CHAPTER 25
Mathematical Statistics
Changes
Regression and a short introduction to correlation have been combined in the same section
(Sec. 25.9).
SECTION 25.1. Introduction. Random Sampling, page 1044
Purpose.
To explain the role of (random) samples from populations.
Main Content, Important Concepts
Population
Sample
Random numbers, random number generator
Sample mean:
x
w
; see (1)
Sample variance
s
2
; see (2)
Comments on Content
Sample mean and sample variance are the two most important parameters of a sample.
x
w
measures the central location of the sample values and
s
2
their spread (their variability).
Small
s
2
may indicate high quality of production, high accuracy of measurement, etc.
Note well that
x
w
and
s
2
will generally vary from sample to sample taken from the same
population, whose mean
m
and variance
s
2
are unique, of course. This is an important
conceptual distinction that should be mentioned explicitly to the students.
SECTION 25.2. Point Estimation of Parameters, page 1046
Purpose.
As a first statistical task we discuss methods for obtaining approximate values
of unknown population parameters from samples; this is called
estimation of parameters.
Main Content, Important Concepts
Point estimate, interval estimate
Method of moments
Maximum likelihood method
SOLUTIONS TO PROBLEM SET 25.2, page 1048
2.
Put
5
0 in Example 1 and proceed with the second equation in (8), as in the example,
to get the estimate

2
5
o
n
i
5
1
x
i
2
.
4.
The likelihood function is (we can drop the binomial factors)
<
5
p
k
1
(1
2
p
)
n
2
k
1
•••
p
k
m
(1
2
p
)
n
2
k
m
5
p
k
1
1
•••
1
k
m
(1
2
p
)
nm
2
(
k
1
1
•••
1
k
m
)
.
1
}
n
392
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2:06 PM
Page 392
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View Full DocumentThe logarithm is
ln
<
5
(
k
1
1
•••
1
k
m
) ln
p
1
[
nm
2
(
k
1
1
1
k
m
)
]
ln (1
2
p
).
Equating the derivative with respect to
p
to zero, we get, with a factor
2
1 from the
chain rule,
(
k
1
1
1
k
m
)
5
[
nm
2
(
k
1
1
1
k
m
)]
.
Multiplication by
p
(1
2
p
) gives
(
k
1
1
••
1
k
m
)(1
2
p
)
5
[
nm
2
(
k
1
1
1
k
m
)
]
p
.
By simplification,
k
1
1
k
m
5
nmp
.
The result is
p
ˆ
5
o
m
i
5
1
k
i
.
6.
We obtain
<
5
ƒ(
x
1
)ƒ(
x
2
)
ƒ(
x
n
)
5
p
(1
2
p
)
x
1
2
1
p
(1
2
p
)
x
2
2
1
p
(1
2
p
)
x
n
2
1
5
p
n
(1
2
p
)
x
1
1
•••
1
x
n
2
n
.
The logarithm is
ln
<
5
n
ln
p
1
(
o
n
m
5
1
x
m
2
n
)
ln (1
2
p
).
Differentiation with respect to
p
gives, with a factor
2
1 from the chain rule,
52
o
n
m
5
1
.
Equating this derivative to zero gives
n
(1
2
p
ˆ
)
5
p
ˆ
(
o
n
m
5
1
x
m
2
n
)
.
Thus
p
ˆ
5
1/
x
w
.
8.
p
ˆ
5
2/(7
1
6)
5
2/13, by Prob. 6.
10.
<
5
1(
b
2
a
)
n
is maximum if
b
2
a
is as small as possible, that is,
a
equal to the
smallest sample value and
b
equal to the largest.
12.
m
5
1/
u
,
ˆ
5
x
w
14.
ˆ
5
1/
x
w
5
2,
F
(
x
)
5
1
2
e
2
2
x
if
x
^
0 and 0 otherwise. A graph shows that the
step function
F
˜
(
x
) (the sample distribution function) approximates
F
(
x
) reasonably
well. (For goodness of fit, see Sec. 25.7.)
SECTION 25.3. Confidence Intervals, page 1049
Purpose.
To obtain interval estimates (“confidence intervals”) for unknown population
parameters for the normal distribution and other distributions.
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