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# More tests for proportions

By Clarence Willis,2014-10-11 20:51
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More tests for proportions

Fisher’s Exact Test

Does not rely on Normality assumption.

Uses “exact” distribution instead of a Normal

approximation.

2 Use in place of χ test when any expected cell

frequency is less than 5.

Example: caries incidence

Caries

yes no total

control 10 26 36

intervention 6 62 68

total 16 88 104

The null-hypothesis of Fisher’s Exact test is that there is no relationship between the two categories. Thus, every possible arrangement of observations in the respective cells is equally likely (we assume the row and column totals are fixed).

The p-value is computed by calculating the number of possible arrangements of observations that produce tables that are more extreme than the observed and then dividing this by the total number of possible arrangements of the observations.

Example: Caries incidence

observed table Caries

p;- p;= ci yes no total 0.190

control 10 26 36

intervention 6 62 68 probability under H 0

total 16 88 104 p= 0.01048 H0

Tables more extreme (result in greater difference in proportions)

p;- p;= p;- p;= cici11 25 36 15 21 36 0.23 0.40

5 63 68 1 67 68

16 88 104 p= = 16 88 104 p0.00236 0.00000 H0H0

p;- p;= p;- p;= cici12 24 36 16 20 36 0.27 0.44

4 64 68 0 68 68

16 88 104 p= 16 88 104 p= 0.00038 0.00000 H0H0

p;- p;= p;- p;= cici13 23 36 0 36 36 0.32 -0.24

3 65 68 16 52 68

16 88 104 p= = 16 88 104 p0.00004 0.00055 H0H0

p;- p;= p;- p;= cici14 22 36 1 35 36 0.36 -0.19

2 66 68 15 53 68

16 88 104 p= = 16 88 104 p0.00000 0.00602 H0H0

Total probability of all as or more extreme tables = 0.01985

Formula for Probability of Table in Fisher’s Exact Test

Table Probability

(a;c)!(c;d)!(a;c)!(b;d)! a b

c d n!a!b!c!d!

SPSS output

treatment group * caries at age two CrosstabulationCount

caries at age two

noyesTotaltreatmentcontrols261036

groupintervention62668

Total8816104

Chi-Square Tests

Asymp. Sig.Exact Sig.Exact Sig.

Valuedf(2-sided)(2-sided)(1-sided)

bPearson Chi-Square6.4961.011

aContinuity Correction5.1221.024

Likelihood Ratio6.1711.013

Fisher's Exact Test.020.013Linear-by-Linear6.4341.011Association

cMcNemar Test.000N of Valid Cases104

a. Computed only for a 2x2 table

b. 0 cells (.0%) have expected count less than 5. The minimum expected count is

5.54.

c. Binomial distribution used.

McNemar’s Test for Proportions (Paired Data)

Use for comparing proportions from paired data

Example: Change in plaque index

Fifty-three study participants assessed twice for plaque index (PI), at baseline and 4 weeks later. We wish to assess whether the proportion of patients with high PI changes.

PI at 4 weeks

PI at baseline low high

low 29 1

high 13 10

Incorrect methods:

1123ˆˆpp121. Comparing with using the Z-test: 5353

This will not give a valid p-value because it does not

compare independent samples. The same 53 people

are used in each proportion.

2. Performing a Chi-square or Fisher’s Exact test on the

above 2×2 table: These would test whether the

proportion of high’s at baseline is related to the

proportion of high’s at 4 weeks. They would not test

whether or not the proportions are different.

PI at 4 weeks

PI at baseline low high

low 29 1

high 13 10

McNemar’s Test assesses the null hypothesis

H: P(PI high at baseline) = P(PI high at 4 week), 0

by noting that it is equivalent to:

P(PI changes high to low) = P(PI changes low to high), H: 0

for all discordant pairs.

The discordant pairs are those that have different values for the two observations. Note that each entry

in the table is the number of pairs.

The latter H can be evaluated using a one-sample 0

test for proportions with,

H: p = 0.50, vs. H: p ? 0.50, 01

where p = proportion of discordant pairs that increase.

PI at 4 weeks

PI at baseline low high

low 29 1

high 13 10

If n > 20 (where n is # of discordant pairs) can use

Z-test for proportions (chapter 9.3).

If n < 20 (as in the current example, n = 14) use the

binomial distribution to compute the exact p-value.

Let X = number of discordant pairs that increase, which, under H, is binomial(n = 14, p = 0.5). 0

The two-sided p-value is the probability that we would see a more unbalanced sample of the discordant pairs than 13 vs 1, which is

P(X < 1) + P(X > 13)

= P(X=0) + P(X=1) + P(X=13) + P(X=14)

= 0.0001 + 0.0009 + 0.0009 + 0.0001

= 0.0020

SPSS output

baseline PI * four week PI CrosstabulationCount

four week PI

lowhighTotal

baselinelow29130

PIhigh131023

Total421153

Chi-Square Tests

Asymp. Sig.Exact Sig.Exact Sig.

Valuedf(2-sided)(2-sided)(1-sided)

bPearson Chi-Square12.7571.000

aContinuity Correction10.4331.001

Likelihood Ratio13.8721.000

Fisher's Exact Test.000.000Linear-by-Linear12.5161.000Association

cMcNemar Test.002N of Valid Cases53

a. Computed only for a 2x2 table

b. 1 cells (25.0%) have expected count less than 5. The minimum expected count is

4.77.

c. Binomial distribution used.

Analysis of Categorical Data Summary

; Proportions from two independent samples

Large samples Z-test for proportions

Small samples Fisher’s Exact Test

; Proportions from > 2 independent samples

Chi-square test

; Proportions from paired data

McNemar’s Test

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