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McNemar test

Charlotte-9
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,

Lou
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Re: McNemar test

Burleson,Joseph A.
Lou:

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:[hidden email]] On Behalf Of
Lou
Sent: Monday, April 02, 2007 11:18 AM
To: [hidden email]
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,

Lou
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Re: McNemar test

swells-2
I have used the McNemar to examine case characteristics and decisions in
matched pairs. For example, given children of two different races who are
matched on gender, county, reason for report to agency and age; are there
other significant differences between the members of the pair, for example,
biological father in the household? After determining the comparability of
the members of each pair, I used the McNemar again to look at whether there
was a difference between the case disposition (accepted or not accepted for
service) for the white child of the pair v. the African American child of
the pair. For reporting it out, I do one table to illustrate how it works
and then use summaries to report ensuing McNemar analyses. I report it as a
McNemar chi square for paired samples.

Susan


On Apr 2 2007, Burleson,Joseph A. wrote:

>Lou:
>
>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:[hidden email]] On Behalf Of
>Lou
>Sent: Monday, April 02, 2007 11:18 AM
>To: [hidden email]
>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,
>
>Lou
>
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Re: McNemar test

Swank, Paul R
In reply to this post by Burleson,Joseph A.
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:[hidden email]] On Behalf Of
Burleson,Joseph A.
Sent: Monday, April 02, 2007 12:27 PM
To: [hidden email]
Subject: Re: McNemar test

Lou:

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:[hidden email]] On Behalf Of
Lou
Sent: Monday, April 02, 2007 11:18 AM
To: [hidden email]
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,

Lou
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Re: McNemar test

Burleson,Joseph A.
In reply to this post by Charlotte-9
Paul:

I think the simple question is why does SPSS not print the value of the
McNemar chi-square (3.333) under the value column? Siegel (p. 64)
identifies the test as a chi-square test with df = 1, and I concur with
your results, with the SPSS 2-tailed McNemar value equal to .099.

Joe Burleson

-----Original Message-----
From: SPSSX(r) Discussion [mailto:[hidden email]] On Behalf Of
Swank, Paul R
Sent: Tuesday, April 03, 2007 4:06 PM
To: [hidden email]
Subject: Re: McNemar test

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:[hidden email]] On Behalf Of
Burleson,Joseph A.
Sent: Monday, April 02, 2007 12:27 PM
To: [hidden email]
Subject: Re: McNemar test

Lou:

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:[hidden email]] On Behalf Of
Lou
Sent: Monday, April 02, 2007 11:18 AM
To: [hidden email]
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,

Lou
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Re: McNemar test

Swank, Paul R
SPSS does the exact test for McNemar which is based upon the binomial
distribution. I imagine, if the sample were large enough, it might
revert to the large sample approximation, ie, the chi-square.

Paul R. Swank, Ph.D.
Professor, Developmental Pediatrics
Director of Research,


University of Texas Health Science Center at Houston

-----Original Message-----
From: SPSSX(r) Discussion [mailto:[hidden email]] On Behalf Of
Burleson,Joseph A.
Sent: Tuesday, April 03, 2007 3:51 PM
To: [hidden email]
Subject: Re: McNemar test

Paul:

I think the simple question is why does SPSS not print the value of the
McNemar chi-square (3.333) under the value column? Siegel (p. 64)
identifies the test as a chi-square test with df = 1, and I concur with
your results, with the SPSS 2-tailed McNemar value equal to .099.

Joe Burleson

-----Original Message-----
From: SPSSX(r) Discussion [mailto:[hidden email]] On Behalf Of
Swank, Paul R
Sent: Tuesday, April 03, 2007 4:06 PM
To: [hidden email]
Subject: Re: McNemar test

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:[hidden email]] On Behalf Of
Burleson,Joseph A.
Sent: Monday, April 02, 2007 12:27 PM
To: [hidden email]
Subject: Re: McNemar test

Lou:

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:[hidden email]] On Behalf Of
Lou
Sent: Monday, April 02, 2007 11:18 AM
To: [hidden email]
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,

Lou
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Re: McNemar test

Margaret MacDougall
In reply to this post by Swank, Paul R
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

  Margaret


"Swank, Paul R" <[hidden email]> 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:[hidden email]] On Behalf Of
Burleson,Joseph A.
Sent: Monday, April 02, 2007 12:27 PM
To: [hidden email]
Subject: Re: McNemar test

Lou:

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:[hidden email]] On Behalf Of
Lou
Sent: Monday, April 02, 2007 11:18 AM
To: [hidden email]
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,

Lou



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Re: McNemar test

Swank, Paul R
McNemar's test is not a test of concordance at all. It is a test of
equality of the marginal distributions in a 2 by 2 table and, therefore,
a test of symmetry . In your example, the result indicates that the
level of agreement (or disagreement) is the same for the two statements.
It is described by Marascuilo and Mc Sweeney (1977) as a test of the
homogeneity of proportions for qualitative variables in correlated
samples. Because, in the 2 by 2 table, the off diagonals must be the
same if the marginal proportions are the same, it can also be used as a
test of symmetry. By homogeneity of proportions, I mean the proportion
of agreements (or disagreements) for one item being equal to the same
proportion for the other item. Because these proportions are on the same
sample, is this not a paired test?

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

________________________________

From: Margaret MacDougall [mailto:[hidden email]]
Sent: Friday, April 06, 2007 6:14 AM
To: Swank, Paul R; [hidden email]
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

Margaret


"Swank, Paul R" <[hidden email]> 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:[hidden email]] On
Behalf Of
        Burleson,Joseph A.
        Sent: Monday, April 02, 2007 12:27 PM
        To: [hidden email]
        Subject: Re: McNemar test

        Lou:

        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:[hidden email]] On
Behalf Of
        Lou
        Sent: Monday, April 02, 2007 11:18 AM
        To: [hidden email]
        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,

        Lou



________________________________

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Re: McNemar test

Burleson,Joseph A.
In reply to this post by Margaret MacDougall
Margaret:

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: [hidden email]
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

  Margaret


"Swank, Paul R" <[hidden email]> 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:[hidden email]] On Behalf Of
Burleson,Joseph A.
Sent: Monday, April 02, 2007 12:27 PM
To: [hidden email]
Subject: Re: McNemar test

Lou:

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:[hidden email]] On Behalf Of
Lou
Sent: Monday, April 02, 2007 11:18 AM
To: [hidden email]
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,

Lou



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