Chapter 6
Confidence Intervals
6.1 Confidence Intervals for the mean
(Large Samples)
Lesson Objectives
• Find a point estimate and a margin of error
• Construct and interpret confidence intervals for
the population mean
• Determine the minimum sample size for
estimating the population mean
Point Estimate for Population μ
A point estimate is a single value estimate for a population parameter. The most
unbiased point estimate of the population mean, , is the sample mean, .
x
Example:
A random sample of 32 textbook prices (rounded to the nearest dollar) is taken from a
mean, .
local college bookstore. Find a point estimate for the population

34
56
79
94
34
65
86
95
38
65
87
96
45
66
87
98
45
67
87
98
45
67
88
101
45
68
90
110
54
74
90
121
x  74.22
The point estimate for the population mean of textbooks in the bookstore is $74.22.
Try-it-yourself #1
• Calculator steps
Interval Estimate
An interval estimate is an interval, or range of values, used to estimate a population
parameter.
Point estimate for
textbooks
•
74.22
interval estimate
How confident do we want to be that the interval estimate contains the population
mean, μ?
Level of Confidence
The level of confidence c is the probability that the interval estimate contains the
population parameter.
c
(1 – c)
c is the area beneath the
normal curve between the
critical values.
(1 – c)
zc
z=0
Critical values
The remaining area in the tails is 1 – c .
zc
z
Use the Standard Normal
Table to find the
corresponding z-scores.
Common Levels of Confidence
If the level of confidence is 90%, this means that we are 90% confident that the interval
contains the population mean, μ.
0.90
0.05
0.05
zc = 1.645
zc
z
z=0
The corresponding z-scores are ± 1.645.
zc = z1.645
c
Common Levels of Confidence
If the level of confidence is 95%, this means that we are 95% confident that the interval
contains the population mean, μ.
0.95
0.025
0.025
zc = z1.96
c
z
z=0
The corresponding z-scores are ± 1.96.
zc =z1.96
c
Common Levels of Confidence
If the level of confidence is 99%, this means that we are 99% confident that the interval
contains the population mean, μ.
0.99
0.005
0.005
zc = 2.575
zc
z
z=0
The corresponding z-scores are ± 2.575.
zc = z2.575
c
Margin of Error
The difference between the point estimate and the actual population parameter value is
called the sampling error.
x
When μ is estimated, the sampling error is the difference
. Since μ is usually
unknown, the maximum value for the error can be calculated using the level of
confidence.
Given a level of confidence, the margin of error (sometimes called the maximum error of
estimate or error tolerance) E is the greatest possible distance between the point estimate
and the value of the parameter it is estimating.

E  z cσ x  z c σ
n
When n  30, the sample standard
deviation, s, can be used for .
Margin of Error
Example:
A random sample of 32 textbook prices is taken from a local college bookstore. The mean
of the sample is
= 74.22, and the sample standard deviation is s = 23.44.
Use a 95% confidence level and find the margin of error for the mean price of all textbooks
in the bookstore.
x

E  z c σ  1.96  23.44
n
32
Since n  30, s can be substituted for σ.
 8.12
We are 95% confident that the margin of error for the population mean (all the textbooks
in the bookstore) is about $8.12.
Try-it-yourself #2
Confidence Intervals for μ
A c-confidence interval for the population mean μ is
E<μ<
+ E.
The probability that the confidence interval contains μ is c.
Example:
A random sample of 32 textbook prices is taken from a local college bookstore. The
mean of the sample is = 74.22, the sample standard deviation is s = 23.44, and the
margin of error is E = 8.12.
Construct a 95% confidence interval for the mean price of all textbooks in the bookstore.
Continued.
Confidence Intervals for μ
Example continued:
Construct a 95% confidence interval for the mean price of all textbooks in the bookstore.
s = 23.44
= 74.22
E = 8.12
Left endpoint = ?
•
 E = 74.22 – 8.12
= 66.1
Right endpoint = ?
•
= 74.22
•
+ E = 74.22 + 8.12
= 82.34
With 95% confidence we can say that the cost for all textbooks in the bookstore is
between $66.10 and $82.34.
Finding Confidence Intervals for μ
Finding a Confidence Interval for a Population Mean
normally distributed population)
In Words
(n  30 or σ known with a
In Symbols
1. Find the sample statistics n and
.
2. Specify , if known. Otherwise, if n  30, find the
sample standard deviation s and use it as an
estimate for .
3. Find the critical value zc that corresponds to the
given level of confidence.
x 
x
n
( x  x )2
s
n 1
Use the Standard
Normal Table.
E  zc σ
4. Find the margin of error E.
n
Left endpoint: E Right
5. Find the left and right endpoints and form theendpoint: +E Interval:
confidence interval.
E < μ < +E
Confidence Intervals for μ ( Known)
Example:
A random sample of 25 students had a grade point average with a mean of 2.86. Past
studies have shown that the standard deviation is 0.15 and the population is normally
distributed.
Construct a 90% confidence interval for the population mean grade point average.
n = 25
zc = 1.645
= 2.86
 = 0.15
E  z c σ  1.645  0.15  0.05
n
25
± E = 2.86 ± 0.05
2.81 < μ < 2.91
With 90% confidence we can say that the mean grade point
students in the population is between 2.81 and 2.91.
average for all
Try-it-yourself #3
Try-it-yourself #4
Using a calculator
Sample Size
Given a c-confidence level and a maximum error of estimate, E, the minimum sample
size n, needed to estimate , the population mean, is
If  is unknown, you can estimate it using s provided you have a preliminary sample with
at least 30 members.
 zc 
n
.

 E 
2
Example:
You want to estimate the mean price of all the textbooks in the college bookstore. How
many books must be included in your sample if you want to be 99% confident that the
sample mean is within $5 of the population mean?
Continued.
Sample Size
Example continued:
You want to estimate the mean price of all the textbooks in the college bookstore. How
many books must be included in your sample if you want to be 99% confident that the
sample mean is within $5 of the population mean?
  s = 23.44
= 74.22
zc = 2.575
 zc   2.575  23.44 
n



E
5


 
 145.7 (Always round up.)
2
2
You should include at least 146 books in your sample.
6.2 Confidence intervals for the mean
(small Samples)
Lesson Objectives
• Interpret and use the t-distribution table
• Construct confidence intervals when n<30
The t-Distribution
When a sample size is less than 30, and the random variable x is
approximately normally distributed, it follow a t-distribution.
x 
t
s
n
Properties of the t-distribution
1. The t-distribution is bell shaped and symmetric about the mean.
2. The t-distribution is a family of curves, each determined by a parameter
called the degrees of freedom. The degrees of freedom are the number of
free choices left after a sample statistic such as is calculated. When you
use a t-distribution to estimate a population mean, the degrees of freedom
are equal to one less than the sample size.
d.f. = n – 1 Degrees of freedom
x

Continued.
The t-Distribution
3.
4.
5.
The total area under a t-curve is 1 or 100%.
The mean, median, and mode of the t-distribution are equal to zero.
As the degrees of freedom increase, the t-distribution approaches the normal
distribution. After 30 d.f., the t-distribution is very close to the standard
normal z-distribution.
The tails in the t-distribution are
“thicker” than those in the
standard normal distribution.
d.f. = 2
d.f. = 5
t
0
Standard normal
curve
Critical Values of t
Example:
Find the critical value tc for a 95% confidence when the sample size is 5.
Appendix B: Table 5: t-Distribution
Level of
confidence, c
One tail, 
d.f. Two tails, 
1
2
3
4
5
0.50
0.25
0.50
1.000
.816
.765
.741
.727
0.80
0.10
0.20
3.078
1.886
1.638
1.533
1.476
0.90
0.95
0.98
0.05
0.025
0.01
0.10
0.05
0.02
6.314 12.706 31.821
2.920 4.303 6.965
2.353 3.182 4.541
2.132 2.776 3.747
2.015 2.571 3.365
d.f. = n – 1 = 5 – 1 = 4
c = 0.95
tc = 2.776
Continued.
Critical Values of t
Example continued:
Find the critical value tc for a 95% confidence when the sample size is 5.
95% of the area under the t-distribution curve with 4 degrees of freedom lies
between t = ±2.776.
c = 0.95
tc =  2.776
t
tc = 2.776
Try-it-yourself #1
Confidence Intervals and t-Distributions
Constructing a Confidence Interval for the Mean:
In Words
t-Distribution
In Symbols
1. Identify the sample statistics n,
and s.
x,
2. Identify the degrees of freedom, the
level of confidence c, and the critical
value tc.

3. Find the margin of error E.
4. Find the left and right endpoints and
form the confidence interval.
x 
x
n
( x  x )2
s
n 1
d.f. = n – 1
E  tc s
n
xE xE
Constructing a Confidence Interval
Example:
In a random sample of 20 customers at a local fast food restaurant, the mean waiting time
to order is 95 seconds, and the standard deviation is 21 seconds. Assume the wait times
are normally distributed and construct a 90% confidence interval for the mean wait time of

all customers.
n = 20
d.f. = 19
x  95
tc = 1.729
x  E  95  8.1
s = 21
21  8.1
E  tc s  1.729 
20
n
86.9 < μ < 103.1
We are 90% confident that the mean wait time for all customers is between 86.9 and
103.1 seconds.
Try-it-yourself #2
Try-it-yourself #3
Steps for using the calculator
Normal or t-Distribution?
Use the normal distribution with
Is n  30?
Yes
E  zc σ .
n
If  is unknown, use s instead.
No
Is the population normally, or
approximately normally,
distributed?
No
You cannot use the normal distribution
or the t-distribution.
Yes
Use the normal distribution with
Is  known?
No
Use the t-distribution with
E  tc s
n
and n – 1 degrees of freedom.
Yes
E  zc σ .
n
Try-it-yourself #4
Normal or t-Distribution?
Example:
Determine whether to use the normal distribution, the
t-distribution, or neither.
a.) n = 50, the distribution is skewed, s = 2.5
The normal distribution would be used because the sample size is 50.
b.) n = 25, the distribution is skewed, s = 52.9
Neither distribution would be used because n < 30 and the distribution is skewed.
c.) n = 25, the distribution is normal,  = 4.12
The normal distribution would be used because although n < 30, the population
standard deviation is known.
§ 6.3
Confidence Intervals for
Population Proportions
Point Estimate for Population p
The probability of success in a single trial of a binomial experiment is p. This probability
is a population proportion.
The point estimate for p, the population proportion of successes, is given by the
proportion of successes in a sample and is denoted by
where x is the number
of successes in the sample and n is the number in the sample.
x
pˆ  for the proportion of failures is = 1 – The symbols and are
The point estimate
n
read as “p hat” and “q hat.”
qˆ
p̂.
p̂
qˆ
Point Estimate for Population p
Example:
In a survey of 1250 US adults, 450 of them said that their favorite sport to watch is
baseball. Find a point estimate for the population proportion of US adults who say their
favorite sport to watch is baseball.
n = 1250
x = 450
pˆ  x  450  0.36
n 1250
The point estimate for the proportion of US adults who say baseball is their favorite
sport to watch is 0.36, or 36%.
Confidence Intervals for p
A c-confidence interval for the population proportion p is
pˆ  E  p  pˆ  E
where
pq
ˆ
ˆ
E  zc
.
The probability that thenconfidence interval contains p is c.
Example:
Construct a 90% confidence interval for the proportion of US adults who say baseball is
their favorite sport to watch.
n = 1250
x = 450
p̂  0.36
Continued.
Confidence Intervals for p
Example continued:
n = 1250
p̂  0.36
x = 450
ˆˆ
E  z c pq
n
qˆ  0.64
Left endpoint = ?
•
p̂  E  0.36  0.022
 0.338
 1.645
(0.36)(0.64)  0.022
1250
Right endpoint = ?
•
p̂  0.36
•
p̂  E  0.36  0.022
 0.382
With 90% confidence we can say that the proportion of all US adults who say baseball is
their favorite sport to watch is between 33.8% and 38.2%.
Finding Confidence Intervals for p
Constructing a Confidence Interval for a Population Proportion
In Words
In Symbols
1. Identify the sample statistics n and x.
x
p

ˆ
2. Find the point estimate
p̂.
n
3. Verify that the sampling distribution can be
approximated by the normal distribution.
4. Find the critical value zc that corresponds to
the given level of confidence.
5. Find the margin of error E.
6. Find the left and right endpoints and form
the confidence interval.
npˆ  5, nqˆ  5
Use the Standard
Normal Table.
ˆˆ
E  z c pq
n
Left endpoint: p̂  E
Right endpoint: Interval:
p̂  E
pˆ  E  p  pˆ  E
Sample Size
Given a c-confidence level and a margin of error, E, the minimum sample size n, needed
to estimate p is
2
 zc 
ˆ ˆ  for and
n  pq
.
This formula assumes you have an estimate
E
If not, use
and
pˆ  0.5
p̂
qˆ.
qˆ  0.5.
Example:
You wish to find out, with 95% confidence and within 2% of the true population, the
proportion of US adults who say that baseball is their favorite sport to watch.
Continued.
Sample Size
Example continued:
You wish to find out, with 95% confidence and within 2% of the true population, the
proportion of US adults who say that baseball is their favorite sport to watch.
n = 1250
x = 450
2
p̂  0.36
1.96 
z 
ˆ ˆ  c   (0.36)(0.64) 
n  pq

 0.02 
E
2
 2212.8 (Always round up.)
You should sample at least 2213 adults to be 95% confident.
§ 6.4
Confidence Intervals for
Variance and Standard
Deviation
The Chi-Square Distribution
The point estimate for 2 is s2, and the point estimate for  is s. s2 is the most unbiased
estimate for 2.
You can use the chi-square distribution to construct a confidence interval for the variance
and standard deviation.
If the random variable x has a normal distribution, then the distribution of
2
(
n

1)
s
2
forms a chi-square distribution for samples of any size
 
2
σ
n > 1.
The Chi-Square Distribution
Four properties of the chi-square distribution are as follows.
1. All chi-square values χ2 are greater than or equal to zero.
2. The chi-square distribution is a family of curves, each determined by
the degrees of freedom. To form a confidence interval for 2, use the
χ2-distribution with degrees of freedom. To form a confidence
interval for 2, use the
χ2-distribution with degrees of freedom
equal to one less than the sample size.
3. The area under each curve of the chi-square distribution equals one.
4. Find the critical value zc that corresponds to the given level of
confidence.
5. Chi-square distributions are positively skewed.
Critical Values for X2
There are two critical values for each level of confidence. The value χ2R represents the
right-tail critical value and χ2L represents the left-tail critical value.
1
1c
2
X2R
X2
1 2 c   1 2 c
X2
X2L
Area to the right of X2R
Area to the right of X2L
1c
2
X2L
c
1c
2
X2R
The area between the left and right
critical values is c.
X2
Critical Values for X2
Example:
Find the critical values χ2R and χ2L for an 80% confidence when the sample size is 18.
Because the sample size is 18, there are
freedom,
d.f. = n – 1 = 18 – 1 = 17 degrees of
Area to the right of χ2R =
1  c 1  0.8

 0.1
2
2
Area to the right of χ2L =
1  c 1  0.8

 0.9
2
2
Use the Chi-square distribution table to find the critical values.
Continued.
Critical Values for X2
Example continued:
Appendix B: Table 6: χ2-Distribution
Degrees of
freedom
1
2
3
16
17
18
χ2R = 24.769
χ2L = 10.085
0.010
0.072
0.975
- 0.001
0.020 0.051
0.115 0.216

0.95
0.004
0.103
0.352
5.142
5.697
6.265
5.812
6.408
7.015
7.962 9.312 23.542 26.296
8.672 10.085 24.769 27.587
9.390 10.865 25.989 28.869
0.995
0.99
6.908
7.564
8.231
0.90
0.016
0.211
0.584
0.10
2.706
4.605
6.251
0.05
3.841
5.991
7.815
Confidence Intervals for 2 and 
A c-confidence interval for a population variance and standard deviation is as follows.
Confidence Interval for 2:
(n  1)s 2
2
XR
2
σ 
(n  1)s 2
X L2
Confidence Interval for :
(n  1)s 2
2
XR
σ 
(n  1)s 2
X L2
The probability that the confidence intervals contain 2 or  is c.
Confidence Intervals for 2 and 
Constructing a Confidence Interval for a Variance and a Standard
Deviation
In Words
In Symbols
1. Verify that the population has a normal
distribution.
2. Identify the sample statistic n and the
degrees of freedom.
d.f. = n  1
3. Find the point estimate s2.
(x  x )2
s 
n 1
4. Find the critical value χ2R and χ2L that
correspond to the given level of
confidence.
2
Use Table 6 in
Appendix B.
Continued.
Confidence Intervals for 2 and 
Constructing a Confidence Interval for a Variance and a Standard
Deviation
In Words
In Symbols
5. Find the left and right endpoints and
form the confidence interval.
6. Find the confidence interval for the
population standard deviation by
taking the square root of each
endpoint.
(n  1)s 2
2
XR
(n  1)s 2
2
XR
2
σ 
σ 
(n  1)s 2
X L2
(n  1)s 2
X L2
Constructing a Confidence Interval
Example:
You randomly select and weigh 41 samples of 16-ounce bags of potato chips. The sample
standard deviation is 0.05 ounce. Assuming the weights are normally distributed, construct a
90% confidence interval for the population standard deviation.
d.f. = n – 1 = 41 – 1 = 40 degrees of freedom,
Area to the right of χ2R =
Area to the right of χ2L =
1  c 1  0.9

 0.05
2
2
1  c 1  0.9

 0.95
2
2
The critical values are χ2R = 55.758 and χ2L = 26.509.
Continued.
Constructing a Confidence Interval
Example continued:
χ2L = 26.509
χ2R = 55.758
Left endpoint = ?
(n  1)s 2
 R2
Right endpoint = ?
•
(41  1)(0.05)2

55.758
 0.04
(n  1)s 2
 L2
•
(41  1)(0.05)2

26.509
 0.06
0.04  σ  0.06
With 90% confidence we can say that the population standard deviation is between 0.04
and 0.06 ounces.