Unformatted text preview: Chapter 9 Sampling
Sampling
Distributions
Distributions 1 9.1 Introduction In real life calculating parameters of
In
populations is prohibitive because
populations are very large.
populations Rather than investigating the whole
Rather
population, we take a sample, calculate a
statistic related to the parameter of
statistic
parameter
interest, and make an inference. The sampling distribution of the statistic
The sampling
statistic
is the tool that tells us how close is the
statistic to the parameter.
statistic
2 9.2 Sampling Distribution of
9.2
the Mean
the An example A die is thrown infinitely many times. Let X
die
represent the number of spots showing on
any throw.
any
The probability distribution of X is
x
p(x) 1 2 1/6 1/6 3 4 5 1/6 1/6 1/6 E(X) = 1(1/6) +
6 2(1/6) + 3(1/6)+
………………….= 3.5
1/6
V(X) = (13.5)2(1/6) +
(23.5)2(1/6) +
3 Throwing a die twice – sample mean
Throwing Suppose we want to estimate µ
Suppose
from the mean x of a sample of
size n = 2.
size What is the distribution of x? 4 Throwing a die twice – sample mean
Throwing 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 Mean
5,1
5,2
5,3
5,4
5,5
5,6
6,1
6,2
6,3
6,4
6,5
6,6 3
3.5
4
4.5
5
5.5
3.5
4
4.5
5
5.5
6
5 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 Mean
5,1
5,2
5,3
5,4
5,5
5,6
6,1
6,2
6,3
6,4
6,5
6,6 3
3.5
4
4.5
5
5.5
3.5
4
4.5
5
5.5
6 The distribution of x when n = 2
The σ2
2
x
Note µ x = µ x and σ x =
:
2 E( x ) =1.0(1/36)+
1.5(2/36)+….=3.5 6/36
5/36 V(X) = (1.03.5)2(1/36)+
(1.53.5)2(2/36)... =
1.46 4/36
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 x
6.0 6 Sampling Distribution of the
Sampling
Mean
n= 5
µ x = 3.5
σ2
σ = .5833 = x )
(
56
2
x n = 10 n = 25 µ x = 3.5 µ x = 3.5 σ2
2
σ x = .2917 = x )
(
10 σ2
σ = .1167 = x )
(
25
2
x 7 Sampling Distribution of the
Sampling
Mean
n = 10 µ x = 3.5
σ2
2
σ x = .5833 = x )
(
5 n = 25 µ x = 3.5 n= 5 µ x = 3.5 σ2
σ = .2917 = x )
(
10 σ2
σ = .1167 = x )
(
25 2
x 2
x 2 Notice that σ x is smaller than σ x.
Notice that 2 is smaller than .
The larger the sample size the
The larger the sample size the
2
smaller
Therefore, tends
smaller σ x .. Therefore, x tends
o fall closer to µ as the sample
tto fall closer to µ ,, as the sample
size increases.
size increases.
8 Sampling Distribution of the
Sampling
Mean
Demonstration: The variance of the sample mean is
smaller than the variance of the population.
Mean = 1.5
Mean = 2.
Mean = 2.5 1.5
2.5
2
3
2
1.5
2.5
2
1.5
2
2.5
1.5
2
2.5
1.5 the variability of the population
2.5
2
Compare
1.5
2.5
1.5
2of the sample mean.
2.5
to the 1.5
variability
2
2.5
1.5
2.5
2
1.5
2.5
1.5
2
2.5
1.5
2
2.5
Let us take samples
1.5
2
2.5
Population 1 of two observations 9 Sampling Distribution of the
Sampling
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
If
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
the sampling distribution of x will resemble a
normal distribution.
normal 11 Sampling Distribution of the Sample
Sampling
Mean
Mean
1. µ x = µ x
2
σx
2
2. σ x =
n
3. If x is normal, x is normal. If x is nonnormal
x is approximately normally distributed for
sufficiently large sample size. 12 Sampling Distribution of the
Sample Mean
Sample Example 9.1
The amount of soda pop in each bottle is normally
The
distributed with a mean of 32.2 ounces and a
standard deviation of .3 ounces.
standard
Find the probability that a bottle bought by a
Find
customer will contain more than 32 ounces.
customer
Solution
0.7486 The random variable X is the
The
amount of soda in a bottle.
amount
P( x > 32) = P( x − µ 32 − 32.2
>
)
σx
.3 = P( z > −.67) = 0.7486 µ
x = 32 = 32.2
13 Sampling Distribution of the
Sample Mean
Sample Find the probability that a carton of four bottles will
Find
have a mean of more than 32 ounces of soda per
bottle.
bottle. Solution Define the random variable as the mean amount of soda per
Define
bottle.
bottle. P( x > 32) = P( x − µ 32 − 32.2
>
)
σx
.3 4 0.9082 = P( z > −1.33) = 0.9082 0.7486
x = 32
x = 32 µ = 32.2
µ x = 322
. 14 Sampling Distribution of the
Sample Mean
Sample Example 9.2 Dean’s claim: The average weekly income of
Dean’s
B.B.A graduates one year after graduation is
$600.
$600.
Suppose the distribution of weekly income has a
Suppose
standard deviation of $100. What is the probability
that 25 randomly selected graduates have an
average weekly income of less than $550?
average
Solution
x − µ 550− 600
P(x < 550 = P(
)
<
)
σx
100 25
= P(z < −2.5) = 0.0062 15 Sampling Distribution of the Sample
Sampling
Mean
Mean Example 9.2– continued If a random sample of 25 graduates actually had
If
an average weekly income of $550, what would
you conclude about the validity of the claim that
the average weekly income is 600?
the
Solution
With µ = 600 the probability of observing a sample mean
With
as low as 550 is very small (0.0062). The claim that the
mean weekly income is $600 is probably unjustified.
mean It will be more reasonable to assume that µ is smaller
It
than $600, because then a sample mean of $550
becomes more probable.
becomes 16 Using Sampling Distributions for
Using
Inference
Inference To make inference about population parameters we use
To
sampling distributions (as in Example 9.2).
sampling The symmetry of the normal distribution along with the
The
sample distribution of the mean lead to:
sample
P( −1.96 ≤ z ≤ 1.96 ) = .95, or P( −1.96 ≤ x −µ
≤ 1.96 ) = .95
σn This can be written as
σ
σ
P( −1.96
≤ x − µ ≤ 1.96
) = .95
n
n
which become
P(µ −1.96 σ
σ
≤ x ≤ µ +1.96
) = .95
n
n 17 Using Sampling Distributions for
Using
Inference
Inference
Standard normal distribution Z Normal distribution of
x
P(600 − 1.96 100
100
≤ x ≤ 600 + 1.96
) = .95
25
25 .95
.025
.025 .025 1.96 0 Z
1.96 µ − 1..96
P(600 − 196 .95 .025 µ
σ100
σ
100 µ= 6 0 0
Pµ600 +96
( + 1. 1.96
25
n 25
n x
18 Using Sampling Distributions for
Using
Inference
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
There
falls within the interval [560.8, 639.2] if the
population mean is 600.
population Since the sample mean was 550, the
Since
population mean is probably not 600.
population
19 9.3 Sampling Distribution of
9.3
a Proportion
Proportion The parameter of interest for nominal data
The
is the proportion of times a particular
proportion
outcome (success) occurs.
outcome To estimate the population proportion ‘p’
To
we use the sample proportion.
we
The number
of successes ^
p=
The estimate of p =
The X
n
20 9.3 Sampling Distribution of
9.3
a Proportion
Proportion
^ Since X is binomial, probabilities about p
Since
can be calculated from the binomial
distribution.
distribution.
^ Yet, for inference about p we prefer to use
Yet,
normal approximation to the binomial.
normal 21 Normal approximation to the
Normal
Binomial
Binomial Normal approximation to
Normal
the binomial works best
when
when the number of
the
experiments (sample
size) is large, and
size)
the probability of success,
the
p, is close to 0.5.
p, For the approximation to
For
provide good results two
conditions should be met:
conditions
np ≥ 5; n(1  p) ≥ 5 22 Normal approximation to the
Normal
Binomial
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:
µ = np; σ 2 = np(1  p) 23 Normal approximation to the
Normal
Binomial
Binomial
Let us build a normal
distribution to
approximate the binomial
P(X = 10).
P(XBinomial = 10) .176
= µ = np = 20(.5) = 10;
σ2 = np(1  p) = 20(.5)(1  .5) = 5
σ = 51/2 = 2.24 P(9.5<YNormal<10.5)
The approximation =~
P(9.5<Y<10.5)
9.5 = P( 10 10.5 9.5− 10
105 − 10
.
≤ Z≤
) = .1742
2.24
2.24 24 Normal approximation to the
Normal
Binomial
Binomial More examples of normal
More
approximation to the binomial
approximation
P(X ≤ 4) ≅ P(Y< 4.5)
P(X
4 P(X ≥ 14) ≅ P(Y > 13.5)
P(X 14) 4.5 13.5 14 25 Approximate Sampling Distribution
Approximate
of a Sample Proportion
of From the laws of expected value and variance,
From
it can be shown that E( p ) = p and V( p )
ˆ
ˆ
=p(1p)/n
=p(1p)/n If both np > 5 and np(1p) > 5, then
ˆ
p −p
z=
p(1−p)
n Z iis approximately standard normally
s
distributed.
distributed. 26 Example 9.3 A state representative received 52% of the
state
votes in the last election.
votes
One year later the representative wanted
One
to study his popularity.
to
If his popularity has not changed, what is
If
the probability that more than half of a
sample of 300 voters would vote for him?
27 Example 9.3 Solution The
The number of respondents who prefer the
representative is binomial with n = 300 and p
= .52. Thus, np = 300(.52) = 156 and
.52.
n(1p) = 300(1.52) = 144 (both greater than 5) ˆ
P( p > .50) = P ˆ
p− p
p (1 − p ) n > = .7549
(.52)(1 − .52) 300 .50 − .52 28 9.4 Sampling Distribution of the
9.4
Difference Between Two Means
Difference Independent samples are drawn from
Independent
each of two normal populations
each We’re interested in the sampling
We’re
distribution of the difference between the
two sample means x1 − x2
two 29 Sampling Distribution of the
Sampling
Difference Between Two Means
Difference The distribution of x1 − x2 is normal if The two samples are independent, and
The parent populations are normally
The
distributed.
distributed. If the two populations are not both
If
not
normally distributed, but the sample
sizes are 30 or more, the distribution of x1 − x2 is approximately normal. 30 Sampling Distribution of the
Sampling
Difference Between Two Means
Difference Applying the laws of expected value
Applying
and variance we have:
and
E(x1 − x2 ) = E(x1) − E(x2 ) = µ1 − µ 2
σ
σ
V(x1 − x2 ) = V(x1 ) + V(x2 ) =
+
n
n
2
1 2
2 We can define:
Z= ( x 1 − x 2 ) − ( µ1 − µ 2 )
2
σ1 σ 2
+2
n1 n2 31 Sampling Distribution of the
Difference Between Two Means
Difference
Example 9.4 The starting salaries of MBA students from
The
two universities (WLU and UWO) are $62,000
(stand.dev. = $14,500), and $60,000 (stand.
dev. = $18,3000).
dev.
What is the probability that a sample mean of
What
WLU students will exceed the sample mean of
UWO students? (nWLU = 50; nUWO = 60)
UWO 32 Sampling Distribution of the
Sampling
Difference Between Two Means
Difference Example 9.4 – Solution
We need to determine P( x1 − x 2 > 0)
µ 1  µ 2 = 62,000  60,000 = $2,000
2
σ 12 σ 2
14,500 2 18,300 2
+
=
+
= $3,128
n
n
50
60 P(x1 − x2 > 0) = P( x1 − x2 − (µ1  µ 2 )
2
σ1 σ2
+2
n1 n2 = P(z > −.64 = .5+ .2389 .7389
)
= 0− 2000
>
)
3128 33 ...
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 Spring '08
 ROCHON
 Normal Distribution, Probability theory, Binomial distribution

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