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Date:         Fri, 6 Apr 2007 12:22:09 -0400
Reply-To:     "Burleson,Joseph A." <>
Sender:       "SPSSX(r) Discussion" <SPSSX-L@LISTSERV.UGA.EDU>
From:         "Burleson,Joseph A." <>
Subject:      Re: McNemar test
Comments: To: Margaret MacDougall <>
Content-Type: text/plain; charset="iso-8859-1"


One has to keep track of the two seemingly related statistics: the original chi-square (1901), and the McNemar test.

The first is the analog to a dichotomous by dichotomous Pearson correlation coefficient [which came after the chi-square]; in fact, the p-level of the chi-square is identical to the p-level to the p-level one gets when simply running the 2 X 2 as the Pearason r.

The second is the analog to a dichotomous paired t-test. Imagine running a paired t-test on the following data: 1st statement Agree Dis Sum

Agr 40 10 50

2ndDis 10 40 50

Sum 50 50 100

If you calculate the difference scores from 1st(pre) to 2nd(post), you will get only 3 types of difference scores from each of the 4 types of data (4 cells, above), diff = 1st - 2nd.

Let Agree=1, Disagree = 0

1st 2nd diff # times occurs from above #times*value 1 1 0 40 0 1 0 1 10 10 0 1 -1 10 -10 0 0 0 40 0 Sum of diffs: 0

If you take this value and plug it into the paired t-test formula, you will get a t = 0, p = 1.0.

So there is perfect logic to the McNemar test, insofar as the logic from the paired t-test. One thing that can be confusing is the value of 0 derived above for the Agree/Agree as well as the Dis/Dis. This is equivalent to a "lack of change" that would also occur if a subject went from 400 to 400 on the SAT verbal from Test1 to Test2, where another subject went from 600 to 600: both subjects would have a change score of zero, and this is not seen as odd.

Note that regardless of the SD of the differences above [not calculated], the t would still be zero.

Try putting some other values than above and observe the value of the paired t, as well as its' p-values.

Joe Burleson ________________________________

From: SPSSX(r) Discussion on behalf of Margaret MacDougall Sent: Fri 4/6/2007 7:13 AM To: SPSSX-L@LISTSERV.UGA.EDU Subject: Re: McNemar test

Dear Paul

Thank you for your lucid description of what is achieved by the McNemar test. I have been watching this discussion with interest, as I have been developing increasing scepticism about the usefulness of this test as a paired test. In particular, it surely cannot be viewed as a good test of difference in individual views over two statements, say, if we wish in particular to see if the individual feels differently according as to whether you present him or her with statement 1 or 2.

Consider the following example:

1st statement 2nd statement agree disagree total agree 10 40 50

disagree 40 10 50

total 50 50 100 If we wished to test the null hypothesis that an individual's view does not change according to the statement they are presented with, we wouldn't have a rather strange result with the McNemar test - a chi-square value of 0, presumably, and a p-value (according to SPSS) of 1, and yet, common sense tells us from this table that concordance is rather poor to say the least. Although I do not have the data to hand, I have witnessed similarly conflicting results occurring in practice where somebody wishes to test for a difference in diagnosis for a particular condition using two tests. Despite the lack of concordance for the results of these tests, the McNemar test provided no evidence to support what was obvious from the corresponding frequency tables (and thus again p greater than 0.05 in each case). I think that there is some confusion about what this test sets out to achieve.

Please feel free to come back to me on this. For example, I would be interested to receive evidence of where the McNemar test can be truly said to work usefully as a paired test.

Yours gratefully

Best wishes


"Swank, Paul R" <> wrote:

I agree that McNemar's test is like a paired (or dependent samples) t test but it works on a 2 by 2 table where the table represents paired frequencies. For example, if 100 people report whether or not they agree or disagree with two political statements, then the results can be reported as a 2 by 2 table:

1st statement 2nd statement agree disagree total agree 30 10 40

disagree 20 40 60

total 50 50 100

The typical chi-squared test for such a table is a measure of association between the statements Indicatig a significant relation (chi-square(df=1, n=100) = 16.67; p < .01). The McNemar test is concerned with the equality of the marginal proportions (p(1.) = P(.1). That is, is the proportion who agree with the first statement (40%) different from the proportion who agree with the second statement (50%)? This is equivalent to the test of symmetry for the table (p(12) = p(21) The large sample test statistic is chi-square(1) = (10 - 20)**2 / (10+20) = 3.33; p > .05.

Paul R. Swank, Ph.D. Professor Director of Reseach Children's Learning Institute University of Texas Health Science Center-Houston

-----Original Message----- From: SPSSX(r) Discussion [mailto:SPSSX-L@LISTSERV.UGA.EDU] On Behalf Of Burleson,Joseph A. Sent: Monday, April 02, 2007 12:27 PM To: SPSSX-L@LISTSERV.UGA.EDU Subject: Re: McNemar test


The McNemar test does not test the difference between the two proportions in a 2 X 2 chi-square. It is used to assess change across time (or some other within-subjects variable): does the change from 0 to 1 differ significantly from the change from 1 to 0, for example (it ignores the frequencies of subjects who go from 0 to 0 and from 1 to 1).

Hence, it is like a "paired t-test" for dichotomous data.

See Siegel, S. (1956). Non parametric statistics for the behavioral sciences. New York: McGraw-Hill. p. 63-67.

Joe Burleson -----Original Message----- From: SPSSX(r) Discussion [mailto:SPSSX-L@LISTSERV.UGA.EDU] On Behalf Of Lou Sent: Monday, April 02, 2007 11:18 AM To: SPSSX-L@LISTSERV.UGA.EDU Subject: McNemar test

Dear all,

Could someone please advise on the best way to report the results of a McNemar test. If I state beforehand that I am using McNemar, do I simply label the test statistic as being 'chi-square'?

For example, I have compared the proportions 82.59% and 76.62%. This gives a value of the McNemar statistic of 172.567 with p < 0.001 and N = 7920. I have calculated a 95% confidence interval for the difference of (5.09%, 6.85%).

I would like to report this (and similar) information in the most accurate and way possible. Suggestions please?

Many thanks,


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