Transcript Chapter 6: Confidence Intervals

Chapter 6
Confidence Intervals
§ 6.1
Confidence Intervals
for the Mean
(Large Samples)
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, .
Example:
A random sample of 32 textbook prices (rounded to the nearest
dollar) is taken from a local college bookstore. Find a point
estimate for the population mean, .
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
 74.22
The point estimate for the population mean of textbooks
in the bookstore is $74.22.
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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?
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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
z
zc
Critical values
Use the Standard
Normal Table to find the
corresponding z-scores.
The remaining area in the tails is 1 – c .
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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 = z1.645
z = 0 zc = z1.645
c
c
z
The corresponding z-scores are ± 1.645.
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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=0
zc =z1.96
c
z
The corresponding z-scores are ± 1.96.
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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 = z2.575
z = 0 zc = z2.575
c
c
z
The corresponding z-scores are ± 2.575.
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Margin of Error
The difference between the point estimate and the actual
population parameter value is called the sampling error.
When μ is estimated, the sampling error is the difference
μ – . 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 .
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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.
E  zc
σ
n
 1 .9 6 
 8 .1 2
2 3 .4 4
32
Since n  30, s can be
substituted for σ.
We are 95% confident that the margin of error for the
population mean (all the textbooks in the bookstore) is
about $8.12.
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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.
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Confidence Intervals for μ
Example continued:
Construct a 95% confidence interval for the mean price of
all textbooks in the bookstore.
= 74.22
s = 23.44
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.
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Finding Confidence Intervals for μ
(n
Finding a Confidence Interval for a Population Mean
 30 or σ known with a normally distributed population)
In Words
1. Find the sample statistics n and
In Symbols
.
x 
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.
4. Find the margin of error E.
5. Find the left and right endpoints and
form the confidence interval.
s 
x
n
( x  x )
n 1
2
Use the Standard
Normal Table.
E  zc
σ
n
Left endpoint: E
Right endpoint: +E
Interval: E < μ <
+E
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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.
= 2.86
 = 0.15
n = 25
zc = 1.645
E  zc
± E = 2.86 ± 0.05
σ
n
 1.645 
0.15
25
 0.05
2.81 < μ < 2.91
With 90% confidence we can say that the mean grade point
average for all students in the population is between 2.81 and 2.91.
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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
2
 z c 
n
 .
 E 
If  is unknown, you can estimate it using s provided you
have a preliminary sample with at least 30 members.
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.
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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
2
zc = 2.575
2
 z c 
 2.575  23.44 
n  

  
E
5




 1 4 5 .7 (Always round up.)
You should include at least 146 books in your sample.
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§ 6.2
Confidence Intervals
for the Mean
(Small Samples)
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.
t 
x  μ
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
Continued.
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The t-Distribution
3. The total area under a t-curve is 1 or 100%.
4. The mean, median, and mode of the t-distribution are equal to zero.
5. 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
0
t
Standard
normal curve
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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
d.f. = n – 1 = 5 – 1 = 4
tc = 2.776
c = 0.95
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
Continued.
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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
tc = 2.776
t
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Confidence Intervals and t-Distributions
t-
Constructing a Confidence Interval for the Mean:
Distribution
In Words
1. Identify the sample statistics n,
, and s.
2. Identify the degrees of freedom,
the level of confidence c, and
the critical value tc.
In Symbols
x 
x
s 
n
( x  x )
n 1
2
d.f. = n – 1
s
3. Find the margin of error E.
E  tc
4. Find the left and right
endpoints and form the
confidence interval.
Left endpoint: E
Right endpoint: +E
Interval: E < μ <
+E
n
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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.
= 95 s = 21
n = 20
d.f. = 19
tc = 1.729
± E = 95 ± 8.1
E  tc
s
n
 1.729 
21
20
 8.1
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.
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Normal or t-Distribution?
Use the normal distribution with
Is n  30?
Yes
No
Yes
Is  known?
No
σ
n
.
If  is unknown, use s instead.
No
Is the population normally, or
approximately normally,
distributed?
E  zc
You cannot use the normal
distribution or the t-distribution.
Use the normal distribution with
Yes
E  zc
σ
n
.
Use the t-distribution with
E  tc
s
n
and n – 1 degrees of freedom.
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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.
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§ 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
x
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
proportion of failures is qˆ = 1 – pˆ . The symbols pˆ and qˆ
are read as “p hat” and “q hat.”
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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
pˆ 
x = 450
x
450

 0 .3 6
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%.
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Confidence Intervals for p
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.
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
pˆ  0.36
Continued.
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Confidence Intervals for p
Example continued:
n = 1250
x = 450
pˆ  0.36
E  zc
qˆ  0.64
Left endpoint = ?
•
pˆ qˆ
n
 1.645
(0.36)(0.64 )
 0.022
1250
Right endpoint = ?
•
pˆ  0.36
•
pˆ  E  0.36  0.022
pˆ  E  0.36  0.022
 0.338
 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%.
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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
n
2. Find the point estimate pˆ .
pˆ 
3. Verify that the sampling distribution
can be approximated by the normal
distribution.
npˆ  5, nqˆ  5
4. Find the critical value zc that
corresponds to the given level of
confidence.
Use the Standard
Normal Table.
5. Find the margin of error E.
E  zc
6. Find the left and right endpoints and
form the confidence interval.
Left endpoint: pˆ  E
Right endpoint: pˆ  E
Interval: pˆ  E  p  pˆ
pˆ qˆ
n
E
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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 
ˆˆ
n  pq
 .
 E 
This formula assumes you have an estimate for pˆ and qˆ .
If not, use pˆ  0.5 and 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.
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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
pˆ  0.36
z 
 1.96 
ˆ ˆ  c   (0.36)(0.64 ) 
n  pq

 0.02 
 E 
 2 2 1 2 .8
2
(Always round up.)
You should sample at least 2213 adults to be 95% confident.
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§ 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
σ
2
2
forms a chi-square distribution for samples of any size
n > 1.
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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
χ2distribution 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.
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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
X2
X2L
Area to the right of X2R
1 2 c   1 2 c
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
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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
d.f. = n – 1 = 18 – 1 = 17 degrees of freedom,
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.
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Critical Values for X2
Example continued:
Appendix B: Table 6: χ2-Distribution
Degrees of
freedom
1
2
3
16
17
18
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
χ2R = 24.769
χ2L = 10.085
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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
σ
2

( n  1)s
2
2
XL
Confidence Interval for :
( n  1)s
2
XR
2
σ 
( n  1)s
2
2
XL
The probability that the confidence intervals contain
2 or  is c.
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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.
3. Find the point estimate s2.
4. Find the critical value χ2R and χ2L
that correspond to the given level of
confidence.
d .f. = n  1
s
2
(x  x )

n 1
2
Use Table 6 in
Appendix B.
Continued.
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Confidence Intervals for 2 and 
Constructing a Confidence Interval for a Variance and a Standard
Deviation
In Words
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.
In Symbols
( n  1)s
2
2
XR
( n  1)s
2
XR
2
σ
2

σ 
( n  1)s
2
2
XL
( n  1)s
2
2
XL
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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 = 1  c  1  0 .9  0 .0 5
2
2
Area to the right of χ2L = 1  c  1  0 .9  0 .9 5
2
2
The critical values are χ2R = 55.758 and χ2L = 26.509.
Continued.
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Constructing a Confidence Interval
Example continued:
χ2L = 26.509
χ2R = 55.758
Left endpoint = ?
Right endpoint = ?
•
•
( n  1)s
2
R
2

(41  1)(0.05)
55.758
2
 0.04
( n  1)s
2
L
2

(41  1)(0.05)
26.509
2
 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.
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