Transcript confidence interval

Chapter 7
Confidence Intervals and
Sample Size
© McGraw-Hill, Bluman, 5th ed., Chapter 7
1
Chapter 7 Overview
Introduction

7-1 Confidence Intervals for the Mean When
 Is Known and Sample Size

7-2 Confidence Intervals for the Mean When
 Is Unknown

7-3 Confidence Intervals and Sample
Size for Proportions

7-4 Confidence Intervals and Sample Size for
Variances and Standard Deviations
Bluman, Chapter 7
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Chapter 7 Objectives
1. Find the confidence interval for the mean when
 is known.
2. Determine the minimum sample size for finding
a confidence interval for the mean.
3. Find the confidence interval for the mean when
 is unknown.
4. Find the confidence interval for a proportion.
Bluman, Chapter 7
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Chapter 7 Objectives
5. Determine the minimum sample size for
finding a confidence interval for a proportion.
6. Find a confidence interval for a variance and a
standard deviation.
Bluman, Chapter 7
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7.1 Confidence Intervals for the
Mean When  Is Known and
Sample Size
 A point estimate is a specific
numerical value estimate of a
parameter.
 The best point estimate of the
population mean µ is the
sample mean X .
Bluman, Chapter 7
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Three Properties of a Good
Estimator
1. The estimator should be an
unbiased estimator. That is,
the expected value or the mean
of the estimates obtained from
samples of a given size is
equal to the parameter being
estimated.
Bluman, Chapter 7
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Three Properties of a Good
Estimator
2. The estimator should be
consistent. For a consistent
estimator, as sample size
increases, the value of the
estimator approaches the value
of the parameter estimated.
Bluman, Chapter 7
7
Three Properties of a Good
Estimator
3. The estimator should be a
relatively efficient estimator;
that is, of all the statistics that
can be used to estimate a
parameter, the relatively
efficient estimator has the
smallest variance.
Bluman, Chapter 7
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Confidence Intervals for the Mean
When  Is Known and Sample Size


An interval estimate of a
parameter is an interval or a range
of values used to estimate the
parameter.
This estimate may or may not
contain the value of the parameter
being estimated.
Bluman, Chapter 7
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Confidence Level of an Interval
Estimate

The confidence level of an interval
estimate of a parameter is the probability
that the interval estimate will contain the
parameter, assuming that a large
number of samples are selected and that
the estimation process on the same
parameter is repeated.
Bluman, Chapter 7
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Confidence Interval
A
confidence interval is a specific
interval estimate of a parameter
determined by using data obtained
from a sample and by using the
specific confidence level of the
estimate.
Bluman, Chapter 7
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Formula for the Confidence Interval
of the Mean for a Specific a
  
  
X  za / 2 



X

z
a /2 


 n
 n
For a 90% confidence interval: za / 2  1.65
For a 95% confidence interval: za / 2  1.96
For a 99% confidence interval: za / 2  2.58
Bluman, Chapter 7
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95% Confidence Interval of the
Mean
Bluman, Chapter 7
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Maximum Error of the Estimate
The maximum error of the estimate is the
maximum likely difference between the point
estimate of a parameter and the actual value
of the parameter.
  
E  za / 2 

 n
Bluman, Chapter 7
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Confidence Interval for a Mean
Rounding Rule
When you are computing a confidence interval for
a population mean by using raw data, round off to
one more decimal place than the number of
decimal places in the original data.
When you are computing a confidence interval for
a population mean by using a sample mean and a
standard deviation, round off to the same number
of decimal places as given for the mean.
Bluman, Chapter 7
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Chapter 7
Confidence Intervals and
Sample Size
Section 7-1
Example 7-1
Page #358
Bluman, Chapter 7
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Example 7-1: Days to Sell an Aveo
A researcher wishes to estimate the number of days it
takes an automobile dealer to sell a Chevrolet Aveo. A
sample of 50 cars had a mean time on the dealer’s lot of
54 days. Assume the population standard deviation to be
6.0 days. Find the best point estimate of the population
mean and the 95% confidence interval of the population
mean.
The best point estimate of the mean is 54 days.
X  54, s  6.0, n  50,95%  z  1.96
  
  
X  za 2 
    X  za 2 

 n
 n
Bluman, Chapter 7
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Example 7-1: Days to Sell an Aveo
X  54, s  6.0, n  50,95%  z  1.96
  
  
X  za 2 
    X  za 2 

 n
 n
 6.0 
 6.0 
54  1.96 
    54  1.96 

 50 
 50 
54  1.7    54  1.7
52.3    55.7
52    56
One can say with 95% confidence that the interval
between 52 and 56 days contains the population mean,
based on a sample of 50 automobiles.
Bluman, Chapter 7
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Chapter 7
Confidence Intervals and
Sample Size
Section 7-1
Example 7-2
Page #358
Bluman, Chapter 7
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Example 7-2: Ages of Automobiles
A survey of 30 adults found that the mean age of a
person’s primary vehicle is 5.6 years. Assuming the
standard deviation of the population is 0.8 year, find the
best point estimate of the population mean and the 99%
confidence interval of the population mean.
The best point estimate of the mean is 5.6 years.
 0.8 
 0.8 
5.6  2.58 
    5.6  2.58 

 50 
 50 
5.2    6.0
One can be 99% confident that the mean age of all
primary vehicles is between 5.2 and 6.0 years, based on a
sample of 30 vehicles.
Bluman, Chapter 7
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95% Confidence Interval of the
Mean
Bluman, Chapter 7
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95% Confidence Interval of the
Mean
One can be
95% confident
that an interval
built around a
specific sample
mean would
contain the
population
mean.
Bluman, Chapter 7
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Finding za
2
for 98% CL.
za
Bluman, Chapter 7
2
 2.33
23
Example 1
A sample of the reading scores of 35 fifth-graders has a
mean of 82. The standard deviation of the sample is 15.
a. Find the 95% confidence
interval of the mean
reading scores of all fifthgraders.
b. Find the 99% confidence
interval of the mean
reading scores of all fifthgraders.
c. Which interval is larger?
Explain why.
n = 35
s = 15
X = 82
a. Find the 95% confidence interval of
mean reading scores of all fifth-graders.
 s 
 s 
X – za 2  n  < < X +za 2  n 




 15 
 15 
82 – (1.96) 
<  < 82 + (1.96) 


35
35




82 – 4.97< < 82 + 4.97
77 <  < 87
n = 35
s = 15
X = 82
a. Find the 95% confidence interval of
mean reading scores of all fifth-graders.
77 <  < 87
Approximately 95% of the sample
means will fall within 1.96
standard errors of the
population mean,
so use za 2 = 1.96.
n = 35
s = 15
X = 82
b. Find the 99% confidence interval of the
mean reading scores of all fifth-graders.
 s 
 s 
X – za 2  n  < < X +za 2  n 




 15 
 15 
82 – (2.58) 
< m < 82 + (2.58) 

 35 
 35 
82 – 6.54<  < 82 + 6.54
75<  < 89
n = 35
s = 15
X = 82
b. Find the 99% confidence interval of the
mean reading scores of all fifth-graders.
75<  < 89
Approximately 99% of the sample
means will fall within 2.58 standard
errors of the population mean,
so use za 2 = 2.58 .
c. Which interval is larger? Explain why.
95% confidence level = 77<  < 87
99% confidence level = 75<  < 89
The 99% confidence interval
is larger because the
confidence level is larger.
Example 2
A study of 40 English composition professors showed
that they spent, on average, 12.6 minutes correcting a
student’s term paper.
a. Find the 90% confidence interval
of the mean time for all composition
papers when  = 2.5 minutes.
b. If a professor stated that he
spent, on average, 30 minutes
correcting a term paper,
what would be your reaction?
n = 40
X = 12.6
a. Find the 90% confidence interval of the
mean time for all composition papers
when  = 2.5 minutes.
 s 
 s 
X – za 2  n  < < X +za 2  n 




 2.5 
 2.5 
12.6 – (1.65) 
<  < 12.6 + (1.65) 

 40 
 40 
12.6 – 0.652<  < 12.6 + 0.652
11.9<  < 13.3
b. If a professor stated that he spent, on average, 30
minutes correcting a term paper, what would be
your reaction?
11.9 <  < 13.3
It would be highly unlikely since this
is far larger than 13.3 minutes.
Chapter 7
Confidence Intervals and
Sample Size
Section 7-1
Example 7-3
Page #360
Bluman, Chapter 7
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Example 7-3: Credit Union Assets
The following data represent a sample of the assets (in
millions of dollars) of 30 credit unions in southwestern
Pennsylvania. Find the 90% confidence interval of the
mean.
12.23 16.56 4.39
2.89 1.24 2.17
13.19 9.16 1.42
73.25 1.91 14.64
11.59 6.69 1.06
8.74 3.17 18.13
7.92 4.78 16.85
40.22 2.42 21.58
5.01 1.47 12.24
2.27 12.77 2.76
Bluman, Chapter 7
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Example 7-3: Credit Union Assets
Step 1: Find the mean and standard deviation. Using
technology, we find X = 11.091 and s = 14.405.
Step 2: Find α/2. 90% CL  α/2 = 0.05.
Step 3: Find zα/2. 90% CL  α/2 = 0.05  z.05 = 1.65
Table E
The Standard Normal Distribution
z
.00
…
.04
.05
0.9495
0.9505
…
.09
0.0
0.1
.
.
.
1.6
Bluman, Chapter 7
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Example 7-3: Credit Union Assets
Step 4: Substitute in the formula.
  
  
X  za 2 
    X  za 2 

 n
 n
 14.405 
 14.405 
11.091  1.65 
    11.091  1.65 

 30 
 30 
11.091  4.339    11.091  4.339
6.752    15.430
One can be 90% confident that the population mean of the
assets of all credit unions is between $6.752 million and
$15.430 million, based on a sample of 30 credit unions.
Bluman, Chapter 7
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Technology Note
This chapter and subsequent chapters include examples
using raw data. If you are using computer or calculator
programs to find the solutions, the answers you get may
vary somewhat from the ones given in the textbook.
This is so because computers and calculators do not
round the answers in the intermediate steps and can use
12 or more decimal places for computation. Also, they
use more exact values than those given in the tables in
the back of this book.
These discrepancies are part and parcel of statistics.
Bluman, Chapter 7
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Formula for Minimum Sample Size
Needed for an Interval Estimate of the
Population Mean
 za 2   
n

 E 
2
where E is the maximum error of estimate. If necessary,
round the answer up to obtain a whole number. That is, if
there is any fraction or decimal portion in the answer, use
the next whole number for sample size n.
Bluman, Chapter 7
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Chapter 7
Confidence Intervals and
Sample Size
Section 7-1
Example 7-4
Page #362
Bluman, Chapter 7
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Example 7-4: Depth of a River
A scientist wishes to estimate the average depth of a river.
He wants to be 99% confident that the estimate is
accurate within 2 feet. From a previous study, the
standard deviation of the depths measured was 4.38 feet.
99%  z  2.58, E  2,   4.38
2
z


 a2
  2.58  4.38 
n
 
  31.92  32
2

 E  
2
Therefore, to be 99% confident that the estimate is within
2 feet of the true mean depth, the scientist needs at least
a sample of 32 measurements.
Bluman, Chapter 7
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Example 3
An insurance company is trying to estimate the average
number of sick days that full-time food service workers
use per year. A pilot study found the standard deviation
to be 2.5 days. How large a sample must
be selected if the company wants to be
95% confident of getting an interval
that contains the true mean with a
maximum error of 1 day?
s = 2.5
confidence level = 95%
maximum error = 1 day
s = 2.5
confidence level = 95%
maximum error = 1 day
2
 za 2    (1.96)(2.5)  2
2
=
n =
=
(4.9)
 

E
1

 
or
n = 25 workers
n = 24.01
7.2 Confidence Intervals for the
Mean When  Is Unknown
The value of , when it is not known, must be estimated
by using s, the standard deviation of the sample.
When s is used, especially when the sample size is small
(less than 30), critical values greater than the values for
za 2 are used in confidence intervals in order to keep the
interval at a given level, such as the 95%.
These values are taken from the Student t distribution,
most often called the t distribution.
Bluman, Chapter 7
43
Characteristics of the t Distribution
The t distribution is similar to the standard
normal distribution in these ways:
1. It is bell-shaped.
2. It is symmetric about the mean.
3. The mean, median, and mode are equal to
0 and are located at the center of the
distribution.
4. The curve never touches the x axis.
Bluman, Chapter 7
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Characteristics of the t Distribution
The t distribution differs from the standard
normal distribution in the following ways:
1. The variance is greater than 1.
2. The t distribution is actually a family of
curves based on the concept of degrees of
freedom, which is related to sample size.
3. As the sample size increases, the t
distribution approaches the standard
normal distribution.
Bluman, Chapter 7
45
Degrees of Freedom



The symbol d.f. will be used for degrees of
freedom.
The degrees of freedom for a confidence
interval for the mean are found by subtracting
1 from the sample size. That is, d.f. = n - 1.
Note: For some statistical tests used later in
this book, the degrees of freedom are not
equal to n - 1.
Bluman, Chapter 7
46
Formula for a Specific Confidence
Interval for the Mean When  Is
Unknown and n < 30
 s 
 s 
X  ta 2 
    X  ta 2 

 n
 n
The degrees of freedom are n - 1.
Bluman, Chapter 7
47
Chapter 7
Confidence Intervals and
Sample Size
Section 7-2
Example 7-5
Page #369
Bluman, Chapter 7
48
Example 7-5: Using Table F
Find the tα/2 value for a 95% confidence interval when the
sample size is 22.
Degrees of freedom are d.f. = 21.
Bluman, Chapter 7
49
Chapter 7
Confidence Intervals and
Sample Size
Section 7-2
Example 7-6
Page #370
Bluman, Chapter 7
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Example 7-6: Sleeping Time
Ten randomly selected people were asked how long they
slept at night. The mean time was 7.1 hours, and the
standard deviation was 0.78 hour. Find the 95%
confidence interval of the mean time. Assume the variable
is normally distributed.
Since  is unknown and s must replace it, the t distribution
(Table F) must be used for the confidence interval. Hence,
with 9 degrees of freedom, tα/2 = 2.262.
 s 
 s 
X  ta 2 
    X  ta 2 

 n
 n
 0.78 
 0.78 
7.1  2.262 
    7.1  2.262 

 10 
 10 
Bluman, Chapter 7
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Example 7-6: Sleeping Time
 0.78 
 0.78 
7.1  2.262 
    7.1  2.262 

 10 
 10 
7.1  0.56    7.1  0.56
6.54    7.66
One can be 95% confident that the population mean is
between 6.54 and 7.66 inches.
Bluman, Chapter 7
52
Chapter 7
Confidence Intervals and
Sample Size
Section 7-2
Example 7-7
Page #370
Bluman, Chapter 7
53
Example 7-7: Home Fires by Candles
The data represent a sample of the number of home fires
started by candles for the past several years. Find the
99% confidence interval for the mean number of home
fires started by candles each year.
5460 5900 6090 6310 7160 8440 9930
Step 1: Find the mean and standard deviation. The mean
is X = 7041.4 and standard deviation s = 1610.3.
Step 2: Find tα/2 in Table F. The confidence level is 99%,
and the degrees of freedom d.f. = 6
t .005 = 3.707.
Bluman, Chapter 7
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Example 7-7: Home Fires by Candles
Step 3: Substitute in the formula.
 s 
 s 
X  ta 2 
    X  ta 2 

 n
 n
 1610.3 
 1610.3 
7041.4  3.707 
    7041.4  3.707 

7
7




7041.4  2256.2    7041.4  2256.2
4785.2    9297.6
One can be 99% confident that the population mean
number of home fires started by candles each year is
between 4785.2 and 9297.6, based on a sample of home
fires occurring over a period of 7 years.
Bluman, Chapter 7
55
Example 4
A state representative wishes to estimate the mean
number of women representatives per state legislature. A
random sample of 17 states is selected, and the number of
women representatives is shown.Based on
the sample, what is the point estimate of the
mean? Find the 90% confidence interval of
the mean population. (Note: The population
mean is actually 31.72, or about 32.)
Compare this value to the point estimate
and the confidence interval. There is
something unusual about the data.
Describe it and state how it would
affect the confidence interval.
5
31
18
58
X = 33.4
33
16
29
132
35
45
15
37
19
39
24
13
18
s = 28.7
 s 
 s 
X – ta 2  n  <  < X + ta 2  n 




 28.7 
 28.7 
33.4 – 1.746 
<  < 33.4 + 1.746 


17
17




33.4 – 12.2 <  < 33.4 + 12.2
21.2<  < 45.6
Example 5
A recent study of 28 employees of XYZ Company
showed that the mean of the distance they traveled to
work was 14.3 miles. The standard deviation of the
sample mean was 2 miles. Find the 95% confidence
interval of the true mean.
 s 
 s 
X – ta 2  n  <  < X + ta 2  n 




14.3 – 2.052
 
2
28
<  < 14.3 + 2.052
14.3 – 0.776<  < 14.3 + 0.776
 
2
28
 s 
 s 
X – ta 2  n  <  < X + ta 2  n 




14.3 – 2.052
 
2
<  < 14.3 + 2.052
28
14.3 – 0.776<  < 14.3 + 0.776
13.5<  < 15.1
 
2
28
If a manager wanted to be sure that most of his
employees would not be late, how much time would he
suggest they allow for the commute if the average
speed is 30 miles per hour?
14.3 mi
30 mi/hr
-
1
2
hour
about 30 minutes
Example 6
The average yearly income for 28 community college
instructors was $56,718. The standard deviation was
$650. Find the 95% confidence interval of the
true mean.
 s 
 s 
X – ta 2  n  <  < X + ta 2  n 




 s 
 s 
X – ta 2  n  <  < X + ta 2  n 





$56, 718 – 2.052 



 <  < $56, 718 + 2.052 

28 
650
$56, 466 <  < $56, 970


28 
650
If a faculty member wishes to see if he or she is being
paid below average, what salary value should he or she
use?
Use the lower bound of the
confidence interval: $56,466
7.3 Confidence Intervals and
Sample Size for Proportions
p = population proportion
p̂ (read p “hat”) = sample proportion
For a sample proportion,
X
pˆ 
n
and
n X
qˆ 
n
or
qˆ  1  pˆ
where X = number of sample units that possess the
characteristics of interest and n = sample size.
Bluman, Chapter 7
64
Chapter 7
Confidence Intervals and
Sample Size
Section 7-3
Example 7-8
Page #376
Bluman, Chapter 7
65
Example 7-8: Air Conditioned
Households
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.
Since X = 54 and n = 150,
X
54
pˆ 

 0.36  36%
n 150
qˆ  1  pˆ  1  0.36  0.64  64%
Bluman, Chapter 7
66
Formula for a Specific Confidence
Interval for a Proportion
pˆ  za 2
ˆˆ
pq
 p  pˆ  za
n
2
ˆˆ
pq
n
when np  5 and nq  5.
Rounding Rule: Round off to three decimal places.
Bluman, Chapter 7
67
Chapter 7
Confidence Intervals and
Sample Size
Section 7-3
Example 7-9
Page #376
Bluman, Chapter 7
68
Example 7-9: Male Nurses
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.
p  X n  60 500  0.12, qˆ  0.88
ˆˆ
ˆˆ
pq
pq
pˆ  za 2
 p  pˆ  za 2
n
n
0.12  0.88 
0.12  0.88 


0.12  1.65
 p  0.12  1.65
500
500
0.12  0.024  p  0.12  0.024
.096  p  0.144
You can be 90% confident that the percentage of
applicants who are men is between 9.6% and 14.4%.
Bluman, Chapter 7
69
Chapter 7
Confidence Intervals and
Sample Size
Section 7-3
Example 7-10
Page #377
Bluman, Chapter 7
70
Example 7-10: Religious Books
A survey of 1721 people found that 15.9% of individuals
purchase religious books at a Christian bookstore. Find
the 95% confidence interval of the true proportion of
people who purchase their religious books at a Christian
bookstore.
ˆˆ
ˆˆ
pq
pq
pˆ  za 2
 p  pˆ  za 2
n
n
0.159  1.96
 0.159  0.841
1721
 p  0.159  1.96
 0.159  0.841
1721
0.142  p  0.176
You can say with 95% confidence that the true
percentage is between 14.2% and 17.6%.
Bluman, Chapter 7
71
Example 7
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%?
p̂ =
X
55
=
= 0.12
n 450

pˆ – za 2

q̂ = 1– 0.12 = 0.88

ˆˆ
pq
< p < pˆ + za 2
n

ˆˆ
pq
n
X
55
p̂ =
=
= 0.12
n 450

pˆ – za 2

q̂ = 1– 0.12 = 0.88

ˆˆ
pq
< p < pˆ + za 2
n

ˆˆ
pq
n
0.12 – 1.96
(0.12)(0.88)
(0.12)(0.88)
< p < 0.12 + 1.96
450
450
0.12 – 0.03 < p < 0.12 + 0.03
0.09 < p < 0.15
11% is contained in the
confidence interval.
or
9% < p < 15%
Example 8
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.
p̂ = 0.84

pˆ – za 2

q̂ = 0.16

ˆˆ
pq
< p < pˆ + za 2
n

ˆˆ
pq
n
p̂ = 0.84

pˆ – za 2

q̂ = 0.16

ˆˆ
pq
< p < pˆ + za 2
n

ˆˆ
pq
n
(0.84)(0.16)
(0.84)(0.16)
0.84 – 1.65
< p < 0.84 + 1.65
200
200
0.84 – 0.043 < p < 0.84 + 0.043
0.797< p < 0.883
Example 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’s study.
Find the 99% confidence interval of the
true proportion.
p̂ = 0.23
n = 200
q̂ = 1– 0.23 = 0.77

pˆ – za 2


ˆˆ
pq
< p < pˆ + za 2
n

ˆˆ
pq
n
(0.23)(0.77)
(0.23)(0.77)
0.23 – 2.58
< p < 0.23 + 2.58
200
200
0.23 – 0.077 < p < 0.23 + 0.077
0.153 < p < 0.307
A study by the University of Michigan found that one in
five 13- and 14-year-olds is a sometime smoker. Find the
99% confidence interval of the true proportion and
compare this with the University of
Michigan’s study.
0.153 < p < 0.307
1
5
= 0.20
The University of Michigan study
falls within this confidence interval.
Formula for Minimum Sample Size
Needed for Interval Estimate of a
Population Proportion
 za 2 
ˆ ˆ
n  pq

 E 
2
If necessary, round up to the next whole number.
Bluman, Chapter 7
81
Chapter 7
Confidence Intervals and
Sample Size
Section 7-3
Example 7-11
Page #378
Bluman, Chapter 7
82
Example 7-11: Home Computers
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.
2
2
 za 2 
 1.96 
ˆ ˆ
n  pq
   0.40  0.60   0.02   2304.96


 E 
The researcher should interview a sample of at least
2305 people.
Bluman, Chapter 7
83
Chapter 7
Confidence Intervals and
Sample Size
Section 7-3
Example 7-12
Page #378
Bluman, Chapter 7
84
Example 7-12: Car Phone Ownership
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.
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.
2
2
z
 a2
 1.65 
ˆ
ˆ
n  pq 
 272.25
   0.50  0.50  

 0.05 
 E 
The researcher should ask at least 273 executives.
Bluman, Chapter 7
85
Example 10
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?
a. How large should the sample size be?
p̂ = 0.25 q̂ = 0.75
 za /2 
n = p̂q̂ 

 E 
2
 2.58 
n =  0.25  0.75 
 .02 
n = 3120.187
or
2
n = 3121
b. If no estimate of the sample proportion is
available, how large should the sample be?
p̂ = 0.5
q̂ = 0.5
 za / 2 
ˆˆ
n = pq
 E 
n = 4160.25
2
2.58 
n =  0.5  0.5  
 0.02 
or
n = 4161
Since there is no prior knowledge of
p or q, assign the values
q̂ = 0.5 and p̂ = 0.5.
2
7-4 Confidence Intervals for
Variances and Standard Deviations



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.
Bluman, Chapter 7
89
Chi-Square Distributions

The chi-square distribution must be used to calculate
confidence intervals for variances and standard
deviations.

The chi-square variable is similar to the t variable in
that its distribution is a family of curves based on the
number of degrees of freedom.

2

The symbol for chi-square is (Greek letter chi,
pronounced “ki”).

A chi-square variable cannot be negative, and the
distributions are skewed to the right.
Bluman, Chapter 7
90
Chi-Square Distributions

At about 100 degrees of freedom, the chi-square
distribution becomes somewhat symmetric.

The area under each chi-square distribution is equal to
1.00, or 100%.
Bluman, Chapter 7
91
Formula for the Confidence Interval for
a Variance
 n  1 s 2   2   n  1 s 2 ,
2
2
 right
d.f. = n  1
left
Formula for the Confidence Interval for
a Standard Deviation
 n  1 s

2
right
2
 
 n  1 s

2
left
Bluman, Chapter 7
2
,
d.f. = n  1
92
Chapter 7
Confidence Intervals and
Sample Size
Section 7-4
Example 7-13
Page #385
Bluman, Chapter 7
93
Example 7-13: Using Table G
Find the values for 
interval when n = 25.
2
right
and
2
 left
for a 90% confidence
2

To find right, subtract 1 - 0.90 = 0.10.
Divide by 2 to
get 0.05. 2
To find  left , subtract 1 - 0.05 to get 0.95.
Bluman, Chapter 7
94
Example 7-13: Using Table G
Use the 0.95 and 0.05 columns and the row
corresponding to 24 d.f. in Table G.
2
2
The right
value is 36.415; the left
value is 13.848.
Bluman, Chapter 7
95
Confidence Interval for a Variance
or Standard Deviation
Rounding Rule
When you are computing a confidence interval for a
population variance or standard deviation by using raw
data, round off to one more decimal places than the
number of decimal places in the original data.
When you are computing a confidence interval for a
population variance or standard deviation by using a
sample variance or standard deviation, round off to the
same number of decimal places as given for the sample
variance or standard deviation.
Bluman, Chapter 7
96
Chapter 7
Confidence Intervals and
Sample Size
Section 7-4
Example 7-14
Page #387
Bluman, Chapter 7
97
Example 7-14: Nicotine Content
Find the 95% confidence interval for the variance and
standard deviation of the nicotine content of cigarettes
manufactured if a sample of 20 cigarettes has a standard
deviation of 1.6 milligrams.
To find 
get 0.025.
2
right,
subtract 1 - 0.95 = 0.05. Divide by 2 to
2

To find left , subtract 1 - 0.025 to get 0.975.
In Table G, the 0.025 and 0.975 columns with the d.f.
19 row yield values of 32.852 and 8.907, respectively.
Bluman, Chapter 7
98
Example 7-14: Nicotine Content
 n  1 s 2   2   n  1 s 2
2
2
 right
left
2
2
19 1.6    2  19 1.6 
32.852
8.907
1.5   2  5.5
You can be 95% confident that the true variance for the
nicotine content is between 1.5 and 5.5 milligrams.
1.5    5.5
1.2    2.3
You can be 95% confident that the true standard
deviation is between 1.2 and 2.3 milligrams.
Bluman, Chapter 7
99
Chapter 7
Confidence Intervals and
Sample Size
Section 7-4
Example 7-15
Page #387
Bluman, Chapter 7
100
Example 7-15: Cost of Ski Lift Tickets
Find the 90% confidence interval for the variance and
standard deviation for the price in dollars of an adult
single-day ski lift ticket. The data represent a selected
sample of nationwide ski resorts. Assume the variable is
normally distributed.
59 54 53 52 51
39 49 46 49 48
Using technology, we find the variance of the data is
s2=28.2.
In Table G, the 0.05 and 0.95 columns with the d.f. 9
row yield values of 16.919 and 3.325, respectively.
Bluman, Chapter 7
101
Example 7-15: Cost of Ski Lift Tickets
 n  1 s 2   2   n  1 s 2
2
2
 right
left
 9  28.2    2   9  28.2 
16.919
3.325
15.0   2  76.3
You can be 95% confident that the true variance for the
cost of ski lift tickets is between 15.0 and 76.3.
15.0    76.3
3.87    8.73
You can be 95% confident that the true standard
deviation is between $3.87 and $8.73.
Bluman, Chapter 7
102
Example 11
Find the 90% confidence interval for the variance and
standard deviation for the time it takes a customer to
place a telephone order with a large catalog company
is a sample of 23 telephone orders has a standard
deviation of 3.8 minutes. Assume the variable is
normally distributed.
2
2
2
2
(n – 1)s
2 (n – 1)s
< <
2
2
 right
 left
22 3.8 
33.924
2
< <
22 3.8 
2
12.338
9.36 <  < 25.75
3.06 <  < 5.07
Example 12
Find the 99% confidence interval for the variance and
standard deviation of the weights of 25 one-gallon
containers of motor oil if a sample of 14 containers has
a variance of 3.2. Assume the variable is normally
distributed.
Find the 99% confidence interval for the
variance and standard deviation.
n = 14
s 2 = 3.2
(n – 1)s 2
(n – 1)s 2
2
< <
2
2
right

left
13(3.2)
13(3.2)
2
<

<
29.819
3.565
1.4 <  2 < 11.7
1.2 <  < 3.4
Example 13
The number of calories in a 1-ounce serving of
various kinds of regular cheese is shown.
Estimate the population variance and standard
deviation with 90% confidence.
110
45
100
95
110
110
100
110
95
120
130
100
80
105
105
90
110
70
125
108
(n – 1)s 2
2
(n – 1)s 2
2
< <
2

right
left
2
19(19.1913) 2
19(19.1913)
< 2 <
30.144
10.117
2
232.1 <  < 691.7
15.2 <  < 26.3
Example 14
A service station advertises that customers will have to
wait no more than 30 minutes for an oil change. A
sample of 28 oil changes has a standard deviation of
5.2 minutes. Find the 95% confidence interval of the
population standard deviation of the time spent waiting
for an oil change.
(n – 1)s 2
2
(n – 1)s 2
2
< <
2
right

left
2
27(5.2)
27(5.2)2
< 2 <
43.194
14.573
16.9 <  2 < 50.1
4.1 <  < 7.1