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A confidence interval for an unknown
parameter consists of an interval of
numbers based on a point estimate.
The level of confidence represents the
expected proportion of intervals that will
contain the parameter if a large number of
different samples is obtained. The level of
confidence is denoted (1 – α)·100%.
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For example, a 95% level of confidence
(α = 0.05) implies that if 100 different
confidence intervals are constructed, each
based on a different sample from the same
population, we will expect 95 of the intervals
to contain the parameter and 5 not to include
the parameter.
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Section
9.2
Estimating a
Population Mean
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A point estimate is the value of a statistic
that estimates the value of a parameter.
For example, the sample mean, x , is a
point estimate of the population mean μ.
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Estimation of a Population Mean, m
100(1-)% Confidence Interval for a Population Mean m using a Large Sample (n>=30)
x  z / 2

where
x

if
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n
is the sample mean in a simple random sample and
is the population standard deviation
is unknown, s may be used instead as an approximation.
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Parallel Example 1: Computing a Point Estimate
Pennies minted after 1982 are made from 97.5% zinc and 2.5%
copper. The following data represent the weights (in grams) of
17 randomly selected pennies minted after 1982.
2.46 2.47 2.49 2.48 2.50 2.44 2.46 2.45 2.49
2.47 2.45 2.46 2.45 2.46 2.47 2.44 2.45
Treat the data as a simple random sample. Estimate the
population mean weight of pennies minted after 1982.
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Solution
The sample mean is
2.46  2.47 
x
17
 2.45
 2.464
The point estimate of μ is 2.464 grams.
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• 1) Suppose 40 teachers from Washington have been randomly
selected and their salaries recorded. The resulting sample average was
$29,000 and the standard deviation was $2,000.
• Construct a 90, 95, and 99% confidence interval for μ, the mean
salary for the population of teachers in Washington. What do you
learn about the width of the intervals as you increase the confidence
level?
• Suppose our sample size is 80 instead of 60. Construct a 95%
confidence interval for μ.
• Now suppose our sample size is 100. Construct a 95% confidence
interval for μ. What do you learn as you increase the sample size?
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• If we have a sample of size smaller than 30,
we use student’s t distribution
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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m
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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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 μ = 50 and σ = 10.
b)
Determine the sample mean and sample standard deviation
for each of the samples.
Compute z 
c)
d)
x m

n
and
x m
for each sample.
t
s
n
Draw a histogram for both z and t.

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
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Histogram for z
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Histogram for t
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CONCLUSION:
• 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.
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CONCLUSION (continued):
• 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 σ. Because the sample
standard deviation is itself a random variable (rather
than a constant such as σ), we have more dispersion
in the distribution of t.
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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, which equals 1/2.
4. As t increases or decreases without bound, the graph
approaches, but never equals, zero.
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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 σ,
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 σ, by the Law
of Large Numbers.
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Objective 3
• Determine t-Values
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Parallel Example 2: Finding t-values
Find the t-value such that the area under the t-distribution
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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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 0.879.
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Objective 4
• Construct and Interpret a Confidence Interval
for the Population Mean
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Constructing a (1 – α)100% Confidence
Interval for μ
A (1 – α)·100% confidence interval for μ is given by
Lower
bound:
where
s
x  t 
n
2
t
Upper
bound:
s
x  t 
n
2
is the critical value with n – 1 df.
2
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Parallel Example 2: Using Simulation to Demonstrate the
Idea of a Confidence Interval
We will use Minitab to simulate obtaining 30 simple
random samples of size n = 8 from a population that is
normally distributed with μ = 50 and σ = 10. Construct
a 95% confidence interval for each sample. How many
of the samples result in intervals that contain μ = 50 ?
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Sample
C1
C2
C3
C4
C5
C6
C7
C8
C9
C10
C11
C12
C13
C14
C15
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Mean
47.07
49.33
50.62
47.91
44.31
51.50
52.47
59.62
43.49
55.45
50.08
56.37
49.05
47.34
50.33
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
95.0% CI
40.14,
54.00)
42.40,
56.26)
43.69,
57.54)
40.98,
54.84)
37.38,
51.24)
44.57,
58.43)
45.54,
59.40)
52.69,
66.54)
36.56,
50.42)
48.52,
62.38)
43.15,
57.01)
49.44,
63.30)
42.12,
55.98)
40.41,
54.27)
43.40,
57.25)
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SAMPLE
C16
C17
C18
C19
C20
C21
C22
C23
C24
C25
C26
C27
C28
C29
C30
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MEAN
44.81
51.05
43.91
46.50
49.79
48.75
51.27
47.80
56.60
47.70
51.58
47.37
61.42
46.89
51.92
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
95%
37.88,
44.12,
36.98,
39.57,
42.86,
41.82,
44.34,
40.87,
49.67,
40.77,
44.65,
40.44,
54.49,
39.96,
44.99,
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CI
51.74)
57.98)
50.84)
53.43)
56.72)
55.68)
58.20)
54.73)
63.52)
54.63)
58.51)
54.30)
68.35)
53.82)
58.85)
Note that 28 out of 30, or 93%, of the confidence
intervals contain the population mean μ = 50.
In general, for a 95% confidence interval, any
sample mean that lies within 1.96 standard errors
of the population mean will result in a
confidence interval that contains μ.
Whether a confidence interval contains μ
depends solely on the sample mean, x .
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Parallel Example 3: Constructing a Confidence Interval
Construct a 99% confidence interval about the
population mean weight (in grams) of pennies minted
after 1982. Assume μ = 0.02 grams.
2.46 2.47 2.49 2.48 2.50 2.44 2.46 2.45 2.49
2.47 2.45 2.46 2.45 2.46 2.47 2.44 2.45
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• t 2  1.746
• Lower bound:
s = 2.464 – 1.746
x  t 2 
n
0.02




17 
= 2.464 – 0.008 = 2.456
• Upper bound:

s = 2.464 + 2.575
x  t 2 
n
0.02




17 
= 2.464 + 0.008 = 2.472
We are 99% confident that the mean weight of pennies

minted after 1982 is between 2.456 and 2.472 grams.
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For a fixed sample size, as the confidence level
increases, the margin of error increases.
As the sample size increases, the margin of error
decreases.
So, how to decide the sample size for a fixed
margin of error and confidence level ??
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Determining the Sample Size n
The sample size required to estimate the
population mean, µ, with a level of confidence
(1– α)·100% with a specified margin of error, E,
is given by
 z  s 
2


n
 E 


2
where n is rounded up to the nearest whole
number.
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Parallel Example 7:
Determining the Sample Size
Back to the pennies. How large a sample would be
required to estimate the mean weight of a penny
manufactured after 1982 within 0.005 grams with 99%
confidence? Assume  = 0.02.
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• z 2  z0.005  2.575
• s = 0.02
• E = 0.005
2
 z  s 
2
2.575(0.02)


2
 
n
 106.09

 0.005 
 E 


Rounding up, we find n = 107.
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