# Sampling Distributions

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```					 Chapter 9

Sampling
Distributions

1
9.1 Introduction

 In real life calculating parameters of
populations is prohibitive because
populations are very large.
 Rather than investigating the whole
population, we take a sample, calculate a
statistic related to the parameter of
interest, and make an inference.
 The sampling distribution of the statistic
is the tool that tells us how close is the
statistic to the parameter.
2
9.2 Sampling Distribution of
the Mean

 An example
   A die is thrown infinitely many times. Let X
represent the number of spots showing on
any throw.
   The probability distribution of X is
E(X) = 1(1/6) +
x    1 2 3 4 5 6               2(1/6) + 3(1/6)+
………………….= 3.5
p(x) 1/6 1/6 1/6 1/6 1/6 1/6
V(X) = (1-3.5)2(1/6) +
(2-3.5)2(1/6) +
…………. …= 2.92
3
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 ?

4
Throwing a die twice – sample mean

Sample         Mean Sample     Mean   Sample         Mean
1     1,1     1    13   3,1   2      25     5,1        3
2     1,2    1.5   14   3,2  2.5     26     5,2      3.5
3     1,3     2    15   3,3   3      27     5,3        4
4     1,4    2.5   16   3,4  3.5     28     5,4      4.5
5     1,5     3    17   3,5   4      29     5,5        5
6     1,6    3.5   18   3,6  4.5     30     5,6      5.5
7     2,1    1.5   19   4,1  2.5     31     6,1      3.5
8     2,2     2    20   4,2   3      32     6,2        4
9     2,3    2.5   21   4,3  3.5     33     6,3      4.5
10     2,4     3    22   4,4   4      34     6,4        5
11     2,5    3.5   23   4,5  4.5     35     6,5      5.5
12     2,6     4    24   4,6   5      36     6,6        6

5
Sample          Mean Sample     Mean          Sample            Mean
1      1,1     1    13   3,1   2             25        5,1        3
The distribution of x when n = 2
2
3
1,2
1,3
1.5
2
14
15
3,2
3,3
2.5
3
26
27
5,2
5,3
3.5
4
4      1,4    2.5   16   3,4  3.5            28        5,4      4.5
2
5      1,5     3    17   3,5   4             29
2         x
5,5        5
6
7     Note : m x  m x and
1,6
2,1
3.5
1.5
18
19
3,6
4,1
4.5
2.5           x
30
31       5,6
6,1
5.5
3.5
8
9
2,2
2,3
2
2.5
20
21
4,2
4,3
3
3.5
32
33
2
6,2
6,3
4
4.5
10      2,4     3    22   4,4   4             34        6,4        5
11      2,5    3.5   23   4,5  4.5            35        6,5      5.5
12      2,6     4    24   4,6   5             36        6,6        6

E( x) =1.0(1/36)+
6/36                                                       1.5(2/36)+….=3.5
5/36
V(X) = (1.0-3.5)2(1/36)+
4/36                                                       (1.5-3.5)2(2/36)... = 1.46
3/36
2/36
1/36
1      1.5    2.0   2.5   3.0   3.5   4.0   4.5   5.0    5.5 6.0       x       6
Sampling Distribution of the
Mean
n5                 n  10                n  25
m x  3.5           m x  3.5             m x  3.5
2                    2                  2
  .5833 (  x )
2
x                   x  .2917 (  x )
2
  .1167 (  x )
2
x
5 6                  10                  25

7
Sampling Distribution of the
Mean
n5                      n  10                n  25
m x  3.5                m x  3.5             m x  3.5
2                  2                    2
 2  .5833 (     x
)      .2917 (  x )
2
x                     .1167 (  x )
2
x
x
5                   10                    25

 2
Notice that  x is smaller than . x.
2

The larger the sample size the
smaller  x . Therefore, x tends
2

to fall closer to m, as the sample
size increases.
8
Sampling Distribution of the
Mean
Demonstration: The variance of the sample mean is
smaller than the variance of the population.

Mean = 1.5 Mean = 2. Mean = 2.5

Population      1      1.5        2
2       2.5     3
1.5
1.5        2
2       2.5
2.5
1.5        2       2.5
Compare                   2.5
1.5 the variability of the population
2
1.5
1.5        2 of the2.5
2.5
1.5
to the variability2       2.5
sample mean.
1.5        2       2.5
1.5                2.5
1.5        2       2.5
Let us take samples     1.5
1.5        2
2       2.5
2.5
of two observations

9
Sampling Distribution of the
Mean

Also,
Expected value of the population =
(1 + 2 + 3)/3 = 2

Expected value of the sample mean =
(1.5 + 2 + 2.5)/3 = 2

10
The Central Limit Theorem

 If a random sample is drawn from any
population, the sampling distribution of the
sample mean 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.

11
Sampling Distribution of the Sample
Mean

1. m x  m x
x
2
2.  
2
x
n
3. If x is normal, x is normal. If x is nonnormal
x is approximately normally distribute d for
sufficient ly large sample size.

12
Sampling Distribution of the
Sample Mean
 Example 9.1
   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.
   Solution
0.7486
 The random variable X is the
amount of soda in a bottle.
x  m 32  32.2
P( x  32)  P(                  )
x      .3         x = 32 m = 32.2
 P( z  .67)  0.7486                       13
Sampling Distribution of the
Sample Mean
 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
P( x  32)  P(                  )                        0.9082
x    .3 4
 P( z  1.33)  0.9082
0.7486
x = 32
x  32 m = 32.2
m x  32.2         14
Sampling Distribution of the
Sample Mean
 Example 9.2
   Dean’s claim: The average weekly income of
\$600.
   Suppose the distribution of weekly income has a
standard deviation of \$100. What is the
probability that 25 randomly selected graduates
have an average weekly income of less than
\$550?
   Solution                   x  m 550  600
P( x  550)  P(                  )
x     100 25
 P( z  2.5)  0.0062      15
Sampling Distribution of the Sample
Mean

 Example 9.2– continued
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 of observing a sample mean
as low as 550 is very small (0.0062). 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.

16
Using Sampling Distributions for
Inference
 To make inference about population parameters we use
sampling distributions (as in Example 9.2).
 The symmetry of the normal distribution along with the
sample distribution of the mean lead to:
xm
P( 1.96  z  1.96 )  .95, or P( 1.96         1.96 )  .95
 n
- Z.025     Z.025
This can be written as
                  
P( 1.96     x  m  1.96     )  .95
n                  n
which become
                 
P(m  1.96      x  m  1.96    )  .95
n                 n
17
Using Sampling Distributions for
Inference

Standard normal distribution Z         Normal distribution of    x
100                  100
P(600  1.96        x  600  1.96     )  .95
25                   25

.95
.95
.025                           .025 .025                                        .025

Z                                                 x
-1.96    0    -1.96                     
100      m             
m600 Pm600 9696 100
m  1..96
P(600  196                (  1. 1.
n
25                     n 25
18
Using Sampling Distributions for
Inference

100                  100
P(600  1.96      x  600  1.96     )  .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.
19
9.3 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 =   X
n

20
9.3 Sampling Distribution of
a Proportion

 Since X is binomial, probabilities about ^
p
can be calculated from the binomial
distribution.
^
 Yet, for inference about p we prefer to use
normal approximation to the binomial.

21
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)      5

22
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:

m = np; 2 = np(1 - p)

23
Normal approximation to the
Binomial

Let us build a normal                          m = np = 20(.5) = 10;
distribution to approximate the                2 = np(1 - p) = 20(.5)(1 - .5) = 5
 = 51/2 = 2.24
binomial P(X = 10).

P(XBinomial = 10) =.176                                  P(9.5<YNormal<10.5)
The approximation
~ P(9.5<Y<10.5)
=

9.5   10   10.5
9.5  10     10.5  10
 P(            Z           )  .1742
2.24         2.24
24
Normal approximation to the
Binomial

 More examples of normal approximation
to the binomial
P(X  4) @ P(Y< 4.5)

4
4.5
P(X 14) @ P(Y > 13.5)

13.5   14   25
Approximate Sampling Distribution
of a Sample Proportion

 From the laws of expected value and variance,
ˆ              ˆ
it can be shown that E( p ) = p and V( p )
=p(1-p)/n
 If both np > 5 and np(1-p) > 5, then

ˆ
pp
z
p(1  p)
n
 Z is approximately standard normally
distributed.                                26
 Example 9.3
   A state representative received 52% of the
   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?

27
 Example 9.3
   Solution
   The number of respondents who prefer the
representative is binomial with n = 300 and p =
.52. Thus, np = 300(.52) = 156 and
n(1-p) = 300(1-.52) = 144 (both greater than 5)

    p p
ˆ                .50  .52     
P( p  .50)  P
ˆ                                                .7549
   p(1  p) n   (.52)(1  .52) 300 
                                   

28
9.4 Sampling Distribution of the
Difference Between Two Means

 Independent samples are drawn from
each of two normal populations
 We’re interested in the sampling
distribution of the difference between the
two sample means x 1  x 2

29
Sampling Distribution of the
Difference Between Two Means

 The distribution of x 1  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
x 1  x 2 is approximately normal.
30
Sampling Distribution of the
Difference Between Two Means

 Applying the laws of expected value and
variance we have:
E( x1  x 2 )  E( x1 )  E( x 2 )  m1  m 2
1  2
2

V( x 1  x 2 )  V( x 1 )  V( x 2 )      2
n    n
 We can define:
( x 1  x 2 )  ( m1  m 2 )
Z
1  2
2
 2
n1 n2                    31
Sampling Distribution of the
Difference Between Two Means

Example 9.4
   The starting salaries of MBA students from
two universities (WLU and UWO) are \$62,000
(stand.dev. = \$14,500), and \$60,000 (stand.
dev. = \$18,3000).
   What is the probability that a sample mean of
WLU students will exceed the sample mean of
UWO students? (nWLU = 50; nUWO = 60)

32
Sampling Distribution of the
Difference Between Two Means
 Example 9.4 – Solution
We need to determine P( x1  x 2  0)

m1 - m2 = 62,000 - 60,000 = \$2,000

 12   2
2
14,5002 18,3002
                                      \$3,128
n       n            50            60
x1  x 2  (m1 - m 2 ) 0  2000
P( x1  x 2  0)  P(                               )
1 2
2     2         3128

n1 n2
 P( z  .64)  .5  .2389  .7389
33

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