Transcript Chapter 8 Section 1 Distributions of the Sample Mean Chapter 8 – Section 1 • Learning objectives – Understand the concept of a sampling distribution– Describe the.

Chapter 8
Section 1
Distributions of the
Sample Mean
Chapter 8 – Section 1
• Learning objectives
1
–
Understand the concept of a sampling distribution
2
– Describe the distribution of the sample mean for
3 samples obtained from normal populations
– Describe the distribution of the sample mean for
samples obtained from a population that is not
normal
Chapter 8 – Section 1
• Learning objectives
1
–
Understand the concept of a sampling distribution
2
– Describe the distribution of the sample mean for
3 samples obtained from normal populations
– Describe the distribution of the sample mean for
samples obtained from a population that is not
normal
Chapter 8 – Section 1
• Lets look at a small population.
2,4,6 8,4,3,7
Find the mean. Is this mu or x bar?
Now lets take all the possible samples of 2 from
this population.
4
2,4,6 8,4,3,7
2,2
2,4
2,6
2,8
2,4
2,3
2,7
4,4
4,6
4,8
4,4
4,3
4,7
6,6
6,8
6,4
6,3
6,7
8,8
8,4
8,3
8,7
4,4
4,3
4,7
3.3
3,7
7,7
5
2,4,6 8,4,3,7
x
2,2
2,4
2,6
2,8
2,4
2,3
2,7
x
4,4
4,6
4,8
4,4
4,3
4,7
x
x
6,6
6,8
6,4
6,3 x
6,7
8,8
8,4
8,3
8,7
x
4,4
4,3
4,7
3.3
3,7
7,7
6
Chapter 8 – Section 1
• Often the population is too large to perform a census
… so we take a sample
• How do the results of the sample apply to the
population?
– What’s the relationship between the sample mean and the
population mean?
– What’s the relationship between the sample standard
deviation and the population standard deviation?
• This is statistical inference
Chapter 8 – Section 1
• We want to use the sample mean x to estimate
the population mean μ
• If we want to estimate the heights of eight year
old girls, we can proceed as follows
– Randomly select 100 eight year old girls
– Compute the sample mean of the 100 heights
– Use that as our estimate
• This is using the sample mean to estimate the
population mean
Chapter 8 – Section 1
• However, if we take a series of different
random samples
–
–
–
–
Sample 1 – we compute sample mean x1
Sample 2 – we compute sample mean x2
Sample 3 – we compute sample mean x3
Etc.
• Each time we sample, we may get a different
result
• The sample mean x is a random variable!
Chapter 8 – Section 1
• Because the sample mean is a random variable
– The sample mean has a mean
– The sample mean has a standard deviation
– The sample mean has a probability distribution
• This is called the sampling distribution of the
sample mean
CENTRAL LIMIT THEOREM
• The Central Limit Theorem is about the
distribution of sample means. x
The Central Limit Theorem is what we
will base most or the statistical concepts
on for the rest of this course.
Definition
Sampling Distribution of the mean
is the probability distribution of
sample means, with all samples
having the same sample size n.
Central Limit Theorem
Given:
1. The random variable x has a distribution (which
may or may not be normal) with mean µ and
standard deviation .
2. Samples all of the same size n are randomly
selected from the population of x values.
http://onlinestatbook.com/stat_sim/sampling_dist/index.html
Central Limit Theorem
Conclusions:
Central Limit Theorem
Conclusions:
1. The distribution of sample x will, as the
sample size increases, approaches a normal
distribution.
Central Limit Theorem
Conclusions:
1. The distribution of sample x will, as the
sample size increases, approach a normal
distribution.
2. The mean of the sample means will be the
population mean µ.
Central Limit Theorem
Conclusions:
1. The distribution of sample x will, as the
sample size increases, approach a normal
distribution.
2. The mean of the sample means will be the
population mean µ.
3. The standard deviation of the sample means
n
will approach 
Practical Rules Commonly Used:
1. For samples of size n larger than 30, the distribution of
the sample means can be approximated reasonably well
by a normal distribution. The approximation gets better
as the sample size n becomes larger.
2. If the original population is itself normally distributed,
then the sample means will be normally distributed for
any sample size n (not just the values of n larger than 30).
http://onlinestatbook.com/stat_sim/sampling_dist/index.html
Notation
Notation
the mean of the sample means
µx = µ
Notation
the mean of the sample means
µx = µ
the standard deviation of sample mean

x = n
Notation
the mean of the sample means
µx = µ
the standard deviation of sample mean

x = n
(often called standard error of the mean)
As the sample size increases, the
sampling distribution of sample
means approaches a normal
distribution.
Example: Given the population of women has normally
distributed weights with a mean of 143 lb and a standard
deviation of 29 lb,
a.) if one woman is randomly selected, find the probability
that her weight is greater than 150 lb.
b.) if 36 different women are randomly selected, find the
probability that their mean weight is greater than 150 lb.
Example: Given the population of women has normally
distributed weights with a mean of 143 lb and a standard
deviation of 29 lb,
a.) if one woman is randomly selected, find the probability
that her weight is greater than 150 lb.
z = 150-143 = 0.24
29
P(x>150) = P(z>.24)
= 1 - 0.5948 = 0.4052
P(z>.24) = 1 - 0.5948 = 0.4052
0.5948
 = 143
= 29
0
150
0.24
Example: Given the population of women has normally
distributed weights with a mean of 143 lb and a standard
deviation of 29 lb,
b.) if 36 different women are randomly selected, find the
probability that their mean weight is greater than 150 lb.
Example: Given the population of women has normally
distributed weights with a mean of 143 lb and a standard
deviation of 29 lb,
b.) if 36 different women are randomly selected, find the
probability that their mean weight is greater than 150 lb.
x = 143
150
x = 29 = 4.83333
36
Example: Given the population of women has normally
distributed weights with a mean of 143 lb and a standard
deviation of 29 lb,
b.) if 36 different women are randomly selected, find the
probability that their mean weight is greater than 150 lb.
z = 150-143 = 1.45
29
36
0.9265
x = 143
x = 4.83333
0
150
1.45
normally distributed weights with a mean of 143 lb and
a standard deviation of 29 lb,
b.) if 36 different women are randomly selected, find
the probability that their mean weight is greater than
150 lb.
P(x36  150)  P(z 
150  143
29
36
 P (z  145
. )
z = 150-143 = 1.45
29
36
 1 0.9265
 0.0735
= 0.0735
0.9265
x = 143
x = 4.83333
0
150
1.45
)
Example: Given the population of women has normally
distributed weights with a mean of 143 lb and a standard
deviation of 29 lb,
b.) if 36 different women are randomly selected, the
probability that their mean weight is greater than 150 lb is
0.0735.
z = 150-143 = 1.45
29
36
1 - 0.9265 = 0.0735
0.4265
x = 143
x = 4.83333
0
150
1.45
Example: Given the population of women has normally
distributed weights with a mean of 143 lb and a standard
deviation of 29 lb,
Example: Given the population of women has normally
distributed weights with a mean of 143 lb and a standard
deviation of 29 lb,
a.) if one woman is randomly selected, find the probability
that her weight is greater than 150 lb.
P(x > 150) = 0.4052
Example: Given the population of women has normally
distributed weights with a mean of 143 lb and a standard
deviation of 29 lb,
a.) if one woman is randomly selected, find the probability
that her weight is greater than 150 lb.
P(x > 150) = 0.4052
b.) if 36 different women are randomly selected, their mean
weight is greater than 150 lb.
P(x > 150) = 0.0735
Example: Given the population of women has normally
distributed weights with a mean of 143 lb and a standard
deviation of 29 lb,
a.) if one woman is randomly selected, find the probability
that her weight is greater than 150 lb.
P(x > 150) = 0.4052
b.) if 36 different women are randomly selected, their mean
weight is greater than 150 lb.
P(x36 > 150) = 0.0735
It is much easier for an individual to deviate from the
mean than it is for a group of 36 to deviate from the mean.
Sampling Without Replacement
If n > 0.05 N
Sampling Without Replacement
If n > 0.05 N
x =

n
N-n
N-1
Sampling Without Replacement
If n > 0.05 N
x =

n
N-n
N-1
finite population
correction factor
38
Frequency
Distribution of 50 Sample Means
for 50 Students
15
10
5
0
0
Figure 5-20
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