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Chapter
9
Estimating the
Value of a
Parameter Using
Confidence
Intervals
© 2010 Pearson Prentice Hall. All rights reserved
Section
9.2
Confidence Intervals
about a Population
Mean When the
Population Standard
Deviation is
Unknown
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Objectives
1. Know the properties of Student’s
t-distribution
2. Determine t-values
3. Construct and interpret a confidence interval
for a population mean
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9-3
Objective 1
• Know the Properties of Student’s
t-Distribution
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9-4
Student’s t-Distribution
Suppose that a simple random sample of size n is
taken from a population. If the population
from which the sample is drawn follows a
normal distribution, the distribution of
x 
t
s
n
follows Student’s t-distribution with n-1
degrees of freedom where x is the sample
mean and s is the sample standard deviation.

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9-5
Parallel Example 1: Comparing the Standard Normal
Distribution to the t-Distribution Using Simulation
a)
Obtain 1,000 simple random samples of size n=5 from a
normal population with =50 and =10.
b)
Determine the sample mean and sample standard deviation
for each of the samples.
c)
d)
Compute z 
x 

n
and
x 
for each sample.
t
s
n
Draw a histogram for both z and t.


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9-6
Histogram for z
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9-7
Histogram for t
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CONCLUSIONS:
• The histogram for z is symmetric and bell-shaped
with the center of the distribution at 0 and virtually all
the rectangles between -3 and 3. In other words, z
follows a standard normal distribution.
• The histogram for t is also symmetric and bell-shaped
with the center of the distribution at 0, but the
distribution of t has longer tails (i.e., t is more
dispersed), so it is unlikely that t follows a standard
normal distribution. The additional spread in the
distribution of t can be attributed to the fact that we
use s to find t instead of . Because the sample
standard deviation is itself a random variable (rather
than a constant such as ), we have more dispersion
in the distribution of t.
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9-9
Properties of the t-Distribution
1. The t-distribution is different for different degrees of
freedom.
2. The t-distribution is centered at 0 and is symmetric
about 0.
3. The area under the curve is 1. The area under the
curve to the right of 0 equals the area under the curve
to the left of 0 equals 1/2.
4. As t increases without bound, the graph approaches,
but never equals, zero. As t decreases without bound,
the graph approaches, but never equals, zero.
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9-10
Properties of the t-Distribution
5. The area in the tails of the t-distribution is a little
greater than the area in the tails of the standard normal
distribution, because we are using s as an estimate of
, thereby introducing further variability into the tstatistic.
6. As the sample size n increases, the density curve of t
gets closer to the standard normal density curve. This
result occurs because, as the sample size n increases,
the values of s get closer to the values of , by the
Law of Large Numbers.
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9-11
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9-12
Objective 2
• Determine t-Values
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9-13
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9-14
Parallel Example 2: Finding t-values
Find the t-value such that the area under the tdistribution to the right of the t-value is 0.2 assuming
10 degrees of freedom. That is, find t0.20 with 10
degrees of freedom.
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9-15
Solution
The figure to the left
shows the graph of the
t-distribution with 10
degrees of freedom.
The unknown value of t is labeled, and the area under
the curve to the right of t is shaded. The value of t0.20
with 10 degrees of freedom is 0.8791.
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Objective 3
• Construct and Interpret a Confidence Interval
for a Population Mean
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Constructing a (1-)100% Confidence
Interval for ,  Unknown
Suppose that a simple random sample of size n is taken
from a population with unknown mean  and
unknown standard deviation . A (1-)100%
confidence interval for  is given by
Lower
bound:
s
x  t 
n
2
Upper
bound:
s
x  t 
n
2
Note: The interval is exact when the population is normally
distributed. It is approximately correct for nonnormal
populations, provided that n is large enough.


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9-18
Parallel Example 3: Constructing a Confidence Interval
about a Population Mean
The pasteurization process reduces the amount of
bacteria found in dairy products, such as milk. The
following data represent the counts of bacteria in
pasteurized milk (in CFU/mL) for a random sample of
12 pasteurized glasses of milk. Data courtesy of Dr.
Michael Lee, Professor, Joliet Junior College.
Construct a 95% confidence interval for the bacteria
count.
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NOTE: Each observation is in tens of thousand.
So, 9.06 represents 9.06 x 104.
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Solution: Checking Normality and
Existence of Outliers
Normal Probability Plot for CFU/ml
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Solution: Checking Normality and
Existence of Outliers
Boxplot of CFU/mL
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• x  6.41 and s  4.55
•   0.05, n  12, so t 0.05  2.201
2
Lower
bound:
4.55
6.41 2.201
 3.52
12
Upper
bound:

4.55
6.41 2.201
 9.30
12
The 95% confidence interval for the mean bacteria
count in pasteurized milk is (3.52, 9.30).

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9-23
Parallel Example 5: The Effect of Outliers
Suppose a student miscalculated the amount of
bacteria and recorded a result of 2.3 x 105. We
would include this value in the data set as 23.0.
What effect does this additional observation have on
the 95% confidence interval?
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Solution: Checking Normality and
Existence of Outliers
Boxplot of CFU/mL
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Solution
• x  7.69 and s  6.34
•   0.05, n 13, so t 0.05  2.179
2
Lower
bound:
6.34
7.69  2.179 
 3.86
13
6.34
Upper
7.69  2.179 
 11.52
bound:
13

The 95% confidence interval for the mean bacteria
count in pasteurized milk, including the outlier

is (3.86, 11.52).
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9-26
CONCLUSIONS:
• With the outlier, the sample mean is larger because
the sample mean is not resistant
• With the outlier, the sample standard deviation is
larger because the sample standard deviation is not
resistant
• Without the outlier, the width of the interval
decreased from 7.66 to 5.78.
Without
Outlier

With
Outlier
x
s
95% CI
6.41
4.55
(3.52, 9.30)
7.69
6.34
(3.86, 11.52)
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9-27