Estimation Procedures

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Estimation Procedures
Confidence Interval Estimation
Point Estimation
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Three Properties of Point Estimators
1. Unbiasedness
2. Consistency
3. Efficiency
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• Estimate Number Error
• 1 6
• 2 8
• 3 -10
• 4 2
• 5 -6

Error 0 0 0 -1 0
The estimates in green are more
efficient (smaller standard error) but the
estimates in red are unbiased
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xMEAN
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The Sampling Distribution of xMEAN for large
samples
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The standard error (s.e.) of estimation for xMEAN
is given by s.e. s/?n where s is the
population standard deviation and n is the
sample size
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s.e. s /?n
Q. Why is the standard error (s.e.) directly
related to s?
A.If the population is more varied (dispersed) it
is more difficult to locate the typical value
In which case are you likely to predict the
population mean more accurately??
1. The age distribution of all students in
English schools, or
2. The age distribution of all students in
English sixth form colleges?
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s.e. s /?n
Q. Why is the s.e. inversely related to the
sample size?
A. The larger the n, the more representative
the sample is of the population and hence the
smaller sampling error
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• Confidence Interval (CI)
• Sometimes, it is possible and convenient to
  predict, with a certain amount of confidence in
  the prediction, that the true value of the
  parameter lies within a specified interval.
• Such an interval is called a Confidence Interval
  (CI)

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• The statement mL, mH is the 95 CI of m is
  to be interpreted that with 95 chance the
  population mean lies within the specified
  interval and with 5 chance it lies outside.

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• Two points to appreciate about the CI

A. The larger the standard error, longer is the
CI, ceteris paribus
B. The higher the level of confidence, the
longer is the CI, ceteris paribus
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The area shaded orange is approximately98 of
the whole
z
-2.33 0 2.33
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The area shaded orange is approximately95 of
the whole
z
-1.96 0 1.96
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• Example1 (Confidence Interval for the population
  mean) Suppose that the result of sampling yields
  the following
• xMEAN 25 n 36. Use this information to
  construct a 95 CI for m, given that s 16

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• Since n gt24, we can say that xMEAN is
  approximately Normal(m, s2/36).
• Standardisation means that (xMEAN - m)/(s/6) is
  approximately z.
• Now find the two symmetric points around 0 in the
  z table such that the area is 0.95. The answer is
  
• z ?1.96.

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• Now solve
•  (xMEAN - m)/(s/6) ?1.96.
•  (25- m)/(16/6) ?1.96 to get two values of m
  19.77 and m 30.23. Thus, the 95 CI for m is
  19.77 30.23

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• Question How is the length of the CI related to
  the standard error?

• Answer Ceteris Paribus, the CI is directly
  related to standard error

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• Example 2 (Confidence Interval for the
  population mean) Suppose that the result of
  sampling yields the following
• xMEAN 25 n 36. Use this information to
  construct a 95 CI for m, given that s 32

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• Now solve
•  (xMEAN - m)/(s/6) ?1.96.
•  (25- m)/(32/6) ?1.96 to get two values of m
  14.55 and m 35.45. Thus, the 95 CI for m is
  14.55 35.45

Compare with the 95 CI for 19.77 30.23 for s
16
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• Question How is the length of the CI related to
  the level of confidence?

• Answer Ceteris Paribus, the CI will be longer
  the higher the level of confidence.

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• Example 3 (Confidence Interval for the
  population mean) Suppose that the result of
  sampling yields the following
• xMEAN 25 n 36. Use this information to
  construct a 90 CI for m, given that s 16

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• Solve
•  (xMEAN - m)/(s/6) ?1.645.
•  (25- m)/(16/6) ?1.645 to get two values of
  m 20.61 and m 29.39. Thus, the 90 CI for m
  is 20.61 29.39

Compare with the 95 CI for 19.77 30.23
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Some Procedural Problems in Parametric Analysis

• 1. The sample size n is small

The CLT does not work! To do any kind of
parametric analysis we need the population to be
normally distributed
Case 1 The population standard deviation s is
known
Theory If X is normal(m, s2 ) then xMEAN is also
normal(m, s2 /n)
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• Example4 (Confidence Interval for the population
  mean with small samples)
• Suppose that the result of sampling from a normal
  population with s 4 yields the following

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• xMEAN 25 n 18. Use this information to
  construct the 90 CI for m,

Since X is normal(m, 42 ) then xMEAN is also
normal(m, 42 /18)
(xMEAN - m)/(4/?18) ?1.645. (25- m)/(4/ ?18)
?1.645
m 26.55, or m 23.45
The required CI is 23.45, 26.55
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• 1. The sample size n is small

Case 2 The population standard deviation s is
unknown
Theory If X is normal(m, s2 ) then xMEAN is also
normal(m, s2 /n) with s unknown
Theory If xMEAN is normal(m, s2 /n) with s
unknown, then (xMEAN m)/s/?n has a
t-distribution with (n-1) degrees of freedom.
s ?(?fi(xi xMEAN)2/(n-1) for grouped
data
s ?(?(xi xMEAN)2/(n-1) for raw data,
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• Example5 (Confidence Interval for the population
  mean)
• Suppose that the result of sampling from a normal
  population yields the following

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• xMEAN 25 n 18. Use this information to
  construct a 95 CI for m, given that s2 16
• First, note that as s is unknown, we use s for s.
• But since n lt 24, we can only say that xMEAN has
  a t-distribution with 17 degrees of freedom.
• Now find from the t-distribution table the two
  symmetric values of t such that the area in
  between them is 0.95.

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• The answer is t ? 2.11. Now solve
• (xMEAN - m)/(s/6) ? 2.11
• (25- m)/(16/6) ?2.11
• to get two values of mL 20.36 and mH 29.63.
  Thus the 95 CI for m is 19.37, 30.63.

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• 2.The population standard deviation(s) is unknown
  but the sample size is large

We estimate s by either of the two estimates, s
or where
s ?(?(xi xMEAN)2/N for raw data, and
s ?(?fi(xi xMEAN)2/N for grouped data
Then we proceed as in Example1 above.
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The Sampling Distribution of the Sample
proportion (p)
Suppose that the population mean p 0.6 and
consider the following statistical process
Sample Number Value of p
1 0.48
2 0.54
3 0.65
- -
100 0.5
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Density
p ? Sample Proportion
p
p
This is the distribution of p provided np and
n(1- p) are ? 5
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Density
p ? Sample Proportion
p
p
This is the distribution of p provided np and
n(1- p) are ? 5
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Density
p ? Sample Proportion
p
p
This is the distribution of p provided np and
n(1- p) are ? 5
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Density
p ? Sample Proportion
p
p
As n gets larger
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Density
p ? Sample Proportion
p
p
and larger.
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Density
p ? Sample Proportion
p
p
and larger.
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Density
p ? Sample Proportion
p
p
and larger.
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Density
p ? Sample Proportion
p
p
The distribution gets more compact around the
mean value (p)
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Density
p ? Sample Proportion
p
p
The distribution gets more compact around the
mean value (p)
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Density
p ? Sample Proportion
p
p
The distribution gets more compact around the
mean value (p)
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Density
Sample Size n3
Sample Size n2
Sample Size n1
p
p
The distribution of the sample proportion (p )
for three sample sizes n1 lt n2 lt n3
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Properties of p

• p is an unbiased estimator of the population mean
  m
• E(p ) p

2. Standard error of p (s.e.p) is given by s.ep
?p(1-p)/n
Therefore, p is a consistent estimator of p
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• Example1 (Confidence Interval for the population
  proportion) Suppose that the result of sampling
  yields the following

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• p 0.4 n 36.
• Use this information to construct a 98 CI for p.
  
• First, we do the validity check.
• This requires np ? 5 as well as n(1-p) ? 5.
• Because we dont know what p is, we use p in the
  place of p.

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• Since p 0.4 and n gt 30, the validity check is
  satisfied.
• We can therefore say that p is approximately
  N(p,s2/36) where s2 p(1-p).
• Standardisation means that (p-p)s/6 is
  approximately z.
• Now find the two symmetric points around 0 in the
  z table such that the area is 0.98. The answer is
  
• z ?2.33.

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• Now solve
• (p-p)/s/6 ?2.33
• (0.4-p)/s/6) ?2.33
• In this expression we do not know what p is, so
  we dont know what s is.
• We use 0.4 as a point estimator for p and
  calculate an estimate for s,s 0.49

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• (0.4- p)/ 0.49/6 ?2.33 to get two values of
  pL 0.21 and pH 0.59.
• Thus the 98 CI for p is 0.21 0.59