Distribution of a population
Frequency
How many

individuals

Individual values of X
Distribution of a sample
Frequency
How many
s
individuals
X
Individual values of X
Distribution of a sample: The way in which the
individual scores within a sample are spread
out in relation to the mean.
Sampling Distribution: The way in which
sample means, drawn from the same
population, spread out from the population
mean.
Sampling Distribution
Frequency
How many
x
samples

Mean values of all possible
samples.
Measure of Variability:
Standard error - the average amount that
sample means differ from the population
mean.
x
X 

n
If:
a) randomly select enough samples
b) all samples were the same size
Then:
The distribution of those sample means
would form a normal distribution.
We can use the normal distribution to
predict the likelihood of randomly selecting
samples with particular means.
Z
X 
X
X 

n
 = 45
 = 10
N = 16
What is the likelihood that a sample,
selected at random from this population,
would have a mean of 43 or less?
Z
X 
X
1) compute the standard error
2) compute Z
3) Use the Z table to determine the probability
 = 45
 = 10
N = 16
X  43
1) compute the standard error
X 

n
10
x 
16
 x = 2.5
10
x 
4
2) compute Z using the standard error
z
x
43  45
z
2 .5

Z = - .8
Reminders about Z table:
a column: Z score
b column: prob. between mean and Z
c column: prob. beyond z into the tail
b column + c column = .5000
b
c
b
c
Z table:
z score
a
b
c
.79
.2852
.2148
.80
.2881
.2119
.81
.2910
.2090
Z table:
z score
a
b
c
.79
.2852
.2148
.80
.2881
.2119
.81
.2910
.2090
The prob. that a sample mean
would below 43 is .2119.
.5000 or 50.0 %
.2881 or
28.81%
.2119 or
21.19%

Raw means
Z scores
43 45
- .8
0
With the probability info we can:
- Determine how often we expect a
particular sample to be drawn.
- Establish the probability that a
particular sample that we have
is a random sample.
You develop an expensive training
program for athletes.
You say that your program reduce
event times for runners.
Typical running time: 50 sec.
After your program the group’s
average running time is 49 sec.
Is the program effective,
or is this random variability?
Assume that :
 = 50 sec.
N = 15
 = 4 sec.
Z
X 
X
X 

n
4
X 
15
15  3.87
4
X 
3.87
standard error = 1.03
49  50
z
1.03
Z = - .97
1
z
1.03
Probability that a sample would
have a mean of 49 or a more extreme
score?
P = .1660
33.40%
16.60%
Raw score
Z score
49
50
- .97
0
Sampling Distribution
Frequency
How many
x
samples

Mean values of all possible
samples.
The convention is that we will
assume that a sample is different
from the population mean for
reasons other than chance when the
probability of randomly selecting
the sample is less than or equal to
.05 (5%)
Statistical Significance:
The outcome could occur by chance alone
less than 5% of the time.
Null hypothesis: Mathematical statement
that describes the population in the absence
of any treatment.
Ho :  = 50
Alternative hypothesis: The
mathematical statement that
describes the population if the
treatment given would be expected to
have an effect and therefore change
the population.
H1 :   50
If an outcome is statistically significant we
reject the null hypothesis and accept the
alternative.
2.5%
Ho
H1
2.5%
Why is the 5% split into two parts?
H1 :   50
This doesn’t suggest the direction of
an unusual outcome.
Samples means much less than
the population mean of 50 as well
as samples means much greater
than 50 would be unusual.
Ho :  = 50
2.5%
2.5%
50
H1 :   50
This is know as a two tailed test.
Both tails of the distribution are included
as possible outcomes in the hypothesis
statement.
Assume a teacher knows that students
generally have an average score on a
test of 85. The standard deviation of
all of the students has been 7 points.
Ho :  = 85
H1 :   85
The teacher uses a new curriculum this
year and finds that the class (N = 20) has
an average on the test of 81.
Has this curriculum made a difference in
the performance of the class?
Z
X 
X
X 

n
 = 85
X 
=7

n
n = 20
X  81
7
X 
20
7
X 
20
7
X 
4 .7
 X  1.56
81 - 85
Z =
1.56
Z = --2.56
From the Z table:
Prob. of a Z score between 0 and
2.56 is .4948 (col. B)
The prob. of a Z score beyond 2.56 is
.0052 (col. C)
.0052 < .05
Therefore this qualifies as a
statistically significant event.
H1
H1
Sample means
Z scores
81
-2.56
85
0
Conclusion:
Reject the null hypothesis.
This curriculum probably caused
the class to perform differently than
the population as a whole.
Could our conclusion be wrong?
It is possible (although unlikely) that a
sample that had a mean of as low as 81
could have been selected at random from
the population.
In other words, it is possible (although
unlikely) that the curriculum had no effect
on the sample.
Type One Error (alpha error,  )
When you reject the null and it is
actually true.
Type two error (beta error,  )
Accept the null hypothesis
(saying that the independent
variable has no effect) when the
null is false.
Ho :  = 85
H1 :   85
One tailed test:
When the alternative hypothesis predicts a
direction.
H1 :  < 85
5%
85
Z score where 2.5% of the
distribution lies in the tail:
Z = + 1.96
Critical value for a two tailed
test.
2.5%
2.5%
-1.96
0
1.96
Z score where 5% of the distribution lies
in the tail:
Z = + 1.65
Critical value for a one tailed test.
5%
- 1.65
0
5%
0
1.65