Download Chapter 9 - Sampling Distributions

Ch 9 實習
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Sampling Distributions…
A sampling distribution is created by, as the name suggests,
sampling.
The method we will employ on the rules of probability and
the laws of expected value and variance to derive the
sampling distribution.
For example, consider the roll of one and two dice…
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9.2
Sampling Distribution of the Mean…
A fair die is thrown infinitely many times,
with the random variable X = # of spots on any throw.
The probability distribution of X is:
x
P(x)
1
2
3
4
5
6
1/6
1/6
1/6
1/6
1/6
1/6
…and the mean and variance are calculated as well:
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9.3
Sampling Distribution of Two Dice
A sampling distribution is created by looking at
all samples of size n=2 (i.e. two dice) and their means…
While there are 36 possible samples of size 2, there are only
11 values for , and some (e.g. =3.5) occur more
frequently than others (e.g.
=1).
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9.4
Sampling Distribution of Two Dice…
The sampling distribution of
1/36
2/36
3/36
4/36
5/36
6/36
5/36
4/36
3/36
2/36
1/36
5/36
)
6/36
4/36
P(
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
6.0
P( )
3/36
2/36
1/36
1.0
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is shown below:
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
6.0
9.5
Compare…
Compare the distribution of X…
1
2
3
4
5
6
1.0
1.5
…with the sampling distribution of
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
6.0
.
As well, note that:
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9.6
Generalize…
We can generalize the mean and variance of the sampling of
two dice:
…to n-dice:
The standard deviation of the
sampling distribution is
called the standard error:
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9.7
Central Limit Theorem…
The sampling distribution of the mean of a random sample
drawn from any population is approximately normal for a
sufficiently large sample size.
The larger the sample size, the more closely the sampling
distribution of X will resemble a normal distribution.
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9.8
Central Limit Theorem…
If the population is normal, then X is normally distributed
for all values of n.
If the population is non-normal, then X is approximately
normal only for larger values of n.
In most practical situations, a sample size of 30 may be
sufficiently large to allow us to use the normal distribution as
an approximation for the sampling distribution of X.
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9.9
Sampling Distribution of the Sample Mean
1.
2.
3. If X is normal, X is normal. If X is nonnormal, X is
approximately normal for sufficiently large sample sizes.
Note: the definition of “sufficiently large” depends on the
extent of nonnormality of x (e.g. heavily skewed;
multimodal)
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9.10
Sampling Distribution of sample mean

Standard Deviation of sample mean
Finite Population

N n
x  ( )
n N 1
Infinite Population
x 

n
• A finite population is treated as being infinite if
n/N < .05.
• ( N  n) / ( N  1) is the finite correction factor.
•  x is referred to as the standard error of the mean.
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Example 1
• The amount of time the university professors devote
to their jobs per week is normally distributed with a
mean of 52 hours and a standard deviation of 6 hours
• a. What is the probability that a professor works for
more than 60 hours per week?
• b. Find the probability that the mean amount of work
per week for three randomly selected professors is
more than 60 hours.
• c. Find the probability that if three professors are
randomly selected, all three work for more than 60
hours per week.
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Solution
• a.
• b.
• c.
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Using the Sampling Distribution for Inference
Here’s another way of expressing the probability calculated from a
sampling distribution.
P(-1.96 < Z < 1.96) = .95
Substituting the formula for the sampling distribution
P(1.96 
X
/ n
 1.96 )  .95
With a little algebra
P(  1.96

n
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 X    1.96

)  .95
n
9.14
Using the Sampling Distribution for Inference
Returning to the chapter-opening example where µ = 800, σ = 100,
and n = 25, we compute
P(800  1.96
100
 X  800  1.96
25
100
)  .95
25
or
P(760 .8  X  839 .2)  .95
This tells us that there is a 95% probability that a sample mean will
fall between 760.8 and 839.2. Because the sample mean was
computed to be $750, we would have to conclude that the dean's
claim is not supported by the statistic.
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9.15
Sampling Distribution of a Proportion
• The parameter of interest for nominal data is the
proportion of times a particular outcome (success)
occurs.
• To estimate the population proportion ‘p’ we use the
sample proportion.
The number
of successes
^ =
The estimate of p = p
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X
n
Sampling Distribution of a Proportion
^ can
• Since X is binomial, probabilities about p
be calculated from the binomial distribution.
^ we prefer to use
• Yet, for inference about p
normal approximation to the binomial.
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Normal approximation to the Binomial
–Normal approximation to the binomial works best when
•the number of experiments (sample size) is large, and
•the probability of success, p, is close to 0.5.
–For the approximation to provide good results two conditions
should be met:
np  5; n(1 - p)
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
5
Normal approximation to the Binomial
Example
Approximate the binomial probability P(x=10)
when n = 20 and p = .5
The parameters of the normal distribution
used to approximate the binomial are:
 = np; 2 = np(1 - p)
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Normal approximation to the Binomial
Let us build a normal
distribution to approximate the
binomial P(X = 10).
 = np = 20(.5) = 10;
2 = np(1 - p) = 20(.5)(1 - .5) = 5
 = 51/2 = 2.24
P(9.5<YNormal<10.5)
The approximation
P(XBinomial = 10) =.176
~= P(9.5<Y<10.5)
9.5
10
10.5
9.5  10
10.5  10
 P(
Z
)  .1742
2.24
2.24
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Normal approximation to the Binomial
 More examples of normal approximation
to the binomial
P(X 4) @ P(Y< 4.5)
4
P(X 14) @ P(Y > 13.5)
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4.5
13.5
14
Sampling Distribution of a Sample Proportion…
Using the laws of expected value and variance, we can
determine the mean, variance, and standard deviation of .
(The standard deviation of is called the standard error of
the proportion.)
Sample proportions can be standardized to a standard normal
distribution using this formulation:
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9.22
Sampling Distribution of
• Standard Deviation of
p
Finite Population
p 
p
p(1  p) N  n
n
N 1
Infinite Population
p 
p(1  p)
n
–  p is referred to as the standard error of the
proportion.
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Example 2
• A psychologist believes that 80% of male drivers
when lost continue to drive, hoping to find the
location they seek rather than ask directions. To
examine this belief, he took a random sample of
350 male drivers and asked each what they did
when lost. If the belief is true, determine the
probability that less than 75% said they continue
driving.
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Solution
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Sampling Distribution:
Difference of two means
The final sampling distribution introduced is that of the
difference between two sample means. This requires:
 independent random samples be drawn from each of two
normal populations
If this condition is met, then the sampling distribution of the
difference between the two sample means, i.e.
will be normally distributed.
(note: if the two populations are not both normally
distributed, but the sample sizes are “large” (>30), the
distribution of
is approximately normal)
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9.26
Sampling Distribution:
Difference of two means
The expected value and variance of the sampling
distribution of
are given by:
mean:
standard deviation:
(also called the standard error if the difference between two
means)
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9.27
Example 9.3…
Since the distribution of
is normal and has a
mean of
and a standard deviation of
We can compute Z (standard normal random variable) in this
way:
Copyright © 2009 Cengage Learning
9.28
Example 3
• The manager of a restaurant believes that waiters and
waitresses who introduce themselves by telling
customers their names will get larger tips than those
who don’t. In fact, she claims that the average tip for the
former group is 18% whereas that of the latter is only
15%. If tips are normally distributed with a standard
deviation of 3 %, what is the probability that in a
random sample of 10 tips recorded from waiters and
waitresses who introduce themselves and 10 tips from
waiters and waitresses who don’t, the mean of the
former will exceed that of the latter?
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Solution
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