using the chi-squared test here. Of course, the fact that McNemar's test provided sensible results doesn't prove that it is valid, it may just have been a coincidence, but the chi-squared test is clearly wrong.

Let's see if we can work through the argument for why McNemar's test might be the right one. I will use a third dataset:

mat3 = as.table(rbind(c(190,  15),
c( 60, 135) ))
colnames(mat3)        = c("US", "UK")
rownames(mat3)        = c("Hi", "Normal")
names(dimnames(mat3)) = c("BP", "Nationality")
mat3
#         Nationality
# BP        US  UK
#   Hi     190  15
#   Normal  60 135
margin.table(mat3, 1)/sum(mat3)
# BP
#     Hi Normal
# 0.5125 0.4875
margin.table(mat3, 2)/sum(mat3)
# Nationality
#    US    UK
# 0.625 0.375


This time we want to compare $51.25\%$ to $62.5\%$ and wonder if in the population the true marginal proportions might have been the same. Because we are comparing two proportions, the most intuitive option would be to use a z-test for the equality of two proportions. We can try that here:

prop.test(x=c(205, 250), n=c(400, 400))
#  2-sample test for equality of proportions with continuity correction
#
# data:  c(205, 250) out of c(400, 400)
# X-squared = 9.8665, df = 1, p-value = 0.001683
# alternative hypothesis: two.sided
# 95 percent confidence interval:
#   -0.18319286 -0.04180714
# sample estimates:
# prop 1 prop 2
# 0.5125 0.6250


(To use prop.test() to test the marginal proportions, I had to enter the numbers of 'successes' and the total number of 'trials' manually, but you can see from the last line of the output that the proportions are correct.) This suggests that it is unlikely to get marginal proportions this far from equality if they were actually equal, given the amount of data we have.

Is this test valid? There are two problems here: The test believes we have 800 data, when we actually have only 400. This test also does not take into account that these two proportions are not independent, in the sense that they were measured on the same people.

Let's see if we can take this apart and find another way. From the contingency table, we can see that the marginal proportions are:
$$\text{% high BP: }\frac{190 + 15}{400} \qquad\qquad\qquad \text{% US: }\frac{190 + 60}{400}$$ What we see here is that the $190$ US statisticians with high blood pressure show up in both marginal proportions. They are both being counted twice and contributing no information about the differences in the marginal proportions. Moreover, the $400$ total shows up in both denominators as well. All of the unique and distinctive information is in the two off-diagonal cell counts ($15$ and $60$). Whether the marginal proportions are the same or different is due only to them. Whether an observation is equally likely to fall into either of those two cells is distributed as a binomial with probability $\pi = .5$ under the null. That was McNemar's insight. In fact, McNemar's test is essentially just a binomial test of whether observations are equally likely to fall into those two cells:

binom.test(x=15, n=(15+60))
#  Exact binomial test
#
# data:  15 and (15 + 60)
# number of successes = 15, number of trials = 75, p-value = 1.588e-07
# alternative hypothesis: true probability of success is not equal to 0.5
# 95 percent confidence interval:
#   0.1164821 0.3083261
# sample estimates:
# probability of success
#                    0.2


In this version, only the informative observations are used and they are not counted twice. The p-value here is much smaller, 0.0000001588, which is often the case when the dependency in the data is taken into account. That is, this test is more powerful than the z-test of difference of proportions. We can further see that the above version is essentially the same as McNemar's test:

mcnemar.test(mat3, correct=FALSE)
#  McNemar's Chi-squared test
#
# data:  mat3
# McNemar's chi-squared = 27, df = 1, p-value = 2.035e-07


If the non-identicallity is confusing, McNemar's test typically, and in R, squares the result and compares it to the chi-squared distribution, which is not an exact test like the binomial above:

(15-60)^2/(15+60)
# [1] 27
1-pchisq(27, df=1)
# [1] 2.034555e-07


Thus, when you want to check the marginal proportions of a contingency table are equal, McNemar's test (or the exact binomial test computed manually) is correct. It uses only the relevant information without illegally using any data twice. It does not just 'happen' to yield results that make sense of the data.

I continue to believe that trying to figure out whether a contingency table is "paired" is unhelpful. I suggest using the test that matches the question you are asking of the data.

• You got my vote. :) – Alexis Mar 12 '15 at 15:28

The question of which test to use, contingency table $\chi^{2}$ versus McNemar's $\chi^{2}$ of a null hypothesis of no association between two binary variables is simply a question of whether your data are paired/dependent, or unpaired/independent:

Binary Data in Two Independent Samples
In this case, you would use a contingency table $\chi^{2}$ test.

For example, you might have a sample of 20 statisticians from the USA, and a separate independent sample of 37 statisticians from the UK, and have a measure of whether these statisticians are hypertensive or normotensive. Your null hypothesis is that both UK and US statisticians have the same underlying probability of being hypertensive (i.e. that knowing whether one is from the USA or from the UK tells one nothing about the probability of hypertension). Of course it is possible that you could have the same sample size in each group, but that does not change the fact of the samples being independent (i.e. unpaired).

Binary Data in Paired Samples
In this case you would use McNemar's $\chi^{2}$ test.

For example, you might have individually-matched case-control study data sampled from an international statistician conference, where 30 statisticians with hypertension (cases) and 30 statisticians without hypertension (controls; who are individually matched by age, sex, BMI & smoking status to particular cases), are retrospectively assessed for professional residency in the UK versus residency elsewhere. The null is that the probability of residing in the UK among cases is the same as the probability of residing in the UK as controls (i.e. that knowing about one's hypertensive status tells one nothing about one's UK residence history).

In fact, McNemar's test analyzes pairs of data. Specifically, it analyzes discordant pairs. So the $r$ and $s$ from $\chi^{2}=\frac{[(r−s)−1]^{2}}{(r+s)}$ are counts of discordant pairs.

Anto, in your example, your data are paired (same variable measured twice in same subject) and therefore McNemar's test is the appropriate choice of test for association.

Quoted References
"Assuming that we are still interested in comparing proportions, what can we do if our data are paired, rather than independent?... In this situation, we use McNemar's test."–Pagano and Gauvreau, Principles of Biostatistics, 2nd edition, page 349. [Emphasis added]

"The expression is better known as the McNemar matched-pair test statistic (McNemar, 1949), and has been a mainstay of matched-pair analysis."—Rothman, Greenland, & Lash. Modern Epidemiology, page 286. [Emphasis added]

"The paired t test and repeated measures of analysis of variance can be used to analyze experiments in which the variable being studied can be measured on an interval scale (and satisfies other assumptions required of parametric methods). What about experiments, analogous to the ones in Chapter 5, where the outcome is measured on a nominal scale? This problem often arises when asking whether or not a an individual responded to a treatment or when comparing the results of two different diagnostic tests that are classified positive or negative in the same individuals. We will develop a procedure to analyze such experiments, Mcnemar's test for changes, in the context of one such study."—Glanz, Primer of Biostatistics, 7th edition, page 200. [Emphasis added. Glanz works through an example of a misapplication the contingency table $\chi^{2}$ test to paired data on page 201.]

"For matched case-control data with one control per case, the resultant analysis is simple, and the appropriate statistical test is McNemar's chi-squared test... note that for the calculation of both the odds ratio and the statistic, the only contributors are the pairs which are disparate in exposure, that is the pairs where the case was exposed but the control was not, and those where the control was exposed but the case was not."—Elwood. Critical Appraisal of Epidemiological Studies and Clinical Trials, 1st edition, pages 189–190. [Emphasis added]

My understanding of McNemar's test is as follows: It is used to see whether an intervention has made a significant difference to a binary outcome. In your example, a group of subjects are checked for infection and the response is recorded as yes or no. All subjects are then given some intervention, say an antibiotic drug. They are then checked again for infection and response is recorded as yes/no again. The (pairs of) responses can be put in the contigency table:

             After
|no  |yes|
Before|No  |1157|35 |
|Yes |220 |13 |


And McNemar's test would be appropriate for this.

It is clear from the table that many more have converted from 'yes' to 'no' (220/(220+13) or 94.4%) than from 'no' to 'yes' (35/(1157+35) or 2.9%). Considering these proportions, McNemar's P value (4.901e-31) appears more correct than chi-square P value (0.04082 ).

If contigency table represents 2 different infections (question 2), then Chi-square would be more appropriate.

Your 3rd question is ambiguous: you first state relating Y at t2 with Y at t1 but in the table you write 'X' at t1 vs Y at t2. Y at t2 vs Y at t1 is same as your first question and hence McNemar's test is needed, while X at t1 and Y at t2 indicates different events are being compared and hence Chi-square will be more appropriate.

Edit: As mentioned by Alexis in the comment, matched case-control data are also analyzed by McNemar's test. For example, 1425 cancer patients are recruited for a study and for each patient a matched control is also recruited. All these (1425*2) are checked for infection. The results of each pair can be shown by similar table:

             Normal
|no  |yes|
Cancer|No  |1157|35 |
|Yes |220 |13 |


More clearly:

                                    Normal:
No infection   Infection
Cancer patient:     No infection    1157            35
Infection       220             13


It shows that it is much more often that cancer patient had infection and control did not, rather than the reverse. Its significance can be tested by McNemar's test.

If these patients and controls were not matched and independent, one can only make following table and do a chisquare test:

            Infection
No    Yes
Cancer  No  1377   48
Yes 1192  233


More clearly:

                No infection        Infection
No cancer       1377                48
Cancer          1192                233


Note that these numbers are same as margins of the first table:

> addmargins(mat)
After
Before   No  Yes  Sum
No  1157   35 1192
Yes  220   13  233
Sum 1377   48 1425


That must be the reason for use of terms like 'marginal frequencies' and 'marginal homogeneity' in McNemar's test.

Interestingly, the addmargins function can also help decide which test to use. If the grand-total is half the number of subjects observed (indicating pairing has been done), then McNemar's test is applicable, else chisquare test is appropriate:

> addmargins(mat)
Normal
Cancer   No  Yes  Sum
No  1157   35 1192
Yes  220   13  233
Sum 1377   48 1425
>
Infection
Cancer   No  Yes  Sum
No  1377   48 1425
Yes 1192  233 1425
Sum 2569  281 2850


The R codes for above tables are as from answers above:

mat = as.table(rbind(c(1157, 35),
c( 220, 13) ))
colnames(mat) <- rownames(mat) <- c("No", "Yes")
names(dimnames(mat)) = c("Cancer", "Normal")

mat3 = as.table(rbind(c(1377, 48),
c(1192, 233) ))
colnames(mat3) <- rownames(mat3) <- c("No", "Yes")
names(dimnames(mat3)) = c("Cancer", "Infection")


Following pseudocode may also help knowing the difference:

subject_id      result_first_observation    result_second_observation
1               no                          yes
2               yes                         no
...

mcnemar.test(table(result_first_observation, result_second_observation))

pair_id     result_case_subject     result_control_subject
1           no                      yes
2           yes                     no
...

mcnemar.test(table(result_case_subject, result_control_subject))

subject_id      result_first_test       result_second_test
1               yes                     no
2               no                      yes
..

chisq.test(table(result_first_test, result_second_test))


Edit:

mid-p variation of peforming McNemar test ( https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3716987/ ) is interesting. It compares b and c of contingency table, i.e. number who changed from yes to no versus number who changed from no to yes (ignoring number of those who remained yes or no through the study). It can be performed using binomial test in python, as shown at https://gist.github.com/kylebgorman/c8b3fb31c1552ecbaafb

It could be equivalent to binom.test(b, b+c, 0.5) since in a random change, one would expect b to be equal to c.

• Not only for intervention analysis: it is used to analyze matched case-control data in an observational sense as well. – Alexis Mar 12 '15 at 15:27
• Given the description / setup prior to the table for Q3, I suspect the "X" is a typo, but that was a good catch & this is a useful contribution to the thread +1. – gung Mar 12 '15 at 18:22
• @mso Edited Q3. it is X at t1! otherwise, as you say it not different from Q1. this Q is over a year old and surprised to see someone coming back to it with the same thoughts that confused me. Following with much interest! – Anto Mar 12 '15 at 19:51
• My apologies, the OP has clarified Q3, evidently it is 2 different diseases at 2 different times. Again, good catch. – gung Mar 12 '15 at 19:53