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Chapter 9
Sampling
Distributions
1
Introduction
 In this chapter we study some
relationships between population and
sample characteristics.
 Generally, we are interested in
population parameters such as



Mean return
Variability of demand
Proportion of defectives in a production line
2
Introduction
 Such parameters are usually unknown
 Therefore, we draw a samples from the
population, and use them to make
inference about the parameters.
 This is done by constructing sample
statistics, that have close relationship to
the population parameters.
3
Introduction
 Samples are random, so the sample
statistic is a random variable.
 As such it has a sample distribution.
 Sample distributions for various
statistics are studied in this chapter
4
9.1 Sampling Distribution of
the Mean
 Example 1


A die is thrown infinitely many times. Let X
represent the number of spots showing on
any throw.
The probability distribution of X is
x
1 2 3 4 5 6
p(x) 1/6 1/6 1/6 1/6 1/6 1/6
E(X) = 1(1/6) +
2(1/6) + 3(1/6)+
………………….= 3.5
V(X) = (1-3.5)2(1/6) +
(2-3.5)2(1/6) +
…………. …= 2.92
5
Throwing a die twice – sample mean
 Suppose we want to estimate m
from the mean x of a sample of
size n = 2.
 What is the distribution of x ?
6
Throwing a die twice – sample mean
these
are the
means
of each
pair2 throws
These are And
all the
possible
pairs
of values
for the
Sample
1
2
3
4
5
6
7
8
9
10
11
12
1,1
1,2
1,3
1,4
1,5
1,6
2,1
2,2
2,3
2,4
2,5
2,6
Mean Sample
Mean
1
13
3,1
2
1.5
14
3,2
2.5
2
15
3,3
3
2.5
16
3,4
3.5
3
17
3,5
4
3.5
18
3,6
4.5
1.5
19
4,1
2.5
2
20
4,2
3
2.5
21
4,3
3.5
3
22
4,4
4
3.5
23
4,5
4.5
4
24
4,6
5
Sample
25
26
27
28
29
30
31
32
33
34
35
36
5,1
5,2
5,3
5,4
5,5
5,6
6,1
6,2
6,3
6,4
6,5
6,6
Mean
3
3.5
4
4.5
5
5.5
3.5
4
4.5
5
5.5
6
7
The distribution of x when n = 2
Calculating the relative frequency of each value
of x we have the following results
x
Frequency
1
1
1.5
2
2.0
3
Relative freq 1/36 2/36 3/36
(1+1)/2 = 1
(1+2)/2 = 1.5
(2+1)/2 = 1.5
2.5
4
4/36
3.0
5
3.5
6
5/36
(1+3)/2 = 2
(2+2)/2 = 2
(3+1)/2 = 2
6/36
4.0
5
5/36
4.5
4
4/36
5.0
3
3/36
5.5 6.0
2
1
2/36
1/36
Notice there are 36 possible
pairs of values:
1,1 1,2 ….. 1,6
2,1 2,2 ….. 2,6
………………..
6,1 6,2 ….. 6,6
8
The Relationship between the
sample size and the sampling
distribution of the sample mean
n5
m x  3.5
n  10
m x  3.5
n  25
m x  3.5
2x
  .5833 (  )
5
2x
2
 x  .2917 (  )
10
2x
  .1167 (  )
25
2
x
As the sample size changes, the
mean of the sample mean does not
change!
2
x
9
The Relationship between the
sample size and the sampling
distribution of the sample mean
n5
m x  3.5
n  10
m x  3.5
n  25
m x  3.5
2x
  .5833 (  )
5
2x
2
 x  .2917 (  )
10
2x
  .1167 (  )
25
2
x
As the sample size increases, the
variance of the sample mean
decreases!
2
x
10
The Relationship between the
sample size and the sampling
distribution of the sample mean
n5
m x  3.5
n  10
m x  3.5
n  25
m x  3.5
2x
  .5833 (  )
5
2x
2
 x  .2917 (  )
10
2x
  .1167 (  )
25
2
x
Also, note the interesting relationship
between the sample size and the
variance of the sample mean.
We’ll formalize this relationship soon.
2
x
11
The Sample Variance
Demonstration: Why is the variance of the sample
mean is smaller than the population variance.
Mean = 1.5 Mean = 2. Mean = 2.5
Population
1
1.5
2
2.5
3
Compare
the
range
of the population
Let us
take
samples
to the
range
of the sample
mean.
of two
observations.
Click
12
The Central Limit Theorem
 If a random sample is drawn from any
population, the sampling distribution of the
sample mean is:


Normal if the parent population is normal,
Approximately normal if the parent population is
not normal, provided the sample size is sufficiently
large.
 The larger the sample size, the more closely
the sampling distribution of x will resemble a
normal distribution.
13
The Parameters of the
Sampling Distribution of X
The mean of X is equal to the mean of the
parent population
μx  μx
The variance of X is equal to the parent
population variance divided by ‘n’.
2
σ
σ 2x  x
n
14
The Sampling Distribution of X
- Example
 Example 2


The amount of soda pop in each bottle is
normally distributed with a mean of 32.2
ounces and a standard deviation of .3
ounces.
Find the probability that a bottle bought by
a customer will contain more than 32
ounces.
15
The Sampling Distribution of X
- Example
 Example 2
Solution
The random variable X is the amount of soda in a bottle.
0.7486
P(x  32)
x = 32 m = 32.2
x  μ 32  32.2
P(x  32)  P(

)  P(z  .67)  0.7486
σx
.3
16
The Sampling Distribution of X
 Find the probability that a carton of four bottles
will have a mean of more than 32 ounces of
soda per bottle.
 Solution

Define the random variable as the mean amount of soda
per bottle.
x  m 32  32.2

)
x
.3 4
 P( z  1.33)  0.9082
P( x  32)  P(
P(x  32)
0.9082
x  32 m x  32.2
17
The Sampling Distribution of X
 Example 3



The average weekly income of B.B.A graduates
one year after graduation is $600.
Suppose the distribution of weekly income has a
standard deviation of $100. What is the
probability that 35 randomly selected graduates
have an average weekly income of less than
$550?
Solution P(x  550)  P( x  μ  550  600 )
σx
100 35
 P(z  2.97)  0.0015
18
The Sampling Distribution of X
 Example 3 – continued


If a random sample of 35 graduates actually had
an average weekly income of $550, what would
you conclude about the validity of the claim that
the average weekly income is 600?
Solution


With m = 600 the probability to have a sample mean as
low as 550 is very small (0.0015). The claim that the
mean weekly income is $600 is probably unjustified.
It will be more reasonable to assume that m is smaller
than $600, because then a sample mean of $550
becomes more probable.
19
<
9.2 Sampling Distribution of
a Sample Proportion (p)
 The parameter of interest for qualitative
(nominal) data is the proportion of times
a particular outcome (success) occurs for
a given population.
 This is the motivation for studying the
distribution of the sample proportion
20
<
9.2 Sampling Distribution of
a Sample Proportion (p)
 Let X be the number of times an event of interest takes
place (we can call such an event a success just like the
definition we used for the binomial experiment)
The number
of successes
<
The sample proportion = p =
X
n
21
<
9.2 Sampling Distribution of
a Sample Proportion (p)
<
<
 Since X is binomial, probabilities for p can
be calculated from the binomial
distribution.
 Yet, for inference about p we prefer to use
normal approximation to the binomial.
22
Approximate Sampling
Distribution
of a Sample Proportion
p̂
 From the laws of expected value and
variance, it can be shown that mp̂= p and
p̂2 = p(1-p)/n
 Z is calculated by:
Z
ˆ p
p
p(1 p)
n
 If both np > 5 and n(1-p) > 5, then Z is
approximately standard normal.
23
Approximate Sampling Distribution
of a Sample Proportion
 Example 5



A state representative received 52% of the
votes in the last election.
One year later the representative wanted
to study his popularity.
If his popularity has not changed, what is
the probability that more than half of a
sample of 300 voters would vote for him?
24
Approximate Sampling Distribution
of a Sample Proportion
 Example 5

Solution

The number of respondents who prefer the
representative is binomial with n = 300 and p =
.52. Thus, np = 300(.52) = 156 > 5
n(1-p) = 300(1-.52) = 144 > 5. The normal
approximation can be applied here:

ˆ p
p
.50  .52 

ˆ
P(p  .50)  P

 .7549
 p(1  p) n

.0288 

25
Using Sampling Distributions for
Inference
 Sampling distributions can be used to make an
inference about population parameters
 For example let us look at an inference about the
population mean
 Generally we’ll compare the actual sample mean
with a hypothesized value of the unknown
population mean, and make an informed decision
about the likelihood of this hypothesis
26
Using Sampling Distributions for
Inference
 Let us guess what the value of m is, and build a symmetrical interval
around m large enough to make it very likely that the sample mean
falls inside it.
 If the sample mean falls outside the interval (although this is very
unlikely), we tend to believe that m is different than the value of m we
guessed.
 The sampling distribution of the sample mean helps in performing
the calculations.
Large
probability
that x falls inside
[mD, m+D]
mD
m
m+D
x
27
Using Sampling Distributions for
Inference
Suppose .95 is considered sufficiently large probability
the sample mean falls inside the interval.
Let us build a symmetrical interval around m.
Using the notation m  D and m + D we have:
P(m  D  x  m + D) = .95.
x
28
Using Sampling Distributions for
Inference
Performing the usual standardization we find that the interval covering
95% of the distribution of the sample mean is:
σ
σ
μ  1.96
 x  μ + 1.96
n
n
0.95
x
μ  1.96
σ
n
μ + 1.96
σ
n
29
Using Sampling Distributions for
Inference
Now let us apply this interval to example 3.
P(m  1.96
P(600  1.96

 x  m + 1.96
n
100
m
n
 x  600 + 1.96
)  .95
100
)  .95
25
25
Which reduces to P(560.8  x  639.2)  .95
 Conclusion
 There is 95% chance that the sample mean falls within
the interval [560.8, 639.2] if the population mean is
600.
 Since the sample mean was 550, the population mean
is probably not 600.
30
Optional: Sampling Distribution of the
Difference Between Two Means
 The difference between two means can
become a parameter of interest when the
comparison between two populations is
studied.
 To make an inference about m1 - m2 we
observe the distribution of x1  x.2
31
9.3 Normal Distribution of the
Difference Between two Sample
Means
 The distribution of x1  x 2 is normal if


The two samples are independent, and
The parent populations are normally
distributed.
 If the two populations are not both
normally distributed, but the sample
sizes are 30 or more, the distribution of
x1  x 2 is approximately normal.
32
9.3 Normal Distribution of the
Difference Between two Sample
Means
 Applying the laws of expected value and
variance we have:
μ x1  x 2  μ1  μ2
σ
2
x1  x 2
σ12 σ 22

+
n
n
 We can define:
Z
( x1  x 2 )  (m1  m 2 )
12 22
+
n1 n2
33
9.3 Normal Distribution of the
Difference Between two Sample
Means
Example 6
The starting salaries of MBA students from
two universities (WLU and UWO) are
$62,000 (stand.dev. = $14,500), and
$60,000 (stand. dev. = $18,300).

What is the probability that a sample mean of
WLU students will exceed the sample mean of
UWO students? (nWLU = 50; nUWO = 60)
34
9.3 Normal Distribution of the
Difference Between two Sample
Means
 Example 6 – Solution
We need to determine P( x1  x2  0)
m1 - m2 = 62,000 - 60,000 = $2,000
12  22
14,500 2 18,3002
+

+
 $3,128
n
n
50
60
x1  x2  (m1 - m2 ) 0  2000
P( x1  x2  0)  P(

)
2
2
3128
1 2
+
n1 n2
 P( z  .64 )  .5 + .2389  .7389
35