19. Inference about a
population proportion
The Practice of Statistics in the Life Sciences
Third Edition
© 2014 W.H. Freeman and Company
Objectives (PSLS Chapter 19)
Inference for a population proportion

Conditions for inference on proportions

The sample proportion p̂ (phat)

The sampling distribution of p̂

Significance test for a proportion

Confidence interval for p

Sample size for a desired margin of error
Conditions for inference on proportions
Assumptions:
1. The data used for the estimate are a random sample from the
population studied.
2. The population is at least 20 times as large as the sample. This
ensures independence of successive trials in the random sampling.
3. The sample size n is large enough that the shape of the sampling
distribution is approximately Normal. How large depends on the type
of inference conducted.
The sample proportion p̂
We now study categorical data and draw inference on the proportion, or
percentage, of the population with a specific characteristic.
If we call a given categorical characteristic in the population “success,”
then the sample proportion of successes,
pˆ 
p̂(phat) is:
count of successes in the sample
count of observatio ns in the sample
We treat a group of 120 Herpes patients with a new drug; 30 get better:
p̂ = (30)/(120) = 0.25 (proportion of patients improving in sample)
Sampling distribution of p̂
The sampling distribution of 𝑝 is never exactly Normal. But for large
enough sample sizes, it can be approximated by a Normal curve.
The mean and standard deviation (width)
of the sampling distribution are both
completely determined by p and n.
Significance test for p
When testing:
H0: p = p0 (a given value we are testing)
p0 (1 p0 )
n
If H0 is true, the sampling distribution is known 
The test statistic is the standardized value of 𝑝
z
pˆ  p0
p0 (1 p0 )
n
p0


pˆ
This is valid when both expected counts — expected successes np0 and

expected failures n(1 − p0) — are each 10 or larger.

P-value for a one or two sided alternative
The P-value is the probability, if H0 was true, of obtaining a test statistic
like the one computed or more extreme in the direction of Ha.
Aphids evade predators (ladybugs) by dropping off the leaf. An
experiment examined the mechanism of aphid drops.
“When dropped upside-down from delicate tweezers, live aphids
landed on their ventral side in 95% of the trials (19 out of 20). In contrast,
dead aphids landed on their ventral side in 52.2% of the trials (12 out of 23).”
Is there evidence (at significance level 5%) that live aphids land right side up
(on their ventral side) more often than chance would predict?
Here, “chance” would be 50% ventral landings. So we test:
H 0 : p  0.5 versus H a : p  0.5
z
pˆ  p0
0.95  0.5

 4.02
p0 (1  p0 ) n
(.5  .5) 20
The expected counts of success and failure are each 10, so the z procedure is
valid. The test P-value is P(z ≥ 4.02). From Table B, P = 1 – P(z < 4.02) < 0.0002,
highly significant. We reject H0. There is very strong evidence (P < 0.0002) that
the righting behavior of live aphids is better than chance.
Mendel’s first law of genetic inheritance states that crossing dominant and
recessive homozygote parents yields a second generation made of 75% of
dominant-trait individuals.
When Mendel crossed pure breeds of plants producing smooth peas and plants
producing wrinkled peas, the second generation (F2), was made of 5474 smooth
peas and 1850 wrinkled peas.
Do these data provide evidence that the proportion of smooth peas in the F2
population is not 75%?
The sample proportion of smooth peas is:
pˆ 
5474
 0.7474
5474  1850
We test: H 0 : p  0.75 versus H a : p  0.75
z
pˆ  p0
0.7474  0.75

 0.513
p0 (1  p0 ) n
(.75  .25) 7324
From Table B, we find P = 2P(z < –0.51) = 2 x 0.3050 = 0.61, not significant. We
fail to reject H0. The data are consistent with a dominant-recessive genetic
model.
Confidence interval for p
When p is unknown, both the center
and the spread of the sampling
distribution are unknown  problem.
We need to “guess” a value for p.
Our options:
* Use the sample proportion p^
This is the “large sample method”.
It performs poorly.
* Use an improved p^ , ~p
This is the “plus four method”. It
is reasonably accurate.
Always use with caution
Large-sample confidence interval for p
Confidence intervals contain the population proportion p in C % of
samples. For an SRS of size n drawn from a large population and with
sample proportion p̂ calculated from the data, an approximate level C
confidence interval for p is
CI : pˆ  m , with
ˆ  z * pˆ (1  pˆ ) n
m  z * SE
Use this method when the number
of successes and the number of
failures are both at least 15.
C
m
-z*
m
z*
C is the area under the standard
normal curve between -z* and z*.
Medication side effects
Arthritis is a painful, chronic inflammation of the joints.
An experiment on the side effects of pain relievers examined arthritis patients to
find the proportion of patients who suffer side effects.
What are some side effects of ibuprofen?
Serious side effects (seek medical attention immediately):
Allergic reactions (difficulty breathing, swelling, or hives)
Muscle cramps, numbness, or tingling
Ulcers (open sores) in the mouth
Rapid weight gain (fluid retention)
Seizures
Black, bloody, or tarry stools
Blood in your urine or vomit
Decreased hearing or ringing in the ears
Jaundice (yellowing of the skin or eyes)
Abdominal cramping, indigestion, or heartburn
Less serious side effects (discuss with your doctor):
Dizziness or headache
Nausea, gaseousness, diarrhea, or constipation
Depression
Fatigue or weakness
Dry mouth
Irregular menstrual periods.
We compute a 90% confidence interval for the population proportion of
arthritis patients who suffer some "adverse symptoms."
What is the sample proportion p̂ ?
pˆ 
23
 0.052
440
For a 90% confidence level, z* = 1.645.
Using the large sample method:
df
m  z * pˆ (1  pˆ ) n
m  1.645* 0.052(1  0.052) / 440
Confidence level C
0.50 0.60 0.70 0.80 0.90
0.95
0.96
z* 0.674 0.841 1.036 1.282 1.645 1.960 2.054
90% CI for p : pˆ  m
0.052  0.017
m  1.645*0.0106  0.017
 With 90% confidence level, between 3.5% and 6.9% of arthritis patients
taking this pain medication experience some adverse symptoms.
“Plus four” confidence interval for p
The “plus four” method gives reasonably accurate confidence
intervals. We act as if we had four additional observations, two
successes and two failures. Thus, the new sample size is n + 4 and the
count of successes is X + 2.
The “plus four” estimate of p is: ~
p
counts of successes  2
count of all observatio ns  4
An approximate level C confidence interval is:
CI : ~
p  m , with
~
m  z * SE  z * ~
p (1  ~
p ) (n  4)
Use this method when C is at least 90% and sample size is at least 10.
We want a 90% CI for the population proportion of arthritis patients
who suffer some “adverse symptoms.”
23  2
25
What is the value of the “plus four” estimate of p? ~
p

 0.056
440  4 444
An approximate 90% confidence interval for p using the “plus four” method is:
m  z* ~
p (1  ~
p ) (n  4)
m  1.645 * 0.056(1  0.056) / 444
m  1.645 * 0.011  0.018
90% CI for p : p  m
0.056  0.018
 With 90% confidence, between 3.8% and 7.4% of the population of arthritis
patients taking this pain medication experience some adverse symptoms.
df
Confidence level C
0.50 0.60 0.70 0.80 0.90
0.95
0.96
z* 0.674 0.841 1.036 1.282 1.645 1.960 2.054
0.98
2.326
0.99
2.576
0.995
2.807
0.998
3.091
0.999
3.291
Sample size for a desired margin of error
You may need to choose a sample size large enough to achieve a
specified margin of error.
Because the sampling distribution of p̂ is a function of the unknown
population proportion p this process requires that you guess a likely
value for p: p*.

p ~ N p, p(1  p ) n

2
 z*
 n    p * (1  p*)
m
Make an educated guess, or use p* = 0.5 (most conservative estimate).
What sample size would we need in order to achieve a margin of error no
more than 0.01 (1 percentage point) with a 90% confidence level?
We could use 0.5 for our guessed p*. However, since the drug has been
approved for sale over the counter, we can safely assume that no more than
10% of patients should suffer “adverse symptoms” (a better guess than 50%).
For a 90% confidence level, z* = 1.645.
2
2
 z*
 1.645 
n    p * (1  p*)  
 (0.1)(0.9)  2434.4
m
 0.01 
 To obtain a margin of error no more than 0.01 we need a sample size n of
at least 2435 arthritis patients.