Chap007.ppt

Copyright © 2015, 2012, 2009 Pearson Education, Inc. Chapter 07, Slide 1
Sampling Distributions
Next chapters:
Ch 8, 9,
13 & 14
Chapter 7

Learning Objectives
In this chapter, you learn:
 The concept of the sampling distribution
 To compute probabilities related to the sample
mean and the sample proportion
 The importance of the Central Limit Theorem

Sampling Distributions
 A sampling distribution is a distribution of all of the
possible values of a sample statistic for a given sample
size selected from a population.
 For example, suppose you sample 50 students from your
college regarding their mean GPA. If you obtained many
different samples of size 50, you will compute a different
mean for each sample. We are interested in the
distribution of all potential mean GPAs we might calculate
for any sample of 50 students.
DCOVA

Developing a Sampling Distribution
 Assume there is a population …
 Population size N=4
 Random variable, X,
is age of individuals
 Values of X: 18, 20,
22, 24 (years)
A B C D
DCOVA

.3
.2
.1
0
18 20 22 24
A B C D
Uniform Distribution
P(x)
x
(continued)
Summary Measures for the Population Distribution:
21
4
24
22
20
18
N
X
μ i







2.236
N
μ)
(X
σ
2
i




DCOVA

16 possible samples
(sampling with
replacement)
Now consider all possible samples of size n=2
1st 2nd Observation
Obs 18 20 22 24
18 18 19 20 21
20 19 20 21 22
22 20 21 22 23
24 21 22 23 24
(continued)
16 Sample
Means
1st
Obs
2nd Observation
18 20 22 24
18 18,18 18,20 18,22 18,24
20 20,18 20,20 20,22 20,24
22 22,18 22,20 22,22 22,24
24 24,18 24,20 24,22 24,24
DCOVA

1st 2nd Observation
Obs 18 20 22 24
18 18 19 20 21
20 19 20 21 22
22 20 21 22 23
24 21 22 23 24
Sampling Distribution of All Sample Means
18 19 20 21 22 23 24
0
.1
.2
.3
P(X)
X
Sample Means
Distribution
16 Sample Means
_
(continued)
(no longer uniform)
_
DCOVA

Summary Measures of this Sampling Distribution:
Developing A Sampling Distribution
(continued)
21
16
24
19
19
18
μX







1.58
16
21)
-
(24
21)
-
(19
21)
-
(18
σ
2
2
2
X






DCOVA
Note: Here we divide by 16 because there are 16
different samples of size 2.

Comparing the Population Distribution
to the Sample Means Distribution
18 19 20 21 22 23 24
0
.1
.2
.3
P(X)
X
18 20 22 24
A B C D
0
.1
.2
.3
Population
N = 4
P(X)
X _
1.58
σ
21
μ X
X


2.236
σ
21
μ 

Sample Means Distribution
n = 2
_
DCOVA

Sample Mean Sampling Distribution:
Standard Error of the Mean
 Different samples of the same size from the same
population will yield different sample means
 A measure of the variability in the mean from sample to
sample is given by the Standard Error of the Mean:
(This assumes that sampling is with replacement or
sampling is without replacement from an infinite population)
 Note that the standard error of the mean decreases as
the sample size increases
n
σ
σX

DCOVA

If the Population is Normal
 If a population is normal with mean μ and
standard deviation σ, the sampling distribution
of is also normally distributed with
and
X
μ
μX

n
σ
σX

DCOVA

Z-value for Sampling Distribution
of the Mean
 Z-value for the sampling distribution of :
where: = sample mean
= population mean
= population standard deviation
n = sample size
X
μ
σ
n
σ
μ)
X
(
σ
)
μ
X
(
Z
X
X 



X
DCOVA

Normal Population
Distribution
Normal Sampling
Distribution
(has the same mean)
Sampling Distribution Properties
(i.e. is unbiased )
x
x
x
μ
μx 
μ
x
μ
DCOVA

Sampling Distribution Properties
As n increases,
decreases
Larger
sample size
Smaller
sample size
x
(continued)
x
σ
μ
DCOVA

Determining An Interval Including A
Fixed Proportion of the Sample Means
Find a symmetrically distributed interval around µ
that will include 95% of the sample means when µ
= 368, σ = 15, and n = 25.
 Since the interval contains 95% of the sample means
5% of the sample means will be outside the interval
 Since the interval is symmetric 2.5% will be above
the upper limit and 2.5% will be below the lower limit.
 From the standardized normal table, the Z score with
2.5% (0.0250) below it is -1.96 and the Z score with
2.5% (0.0250) above it is 1.96.
DCOVA

Determining An Interval Including A
Fixed Proportion of the Sample Means
 Calculating the lower limit of the interval
 Calculating the upper limit of the interval
 95% of all sample means of sample size 25 are
between 362.12 and 373.88
12
.
362
25
15
)
96
.
1
(
368 





n
Z
X L
σ
μ
(continued)
88
.
373
25
15
)
96
.
1
(
368 




n
Z
XU
σ
μ
DCOVA

If the Population is not Normal
 We can apply the Central Limit Theorem:
 Even if the population is not normal,
 …sample means from the population will be
approximately normal as long as the sample size is
large enough.
Properties of the sampling distribution:
and
μ
μx 
n
σ
σx 
DCOVA

n↑
Central Limit Theorem
As the
sample
size gets
large
enough…
the sampling
distribution of
the sample
mean becomes
almost normal
regardless of
shape of
population
x
DCOVA

Population Distribution
Sampling Distribution
(becomes normal as n increases)
Central Tendency
Variation
x
x
Larger
sample
size
Smaller
sample size
If the Population is not Normal
(continued)
Sampling distribution
properties:
μ
μx 
n
σ
σx 
x
μ
μ
DCOVA

How Large is Large Enough?
 For most distributions, n > 30 will give a
sampling distribution that is nearly normal
 For fairly symmetric distributions, n > 15
 For a normal population distribution, the
sampling distribution of the mean is always
normally distributed
DCOVA

Example
 Suppose a population has mean μ = 8 and
standard deviation σ = 3. Suppose a random
sample of size n = 36 is selected.
 What is the probability that the sample mean is
between 7.8 and 8.2?
DCOVA

Example
Solution:
 Even if the population is not normally
distributed, the central limit theorem can be
used (n > 30)
 … so the sampling distribution of is
approximately normal
 … with mean = 8
 …and standard deviation
(continued)
x
x
μ
0.5
36
3
n
σ
σx 


DCOVA

Example
Solution (continued):
(continued)
0.3108
0.3446
-
0.6554
0.4)
Z
P(-0.4
36
3
8
-
8.2
n
σ
μ
-
X
36
3
8
-
7.8
P
8.2)
X
P(7.8




















Z
7.8 8.2 -0.4 0.4
Sampling
Distribution
Standard Normal
Distribution
Population
Distribution
?
?
?
?
?
?
?
?
?
?
?
?
Sample Standardize
8
μ  8
μX
 0
μz 
x
X
DCOVA

Population Proportions
π = the proportion of the population having
some characteristic
 Sample proportion (p) provides an estimate
of π:
 0 ≤ p ≤ 1
 p is approximately distributed as a normal distribution
when n is large
(assuming sampling with replacement from a finite population or
without replacement from an infinite population)
size
sample
interest
of
stic
characteri
the
having
sample
in the
items
of
number
n
X
p 

DCOVA

Sampling Distribution of p
 Approximated by a
normal distribution if:

where
and
(where π = population proportion)
P(ps)
.3
.2
.1
0
0 . 2 .4 .6 8 1 p
π

p
μ
n
)
(1
σp
π
π 

5
)
n(1
5
n
and





DCOVA

Z-Value for Proportions
n
)
(1
p
σ
p
Z
p 








Standardize p to a Z value with the formula:
DCOVA

Example
 If the true proportion of voters who support
Proposition A is π = 0.4, what is the probability
that a sample of size 200 yields a sample
proportion between 0.40 and 0.45?
 i.e.: if π = 0.4 and n = 200, what is
P(0.40 ≤ p ≤ 0.45) ?
DCOVA

Example
 if π = 0.4 and n = 200, what is
P(0.40 ≤ p ≤ 0.45) ?
(continued)
0.03464
200
0.4)
0.4(1
n
)
(1
σp 






1.44)
Z
P(0
0.03464
0.40
0.45
Z
0.03464
0.40
0.40
P
0.45)
p
P(0.40








 






Find :
Convert to
standardized
normal:
p
σ
DCOVA

Example
Z
0.45 1.44
0.4251
Standardize
Standardized
Normal Distribution
 if π = 0.4 and n = 200, what is
P(0.40 ≤ p ≤ 0.45) ?
(continued)
Utilize the cumulative normal table:
P(0 ≤ Z ≤ 1.44) = 0.9251 – 0.5000 = 0.4251
0.40 0
p
DCOVA

Chapter Summary
In this chapter we discussed
 Sampling distributions
 The sampling distribution of the mean
 For normal populations
 Using the Central Limit Theorem
 The sampling distribution of a proportion
 Calculating probabilities using sampling distributions

On-Line Topic:
Sampling from Finite
Populations
Chapter 7

Chap007.ppt

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