Topic 9.
Inference on Proportions (Ch. 14)
1) Introduction.
• In Chapter 7, we studied the binomial distribution. Suppose one has an experiment
which
1) is repeated n times, with each trial having one of two possible outcomes,
success (S) or failure (F),
2) has independent trials, and
3) has constant probability of success on each trial,
i.e.
P(S) = p.
Then the number of successes, x, has the binomial distribution, specifically
P ( x ) = ( nx ) p x q n − x
(1)
where q = 1-p.
• Often, we would like to make statistical inferences concerning p from sample data, i.e.
from x and n.
2) Normal Approximation to the Binomial.
• The binomial calculations can be tedious. In Chapter 7, we learned to approximate
binomial probabilities for rare events (with large n and small p) using the Poisson
distribution. We can also use a Normal approximation.
• Consider the previous problem on the number of smokers in a random sample of n = 30
people. If p = 0.29, one could find the probability of 6 or fewer smokers by calculating
P(0) through P(6), and adding. One would find
P ( X ≤ 6) = 0.190.
• Let’s consider using a normal approximation for this. If
1) np ≥ 5 and 2) n(1 − p ) ≥ 5,
then the binomial distribution is fairly symmetric. (Do we need this condition for the
Poisson approximation?) Recall that the mean and variance of the binomial were
μ = np and
σ 2 = npq.
The logic is to approximate some binomial probability, e.g.
P ( X ≤ x)
by a normal distribution with the same mean and variance as with the binomial. We
indicated previously the possibility of this in Chapter 7, e.g. with Figure 7.7.
• For example, on the problem relating to the number of smokers in a sample of n = 30
people, one has
μ = np = 8.7
σ 2 = npq = 6.177
σ = 2.419
hence using the normal approximation (because both np > 5 and n(1-p) > 5)
6 − 8.7 ⎞
⎛
P ( X ≤ 6) = P ⎜ z ≤
⎟
2.49 ⎠
⎝
= P ( z < −1.09)
= 0.138.
• The above approximation isn’t very accurate, because the true value is 0.190. The
principal reason is that we are approximating a discrete distribution using a continuous
one. Particularly in cases with a relatively small n, one should use the continuity
correction. Note that the binomial probability
P ( X ≤ 6)
corresponds better using the normal distribution to the probability
P ( X ≤ 6.5).
Why? In this case, note the normal probability approximation is
6.5 − 8.7 ⎞
⎛
P ( X ≤ 6.5) = P ⎜ z ≤
⎟
2.49 ⎠
⎝
= P ( z < −0.89)
= 0.187
which is a close approximation.
3) Sampling Distribution of a Proportion.
• Often, the parameter p is unknown in a binomial experiment, and one problem of
interest is statistical inference concerning p. Suppose one has n trials with x successes.
One can show that the best estimate of p is the observed proportion, pˆ , i.e.
pˆ = x / n .
For example, if one sampled n = 50 M&M’s and found x = 6 greens, then our best
estimate of p, the proportion of greens in the population is pˆ = 0.12.
• Clearly, p̂ is a statistic, and hence has a sampling distribution (as did x before). This
sampling distribution has the following properties:
1) The mean (center) is
μ p̂ = p
2) The standard error (spread) is
σ pˆ = p (1 − p / n
3) The shape of the distribution is approximately normal for large n.
In summary,
(
pˆ ~ N p, p (1 − p ) / n
)
(2)
Property 3 is not surprising in light of our previous normal approximation claim.
Moreover, suppose one codes the data so that S = 1 and F = 0. Then averaging 0’s and 1’s
one can show
2
p̂ = x
and hence use the central limit theorem to show that p̂ has a normal distribution for large
n.
• Consider the following example. Suppose the 5-year survival rate of lung cancer
patients is thought to be p = .1. Suppose one has n = 50 patients in a year, and that 10
survive, for pˆ = .20. What is the probability of this occurring given p = .1?
Note that
⎛
⎞
.2 − .1
P ( pˆ ≥ .2) = P ⎜ z ≥
⎟
⎜
(.1)(.9) / 50 ⎟⎠
⎝
= P ( z ≥ 2.36)
= 0.009
Hence, if p = .1, there is less than a 1% chance that pˆ = .2. What might you conclude
based on your observed pˆ = .2 ? This is the essence of a hypothesis test.
4) Confidence Intervals for p.
• Recall, from (2)
(
)
pˆ ~ N p, p (1 − p ) / n .
It follows that
z=
pˆ − p
p(1 − p) / n
(3)
is a standard normal random variable.
Recall,
P (− zα / 2 < z < + zα / 2 ) = 1 − α
Substituting (3) into (4), and using a little algebra on can show
(
(4)
)
P pˆ − zα / 2 p (1 − p ) / n < p < pˆ + zα / 2 p (1 − p ) / n = 1 − α
Unfortunately, we don’t know p for the upper and lower limits, so we replace it by pˆ . It
has been shown that
(
)
P pˆ − zα / 2 pˆ (1 − pˆ ) / n < p < pˆ + zα / 2 pˆ (1 − pˆ ) / n = 1 − α .
This is a 100(1 − α )% confidence interval for p. One could also find a one-sided
confidence interval as before.
• For example, a national survey of n = 1000 males find that x = 80 feel confident that
they would win a dispute with an IRS agent. Hence
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pˆ = 80 /1000 = .08
which is the best estimate of p. A 95% confidence interval for p is
pˆ ± zα / 2 pˆ (1 − pˆ ) / n
= .08 ± 1.96 .08(.92) /1000
= .08 ± .017
= (.063, .097).
5) Testing Hypotheses about p.
• Hypothesis tests follow immediately because p̂ has a normal distribution. Recall from
(3) that
pˆ − p
z=
.
p(1 − p) / n
Therefore one could test a hypothesis such as
H 0 : p = p0
using
pˆ − p0
z=
.
p0 (1 − p0 ) / n
• For example, suppose the 5-year survival rate for lung cancer patients over 40 is p =
0.082. We wish to test whether this same rate holds for men under 40. A sample of n = 52
such men yields pˆ = 0.115. Note our hypothesis testing framework is
1) H 0 : p = 0.082
2) H A : p ≠ 0.082
pˆ − 0.082
3) TS :
z=
0.082(0.918) / 52
=
pˆ − 0.082
.038
4) RR: reject if
|z| > 1.96, for α = .05
5) with pˆ = 0.115, one has
0.115 − 0.082
z=
= 0.87.
0.038
Hence, one would not reject H 0 , indeed the p-value is p = 0.384.
6) Sample Size Estimation.
• Consider a one-sided test, such as
H 0 : p ≤ p0
vs.
H A : p > p0 .
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Suppose we fix α, and want to set β for some specific value p, under H A . In other words,
we want to reject H 0 with power 1 - β when p = p1. Using some algebra, the formula for
n is
2
⎡ z p (1 − p0 ) + z β p1 (1 − p1 ) ⎤
n=⎢ α 0
⎥ .
p1 − p0
⎣⎢
⎦⎥
The corresponding formula for a two-sided test is
⎡z
p (1 − p0 ) + z β
n = ⎢ α /2 0
p1 − p0
⎣⎢
2
p1 (1 − p1 ) ⎤
⎥ .
⎦⎥
• For example, let’s reconsider our previous test in slightly revised form. Suppose we
wish to test
H 0 : p ≤ .082
at α = .05. Our goal is to reject H 0 with power 0.95 if p = .2. Then
⎡1.645 0.082(0.918) + 1.645 .2(.8) ⎤
n=⎢
⎥
.2 − 0.082
⎥⎦
⎣⎢
= (9.4) 2 = 88.36
Hence n = 90 subjects would satisfy the criteria.
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7) Comparing Two Populations.
• This is a very common and hence very important problem. Suppose one has two
populations and takes a sample from each. The data are
Pop
1
2
Sample Size
n1
n2
Successes
x1
x2
Proportion
p̂1
p̂2
We desire to test
H 0 : p1 = p2 .
(5)
Note that if p1 = p2 as specified by H 0 , the best common estimate is
x +x
pˆ = 1 2 ,
n1 + n2
a weighted average of pˆ1 and pˆ 2 . If n1 and n2 are large enough, i.e. n1 pˆ , n2 pˆ ,
n1 (1 − pˆ ) and n2 (1 − pˆ ) > 5, then
pˆ1 − pˆ 2 ~ N ,
and the test statistic for (5) is
( pˆ1 − pˆ 2 ) − ( p1 − p2 )
(6)
z=
.
pˆ (1 − pˆ )(1/ n1 + 1/ n2 )
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• For example, suppose one investigates seat belt effectiveness from motor vehicle
accident data. A sample of n1 = 123 children who were wearing seat belts had x1 = 3
fatalities, whereas a sample of n2 = 290 children who were not wearing a belt had
x2 = 13 fatalities. Clearly, pˆ1 = 0.024 and pˆ 2 = 0.045, for a near doubling of the fatality
rate. Is this sufficient evidence to conclude that seat belts reduce the fatality rate? Note
1) H 0 : p1 = p2
H A : p1 ≠ p2
TS: as given in (6)
RR: with α = .05, reject if |z| > 1.96
Data show
3 + 13
pˆ =
= 0.039.
123 + 290
0.024 − 0.045
z=
Hence
0.039(0.961)(1/123 + 1/ 290)
2)
3)
4)
5)
= −1.01.
Therefore, we do not have sufficient evidence to reject H 0 , indeed the p-value is p =
0.312.
• One way to illustrate these results graphically would be to give separate confidence
intervals for p1 and p2 . They would be overlapping, thus illustrating visually why one
does not have sufficient evidence to conclude p1 ≠ p2 .
• A confidence interval on the difference ( p1 − p2 ) is
pˆ1 (1− pˆ1 )
n1
( pˆ1 − pˆ 2 ) ± zα / 2
+
pˆ 2 (1− pˆ 2 )
n2
.
For example, using the previous data on seat belt effectiveness, a 95% confidence
interval for ( p1 − p2 ) is
(0.024 − 0.045) ± 1.96
0.024(0.976)
123
+ 0.045(0.955)
290
= −0.021 ± 0.036
= (−0.057, 0.015).
In other words, we are 95% confident that the use of seat belts yield anywhere from a
5.7% reduction to a 1.5% increase in the fatality rate. Clearly, more data are needed to
conclude statistically whether or not seat belts increase or decrease fatalities among
children.
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