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```Where We Left Off
• What is the probability of randomly
selecting a sample of three individuals, all
of whom have an I.Q. of 135 or more?
So while the odds chance selection of a single person this far
• Find the z-score of 135, compute the tail
above the mean is not all that unlikely, the odds of a sample this
region
it to the 3rd power.
far above
theand
meanraise
are astronomical
z = 2.19
P = 0.0143
0.01433 = 0.0000029
X
X
Anthony J Greene
1
Sampling Distributions
I What is a Sampling Distribution?
A If all possible samples were drawn from
a population
B A distribution described with
Central Tendency µM
And dispersion σM ,the standard error
II The Central Limit Theorem
Anthony J Greene
2
Sampling Distributions
• What you’ve done so far is to determine the
position of a given single score, x,
compared to all other possible x scores
x
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3
Sampling Distributions
• The task now is to find the position of a
group score, M, relative to all other possible
sample means that could be drawn
M
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4
Sampling Distributions
• The reason for this is to find the probability
of a random sample having the properties
you observe.
M
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5
Sampling Distributions
1. Any time you draw a sample from a population,
the mean of the sample, M , it estimates the
population mean μ, with an average error of:

n
2. We are interested in understanding the probability
of drawing certain samples and we do this with
our knowledge of the normal distribution applied
to the distribution of samples, or Sampling
Distribution
3. We will consider a normal distribution that
consists of all possible samples of size n from a
given population Anthony J Greene
6
Sampling Error
Sampling error is the error
resulting from using a sample
to estimate a population
characteristic.
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7
Sampling Distribution of the Mean
For a variable x and a given sample size, the
distribution of the variable M (i.e., of all possible
sample means) is called the
sampling distribution of the mean.
The sampling distribution is purely theoretical
derived by the laws of probability.
A given score x is part of a distribution for that
variable which can be used to assess probability
A given mean M is part of a sampling
distribution for that variable which can be used
to determine the probability
of a given sample
Anthony J Greene
being drawn
8
The Basic Concept
• Extreme events are unlikely -- single events
• For samples, the likelihood of randomly
selecting an extreme sample is more
unlikely
• The larger the sample size, the more
unlikely it is to draw an extreme sample
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9
The original distribution
of x: 2, 4, 6, 8
Now consider all possible
samples of size n = 2
What is the distribution
of sample means M
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10
The Sampling Distribution For n=2
Notice that it’s a normal distribution with μ = 5
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11
Heights of the five starting players
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12
Possible samples and sample means
for samples of size two
M
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Dotplot for the sampling
distribution of the mean for
samples of size two (n = 2)
M
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14
Possible samples and sample means
for samples of size four
M
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15
Dotplot for the sampling
distribution of the mean for
samples of size four (n = 4)
M
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16
Sample size and sampling error
illustrations for the heights of the
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17
Dotplots for
the
sampling
distributions
of the mean
for samples
of sizes one,
two, three,
four, and
five
M
M
M
M
M
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18
Sample Size and Standard Error
The possible sample means cluster closer around the
population mean as the sample size increases. Thus
the larger the sample size, the smaller the sampling
error tends to be in estimating a population mean,
m, by a sample mean, M.
For sampling distributions, the dispersion is called
Standard Error. It works much like standard
deviation.
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19
Standard Error of M
For samples of size n, the standard error of the variable x equals
the standard deviation of x divided by the square root of the
sample size:
M 

n
In other words, for each sample size, the standard error of all
possible sample means equals the population standard deviation
divided by the square root of the sample size.
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20
The Effect of Sample Size on Standard Error
The distribution of sample means for random samples of
size (a) n = 1, (b) n = 4, and (c) n = 100 obtained from a
normal population with µ = 80 and σ = 20. Notice that the
size of the standard error decreases as the sample size
increases.
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21
Mean of the Variable M
For samples of size n, the mean of the variable M
equals the mean of the variable under consideration:
mM  m .
In other words, for each sample size, the mean of all
possible sample means equals the population mean.
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22
The standard error of M for sample
sizes one, two, three, four, and five
Standard error = dispersion of M
σM
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The sample means for 1000
samples of four IQs. The normal
curve for x is superimposed
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24
Sampling Distribution of the Mean
for a Normally Distributed Variable
Suppose a variable x of a population is normally distributed
with mean m and standard deviation . Then, for samples of
size n, the sampling distribution of M is also normally
distributed and has mean
mM = m
and standard error of
M 

n
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25
(a) Normal distribution for IQs
(b) Sampling distribution
of the mean for n = 4
(c) Sampling distribution
of the mean for n = 16
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26
Samples Versus Individual Scores
Distribution
Distribution
of Individual
of Samples
Scores
Distribution
Consists of
Observed Central
Tendency
Theoretical
Central Tendency
Observed
Dispersion
Theoretical
Dispersion
M
x
MM
M
mM
m
sM
s
M

Frequency distribution for U.S.
household size
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Relative-frequency histogram for
household size
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29
Sample means n = 3,
for 1000 samples of household sizes.
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30
The Central Limit Theorem
For a relatively large sample size, the variable M is
approximately normally distributed, regardless
of the distribution of the underlying variable x.
The approximation becomes better and better with increasing
sample size.
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31
Sampling distributions for
normal, J-shaped, uniform variable
M
M
M
M
M
M
M
M
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M
M
32
APA Style: Tables
The mean self-consciousness scores for participants who were
working in front of a video camera and those who were not
(controls).
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APA Style: Bar Graphs
The mean (±SE) score for treatment groups A and B.
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34
APA Style: Line Graphs
The mean (±SE) number of mistakes made for
groups A and B on each trial.
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35
Summary
• We already knew how to determine the position
of an individual score in a normal distribution
• Now we know how to determine the position of
a sample of scores within the sampling
distribution
• By the Central Limit Theorem, all sampling
distributions are normal with
mean m and dispersion of  M 
Anthony J Greene

n
36
Sample Problem 1
• Given a distribution with μ = 32 and σ = 12
what is the probability of drawing a sample
of size 36 where M > 48
12
M 
2
36
M  m 48  32
z

8
M
2
Anthony J Greene
Does it seem
likely that M is
just a chance
difference?
37
Sample Problem 2
• In a distribution with µ = 45 and σ = 45
what is the probability of drawing a sample
of 25 with M >50?
45
M 
9
25
50  45
z
 0.55
9
Anthony J Greene
38
Sample problem 3
• In a distribution with µ = 90 and σ = 18, for a
sample of n = 36, what sample mean M
would constitute the boundary of the most
extreme 5% of scores?
zcrit = ± 1.96
M m
18
 1.96 
M 
3
M
36
-1.96 M
 1.96 * 3z 90
OR
 1.96 *+1.96
3  90  M
M 84.12
95.88
M  98.55
OR Anthony
MJ Greene
 84.12
39
Sample Problem 4
• In a distribution with µ = 90 and σ = 18,
what is the probability of drawing a sample
whose mean M > 93?
What
n = 9information are we missing?
M
z

18 18


 6
n
9 3
M m
M
93  90

 0.50
Anthony J Greene
6
P  0.308540
Sample Problem 5
• In a distribution with µ = 90 and σ = 18,
what is the probability of drawing a sample
whose mean M > 93?
n = 16

18 18
M 

  4.5
n
16 4
M  m 93  90
z

 0.67
M
4.5
Anthony J Greene
P  0.2514 41
Sample Problem 6
• In a distribution with µ = 90 and σ = 18,
what is the probability of drawing a sample
whose mean M > 93?
n = 25

18 18
M 

  3.6
n
25 5
M  m 93  90
z

 0.83
M
3.6
Anthony J Greene
P  0.2033 42
Sample Problem 7
• In a distribution with µ = 90 and σ = 18,
what is the probability of drawing a sample
whose mean M > 93?
n = 36

18 18
M 

  3.0
n
36 6
M  m 93  90
z

 1.00
M
3.0
Anthony J Greene
P  0.1587 43
Sample Problem 8
• In a distribution with µ = 90 and σ = 18,
what is the probability of drawing a sample
whose mean M > 93?
n = 81
M
z

18 18


  2.0
n
81 9
M m
M
93  90

 1.50
Anthony J Greene
2.0
P  0.0668 44
Sample Problem 9
• In a distribution with µ = 90 and σ = 18,
what is the probability of drawing a sample
whose mean M > 93?
n = 169
M
z

18
18


  1.38
n
169 13
M m
M
93  90

 2.17
Anthony J Greene
1.38
P  0.0146 45
Sample Problem 10
• In a distribution with µ = 90 and σ = 18,
what is the probability of drawing a sample
whose mean M > 93?
n = 625
M
z

18
18



 0.72
n
625 25
M m
M
93  90

 4.17
Anthony J Greene
0.72
P  0.00001 46
Sample Problem 10
• In a distribution with µ = 90 and σ = 18,
what is the probability of drawing a sample
whose mean M > 93?
n=1
M
z

18 18


  18.0
n
1 1
M m
M
93  90

 0.17
Anthony J Greene
18.0
P  0.5675
47
0.60
0.50
P-value
0.40
0.30
0.20
0.10
0.00
0
100
200
300
400
500
600
700
Sample Size
Anthony J Greene
48
z = ±2.58
Sample Problem 11
• In a distribution with µ = 200 and σ = 20,
what sample mean M corresponds to the
most extreme 1% ? 
20 20
M 


 20.0
n=1
n
1 1
M m
z
M  z * M  m
M
M  2.58 * 20  200
M  2.58 * 20  200
M  251.6, 148.4 Anthony J Greene
49
z = ±2.58
Sample Problem 12
• In a distribution with µ = 200 and σ = 20,
what sample mean M corresponds to the
most extreme 1% ? 
20 20
M 


 10.0
n=4
z
M m
M
n
4
2
M  z * M  m
M  2.58 *10  200
M  2.58 *10  200
M  225.8, 174.2 Anthony J Greene
50
z = ±2.58
Sample Problem 13
• In a distribution with µ = 200 and σ = 20,
what sample mean M corresponds to the
most extreme 1% ? 
20 20
M 


 5.0
n = 16
z
M m
M
n
16
4
M  z * M  m
M  2.58 * 5  200
M  2.58 * 5  200
M  212.9, 187.1 Anthony J Greene
51
z = ±2.58
Sample Problem 14
• In a distribution with µ = 200 and σ = 20,
what sample mean M corresponds to the
most extreme 1% ? 
20
20
M 


 2.5
n = 64
n
z
M m
M
64
8
M  z * M  m
M  2.58 * 2.5  200
M  206.4, 193.6
M  2.58 * 2.5  200
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52
z = ±2.58
Sample Problem 15
• In a distribution with µ = 200 and σ = 20,
what sample mean M corresponds to the
most extreme 1% ?

20
20
M 


 1.25
n = 258
n
256 16
z
M m
M
M  z * M  m
M  2.58 *1.25  200
M  203.2, 196.8
M  2.58 *1.25  200
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53
270
250
Score
230
210
190
170
150
0
50
100
150
200
250
300
Sample Size
Anthony J Greene
54
n=1
150
z
M
-2.58
148 .4
175
200
Anthony J Greene
225
250
+2.58
251.6
55
n=4
150
175
z
M
-2.58
174.2
200
Anthony J Greene
225
+2.58
225.8
250
56
n = 16
150
175
200
z
-2.58 +2.58
187.1
212.9
Anthony J Greene
M
225
250
57
n = 64
150
175
200
z
-2.58 +2.58
193.6
206.4
Anthony J Greene
M
225
250
58
n = 258
150
175
200
z
-2.58 +2.58
196.8
203.2
Anthony J Greene
M
225
250
59
n=∞
150
175
200
Anthony J Greene
225
250
60
```
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