# 3. 4. Conﬁdence interval - mean n,

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```Section 7–4 Conﬁdence Intervals and Sample Size for Proportions
369
3. Enter 563.2 for the mean of the data, and then select t. Enter 87.9 for the standard
deviation and 10 for n, the sample size.
4. Either type in or scroll to 95% for the conﬁdence level, and then click [OK].
The result of the procedure will show the output in a new chart as indicated.
Conﬁdence interval - mean
95%
563.2
87.9
10
2.262
62.880
626.080
500.320
7–4
Objective
4
Find the conﬁdence
interval for a
proportion.
confidence level
mean
std. dev.
n
t (df = 9)
half-width
upper confidence limit
lower confidence limit
Conﬁdence Intervals and Sample Size
for Proportions
A USA TODAY Snapshots feature stated that 12% of the pleasure boats in the United States
were named Serenity. The parameter 12% is called a proportion. It means that of all the
pleasure boats in the United States, 12 out of every 100 are named Serenity. A proportion
represents a part of a whole. It can be expressed as a fraction, decimal, or percentage. In
12
this case, 12% 0.12 100
or 253 . Proportions can also represent probabilities. In this case,
if a pleasure boat is selected at random, the probability that it is called Serenity is 0.12.
Proportions can be obtained from samples or populations. The following symbols
will be used.
Symbols Used in Proportion Notation
p population proportion
p̂ (read “p hat”) sample proportion
For a sample proportion,
X
nX
and
q̂ or
1 p̂
n
n
where X number of sample units that possess the characteristics of interest and n sample
size.
p̂ For example, in a study, 200 people were asked if they were satisﬁed with their job
or profession; 162 said that they were. In this case, n 200, X 162, and p̂ X兾n 162兾200 0.81. It can be said that for this sample, 0.81, or 81%, of those surveyed were
satisﬁed with their job or profession. The sample proportion is p̂ 0.81.
The proportion of people who did not respond favorably when asked if they were
satisﬁed with their job or profession constituted q̂, where q̂ (n X)兾n. For this survey,
q̂ (200 162)兾200 38兾200, or 0.19, or 19%.
When p̂ and q̂ are given in decimals or fractions, p̂ q̂ 1. When p̂ and q̂ are given
in percentages, p̂ q̂ 100%. It follows, then, that q̂ 1 p̂, or p̂ 1 q̂, when p̂
and q̂ are in decimal or fraction form. For the sample survey on job satisfaction, q̂ can
also be found by using q̂ 1 p̂, or 1 0.81 0.19.
7–23
370
Chapter 7 Conﬁdence Intervals and Sample Size
Similar reasoning applies to population proportions; that is, p 1 q, q 1 p,
and p q 1, when p and q are expressed in decimal or fraction form. When p and q
are expressed as percentages, p q 100%, p 100% q, and q 100% p.
Example 7–8
In a recent survey of 150 households, 54 had central air conditioning. Find p̂ and q̂ ,
where p̂ is the proportion of households that have central air conditioning.
Solution
Since X 54 and n 150,
X
54
p̂ 0.36 36%
n 150
n X 150 54
96
q̂ 0.64 64%
n
150
150
One can also ﬁnd q̂ by using the formula q̂ 1 p̂. In this case, q̂ 1 0.36 0.64.
As with means, the statistician, given the sample proportion, tries to estimate the
population proportion. Point and interval estimates for a population proportion can be
made by using the sample proportion. For a point estimate of p (the population proportion), p̂ (the sample proportion) is used. On the basis of the three properties of a good
estimator, p̂ is unbiased, consistent, and relatively efﬁcient. But as with means, one is not
able to decide how good the point estimate of p is. Therefore, statisticians also use an
interval estimate for a proportion, and they can assign a probability that the interval will
contain the population proportion.
The conﬁdence interval for a particular p is based on the sampling distribution of p̂.
When the sample size n is no more than 5% of the population size, the sampling distribution of p̂ is approximately normally distributed with a mean of p and a standard deviation of 2pqn, where q 1 p.
Conﬁdence Intervals
To construct a conﬁdence interval about a proportion, one must use the maximum error
of estimate, which is

E z Ⲑ2
p̂q̂
n
Conﬁdence intervals about proportions must meet the criteria that np 5 and nq 5.
Formula for a Speciﬁc Conﬁdence Interval for a Proportion

p̂q̂
p̂q̂
p p̂ zⲐ2
n
n
when np and nq are each greater than or equal to 5.
p̂ zⲐ2
Rounding Rule for a Conﬁdence Interval for a Proportion Round off to
three decimal places.
Example 7–9
7–24
A sample of 500 nursing applications included 60 from men. Find the 90% conﬁdence
interval of the true proportion of men who applied to the nursing program.
Section 7–4 Conﬁdence Intervals and Sample Size for Proportions
371
Solution
Since a 1 0.90 0.10 and za兾2 1.65, substituting in the formula

p̂ z Ⲑ2

p̂q̂
p p̂ z Ⲑ2
n
p̂q̂
n
when p̂ 60兾500 0.12 and q̂ 1 0.12 0.88, one gets

0.12 1.65

p 0.12 1.65
500
0.12 0.024 p 0.12 0.024

500
0.096 p 0.144
9.6% p 14.4%
or
Hence, one can be 90% conﬁdent that the percentage of applicants who are men is
between 9.6 and 14.4%.
When a speciﬁc percentage is given, the percentage becomes p̂ when it is changed
to a decimal. For example, if the problem states that 12% of the applicants were men,
then p̂ 0.12.
Example 7–10
A survey of 200,000 boat owners found that 12% of the pleasure boats were named
Serenity. Find the 95% conﬁdence interval of the true proportion of boats named Serenity.
Source: USA TODAY Snapshot.
Solution
From the Snapshot, p̂ 0.12 (i.e., 12%), and n 200,000. Since za兾2 1.96,
substituting in the formula

p̂ z Ⲑ2
yields

0.12 1.96

p̂q̂
p p̂ z Ⲑ2
n

p̂q̂
n

p 0.12 1.96
200,000
0.119 p 0.121

200,000
Hence, one can say with 95% conﬁdence that the true percentage of boats named
Serenity is between 11.9 and 12.1%.
Sample Size for Proportions
To ﬁnd the sample size needed to determine a conﬁdence interval about a proportion, use
this formula:
Objective
5
Determine the
minimum sample
size for ﬁnding a
conﬁdence interval
for a proportion.
Formula for Minimum Sample Size Needed for Interval Estimate of a
Population Proportion
n p̂q̂

2
2
If necessary, round up to obtain a whole number.
7–25
372
Chapter 7 Conﬁdence Intervals and Sample Size
This formula can be found by solving the maximum error of estimate value for n:

p̂q̂
n
There are two situations to consider. First, if some approximation of p̂ is known
(e.g., from a previous study), that value can be used in the formula.
Second, if no approximation of p̂ is known, one should use p̂ 0.5. This value will
give a sample size sufﬁciently large to guarantee an accurate prediction, given the conﬁdence interval and the error of estimate. The reason is that when p̂ and q̂ are each 0.5, the
product p̂q̂ is at maximum, as shown here.
E z Ⲑ2
Example 7–11
p̂
q̂
p̂q̂
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.09
0.16
0.21
0.24
0.25
0.24
0.21
0.16
0.09
A researcher wishes to estimate, with 95% conﬁdence, the proportion of people who
own a home computer. A previous study shows that 40% of those interviewed had a
computer at home. The researcher wishes to be accurate within 2% of the true
proportion. Find the minimum sample size necessary.
Solution
Since za兾2 1.96, E 0.02, p̂ 0.40, and q̂ 0.60, then
n p̂q̂

z Ⲑ2 2
1.96 2

2304.96
E
0.02
which, when rounded up, is 2305 people to interview.
Example 7–12
The same researcher wishes to estimate the proportion of executives who own a car
phone. She wants to be 90% conﬁdent and be accurate within 5% of the true proportion.
Find the minimum sample size necessary.
Solution
Since there is no prior knowledge of p̂, statisticians assign the values p̂ 0.5 and
q̂ 0.5. The sample size obtained by using these values will be large enough to ensure
the speciﬁed degree of conﬁdence. Hence,
n p̂q̂

2
2

0.05 冹
2
272.25
which, when rounded up, is 273 executives to ask.
In determining the sample size, the size of the population is irrelevant. Only the
degree of conﬁdence and the maximum error are necessary to make the determination.
7–26
Section 7–4 Conﬁdence Intervals and Sample Size for Proportions
373
Speaking of
Statistics
Does Success Bring Happiness?
W. C. Fields said, “Start every day off
with a smile and get it over with.”
Do you think people are happy
because they are successful, or are they
successful because they are just happy
people? A recent survey conducted by
Money magazine showed that 34% of the
people surveyed said that they were happy
because they were successful; however,
63% said that they were successful
because they were happy individuals.
The people surveyed had an average
household income of \$75,000 or more.
The margin of error was 2.5%. Based
would be the conﬁdence interval for each
percent?
Applying the Concepts 7–4
Contracting Inﬂuenza
To answer the questions, use the following table describing the percentage of people who
reported contracting inﬂuenza by gender and race/ethnicity.
Inﬂuenza
Characteristic
Gender
Men
Women
Race/ethnicity
Caucasian
African American
Hispanic
Other
Total
Percent
(95% CI)
48.8
51.5
(47.1–50.5%)
(50.2–52.8%)
52.2
33.1
47.6
39.7
50.4
(51.1–53.3%)
(29.5–36.7%)
(40.9–54.3%)
(30.8–48.5%)
(49.3–51.5%)
Forty-nine states and the District of Columbia participated in the study. Weighted means were
used. The sample size was 19,774. There were 12,774 women and 7000 men.
1.
2.
3.
4.
5.
6.
Explain what (95% CI) means.
How large is the error for men reporting inﬂuenza?
What is the sample size?
How does sample size affect the size of the conﬁdence interval?
Would the conﬁdence intervals be larger or smaller for a 90% CI, using the same data?
Where does the 51.5% under inﬂuenza for women ﬁt into its associated 95% CI?
See page 389 for the answers.
7–27
374
Chapter 7 Conﬁdence Intervals and Sample Size
Exercises 7–4
1. In each case, ﬁnd p̂ and q̂.
a.
b.
c.
d.
e.
n 80 and X 40
n 200 and X 90
n 130 and X 60
n 60 and X 35
n 95 and X 43
2. (ans) Find p̂ and q̂ for each percentage. (Use each
percentage for p̂.)
a.
b.
c.
d.
e.
15%
37%
71%
51%
79%
3. A U.S. Travel Data Center survey conducted for Better
Homes and Gardens of 1500 adults found that 39% said
that they would take more vacations this year than
last year. Find the 95% conﬁdence interval for the true
proportion of adults who said that they will travel more
this year.
Source: USA TODAY.
4. A recent study of 100 people in Miami found 27 were
obese. Find the 90% conﬁdence interval of the
population proportion of individuals living in Miami
who are obese.
Source: Based on information from the Center for Disease Control and
Prevention, USA TODAY.
5. The proportion of students in private schools is around
11%. A random sample of 450 students from a wide
geographic area indicated that 55 attended private
schools. Estimate the true proportion of students
attending private schools with 95% conﬁdence. How
does your estimate compare to 11%?
Source: National Center for Education Statistics (www.nces.ed.gov).
6. The Gallup Poll found that 27% of adults surveyed
How many adults should be surveyed to estimate the
with a 95% conﬁdence interval 5% wide?
Source: www.pollingreport.com.
7. A survey found that out of 200 workers, 168 said they
were interrupted three or more times an hour by phone
messages, faxes, etc. Find the 90% conﬁdence interval
of the population proportion of workers who are interrupted three or more times an hour.
Source: Based on information from USA TODAY Snapshot.
8. A CBS News/New York Times poll found that 329 out of
763 adults said they would travel to outer space in their
7–28
lifetime, given the chance. Estimate the true proportion
of adults who would like to travel to outer space with
92% conﬁdence.
Source: www.pollingreport.com.
9. A study by the University of Michigan found that
one in ﬁve 13- and 14-year-olds is a sometime
smoker. To see how the smoking rate of the students
at a large school district compared to the national rate,
the superintendent surveyed two hundred 13- and
14-year-old students and found that 23% said they
were sometime smokers. Find the 99% conﬁdence
interval of the true proportion, and compare this with
the University of Michigan study.
Source: USA TODAY.
10. A survey of 50 ﬁrst-time white-water canoers showed
that 23 did not want to repeat the experience. Find the
90% conﬁdence interval of the true proportion of
canoers who did not wish to canoe the rapids a second
time. If a rafting company wants to distribute brochures
for repeat trips, what is the minimum number it should
print?
11. A survey of 85 families showed that 36 owned at least
one DVD player. Find the 99% conﬁdence interval of the
true proportion of families who own at least one DVD
player. If another survey in a different location found that
the proportion of families who owned at least one DVD
player was 0.52, would you consider that the proportion
of families in this area was larger than in the area where
the original survey was done?
12. In a certain countrywide school district, a survey of 350
students showed that 28% carried their lunches to school.
Find the 95% conﬁdence interval of the true proportion of
students who carried their lunches to school. If the cafeteria manager wanted to be reasonably sure that all the
children who didn’t bring their lunches could purchase a
lunch, how many lunches should she plan to make each
13. In a Gallup Poll of 1005 individuals, 452 thought they
were worse off ﬁnancially than a year ago. Find the
95% conﬁdence interval for the true proportion of
individuals that feel they are worse off ﬁnancially.
Source: Gallup Poll.
14. In a poll of 1000 likely voters, 560 say that the United
States spends too little on ﬁghting hunger at home. Find
a 95% conﬁdence interval for the true proportion of
voters who feel this way.
Source: Alliance to End Hunger.
Section 7–4 Conﬁdence Intervals and Sample Size for Proportions
15. A medical researcher wishes to determine the
percentage of females who take vitamins. He wishes
to be 99% conﬁdent that the estimate is within
2 percentage points of the true proportion. A recent
study of 180 females showed that 25% took vitamins.
a. How large should the sample size be?
b. If no estimate of the sample proportion is available,
how large should the sample be?
16. A recent study indicated that 29% of the 100 women
over age 55 in the study were widows.
a. How large a sample must one take to be 90%
conﬁdent that the estimate is within 0.05 of the
true proportion of women over age 55 who are
widows?
b. If no estimate of the sample proportion is available,
how large should the sample be?
17. A researcher wishes to estimate the proportion of adult
males who are under 5 feet 5 inches tall. She wants to
be 90% conﬁdent that her estimate is within 5% of the
true proportion.
375
a. How large a sample should be taken if in a sample
of 300 males, 30 were under 5 feet 5 inches tall?
b. If no estimate of the sample proportion is available,
how large should the sample be?
18. Obesity is deﬁned as a body mass index (BMI)
of 3 kg/m2 or more. A 95% conﬁdence interval for the
percentage of U.S. adults aged 20 years and over who
were obese was found to be 22.4% to 23.5%. What was
the sample size?
Source: National Center for Health Statistics (www.cdc.gov/nchs).
19. How large a sample should be surveyed to estimate the
true proportion of college students who do laundry once
a week within 3% with 95% conﬁdence?
20. A federal report indicated that 27% of children ages
2 to 5 years had a good diet—an increase over previous
years. How large a sample is needed to estimate the true
proportion of children with good diets within 2% with
95% conﬁdence?
Source: Federal Interagency Forum on Child and Family Statistics,
Washington Observer-Reporter.
Extending the Concepts
21. If a sample of 600 people is selected and the researcher
decides to have a maximum error of estimate of 4% on
the speciﬁc proportion who favor gun control, ﬁnd the
degree of conﬁdence. A recent study showed that 50%
were in favor of some form of gun control.
22. In a study, 68% of 1015 adults said that they believe the
Republicans favor the rich. If the margin of error was
3 percentage points, what was the conﬁdence interval
used for the proportion?
Source: USA TODAY.
Technology Step by Step
MINITAB
Step by Step
Find a Conﬁdence Interval for a Proportion
MINITAB will calculate a conﬁdence interval, given the statistics from a sample or given the
raw data. In Example 7–9, in a sample of 500 nursing applications 60 were from men. Find the
90% conﬁdence interval estimate for the true proportion of male applicants.
1. Select Stat >Basic Statistics>1 Proportion.
2. Click on the button for Summarized data. No data will be entered in the worksheet.
3. Click in the box for Number of trials and enter 500.
4. In the Number of events box, enter 60.
5. Click on [Options].
6. Type 90 for the conﬁdence level.
7–29
376
Chapter 7 Conﬁdence Intervals and Sample Size
7. Check the box for Use test and interval based on normal distribution.
8. Click [OK] twice.
The results for the conﬁdence interval will be displayed in the session window.
Test and CI for One Proportion
Test of p = 0.5 vs p not = 0.
Sample
X
N
Sample p
90% CI
1
60
500
0.120000
(0.096096, 0.143904)
TI-83 Plus or
TI-84 Plus
Step by Step
Z-Value
-16.99
P-Value
0.000
Finding a Conﬁdence Interval for a Proportion
1. Press STAT and move the cursor to TESTS.
Input
2. Press A (ALPHA, MATH) for 1-PropZlnt.
3. Type in the appropriate values.
4. Move the cursor to Calculate and press ENTER.
Example TI7–3
Find the 95% conﬁdence interval of p when X 60 and n 500,
as in Example 7–9.
The 95% conﬁdence level for p is 0.09152 p 0.14848.
Also p̂ is given.
Excel
Step by Step
Output
Finding a Conﬁdence Interval for a Proportion
Excel does not produce conﬁdence intervals for a proportion. However, you may determine
conﬁdence intervals for a proportion by using the MegaStat Add-in available on your CD
and Online Learning Center. If you have not installed this add-in, do so by following the
instructions on page 24.
There were 500 nursing applications in a sample, including 60 from men. Find the 90%
conﬁdence interval for the true proportion of male applicants.
1. From the toolbar, select MegaStat>Conﬁdence Intervals/Sample Size.
2. In the dialog box, select the Conﬁdence interval—p.
3. Enter 60 in the ﬁrst box; p will automatically switch to x.
4. Enter 500 in the second box for n.
5. Either type in or scroll to 90% for the conﬁdence level, and then click [OK].
7–30
Section 7–5 Conﬁdence Intervals for Variances and Standard Deviations
Speaking of
Statistics
377
OTHER PEOPLE’S MONEY
Here is a survey about college students’
credit card usage. Suggest several ways
that the study could have been more
been used.
means—you guessed it—students are
learning to become debtors. According to
the Public Interest Research Groups, only
half of all students pay off card balances in
full each month, 36% sometimes do and
14% never do. Meanwhile, 48% have paid
a late fee. Here's how undergrads stack up,
according to Nellie Mae, a provider of
college loans:
78%
Average number of cards owned
3
Average student card debt
\$1236
Students with 4 or more cards
32%
Balances of \$3000 to \$7000
13%
Balances over \$7000
9%
Reprinted with permission from the January 2002 Reader’s Digest.
The result of the procedure will show the output in a new chart, as indicated here.
Conﬁdence interval - proportion
90%
0.12
500
1.645
0.024
0.144
0.096
7–5
Objective
6
Find a conﬁdence
interval for a variance
and a standard
deviation.
confidence level
proportion
n
z
half-width
upper confidence limit
lower confidence limit
Conﬁdence Intervals for Variances
and Standard Deviations
In Sections 7–2 through 7–4, conﬁdence intervals were calculated for means and proportions. This section will explain how to ﬁnd conﬁdence intervals for variances and
standard deviations. In statistics, the variance and standard deviation of a variable are as
important as the mean. For example, when products that ﬁt together (such as pipes) are
manufactured, it is important to keep the variations of the diameters of the products as
small as possible; otherwise, they will not ﬁt together properly and will have to be
scrapped. In the manufacture of medicines, the variance and standard deviation of the
medication in the pills play an important role in making sure patients receive the proper
dosage. For these reasons, conﬁdence intervals for variances and standard deviations are
necessary.
7–31
```