6.3 Confidence Intervals for
Population Proportions
Statistics
Mrs. Spitz
Spring 2009
Objectives/Assignment
 How to find a sample proportion
 How to construct a confidence interval for a
population proportion
 How to determine a minimum sample size
when estimating a population proportion.
Assignment: pp. 280-282 #1-27 all
Schedule for coming weeks:
 Today – Notes 6.3. Homework due BOC on
Friday.
 Friday, 1/16/09 – Notes 6.4. Assignment due
Tuesday on our return.
 Monday – 1/19/09 – No school
 Tuesday – 1/20/09 – Chapter Review
 Thursday-Chapter Review 6 DUE – Test –
Chapter 6
 Friday – 1/23/09 – 7.1 Hypothesis Testing
Sample Proportions
 Recall from section 4.2 that the probability of
success in a single trial of a binomial
experiment is p. This probability is a
population proportion. In this section, you will
learn how to estimate a population proportion,
p using a confidence interval. As with
confidence intervals for µ, you will start with a
point estimate (6.1)
Definition:
 The point estimate for p, the population
proportion of successes, is given by the
proportion of successes in a sample and is
x
denoted by:
pˆ 
n
 where x is the number of successes in the
sample and n is the number in the sample.
The point estimate for the number of failures
is qˆ  1  pˆ .The symbols q̂ and p̂ are read as
“p hat” and “q hat”
Ex. 1: Finding a point estimate for p
 In a survey of 883 American adults, 380 said
that their favorite sport is football. Find a
point estimate for the population proportion of
adults who say their favorite sport is football.
 SOLUTION: Using n =883 and x = 380
x 380
pˆ  
 0.43  43%
n 883
Insight
 In the first two sections, estimates were made
for the quantitative data. In this section,
sample proportions are used to make
estimates for qualitative data.
Confidence Intervals for a Population P
 Constructing a confidence interval for a
population proportion p is similar to
constructing a confidence interval for a
population mean. You start with a point
estimate and calculate a maximum error of
estimate.
pˆ  E  p  pˆ  E
Definition:
 A c-confidence interval for the population
proportion p is
pˆ  E  p  pˆ  E
where
E  zc
pˆ qˆ
n
The probability that the confidence interval contains p is c.
Notes
 In section 5.5, you learned that a binomial
can be approximated by the normal
distribution if np  5 and nq  5. When npˆ  5
and nqˆ  5, the sampling distribution for p̂ is
approximately normal with a mean of p = p
and a standard error of
p 
pq
n
Guidelines: Constructing a Confidence Interval for a
Population Proportion
In words
1.
ID the sample stats, n and x
2.
Find the point estimate
3.
Verify the sampling
distribution of p(hat) can be
approximated by the normal
distribution.
4.
Find the critical zc that
corresponds to the given level
of confidence, c.
5.
Find the maximum error of
estimate, E.
6.
Find the left and the right
endpoints and form the
confidence interval.
pˆ 
x
n
Is npˆ  5 and is nqˆ  5 ?
Use a standard normal table.
E  zc
pˆ qˆ
n
pˆ  E
ˆE
Right endpoint: p
Interval: p
ˆ  E  p  pˆ  E
Left endpoint:
Ex. 2: Constructing a Confidence
interval for p
 Construct a 95% confidence interval for the
proportion of American adults who say that
their favorite sport is football.
 SOLUTION: Form example 1, pˆ  0.43 , So,
qˆ  1  0.43  0.57 . Using n = 883, you can
verify that the sampling distribution of p̂ can
be approximated by the normal distribution.
and
npˆ  883  0.43  380  5
nqˆ  883  0.57  503  5
Ex. 2: Constructing a Confidence
interval for p
Using zc = 1.96, the maximum error of estimate is:
pˆ qˆ
(0.43)(0.57)
E  zc
 1.96
 0.033
n
883
The 95% confidence interval is as follows:
Left Endpoint
Right Endpoint
pˆ  E  0.43  0.033  0.397
pˆ  E  0.43  0.033  0.463
0.397  p  0.463
So, with 95% confidence, you can say that the proportion of adults who say that
footbal is their favorite sport is between 39.7% and 46.3%.
Opinion Polls
 The confidence level of 95% used in Example
2 is typical of opinion polls. The result;
however, is usually not stated as a
confidence interval. Instead the result of
Example 2 would usually be stated as 43%
with a margin of error of 3.3%.”
Ex. 3: Constructing a Confidence
Interval for p
 The graph shown below is from a survey of 935
adults. Construct a 99% confidence interval for the
proportion of adults who think that airplanes are the
safest mode of transportation.
Solution:
So with 99% confidence, you can say that the
proportion of adults who think that airplanes are
the safest mode of transportation is between
40.8% and 49.2%
 From the graph pˆ  0.45. So, qˆ  1  0.45  0.55
Using these values and the values n = 935
and zc = 2.575, the maximum error of
estimate is:
E  zc
pˆ qˆ
(0.45)(0.55)
 2.575
 0.042
n
935
The 99% confidence inteval is as follows:
Left Endpoint
Right Endpoint
pˆ  E  0.45  0.042  0.408
pˆ  E  0.45  0.042  0.492
0.408  p  0.492
Increasing Sample Size to Increase
Precision
 One way to increase the precision of the confidence
interval without decreasing the level of confidence is
to increase the sample size.
Insight – why 0.5?
 The reason for using 0.5 as values for p hat
and q hat when no preliminary estimate is
available is that these values yield a
maximum value for the product
pˆ qˆ  pˆ (1  pˆ )
 In other words, if you don’t estimate thevalues
of p hat and q hat, you must pay the penalty
of using a larger sample.
Ex. 4: Determining a Minimum Sample
Size
 You are running a political campaign and wish
to estimate with 95% confidence, the
proportion of registered voters, who will vote
for your candidate. What is the minimum
sample size needed if you are to accurately
within 3% of the population proportion?
SOLUTION
 Because you do not have a preliminary
estimate for p, use p
ˆ  0.5 and qˆ  0.5 .
Using zc = 1.96, and E = 0.03, you can solve
for n.
2
2
 zc 
 1.96 
n  pˆ qˆ    (0.5)(0.5)
  1067.11
 0.03 
E
Because n is a decimal, round up to the nearest
whole number. So, at least 1068 registered voters
should be included in the sample.
Assignment due Friday BOC.
Assignment: pp. 280-282 #1-22 all