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  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.
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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
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
(n
In Symbols
.
x 
x
n
2. Specify , if known. Otherwise, if n  30,
find the sample standard deviation s and
use it as an estimate for .
( x  x )2
s
n 1
3. Find the critical value zc that corresponds
to the given level of confidence.
Use the Standard
Normal Table.
E  zc σ
4. Find the margin of error E.
n
Left endpoint: E
5. Find the left and right endpoints and
Right endpoint: +E
form the confidence interval.
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
σ  1.645  0.15
 0.05
zc = 1.645 E  z c
n
25
2.81 < μ < 2.91
± E = 2.86 ± 0.05
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
 zc 
n
.

 E 
2
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
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.
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