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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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