Statistical Analysis for Education and Psychology Researchers

The null hypothesis is that there is no difference in the probability of upper limb injury
between the two age categories: H 0 : π (population proportion for >6-years-old)=0.5 and,
H 1 : P≠Q. A two-tailed test is required.
For the binomial test when π=0.5, (sample proportion P=Q=0.5) and n >25, the
following formula can be used to evaluate the binomial approximation to the normal
distribution (with continuity correction)

Binomial

approximat

ion to

normal

distribution

with

continuity

correction

—6.4

where X is the smaller of the two frequency counts (one for each category), n is the total
sample size, and P is 0.5.
The correction for continuity is +/−0.5 depending upon the expected value for X which
is evaluated as nP or simply half of the sample size. If X is <nP we add 0.5 to X and if X
is >nP we subtract 0.5 from X. In this example, X is <nP (7<33) and we therefore add 0.5
to X when evaluating formulae 6.4:

Interpretation

This Z-value of −6.2777 is so extreme that it is not even tabulated in the table of Z-
values, (see Table 1, Appendix A4). We can say from this table that the one-tailed
probability associated with this Z-value is p<0.0000 (probability associated with the most
extreme tabulated Z-value of 4.0). Since our alternative hypothesis was simply that the
two frequencies would differ, a two-sided test is appropriate. For a two-sided test the
probability is doubled, here the value of P remains as p< 0.0000. The null hypothesis of
no difference in the probability of upper limb injury between the two age categories is
therefore rejected. Proportionately, significantly more injuries were incurred among left-
handed males when they were >6-years-old than when they were ≤6-years-old.

Computer Analysis

The exact probability can be evaluated using the PROBBNML function. The appropriate
SAS code is,

data a;

Inferences involving binomial and nominal count data 177