Transcript Confinence Intervals

Chapter 8
Confidence Intervals for
One Population Mean
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Definition 8.1
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Definition 8.2
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Example 8.3
Consider the prices of new mobile homes, 95.44% of all
samples of 36 new mobile homes have the property that
the interval from x – 2.4 to x + 2.4 contains μ.
To illustrate that the mean price, μ, of all new mobile
homes may or may not lie in the 95.44% confidence
interval obtained, we used a computer to simulate 20
samples of 36 new mobile home prices each. For the
simulation, we assumed that μ = 61 (i.e., $61 thousand)
and σ = 7.2 (i.e., $7.2 thousand). In reality, we don’t know
μ; we are assuming a value for μ to illustrate a point. For
each of the 20 samples of 36 new mobile home prices,
we did three things: computed the sample mean price, x ;
used Equation (8.1) to obtain the 95.44% confidence
interval; and noted whether the population mean, μ = 61,
actually lies in the confidence interval.
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Example 8.3
Figure 8.2 summarizes our results. For each sample, we
have drawn a graph on the right-hand side of Fig. 8.2.
The dot represents the sample mean, x , in thousands of
dollars, and the horizontal line represents the
corresponding 95.44% confidence interval. Note that the
population mean, μ, lies in the confidence interval only
when the horizontal line crosses the dashed line. Figure
8.2 reveals that μ lies in the 95.44% confidence interval in
19 of the 20 samples, that is, in 95% of the samples. If,
instead of 20 samples, we simulated 1000, we would
probably find that the percentage of those 1000 samples
for which μ lies in the 95.44% confidence interval would
be even closer to 95.44%. Hence we can be 95.44%
confident that any computed 95.44% confidence interval
will contain μ.
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Figure 8.2
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Confidence Intervals for One Population Mean When
σ is Known
If x is a normally distributed variable with mean μ and
standard deviation σ, then, for samples of size n, the
variable x is also normally distributed and has mean μ
and standard deviation s
n . And 95.44% of all
samples of size n have means within 2 × s n of μ, Fig.
8.3(a). We can say that 100(1 − α)% of all samples of
size n have means within
of μ, Fig. 8.3(b).
za 2 × s n
Figure 8.3
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Procedure 8.1
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Key Fact 8.1
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Key Fact 8.2
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Example 8.4
The U.S. Bureau of Labor
Statistics collects information on
the ages of people in the civilian
labor force and publishes the
results in Employment and
Earnings. Fifty people in the
civilian labor force are randomly
selected; their ages are
displayed in Table 8.3. Find a
95% confidence interval for the
mean age, μ, of all people in the
civilian labor force. Assume that
the population standard
deviation of the ages is 12.1
years.
Table 8.3
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Solution Example 8.4
Because the sample size is 50, which is large, and the
population standard deviation is known, we can use
Procedure 8.1 to find the required confidence interval.
Figure 8.4
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Definition 8.3
Figure 8.6
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Key Fact 8.4
Figure 8.6
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Formula 8.1
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Key Fact 8.5
Figure 8.6
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Key Fact 8.6
Figure 8.6
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Although there is a different t-curve for each number of
degrees of freedom, all t-curves are similar and resemble
the standard normal curve, as illustrated in Fig. 8.7. That
figure also illustrates the basic properties of t-curves,
listed in Key Fact 8.6. Note that Properties 1–3 of t-curves
are identical to those of the standard normal curve.
Figure 8.7
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Example 8.8
For a t-curve with 13 degrees of freedom, determine t 0.05 ;
that is, find the t-value having area 0.05 to its right, as
shown in Fig. 8.8(a).
Figure 8.8
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Solution Example 8.8
To find the t-value in question, we use Table IV, a portion
of which is given in Table 8.4. The number of degrees of
freedom is 13, so we first go down the out- side columns,
labeled df, to “13.” Then, going across that row to the
column labeled t0.05 , we reach 1.771. This number is the
t-value having
area 0.05 to its
right, as shown
in Fig. 8.8(b). In
other words, for
a t-curve with
df = 13,
t0.05 = 1.771.
Table 8.4
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Procedure 8.2
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Example 8.9
The U.S. Federal Bureau of Investigation (FBI) compiles
data on robbery and property crimes and publishes the
information in Population-at-Risk Rates and Selected
Crime Indicators. A simple random sample of pickpocket
offenses yielded the losses, in dollars, shown in Table 8.5.
Use the data to find a 95% confidence interval for the
mean loss, μ, of all pickpocket offenses.
Table 8.5
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Solution Example 8.9
Because the sample size, n = 25, is moderate, we first
need to consider questions of normality and outliers. To do
that, we constructed the normal probability plot in Fig. 8.9,
which reveals no outliers and falls roughly in a straight
line.
So, we can
apply
Procedure 8.2
to find the
confidence
interval.
Figure 8.9
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Solution Example 8.9
STEP 1 For a confidence level of 1 −α, use Table IV to
find tα/2 with df = n−1,where n is the sample size.
We want a 95% confidence interval, so α = 1−0.95 = 0.05.
For n = 25, we have df = 25−1 = 24. From Table IV,
tα/2 = t0.05/2 = t0.025 = 2.064
STEP 2 Applying the usual formulas for x and s to the
data in Table 8.5 gives x = 513.32 and s = 262.23. So a
95% confidence interval for μ is from
262.23
262.23
513.32 - 2.064
to 513.32 + 2.064
25
25
or 405.07 to 621.57.
STEP 3 We can be 95% confident that the mean loss of
all pickpocket offenses is somewhere between $405.07
and $621.57.
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Example 8.10
The U.S. Department of Agriculture publishes data on
chicken consumption in Food Consumption, Prices, and
Expenditures. Table 8.6 shows a year’s chicken
consumption, in pounds, for 17 randomly selected people.
Find a 90% confidence interval for the year’s mean
chicken consumption, μ.
Table 8.6
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Solution Example 8.10
A normal probability plot of the data, shown in Fig.8.10(a),
reveals an outlier (0 lb). Because the sample size is only
moderate, applying Procedure 8.2 here is inappropriate.
Figure 8.10
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Solution Example 8.10
The outlier of 0 lb might be are cording error or it might
reflect a person in the sample who does not eat chicken
(e.g., a vegetarian). If were move the outlier from the data,
the normal probability plot for the abridged data shows no
outliers and is roughly linear, as seen in Fig.8.10(b).
Thus, if we are willing to take as our population only
people who eat chicken, we can use Procedure 8.2 to
obtain a confidence interval. Doing so yields a 90%
confidence interval of 51.2 to 59.2.
We can be 90% confident that the year’s mean chicken
consumption, among people who eat chicken, is
somewhere between 51.2 lb and 59.2 lb.
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Homework from chapter 8
8.21, 27, 33
8.57, 58, 62
8.78, 94 (Use Minitab)
Due Wed 9/12
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