Comparison of 2 proportions (not continuous outcomes)

We can use a procedure similar to the comparison of 2 independent means (2-sample t test) to compare the proportions of 2 independent samples.

Example

We want to compare the risk (proportion) of caries between girls and boys in a sample of 74 adolescent patients. We count how many participants in each group have caries. Please note that for this example, we do not count multiple caries per patient; rather, with XF
X F
and XM
X M
we could denote the number of caries in girls and boys and, with nF
n F
and nM,
n M ,
the numbers of adolescent girls and boys, respectively, in the sample. Then we can estimate the proportion of caries (or risk of caries) in the population of adolescents as follows:

pF=XF/nFandpM=XM/nM
p F = X F / n F and p M = X M / n M

The Table summarizes the data by sex and indicates that 14% of female patients and 19% of male patients have caries.

Table
Proportion of caries for male and female patients
Sex n Proportion of caries
Female 52 0.14
Male 22 0.19

We can calculate the 95% confidence interval (CI) of the difference of the 2 proportions; if the CI includes the value of zero, then we infer that there is no difference in caries risk between male and female patients.

We need to do the following steps to accomplish our comparison; these are similar to the comparison of 2 independent means.

  • 1.

    Calculate the difference between the 2 proportions: difference = p F − p M = 0.14 − 0.19 = −0.05, where

  • pF=proportion of caries in females,andpM=proportion of caries in males
    p F = proportion of caries in females , and p M = proportion of caries in males

We observed a reduction in the risk of caries by 5% in the girls compared with the boys. This difference is usually called the risk difference, and it is a measure of treatment effect as described earlier.

  • 2.

    Calculate the standard deviations and standard errors from the sample data set for female and male patients and then the standard deviation of the risk difference.

SDF=StandaradDeviationFemale=pF(1pF)=0.14(10.14)=0.12=0.35
SD F = Standarad Deviation Female = p F ∗ ( 1 − p F ) = 0.14 ∗ ( 1 − 0.14 ) = 0.12 = 0.35
SEF=StandaradErrorFemale=pF(1pF)nF=0.14∗(1−0.14)52=0.002=0.048
SE F = Standarad Error Female = p F ∗ ( 1 − p F ) n F = 0 .14∗ ( 1−0 .14 ) 52 = 0.002 = 0.048
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Apr 4, 2017 | Posted by in Orthodontics | Comments Off on Comparison of 2 proportions (not continuous outcomes)
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