Transcript Chapter 6 - Pegasus @ UCF

CHAPTER 6 Sampling Distributions
Homework:
1abcd,3acd,9,15,19,25,29,33,43
Sec 6.0: Introduction
• Parameter
The "unknown" numerical values that is used
to describe the properties of a population.
• Sample Statistics
The computed numerical values from the
measurements in a sample.
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Sec 6.1: What is a Sampling Distribution?
• Sampling Error
The error results from using a sample instead
of censusing the population to estimate a
population quantity.
• Sampling Distributions
The sample statistics vary from one sample to
another. Therefore, there is a distribution function
associated with each sampling statistic. This is
called the sampling distribution.
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Example 6.1:
The three most popular numbers which were
picked up by a group of students are 7, 17, and 24.
Assume that the population mean and standard
deviation are 16 and 8, respectively.
(a). List all the possible samples of two numbers
that can be obtained from this three numbers.
(b). Find the sample mean for each sample.
(c). Compute the sample error for each sample.
(Solutions in the note page)
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Example 6.2 (Basic)
The population of average points per game
(ppg) for the top five players of Orlando Magic is
presented in the following table. (First eight games
in 1996 playoff)
Player
ppg
O'Neal(O)
25.3
Anderson(A)
16.4
Grant(G)
16.9
Hardaway(H)
22.3
Scott(S)
13.4
(a) Find the sampling distribution of the mean of the
"ppg" of three players from the population of five
players. (Part (a) solution is in note page)
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Example 6.2 Continue:
(b) Compute the mean and standard deviation of
the sample mean.
<Solution to part (b)>:
 x  E x   x  p x  18.86
 x    x   x  p x  176
.
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Note: If you forget how to do it, you need to
review Chapter 4.
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Example 6.2 Continue:
(c) Find the sampling distribution of the sample
median of the "ppg" of three players from the
population of five players. (solution in note page)
(d) Compute the mean and mean square error of the
sample median.
<Solution to part (d)>:
 median  E median   median  p median  18.37
 median    median   median  p median  2.58
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Sec 6.2: Some Criterions for choosing a
Statistics
• Point Estimator
A point estimator of a population parameter is
a rule that tell us how to obtain a single number
based on the sample data. The resulting number is
called a point estimator of this unknown
population parameter.
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• Unbiasedness
The mean of the sampling distribution of a
statistic is called the expectation of this statistic.
If the expectation of a statistics is equal to the
population parameter this statistics is intended to
estimate, then the statistics is called an unbiased
estimator of this population parameter. If the
expectation of a statistics is not equal to the
population parameter, the statistics is a biased
estimator.
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NOTES:
(1). If the population is normally distributed and the
sample are randomly selected from this
population, the sample mean is the best unbiased
estimator of the population mean.
(2). If the population has an extremely skewed
distribution and the sample size of the random
sample is small, the sample mean may not be best
estimator of the population mean.
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Example 6.3:
Consider the probability distribution shown here.
x
p(x)
2
1/3
4
1/3
9
1/3
(a) Find .
<Solution to (a)>:
1
1
1
   xp( x)  2   4   9   5
3
3
3
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Example 6.3 Continue:
(b) List all possible samples of three observations.
<solution for part (b)>:
(2,2,2)
(2,,2,4) (2,4,2) (4,2,2)
(2,4,4) (4,2,4) (4,4,2)
(4,4,4)
(2,2,9) (2,9,2) (9,2,2)
(2,4,9) (2,9,4) (4,2,9) (4,9,2) (9,2,4) (9,4,2)
(4,4,9) (4,9,4) (9,4,4)
(2,9,9) (9,2,9) (9,9,2)
(4,9,9) (9,4,9) (9,9,4)
(9,9,9)
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Example 6.3 Continue:
(c) For a sample of 3 observations, find the sampling
distribution of the sample mean and its mean value
<Solution to part (c)>:
mean
P(mean)
2
1/27
8/3
3/27
10/3
3/27
4
1/27
13/3
3/27
5
6/27
17/3
3/27
20/3
3/27
22/3
3/27
9
1/27
1
1
 x  E x   xp x  2  9 
5
27
27
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Example 6.3 Continue:
(d) For a sample of 3 observations, find the sampling
distribution of the sample median and E(median).
<Solution to part (d)>:
median
P(median)
2
7/27
4
13/27
9
7/27
E(median) = Smedian * P(median)
= 2 * 7/27 + 4*13/27  9*7/27 = 14.33
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Sec 6.3: Central Limit Theorem
Some properties of the sample mean
(1). The expectation of the sampling distribution of
x is equal to the population mean , i.e.
 x  E  x   .
(2). The sampling distribution of sample mean is
approximately normal for sufficiently large sample
size even if the sampled population is not
normally distributed.
(3). The sampling distribution of sample mean is
exactly normal if the sampled population is
normally distributed no matter how small the
sample size is.
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• Central Limiting Theorem
Suppose a random sample of size n is drawn from
a population. If the sample size n is sufficiently
large, then the sample mean has at least
approximately a normal distribution.
The mean value of the sample mean is always
equal to the population mean, and the standard
error of the sample mean is always equal to

x 
n
where  is the population standard deviation.
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Note: Each of the following figure shows
the population distribution (n=1) and three
sampling distributions of the sample mean
for different populations. We can see that
the sampling distribution of the sample
mean is approximately normal distributed
when the sample size getting larger.
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0
n=1
2
4
-2
0
n=4
2
4
-4
0
n=2
2
4
-2
0
n = 30
2
4
0.0
1.0
f(x)
f(x)
1.0
0.0
-4
-2
2.0
-2
2.0
-4
f(x)
0.0 1.0 2.0
f(x)
0.0 1.0 2.0
Figure 6.1 Normal Population
-4
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2.0
0.0
n=1
n=2
1.0
0.0
1.0
f(x)
2.0
0.0 0.5 1.0 1.5 2.0 2.5 3.0
2.0
0.0 0.5 1.0 1.5 2.0 2.5 3.0
0.0
f(x)
1.0
f(x)
1.0
0.0
f(x)
2.0
Figure 6.2 Chi-Square Population
0.0 0.5 1.0 1.5 2.0 2.5 3.0
n=4
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0.0 0.5 1.0 1.5 2.0 2.5 3.0
n = 30
f(x)
0 2 4 6 8
f(x)
0 2 4 6 8
Figure 6.1 Uniform Population
0.0
0.2
0.4
0.6
0.8
1.0
0.0
0.2
0.6
0.8
1.0
0.8
1.0
f(x)
0 2 4 6 8
n=2
f(x)
0 2 4 6 8
n=1
0.4
0.0
0.2
0.4
0.6
n=4
0.8
1.0
19
0.0
0.2
0.4
0.6
n = 30
Example 6.4:
Suppose that you selected a random sample
of size 4 from a normal population with mean 8
and standard deviation 2.
(a) Is the sample mean normally distributed?
Explain.
(b) Find P( x > 10).
(c) Find P( x < 7).
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<Solutions to EX 6.4>:
(a) Yes. The sampling distribution of sample mean is exactly
normal if the sampled population is normally distributed
no matter how small the sample size is.

2

1
standard error =  x 
n
4
x   10  
10  8

)  P( z 
)
(b) P(x > 10) = P(
x
x
1
= P(z > 2) = 0.0228
(c) P(x < 7) =
x   10  
78
P(
<
)  P( z <
)
x
x
1
= P(z < -1) = 0.1587.
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Example 6.5:
As reported in the National Center for Health
Statistics, males who are six feet tall and between
18 and 24 years old have a mean weight of 175
pounds with a 14 pound standard deviation.
(a). Find the probability that a random sample of
196 males who are six feet tall and between 18
and 24 years of age has a mean weight greater
than 176 pounds.
(b). Find the probability that a random sample of 4
males who are six feet tall and between 18 and 24
years of age has a mean weight greater than 168
pounds. State the necessary assumptions.
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<Solutions to EX 6.5>:

14
(a) standard error =  x 

1
n
196
x   176  
176  175
P( x > 176) = P(

)  P( z 
)
x
x
1
= P( z > 1) = 0.1587

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x 

7
n
4
x   168  
168  175
P( x > 168) = P(

)  P( z 
)
x
x
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(b) standard error =
= P( z > -1) = 0.8413.
The population needs to have normal distribution.
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