Tests about binomial proportions
Who is going to win an election? Do observed hiring rates of minorities reflect their
representation in the population? These questions relate to claims about a binomial proportion p. For example, in an election between two candidates, if p is the probability of
candidate A winning, p being greater than or less than .5 corresponds to candidate A
winning or losing, respectively.
Generally, we are interested in the hypotheses
p = p0
versus
p 6= p0 ,
where p is the probability of a success. As we’ve already discussed, if p is the observed
sample proportion of successes, E(p) = p and V (p) = p(1 − p)/n; the Central Limit
Theorem then gives the z–statistic,
z= p
p − p0
.
p0 (1 − p0 )/n
Note that this differs from the confidence interval for a binomial proportion in
that p0 is used in the denominator instead of p since p0 is specified by the null.
Example. According to the Centers for Disease Control and Prevention (CDC), 17% of
school-age children in the United States are obese, while 33.8% of adults in the U.S. are
obese (having a body mass index (BMI) of at least 30). In 2005 the Marion County
(Indiana) Health Department conducted a program wherein a sample of 90147 school-age
children had their heights and weights measured, allowing exact determination of their
BMIs. In this survey 22% of the children measured were obese. Does this indicate that
the true obesity rate for children in Marion County is different from the national average?
The hypotheses being tested are
p = .17
versus
p 6= .17,
where p is the probability of a school-age child being obese. The test statistic is
z=p
c 2011, Jeffrey S. Simonoff
.22 − .17
(.17)(.83)/90147
= 40.0,
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which obviously has a tail probability that is 0 to many digits. That is, there is overwhelming evidence that the obesity rate for school-age children is higher than the national
average.
At the same time the Marion County Health Department also conducted a telephone
survey of 4784 adults, and in that survey 25% reported being obese. Does this indicate
that the true adult obesity rate in Marion County is different from the national average?
The hypotheses being tested here are
p = .338
versus
p 6= .338,
where p is the probability of an adult being obese. The test statistic is
z=p
.25 − .338
(.338)(.662)/4784
= −12.87,
which obviously has a tail probability that is 0 to many digits. That is, there is extremely
strong evidence that the adult obesity rate is lower than the national average.
At first glance, the difference in directions of these results seems very surprising. Why
might they not be as surprising as they first appear to be?
Minitab commands
To obtain a hypothesis test for a Binomial proportion, click on Stat → Basic Statistics → 1 Proportion. Enter in the number of trials (n) and the number of successes (X)
under Summarized data. The test that comes out is exact (based on the Binomial distribution). To get the standard Central Limit Theorem–based test, click on Options, and
then mark the box next to Use test and interval based on normal distribution.
c 2011, Jeffrey S. Simonoff
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