Math 221: Confidence Intervals for a Population Proportion
S. K. Hyde
Chapter 19, (Moore, 5th Ed.)
The tobacco industry closely monitors all surveys that involve smoking. One survey showed that among
785 randomly selected subjects who completed four years of college, 144 smoke (based on data from the
American Medical Association).
1. Large-Sample Confidence Interval for a Population Proportion
An approximate level C confidence interval for p is
p̂ ± m
where
x
p̂ = ,
n
r
∗
m = z SEp̂ ,
and
SEp̂ =
p̂(1 − p̂)
n
Use this interval only when the number of successes and failures in the sample are both at
least 15.
Find a 99% confidence interval for the proportion of smokers among the population of people who
have completed four years of college.
2. The “Plus-Four” Confidence Interval for a Population Proportion
A more accurate level C confidence interval for p is
p̃ ± m
where
p̃ =
x+2
,
n+4
r
m = z ∗ SEp̃ ,
and
SEp̃ =
p̃(1 − p̃)
n+4
This is the same as the above method, with the addition of adding two imaginary successes
and two imaginary failures (four overall) to your sample. Hence, the x + 2 and n + 4 in the
definition of p̃. Use this interval when the confidence level is at least 90% and the sample size
n is at least 10.
Find a 99% “plus four” confidence interval for the proportion of smokers among the population of
people who have completed four years of college.
Confidence Interval for a Population Proportion, page 2
3. Sample Size for a Desired Margin of Error
The level C confidence interval for a population proportion p will have margin of error approximately equal to a specified value m when the sample size is
n=
z ∗ 2
m
p∗ (1 − p∗ ),
where p∗ is a guessed value for the sample proportion. The margin of error will be less than
or equal to m if you take the guess p∗ to be 0.5.
Suppose the American Medical Association wants to determine the sample size needed to have a
margin of error of ±1%?
4. Suppose the American Medical Association wants to determine the sample size needed to have a
margin of error of ±1%? What would be the sample size required if you assume that an estimate
for p is not known?
Confidence Interval for a Population Proportion, page 3
Solution
1. Find a 99% confidence interval for the proportion of smokers among the population of people who
have completed four years of college.
The estimate for p is p̂ =
144
785
= 0.1834395. Thus, the standard error is
s
r
144
144
p̂(1 − p̂)
785 1 − 785
SEp̂ =
=
= 0.01381357.
n
785
Thus, the margin of error is m = z ∗ SEp̂ = 2.576(0.01381357) = 0.03558374.
So a 99% confidence interval for the proportion of smokers among the population of people who
have completed four years of college is
p̂ ± m =⇒ 0.1834395 ± 0.03558374 =⇒ (0.148, 0.219)
Note: When using a TI-83 or TI-84
calculator, select STAT −→ TESTS −→ 1-PropZInt , and enter the appropriate data or statistics.
2. Find a 99% “plus four” confidence interval for the proportion of smokers among the population of
people who have completed four years of college.
Here we repeat the above procedure, but use x + 2 for x and n + 4 for n.
The estimate for p is p̃ =
144+2
785+4
=
146
789
= 0.1850444. Thus, the standard error is
s
r
146
146
p̃(1 − p̃)
789 1 − 789
SEp̃ =
=
= 0.01382504.
n+4
789
Thus, the margin of error is m = z ∗ SEp̂ = 2.576(0.01382504) = 0.03561330.
So a 99% confidence interval for the proportion of smokers among the population of people who
have completed four years of college is
p̂ ± m =⇒ 0.1850444 ± 0.03561330 =⇒ (0.149, 0.221)
To use the TI-83 or TI-84 calculator for the “plus four” intervals, select STAT −→ TESTS −→
1-PropZInt , and you should enter x + 2 for “x” and n + 4 for “n”.
3. Suppose the American Medical Association wants to determine the sample size needed to have a
margin of error of ±1%? Here, we will choose the estimate for p to be p∗ = p̂ = 144
785 = 0.1834395.
n=
z ∗ 2
m
p∗ (1 − p∗ ) =
2.576
.01
2
144
785
144
1−
= 9939.692
785
The American Medical Association needs to have a sample size of 9,940 people to reduce the
margin of error to ±1%.
4. Suppose the American Medical Association wants to determine the sample size needed to have a
margin of error of ±1%? Assume that an estimate for p is not known. Here, we will choose the
estimate for p to be p∗ = p̂ = 21 .
n=
z ∗ 2
m
p∗ (1 − p∗ ) =
2.576
.01
2
1
2
1
1−
= 16589.44
2
If no estimate for p is known, then the American Medical Association needs to have a sample size
of 16,590 people to reduce the margin of error to ±1%.