Confidence Intervals for a Population Proportion by hkksew3563rd

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```									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

x                                            p(1 − p)
ˆ     ˆ
p±m
ˆ          where       p=
ˆ      ,     m = z ∗ SEp ,
ˆ       and       SEp =
ˆ
n                                               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

x+2                                            p(1 − p)
˜     ˜
p±m
˜          where       p=
˜        ,      m = z ∗ SEp ,
˜      and      SEp =
˜
n+4                                              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 p. 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 approx-
imately equal to a speciﬁed value m when the sample size is
z∗   2
n=             p∗ (1 − p∗ ),
m
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.
144
ˆ
The estimate for p is p =   785   = 0.1834395. Thus, the standard error is

144
p(1 − p)
ˆ     ˆ           785   1 − 144
785
SEp =
ˆ                       =                      = 0.01381357.
n                    785
Thus, the margin of error is m = z ∗ SEp = 2.576(0.01381357) = 0.03558374.
ˆ

So a 99% conﬁdence interval for the proportion of smokers among the population of people who
have completed four years of college is

p ± m =⇒ 0.1834395 ± 0.03558374 =⇒ (0.148, 0.219)
ˆ

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.
144+2          146
˜
The estimate for p is p =   785+4      =   789   = 0.1850444. Thus, the standard error is

146
p(1 − p)
˜     ˜           789   1 − 146
789
SEp =
˜                       =                      = 0.01382504.
n+4                   789
Thus, the margin of error is m = z ∗ SEp = 2.576(0.01382504) = 0.03561330.
ˆ

So a 99% conﬁdence interval for the proportion of smokers among the population of people who
have completed four years of college is

p ± m =⇒ 0.1850444 ± 0.03561330 =⇒ (0.149, 0.221)
ˆ

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∗ = p = 144 = 0.1834395.
ˆ 785
2
z∗    2                         2.576        144          144
n=              p∗ (1 − p∗ ) =                             1−           = 9939.692
m                                .01         785          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∗ = p = 2 .
ˆ 1
2
z∗    2                      2.576          1        1
n=                 p∗ (1 − p∗ ) =                        1−         = 16589.44
m                             .01           2        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%.

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