CABT SHS Statistics & Probability - Mean and Variance of Sampling Distributions of Sample Means

CABT Statistics & Probability – Grade 11 Lecture Presentation

Mean & Variance of
Sampling
Distributions of
Sample Means
A Grade 11 Statistics and Probability Lecture

At the end of this lesson, you are
expected to:
 find the mean and variance of
the sampling distribution of the
sample means.
 state and explain the Central
Limit Theorem.
 use the Central Limit Theorem to
solve problems involving means
and variances of sampling
distribution of means.
Lesson Objectives
Mean and Variance of Sampling Distributions of Sample Means

Review: Finding the Mean and Variance of Discrete
Probability Distributions
X P(X)
1 0.1
2 0.3
3 0.2
4 0.4
X P(X)
0.1
0.6
0.6
1.6
X2P(X)
0.1
1.2
1.8
6.4
2.9 9.5SUMS
Mean: 2.9 
Variance:
2 2
9.5 2.9  
1.09
Standard deviation:
1.09  1.04

Recall:
Sampling Distribution of the Mean
A sampling distribution of
sample means is a frequency
distribution using the means
computed from all possible
random samples of a specific
size taken from a population.
The probability distribution of the sample
means is also called the sampling distribution
of the sample means, with the sample mean as
the random variable.
http://www.philender.com/courses/intro/notes2/sam.gif

Recall: Constructing Sampling Distribution
of Sample Means
How to Construct a Sampling
Distribution of Sample Means from a
Given PopulationSTEP 1 – Determine the number of
samples of size n from the population of
size N.
STEP 2 – List all the possible samples
and compute the mean of each sample.
STEP 3 – Construct a frequency
distribution of the sample means
obtained in Step 2.
STEP 4 – Construct the probability

The Mean and Variance of Sampling
Distribution of Sample Means
Because the sampling distribution P(X)
of
sample means X is essentially a
discrete probability distribution, the
computation of the mean and variance
of P(X) is the SAME as for any discrete
probability distribution P(X).
Ows,
talaga?
Mismo!

For the sampling distribution of
the sample means :
Mean:  X
X P X   g
Varianc
e:
 2 2 2
X X
X P X   g
Standard
deviation:
2
X X
  
 P X
X

Consider a group of N = 4 people
with the following ages: 18, 20, 22,
24. Consider samples of size n = 2
from the group without
replacement.
If X is the age of one person in
the group and X is the average
age of the two people in a
sample, find the mean and
variance of the sampling
distribution of X.
The Mean and Variance of Sampling Distribution
of Sample Means

Sampling without replacement
Number of samples of size 2:4 2
6C 
Sample
Sample
Mean
18, 20 19
18, 22 20
18, 24 21
20, 22 21
20, 24 22
22, 24 23
Sample
Mean
Frequency Probability
19 1 1/6
20 1 1/6
21 2 1/3
22 1 1/6
23 1 1/6
TOTAL 6 1
of Sample Means

Sample
Mean
19 1 1/6
20 1 1/6
21 2 1/3
22 1 1/6
23 1 1/6
TOTAL 6 1
X P(X)
19 1/6
20 1/6
21 1/3
22 1/6
23 1/6
Sampling without replacement
Number of samples of size 2:4 2
6C 
of Sample Means

Computing the mean and
variance:
X P(X)
19 1/6
20 1/6
21 1/3
22 1/6
23 1/6
X P(X) X2 P(X)
3.17 60.17
3.33 66.67
7.00 147.00
3.67 80.67
3.83 88.17
21.00 442.67SUMS
Mean: 21X
 
Variance:
2 2
442.67 21X
  
1.67
of Sample Means

Redo Example 1 if sampling is with
replacement.
Sampling with replacement
Number of samples of size 2: 2
4 16
Sample Mean Sample Mean Sample Mean Sample Mean
18,18 18 20, 18 19 22,18 20 24,18 21
18, 20 19 20, 20 20 22, 20 21 24, 20 22
18, 22 20 20, 22 21 22, 22 22 24, 22 23
18, 24 21 20, 24 22 22, 24 23 24, 24 24

Sampling with replacement
Number of samples of size 2: 2
4 16
Sample
Mean
18 1 1/16
19 2 1/8
20 3 3/16
21 4 1/4
22 3 3/16
23 2 1/8
24 1 1/16
TOTAL 16 1
X P(X)
18 1/16
19 1/8
20 3/16
21 1/4
22 3/16
23 1/8
24 1/16
of Sample Means

X P(X)
18 1/16
19 1/8
20 3/16
21 1/4
22 3/16
23 1/8
24 1/16
Computing the mean and variance:
X P(X) X2 P(X)
1.13 20.25
2.38 45.13
3.75 75.00
5.25 110.25
4.13 90.75
2.88 66.13
1.50 36.00
21 443.5SUMS
Mean: 21X
 
Variance:
2 2
443.5 21X
  
2.5

Consider a population with values
1, 1, 2, 3. Samples of size 3 are drawn
from the population without
replacement.
a. Construct a sampling distribution
of the sample means for this
population.
b. Find the mean, variance, and
standard deviation of the
sampling distribution.SOLUTION HERE!
of Sample Means

QUESTION!
How are the ACTUAL mean and
variance of the population
related to the mean and
variance of the sampling
distribution?
of Sample Means

Example 1 and 2 revisited:
Population: (18, 20, 22, 24)
of Sample Means
Mean of the population:
18 20 22 24
21
4
  
  
X X -  (X - )2
18 3 9
20 1 1
22 1 1
24 3 9
sum 20
Variance of the population:
 
2
2
X
N
 
 
 20
5
4
 

Let’s compare!
Mean Variance
Population
Sampling
Distribution
(samples of size 2 without replacement)
21 
21X
 
2
5 
2
1.67X
 
Population: (18, 20, 22, 24)
Sampling: n = 2, without replacement
of Sample Means

Let’s compare!
Mean Variance
Population
Sampling
Distribution
(samples of size 2 with replacement)
21 
21X
 
2
5 
2
2.5X
 
Population: (18, 20, 22, 24)
Sampling: n = 2, with replacement
of Sample Means

Let’s compare!
Values of variances (N = 4, n = 2)
of Sample Means
Without replacement:
2
2 5 4 2
1.67
2 4 1 1X
N n
n N
      
          
With replacement:
2
2 5
2.5
2X
n

   

Properties of the Mean and Variance of Sampling
Distributions of Sample Means
of Sample Means
Condition MEAN VARIANCE
For infinite
populations or for
sampling with
replacement
For finite
populations or for
sampling without
replacement
X
  
X
  
2
2
X
n

 
2
2
1X
N n
n N
   
    

Properties of the Mean and Variance of
Sampling Distributions of Sample Means
of Sample Means
1. The mean of the sample means will
be the same as the population mean.
2. The variance or standard deviation of
the sample means will be smaller
than the variance or standard
deviation of the population.

If a population is normal with mean μ and
standard deviation σ, the sampling distribution
of
is also normally distributed with the
following:
X
of Sample Means
mean standard deviation*variance
X
  
2
2
X
n

  X
n

 
*Note: For , you can also useX
 2
X X
  

If the population is infinite with mean μ and
standard deviation σ or the sampling is with
replacement, use the following:
X
  
2
2
X
n

  X
n

 
of Sample Means
 2
X X
  

If the population is finite with mean μ and
standard deviation σ or the sampling is
without replacement, use the following:
X
  
2
2
1X
N n
n N
   
    
of Sample Means
1X
N n
Nn
 
  

 2
X X
  

Normal
Population
Distribution
Normal Sampling
Distribution
(has the same mean)
x
  
of Sample Means
μ X
xμ X
For a normal distribution:

For sampling with replacement:
Larger
sample size
Smaller
sample size
xμ
of Sample Means
As the sample size n increases,
decreases.
X


The Standard Error of the Mean
A measure of the variability in the mean from
sample to sample is given by the Standard Error
of the Mean, which is simply equal to the standard
deviation of the sampling distribution of the sample
means :
Note that the standard error of the mean
decreases as the sample size increases.
X

X
n

 
1X
N n
Nn
 
  


The Finite Population
Correction Factor
1
N n
N


The FINITE POPULATION CORRECTION
FACTOR is used for computing the standard
error of the mean from a finite population. It is
given by the expression
Wow!

Correction Factor
The correction factor is necessary if
relatively large samples are taken from a
small population, because the sample mean
will then more accurately estimate the
population mean and there will be less error
in the estimation.
Why the need for the
correction factor? 1
N n
N



Correction Factor
The finite population correction factor is
also used when the sample size n is
LARGE relative to the population N.
This means that n is larger than 5% of
N:
0.05n N
For example, if N = 10 and n = 6,
 6 0.05 10 5 
so the correction factor is needed in this case.

Consider an infinite population
with mean 20 and standard
deviation 2. Samples of size 4 are
obtained from the population.
Determine the following:
a. Mean of the sampling
distribution
b. Variance of the standard
distribution
c. Standard error of the mean
of Sample Means

of Sample Means
Given:  = 20,  = 2, n = 4
Since the population is normal and
infinite, we use the following formulas:
mean
sampling error of
the mean
(standard deviation)
variance
X
  
2
2
X
n

 
X
n

 
20
2
2
4
 1
2
4
 1

Consider a finite population with
size 10 with mean 20 and standard
deviation 2. Samples of size 4 are
obtained from the population
without replacement. Determine the
following:
a. Mean of the sampling distribution
b. Variance of the standard
distribution
c. Standard error of the mean

The Mean and Variance of Sampling Distributions
of Sample Means
Given:  = 20,  = 2, n = 4, N = 10
Since the population is finite and sampling is
without replacement, we use the following
formulas:
mean
sampling error of
the mean
(standard deviation)
variance
X
  
2
2
1X
N n
n N
   
    
2
X X
  
20
2
2 10 4
4 10 1
  
   
2
0.67X
 
0.67 0.82 

If a population is normal with mean μ and standard
deviation σ, the sampling distribution of the sample
means is also normally distributed, regardless of the
sample size.
What can we say about the shape of the
sampling distribution of the sample means
when the population from which the sample is
selected is not normal?
QUESTION!
The Mean and Variance of Sampling Distributions
of Sample Means

The Central Limit
Theorem
“The World is
Normal”
Theorem

X
   X
n

 
The mean and standard deviation of the
distribution are, respectively,
If random samples of size n are drawn from a
population with replacement, then as n becomes
larger, the sampling distribution of the mean
approaches the normal distribution,
regardless of the shape of the population
distribution.
The Central Limit Theorem

What does this
theorem mean?
If the sample size n drawn from a finite
population of any shape is LARGE enough,
then it is safe to ASSUME that the distribution
is APPROXIMATELY NORMAL, and thus the
following can be used:
X
   X
n

 

How largeis
“largeenough”?
According to most statistics books,
the sample size n is LARGE
ENOUGH when…
 30n
Wow!

From Statistics for Business and Economics, 6e © 2007 Pearson Education, Inc.
Population Distribution
(not normal)
Sampling Distribution
(becomes normal as nincreases) Larger
sample
sizeSmaller
sample
size
Sampling distribution
properties:
μμx 
n
σ
σx 
x
μ
x
xμ

The Central
Limit Theorem
ILLUSTRATION
OF THE CENTRAL
LIMIT THEOREM
FOR THREE
POPULATIONS
(from Statistics for Business and
Economics by Anderson, Sweeney and
Williams, 10th ed)

Two things to remember in using the Central Limit
Theorem:
1. When the original variable is normally
distributed, the distribution of the sample means
will be normally distributed for any sample size
n.
2. When the distribution of the original variable
might not be normal, a sample size of 30 or
more is needed to use a normal distribution to
approximate the distribution of the sample
means. The larger the sample, the better the
approximation will be.

RECALL: If the distribution of a random
variable X is normal with mean  and standard
deviation , the equivalent z-scores is obtained
by using
x
z




This formula transforms the values of the
variable x into standard units or z values.
Application of the
Central Limit Theorem

For a normally (or approximately normally)
distributed population with mean  and
standard deviation , the equivalent z-score of
a sample mean from the sampling
distribution of means is given by the formula
x


x
x
z




x
z
n


or
Application of the

Distribution of Sample Means for a Large Number of
Samples
Application of the

A population is normally distributed
with mean 20 and standard deviation
6. Samples of size 9 are drawn from
the population. What are the
equivalent z-scores of the following
sample means?a. b.22x  25x 
Application of the

A population is normally
distributed with mean 20 and
standard deviation 6. Samples of
size 9 are drawn from the
population. What is the
probability that the sample mean
is
a. greater than 22?
b. between 22 and 25?
Application of the

A population has mean 50 and
standard deviation 12. Samples
of size 36 are drawn from the
population. What is the
probability that the sample mean
is
a. less than 48?
b. between 42 and 50?
c. greater than 53.5?
Application of the

A recent study of a company regarding the
lifetimes of cell phones it issues to its
employees revealed that the average is
24.3 months and the standard deviation is
2.6 months.
a. If a company phone is chosen at
random, what is the probability that the
phone will last for more than 3 years?
b. If a company randomly samples 40
phones, what is the probability that the
average lifetime of the sample is more
than 3 years?
Application of the

Exercise 4 on page 133, edited
In a study of life expectancy of 400 people
in a certain geographic region, the mean
age of death is 70 years and the standard
deviation is 5.1 years. If a sample of 50
people from this region is selected, what
is the probability that the mean life
expectancy is less than 68 years?
Application of the

Check your
understandingDo Exercise
2 on page
132 and
Extension
A#1 on p.
133 of your

Assignment
Do Exercise #1 on
p. 132, due on
Thursday,
August 25.
Quiz on the Central
Limit Theorem on
Tuesday, September

Quiz #1
A population has a mean of 90 and
standard deviation 3.5. Samples of size
49 are taken from the population and the
sample means are taken.
1. Determine the z-scores of the
following sample means:
    88.5 90.15a x b x
2. What is the probability that the
sample mean will lie between 88.5 and
90.15?

CABT SHS Statistics & Probability - Mean and Variance of Sampling Distributions of Sample Means

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CABT SHS Statistics & Probability - Mean and Variance of Sampling Distributions of Sample Means