## Summary and Analysis of Extension Program Evaluation in R

Salvatore S. Mangiafico

# Tests for Paired Nominal Data

Tests of symmetric margins—or marginal homogeneity—for nominal data are used when the counts on a contingency table represent values that are paired or repeated in time.

As an example, consider a question repeated on a pre-test and a post-test.  We may want to know if the number of correct responses changed from the pre-test to the post-test.

Did students have the correct answer to the question?

After
Before      Correct   Incorrect
Correct      2        0
Incorrect   21        7

Note that the row names and column names have the same levels, and that counts represent paired responses.  That is, for each observation you must know the individual’s response before and after.

Also note that the number of students included in the table are 30, or the sum of the cell counts.

In essence, those students with the same response before and after don’t affect the assessment of the change in responses.  We would focus on the “discordant” counts.  That is, how many students had incorrect answers before and correct answers after, in contrast to those who had the reverse trend.  Here, because 21 changed from incorrect to correct, and 0 changed from correct to incorrect, we might suspect that there was a significant change in responses from incorrect to correct.

To grasp the difference between nominal tests of association and nominal tests of symmetry, be sure to visit the coffee and tea example below in the section “An example without repeated measures, comparing test of symmetry with test of association”.

##### Appropriate data

•  Two nominal variables with two or more levels each, and each with the same levels.

•  Observations are paired or matched between the two variables.

•  McNemar and McNemar–Bowker tests may not be appropriate if discordant cells have low counts.

##### Hypotheses

•  Null hypothesis:  The contingency table is symmetric.  That is, the probability of cell [i, j] is equal to the probability of cell [j, i].

•  Alternative hypothesis (two-sided): The contingency table is not symmetric.

##### Interpretation

Depending on the context, significant results can be reported as e.g. “There was an increase in correct answers from the pre-test to the post-test.” Or, “There was a significant change from answer A to answer B.”

##### Post-hoc analysis

Post-hoc analysis for tests on a contingency table larger than 2 x 2 can be conducted by conducting tests for the component 2 x 2 tables.  A correction for multiple tests could be applied.

##### Other notes and alternative tests

•  For unpaired data, see the tests in the chapter Association Tests for Nominal Data.

•  For multiple times or groups, Cochran’s Q test can be used if the response is binary.

### Packages used in this chapter

The packages used in this chapter include:

•  rcompanion

The following commands will install these packages if they are not already installed:

if(!require(rcompanion)){install.packages("rcompanion")}

#### McNemar and McNemar–Bowker tests

For a 2 x 2 table, the most common test for symmetry is McNemar’s test.  For larger tables, McNemar’s test is generalized as the McNemar–Bowker symmetry test.  One drawback to the latter test is that it may fail if there are 0’s in certain locations in the matrix.

McNemar’s test may not be reliable if there are low counts in the “discordant” cells.  Authors recommend that these cells to sum to at least 5 or 10 or 25.

#### Exact tests

Exact tests of symmetry reduce to exact tests for goodness-of-fit.  A 2 x 2 table is analyzed with a binomial exact test.  Examples of this are shown in this chapter in the “Optional analyses: conducting exact tests for symmetry” section.

The nominalSymmetryTest function can conduct these exact tests easily.

### Example of tests for paired data nominal data

Alucard teaches a Master Gardener training on rain gardens for stormwater management and one on rain barrels.  He wishes to assess if people are more willing to install these green infrastructure practices after attending the training.  His data follow.  Note that there are 46 attendees answering each question, and the counts in each table sum to 46.

Are you planning to install a rain barrel?

After
Before   Yes   No
Yes        9    5
No        17   15

Are you planning to install a rain garden?

After
Before   Yes   No   Maybe
Yes        6    0   1
No         5    3   7
Maybe     11    1   12

##### Rain barrel

Before       After.yes   After.no
Before.yes     9          5
Before.no     17         15
"))

Matrix.1

After.yes After.no
Before.yes         9        5
Before.no         17       15

sum(Matrix.1)

[1] 46

##### Rain garden

Before         Yes.after   No.after   Maybe.after
Yes.before      6          0           1
No.before       5          3           7
Maybe.before   11          1          12
"))

Matrix.2

Yes.after No.after Maybe.after
Yes.before           6        0           1
No.before            5        3           7
Maybe.before        11        1          12

sum(Matrix.2)

[1] 46

#### McNemar and McNemar –Bowker tests

##### Rain barrel

mcnemar.test(Matrix.1)

McNemar's Chi-squared test with continuity correction

McNemar's chi-squared = 5.5, df = 1, p-value = 0.01902

##### Rain garden

mcnemar.test(Matrix.2)

McNemar's Chi-squared test

McNemar's chi-squared = 17.833, df = 3, p-value = 0.0004761

##### Rain garden example with post-hoc tests

Here, the global test result is the same as the mcnemar.test result.  Looking at the post-hoc tests, we see that there was a significant change from “maybe” to “yes”.

library(rcompanion)

nominalSymmetryTest (Matrix.2)

\$Global.test.for.symmetry
Dimensions  p.value
1      3 x 3 0.000476

\$Pairwise.symmetry.tests
1       Yes.before/Yes.after : No.before/No.after  0.0736   0.0771
2 Yes.before/Yes.after : Maybe.before/Maybe.after 0.00937   0.0281
3   No.before/No.after : Maybe.before/Maybe.after  0.0771   0.0771

Method
1    fdr

\$statistical.method
Method
1 McNemar test

Maybe to Yes, adjusted p value = 0.0281, with 11 changing from Maybe to Yes, and 1 changing from Yes to Maybe.

Before         Yes.after   No.after   Maybe.after
Yes.before
6          0           1
No.before       5          3           7
Maybe.before
11          1          12

#### Exact tests for symmetry

##### Rain barrel

library(rcompanion)

nominalSymmetryTest(Matrix.1,
exact=TRUE)

\$Global.test.for.symmetry
Dimensions p.value
1      2 x 2  0.0169

\$Statistical.method
Method
1 binomial test

##### Rain garden

Note that the nominalSymmetryTest function won’t return a global exact test result for tables larger than 2 x 2, but will return exact tests for the post-hoc tests.

library(rcompanion)

nominalSymmetryTest (Matrix.2,
exact=TRUE)

\$Global.test.for.symmetry
Dimensions p.value
1      3 x 3      NA

\$Pairwise.symmetry.tests
1       Yes.before/Yes.after : No.before/No.after  0.0625   0.0703
2 Yes.before/Yes.after : Maybe.before/Maybe.after 0.00635   0.0190
3   No.before/No.after : Maybe.before/Maybe.after  0.0703   0.0703

Method
1    fdr

\$statistical.method
Method
1 binomial test

Maybe to Yes, adjusted p value = 0.0190, with 11 changing from Maybe to Yes, and 1 changing from Yes to Maybe.

Before         Yes.after   No.after   Maybe.after
Yes.before
6          0           1
No.before       5          3           7
Maybe.before
11          1          12

#### Effect size

Appropriate effect sizes for data subjected to McNemar, McNemar–Bowker, or equivalent exact tests include Cohen’s g and odds ratio.

Considering a 2 x 2 table, with a and d being the concordant cells and b and c being the discordant cells, the odds ratio is simply the greater of (b/c) or (c/b), and P is the greater of (b/(b+c)) or (c/ b+c)).  Cohen’s g is P – 0.5.  These statistics can be extended to larger tables.

Cohen (1988) lists interpretations for Cohen’s g.  These can easily be translated into interpretations for the odds ratio.  For example, the cutoff for the “small” interpretation is g ≥ 0.05, which would be 0.55 : 0.45 odds, or 0.55/0.45, or 1.22.

Because odds ratio is ratio of values, a large effect for odds ratio is far from 1.  It may be useful to use the reciprocal of the odds ratio for odds ratios less than 1.

 Small Medium Large Cohen’s g (absolute value) 0.05 – < 0.15 0.15 – < 0.25 ≥ 0.25 Odds ratio (OR) 1.22 – < 1.86 0.820 – < 0.538 1.86 – < 3.00 0.538 – < 0.333 ≥ 3.00 ≤ 0.333

##### Rain barrel

In the output, OR is the odds ratio, and g is Cohen’s g.

library(rcompanion)

cohenG(Matrix.1)

\$Global.statistics
Dimensions    OR     P      g
1      2 x 2 0.294 0.227 -0.273

1 / 0.294

[1] 3.401361

### So, the odds ratio can be interpreted as either 0.294 or 3.4.

Confidence intervals for odds ratio can be requested also.

cohenG(Matrix.1,
ci=TRUE)

\$Global.statistics
Dimensions Statistic  Value lower.ci upper.ci
1      2 x 2        OR  0.294   0.0588    0.733
2      2 x 2         P  0.227   0.0526    0.428
3      2 x 2         g -0.273  -0.4474   -0.072

##### Rain garden

library(rcompanion)

cohenG(Matrix.2)

\$Global.statistics
Dimensions   OR    P    g
1      3 x 3 11.5 0.92 0.42

\$Pairwise.statistics
Comparison  OR     P     g
1       Yes.before/Yes.after : No.before/No.after Inf     1   0.5
2 Yes.before/Yes.after : Maybe.before/Maybe.after  11 0.917 0.417
3   No.before/No.after : Maybe.before/Maybe.after   7 0.875 0.375

### An example without repeated measures, comparing test of symmetry with test of association

As another example, consider a survey of tea and coffee drinking, in which each respondent is asked both if they drink coffee, and if they drink tea.

Tea
Coffee   Yes   No
Yes      37    17
No        9    25

We could use a test of symmetry in this case if the question we wanted to answer was, Is coffee more popular than tea?  That is, is it more common for someone to drink coffee and not tea than to drink tea and not coffee?  (Those who drink both or drink neither are not relevant to this question.)

This is an example of using a test of symmetry to test the relative frequency of two dichotomous variables when the same subjects are surveyed.

Coffee   Yes   No
Yes      37    17
No        9    25
"))

mcnemar.test(Matrix.3)

McNemar's Chi-squared test with continuity correction

McNemar's chi-squared = 1.8846, df = 1, p-value = 0.1698

###  Neither coffee nor tea is more popular, specifically because
###    neither the 9 nor the 17 in the table are large relative to
###    the other.

A test of association answers a very different question.  Namely, Is coffee drinking associated with tea drinking?  That is, is someone more likely to drink tea if they drink coffee?

chisq.test(Matrix.3)

Pearson's Chi-squared test with Yates' continuity correction

X-squared = 13.148, df = 1, p-value = 0.0002878

###  Coffee drinking and tea drinking are associated, in this case people who
###    drink coffee are likely to drink tea. This is a positive association.
###    A negative association could also be significant.

### Optional analysis: a 4 x 4 example with several 0’s

As an additional example, imagine a religious caucusing event in which advocates try to sway attendees to switch their religions.

Matrix row names are the attendees’ original religions, and the column names, with a “2” added, are attendees’ new religions after the caucus.  Note that there are several 0 counts in the matrix.

The mcnemar.test function fails in this case because of the position of some 0 counts in the matrix.

Before        Pastafarian.after  Discordiant.after  Dudist.after  Jedi.after
Pastafarian    7                  0                  23             0
Discordiant    0                  7                   0            33
Dudist         3                  0                   7             1
Jedi           0                  1                   0             7
"))

Matrix.4

Pastafarian.after Discordiant.after Dudist.after Jedi.after
Pastafarian                 7                 0           23          0
Discordiant                 0                 7            0         33
Dudist                      3                 0            7          1
Jedi                        0                 1            0          7

##### McNemar –Bowker test

mcnemar.test(Matrix.4)

McNemar's Chi-squared test

McNemar's chi-squared = NaN, df = 6, p-value = NA

##### Post-hoc tests

library(rcompanion)

nominalSymmetryTest(Matrix.4)

\$Global.test.for.symmetry
Dimensions p.value
1      4 x 4     NaN

\$Pairwise.symmetry.tests
1 Pastafarian/Pastafarian.after : Discordiant/Discordiant.after     <NA>       NA
2           Pastafarian/Pastafarian.after : Dudist/Dudist.after 0.000194 2.91e-04
3               Pastafarian/Pastafarian.after : Jedi/Jedi.after     <NA>       NA
4           Discordiant/Discordiant.after : Dudist/Dudist.after     <NA>       NA
5               Discordiant/Discordiant.after : Jedi/Jedi.after 1.06e-07 3.18e-07
6                         Dudist/Dudist.after : Jedi/Jedi.after        1 1.00e+00

Method
1    fdr

\$statistical.method
Method
1 McNemar test

A look at the significant results

Pastafarian to Dudist, adjusted p value = 0.000291

Before        Pastafarian.after   Discordiant.after   Dudist.after   Jedi2.after
Pastafarian
7                   0                   23              0
Discordiant   0                   7                    0             33
Dudist
3                   0                    7              1
Jedi          0                   1                    0              7

Discordiant to Jedi, adjusted p value < 0.0001

Before        Pastafarian.after   Discordiant.after   Dudist.after   Jedi.after
Pastafarian   7                   0                   23              0
Discordiant   0
7                    0             33
Dudist        3                   0                    7              1
Jedi          0
1                    0              7

### Optional analyses: conducting exact tests for symmetry

The exact symmetry tests for a 2 x 2 table can be conducted directly with the binom.test function.  These results match the results of the nominalSymmetryTest function with the exact=TRUE option.

Rain barrel

After
Before   Yes   No
Yes        9    5
No        17   15

##### Rain barrel

For a 2 x 2 matrix, x = the count in one of the discordant cells, and n = the sum of the counts in discordant cells.  The expected proportion is 0.50.

You could also follow the method for an n x n matrix below.

x =  5
n =  5 + 17
expected = 0.50

binom.test(x, n, expected)

Exact binomial test

number of successes = 5, number of trials = 22, p-value = 0.0169

### Optional:  Comparing odds ratio and Cohen’s g

For a 2 x 2 table of matched counts, odds ratio and Cohen’s g are precisely related, and their interpretations according to Cohen will always coincide.  An odds ratio value of 1 corresponds to a Cohen's g of 0.  Cohen’s g approaches 0.5 as odds ratio approaches infinity.  In the second figure below, the colors indicate Cohen’s interpretation of less-than-small, small, medium, and large as the blue becomes darker.

### References

Cohen, J. 1988. Statistical Power Analysis for the Behavioral Sciences, 2nd Edition. Routledge.

### Exercises N

1. Considering Alucard’s data,

a.  How many students responded to the rain barrel question?

b.  How many students changed their answer on the rain barrel question from no to yes?

c.  What do you conclude about the results on the rain barrel question?  Include:

•  The statistical test you are using, p-value from that test, conclusion of the test, and any other comments on the test.

•  The effect size and interpretation.

•  Description of the observed counts or proportions.  Is there anything notable?

•  Any other practical conclusions you wish to note.

d.  What do you conclude about the results of the global test of the rain garden question?

e.  What do you conclude about the results of the post-hoc analysis of the rain garden question?

2. Considering the coffee and tea example, Do you understand the difference between the hypotheses for tests of association and tests of symmetry?  Would you be comfortable choosing the correct approach?

3. Ryuk and Rem held a workshop on planting habitat for pollinators like bees and butterflies.  They wish to know if attendees were more likely to do a planting after the workshop than before.

Will plant?

After
Before   Yes.after   No.after   Maybe.after
Yes       17         0           0
No         5         9          13
Maybe     15         0           7

For each of the following, answer the question, and show the output from the analyses you used to answer the question.

a.  How many students responded to this question?

b.  How many students changed their answer from no to yes?

c.  What do you conclude about the results of the global test?

d.  What do you conclude about the results of the post-hoc analysis?