Transcript Chapter Estimating the Value of a Parameter Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc.

Chapter
9
Estimating the
Value of a
Parameter
Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc.
Section
9.1
Estimating a
Population
Proportion
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Objectives
1. Obtain a point estimate for the population
proportion
2. Construct and interpret a confidence interval
for the population proportion
3. Determine the sample size necessary for
estimating the population proportion within a
specified margin of error
9-3
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Objective 1
• Obtain a Point Estimate for the Population
Proportion
9-4
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A point estimate is the value of a statistic
that estimates the value of a parameter.
For example, the point estimate for the
x
population proportion is pˆ  , where x is
n
the number of individuals in the sample
with a specified characteristic and n is the
sample size.
9-5
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Parallel Example 1: Calculating a Point Estimate for the
Population Proportion
In July of 2008, a Quinnipiac University Poll asked
1783 registered voters nationwide whether they
favored or opposed the death penalty for persons
convicted of murder. 1123 were in favor.
Obtain a point estimate for the proportion of registered
voters nationwide who are in favor of the death
penalty for persons convicted of murder.
9-6
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Solution
Obtain a point estimate for the proportion of registered
voters nationwide who are in favor of the death
penalty for persons convicted of murder.
1123
pˆ 
 0.63
1783
9-7

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Objective 2
• Construct and Interpret a Confidence Interval
for the Population Proportion
9-8
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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%.
9-9
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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.
9-10
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Confidence interval estimates for the population
proportion are of the form
Point estimate ± margin of error.
The margin of error of a confidence interval
estimate of a parameter is a measure of how
accurate the point estimate is.
9-11
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The margin of error depends on three factors:
Level of confidence: As the level of confidence
increases, the margin of error also increases.
Sample size: As the size of the random sample
increases, the margin of error decreases.
Standard deviation of the population: The more
spread there is in the population, the wider our
interval will be for a given level of confidence.
9-12
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Sampling Distribution of pˆ
For a simple random sample of size n, the
sampling distribution of pˆ is approximately
normal with mean  pˆ  p and standard deviation
, provided that  pˆ  p(1 p)

n
np(1 – p) ≥ 10.
NOTE:We also require that each trial be independent;
when sampling from finite populations (n ≤ 0.05N).
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Constructing a (1 – α)·100% Confidence
Interval for a Population Proportion
Suppose that a simple random sample of size n is taken
from a population. A (1 – α)·100% confidence interval
for p is given by the following quantities
pˆ (1 pˆ )
pˆ  z 2 
n
pˆ (1 pˆ )
Upper bound: ˆ
p  z 2 
n
 be the case that npˆ (1 pˆ )  10 and
Note: It must
Lower bound:
n ≤ 0.05N to construct this interval.
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
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Interpretation of a Confidence Interval
A (1 – α)·100% confidence interval
indicates that (1 – α)·100% of all simple
random samples of size n from the
population whose parameter is unknown
will result in an interval that contains the
parameter.
9-15
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Margin of Error
The margin of error, E, in a (1 – α) 100%
confidence interval for a population proportion is
given by
pˆ (1  pˆ )
E  z 
n
2
9-16
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Parallel Example 2: Constructing a Confidence Interval for a
Population Proportion
In July of 2008, a Quinnipiac University Poll asked
1783 registered voters nationwide whether they
favored or opposed the death penalty for persons
convicted of murder. 1123 were in favor.
Obtain a 90% confidence interval for the proportion of
registered voters nationwide who are in favor of the
death penalty for persons convicted of murder.
9-17
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Solution
•
•
pˆ  0.63
npˆ (1 pˆ )  1783(0.63)(1 0.63)  415.6  10 and the
sample size is definitely less than 5% of the
population size
•
α = 0.10 so zα/2 = z0.05 = 1.645
•
0.63(1
0.63)
Lower bound: 0.63 1.645
 0.61
1783
Upper bound: 0.63 1.645 0.63(1 0.63)  0.65
1783
•
9-18
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Solution
We are 90% confident that the proportion of
registered voters who are in favor of the death
penalty for those convicted of murder is between
0.61and 0.65.
9-19
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Objective 3
• Determine the Sample Size Necessary for
Estimating a Population Proportion within a
Specified Margin of Error
9-20
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Sample size needed for a specified margin of
error, E, and level of confidence (1 – α):
2
z 2 
n  pˆ (1 pˆ ) 
 E 
Problem: The formula uses pˆ which depends
the quantity we are trying to determine!
on n,
9-21
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Two possible solutions:
1. Use an estimate of pö based on a pilot study or an
earlier study.
2. Let pö = 0.5 which gives the largest possible value of
n for a given level of
confidence and a given
margin of error.
9-22
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Sample Size Needed for Estimating the
Population Proportion p
The sample size required to obtain a (1 – α)·100%
confidence interval for p with a margin of error E is
given by
2
z 2 
n  pˆ (1 pˆ ) 
 E 
(rounded up to the next integer), where pˆ is a prior
estimate of p.

9-23

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Sample Size Needed for Estimating the
Population Proportion p
If a prior estimate of p is unavailable, the sample size
required is
2
z 2 
n  0.25 
 E 

9-24
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Parallel Example 4: Determining Sample Size
A sociologist wanted to determine the percentage of
residents of America that only speak English at
home. What size sample should be obtained if she
wishes her estimate to be within 3 percentage points
with 90% confidence assuming she uses the 2000
estimate obtained from the Census 2000
Supplementary Survey of 82.4%?
9-25
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Solution
•
E = 0.03
•
z 2  z0.05 1.645
•
pˆ  0.824
•
1.6452
n  0.824(1 0.824)
  436.04
 0.03 
We round this value up to 437. The sociologist must
survey 437 randomly selected American residents.
9-26
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Section
9.2
Estimating a
Population Mean
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Objectives
1.
2.
3.
4.
Obtain a point estimate for the population mean
State properties of Student’s t-distribution
Determine t-values
Construct and interpret a confidence interval for
a population mean
5. Find the sample size needed to estimate the
population mean within a given margin or error
9-28
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Objective 1
• Obtain a Point Estimate for the Population
Mean
9-29
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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.
9-30
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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.
9-31
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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.
9-32
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Objective 2
• State Properties of Student’s t-Distribution
9-33
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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
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.
(population SD is not known)
9-34
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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 

n
and
x 
t
s
n
Draw a histogram for both z and t.

9-35

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for each sample.
Histogram for z
9-36
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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.
9-38
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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.
9-39
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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.
9-40
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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.
9-41
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9-42
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Objective 3
• Determine t-Values
9-43
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9-44
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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.
9-45
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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 0.8791.
9-46
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Objective 4
• Construct and Interpret a Confidence Interval
for the Population Mean
9-47
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Constructing a (1 – α)100% Confidence
Interval for μ
Provided
• Sample data come from a simple random sample or
randomized experiment
• Sample size is small relative to the population size
(n ≤ 0.05N)
• The data come from a population that is normally
distributed, or the sample size is large
A (1 – α)·100% confidence interval for μ is given by
Lower
Upper
s
s
bound: x  t 
bound: x  t  

where
t
2
n
is the critical value with n – 1 df.
2
9-48
2
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n
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
9-49
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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 ?
9-50
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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
9-52
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(area to the left is 0.975) 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 .
9-53
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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
9-54
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9-55
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Weight (in grams) of Pennies
9-56
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• t 2  2.575
• Lower bound:
 = 2.464 – 2.575 0.02
x  z 2 

n
• Upper bound:
x  z 2 

n


17 
= 2.464 – 0.012 = 2.452

0.02


= 2.464 + 2.575 

17 
= 2.464 + 0.012 = 2.476
We are 99% confident that the mean weight of pennies

minted after 1982 is between 2.452 and 2.476 grams.
9-57
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Objective 5
• Find the Sample Size Needed to Estimate the
Population Mean within a Given Margin of Error
9-58
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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.
9-59
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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.
9-60
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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.
9-61
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