What single samples t test (2)?

Welcome to a presentation explaining the concepts
behind the use of a single sample t-test

Welcome to a presentation explaining the concepts
behind the use of a single sample t-test in determining
the probability that a sample and a population are
similar to or different from one another statistically.

We will follow an example where researchers attempt
to determine if the sample they have collected is
statistically significantly similar or different from a
population.

Their hope is that the sample and population are
statistically similar to one another, so they can claim
that results of experiments done to the sample are
generalizable to the population.

Let’s imagine that this is the population distribution for
IQ scores in the country:

It has a population
mean of 100

m = 100
This Greek symbol
represents the mean
of a population

We decide to select a random sample to do
experiments on.
m = 100

So, we randomly
select 20 persons
m = 100

Let’s say that sample
of 20 has an IQ score
mean of 70
m = 100

Let’s say that sample
of 20 has an IQ score
풙 = 70 m = 100
mean of 70

풙 = 70 m = 100
Note, this x with a bar
over it is the symbol
for a sample mean.

풙 = 70 m = 100
Again this Greek
symbol m is the symbol
for a population mean.

Along with a mean of
70 this sample has a
distribution that looks
풙 = 70 m = 100
like this

풙 = 70 m = 100
So, here’s the question:

풙 = 70 m = 100
Is this randomly selected sample of 20 IQ scores
representative of the population?

The Single Sample t-test is a tool used to determine the
probability that it is or is not.
풙 = 70 m = 100

So, how do we determine if the sample is a good
representative of the population?

Let’s look at the population distribution of IQ scores
first:

One thing we
notice right off is
that it has a normal
distribution

Normal
Distributions have
some important
properties or
attributes that
make it possible to
consider rare or
common
occurrences

Also, Normal
Distributions have
some constant
percentages that
are true across all
normal
distributions

Attribute #1:
50% of all of the
scores are above
the mean and the
other 50% of the
scores are below
the mean

Attribute #1:
50% of all of the
scores are above
the mean and the
other 50% of the
scores are below
the mean
The Mean

Attribute #1:
50% of all of the
scores are above
the mean and the
other 50% of the
scores are below
the mean
The Mean 50% of all scores

Attribute #1:
50% of all of the
scores are above
the mean and the
other 50% of the
scores are below
the mean
50% of all scores The Mean

Before going on let’s
take a brief time out

The next section requires an understanding of
the concept of standard deviation.

If you are unfamiliar with this concept do a
search for standard deviation in this software.
After viewing it return to slide 36 of this
presentation.

Time in – let’s get back
to the instruction

The
Mean
Attribute #2:
68% of all of the scores
are between +1 standard
deviation and -1 standard
deviation

The
Mean
Attribute #2:
deviation from the mean

The
Mean
Attribute #2:
+1 sd

The
Mean
Attribute #2:
-1 sd +1 sd

68% of all scores
The
Mean
Attribute #2:
-1 sd +1 sd

68% of all scores
The
Mean
So what this means is – if
you were randomly
selecting samples from
this population you have
a 68% chance or .68
probability of pulling
that sample from this
part of the distribution.
-1 sd +1 sd

68% of all scores
The
Mean
-1 sd +1 sd
Let’s put some
numbers to this
idea.

68% of all scores
The
Mean
-1 sd +1 sd
The mean of IQ scores across
the population is 100

-1 sd +1 sd
m = 100
With a population standard
deviation (s) of 15:
+1 standard deviation
would at an IQ score of 115
and
-1 standard deviation
would be at 85
68% of all scores

m = 100
+1 sd
-1s=85
-1 sd
68% of all scores
and
-1 standard deviation
would be at 85

68% of all scores
-1s=85 m = 100
+1s=115
-1 sd +1 sd
and
-1 standard deviation would
be at 85

68% of all scores So, there is a 68%
m = 100
-1 sd +1 sd
chance or .68
probability that a
sample was
collected between
IQ scores of 85 and
115
-1s=85 +1s=115

68% of all scores
m = 100
-1 sd +1 sd
Attribute 2:
2 standard deviation
units above and
below the mean
constitute 95% of all
scores.
-1s=85 +1s=115

68% of all scores
m = 100
-1 sd +1 sd
Attribute 2:
units above and
below the mean
constitute 95% of all
scores.
-1s=85 +1s=115
+2 sd +2 sd

68% of all scores
m = 100
-1 sd +1 sd
units above the
mean would be an
IQ score of 130 or
100 + 2*15(sd))
-1s=85 +1s=115
+2 sd +2 sd

68% of all scores
m = 100
-1 sd +1 sd
units above the
mean would be an
IQ score of 130 or
100 + 2*15(sd))
-1s=85 +1s=115
+1s=130
+2 sd
-2 sd

68% of all scores
m = 100
-1 sd +1 sd
units below the
mean would be an
IQ score of 70 or
100 - 2*15(sd))
-1s=85 +1s=115
+1s=115
+2 sd
-2 sd

68% of all scores
-2s=70 m = 100
+1s=115
-1 sd +1 sd
units below the
mean would be an
IQ score of 70 or
100 - 2*15(sd))
-1s=85 +1s=115
+2 sd
-2 sd

68% of all scores
-2s=70 m = 100
+1s=115
-1 sd +1 sd
Now, it just so
happens in nature
that 95% of all
scores are between
+2 and -2 standard
deviations in a
normal distribution.
-1s=85 +1s=115
+2 sd
-2 sd

95% of all scores
-2s=70 -1s=85 m = 100
+1s=115
+1s=130
-1 sd +1 sd
+2 sd
-2 sd
Now, it just so
happens in nature
that 95% of all
scores are between
+2 and -2 standard
deviations in a
normal distribution.

95% of all scores
-2s=70 m = 100
+1s=130
-1 sd +1 sd
This means that
there is a .95 chance
that a sample we
select would come
from between these
two points.
-1s=85 +1s=115
+2 sd
-2 sd

Attribute #3: 99% of all scores are between
95% of all scores
-2s=70 -1s=85 m = 100
+1s=115
+1s=130
-1 sd +1 sd
+2 sd
-2 sd
+3 and -3 standard deviations.

99% of all scores
-2s=70 -1s=85 m = 100
+1s=115
+1s=130
-1 sd +1 sd
+2 sd
-2 sd
-3s=55
-3 sd
+3s=55
+3 sd
Attribute #3: 99% of all scores are between
+3 and -3 standard deviations.

99% of all scores
-2s=70 -1s=85 m = 100
+1s=115
+1s=130
-1 sd +1 sd
+2 sd
-2 sd
-3s=55
-3 sd
-3s=55
+3 sd
These standard deviations are only
approximates.

99% of all scores
-2s=70 -1s=85 m = 100
+1s=115
+1s=130
-3s=55 -3s=55
-3 sd +3 sd
-1 sd +1 sd
+2 sd
-2 sd
Here are the actual values

99% of all scores
-2s=70 -1s=85 m = 100
+1s=115
+1s=130
-1 sd +1 sd
+2 sd
-2 sd
-2.58 s
=55
-2.58 sd
-3s=55
+3 sd

99% of all scores
-1.96 s
-1s=85 m = 100
+1s=115
+1s=130 =70
-1 sd +1 sd
+2 sd
-2.58 sd -1.96 sd
-3s=55
+3 sd
-2.58 s
=55

99% of all scores
m = 100
-1 sd +1 sd
+1s=130
+2 sd
-1.96 s
=70
-2.58 sd -1.96 sd
-3s=55
+3 sd
-2.58 s
=55
+1s=115
-1 s
=85

99% of all scores
m = 100
-1 sd +1 sd
+1s=130
+2 sd
-1.96 s
=70
-2.58 sd -1.96 sd
-3s=55
+3 sd
-2.58 s
=55
+1 s
=115
-1 s
=85

99% of all scores
m = 100
-1 sd +1 sd
+1.96 s
=130
-1.96 s
=70
-2.58 sd -1.96 sd
-3s=55
+3 sd
+1.96 sd
-2.58 s
=55
+1 s
=115
-1 s
=85

99% of all scores
m = 100
+1 s
=115
-1 s
=85
-1 sd +1 sd
+1.96 s
=130
-1.96 s
=70
-2.58 sd -1.96 sd
+2.58 s
=145
+2.58 sd
+1.96 sd
-2.58 s
=55

99% of all scores
m = 100
-2.58 sd -1.96 sd -1 sd +1 sd
+1.96 sd +2.58 sd
Based on the percentages of a normal distribution, we can
insert the percentage of scores below each standard deviation
point.
+1 s
=115
-1 s
=85
+1.96 s
=130
+2.58 s
=145
-1.96 s
=70
-2.58 s
=55

99% of all scores
-2.58 sd -1.96 sd -1 sd +1 sd
+1.96 sd +2.58 sd
Attribute #4:
percentages
can be
calculated
below or
above each
standard
deviation
point in the
distribution.
m = 100
+1 s
=115
-1 s
=85
+1.96 s
=130
+2.58 s
=145
-1.96 s
=70
-2.58 s
=55

99% of all scores
m = 100
+1 s
=115
-1 s
=85
+1.96 s
=130
+2.58 s
=145
-1.96 s
=70
-2.58 s
=55
-2.58 sd -1.96 sd -1 sd +1 sd +1.96 sd +2.58 sd

99% of all scores
95% of all scores
m = 100
+1 s
=115
-1 s
=85
+1.96 s
=130
+2.58 s
=145
-1.96 s
=70
-2.58 s
=55
-2.58 sd -1.96 sd -1 sd +1 sd +1.96 sd +2.58 sd

99% of all scores
95% of all scores
68% of all scores
m = 100
+1 s
=115
-1 s
=85
+1.96 s
=130
+2.58 s
=145
-1.96 s
=70
-2.58 s
=55
-2.58 sd -1.96 sd -1 sd +1 sd +1.96 sd +2.58 sd

99% of all scores
95% of all scores
68% of all scores
m = 100
+1 s
=115
-1 s
=85
+1.96 s
=130
-2.58 sd -1.96 sd -1 sd +1 sd +1.96 sd +2.58 sd
With this information we can determine the probability that scores will
fall into a number portions of the distribution.
+2.58 s
=145
-1.96 s
=70
-2.58 s
=55

There is a 0.5%
chance that if
you randomly
selected a
person that their
IQ score would
be below a 55 or
a -2.58 SD
0.5%
m = 100
+1 s
=115
-1 s
=85
+1.96 s
=130
+2.58 s
=145
-1.96 s
=70
-2.58 s
=55
-2.58 sd -1.96 sd -1 sd +1 sd +1.96 sd +2.58 sd

There is a 2.5%
chance that if
you randomly
selected a
person that their
IQ score would
be below a 70 or
a -1.96 SD
2.5%
m = 100
+1 s
=115
-1 s
=85
+1.96 s
=130
+2.58 s
=145
-1.96 s
=70
-2.58 s
=55
-2.58 sd -1.96 sd -1 sd +1 sd +1.96 sd +2.58 sd

There is a 16.5%
chance that if
you randomly
selected a
person that their
IQ score would
be below a 85 or
a -1 SD
16.5%
m = 100
+1 s
=115
-1 s
=85
+1.96 s
=130
+2.58 s
=145
-1.96 s
=70
-2.58 s
=55
-2.58 sd -1.96 sd -1 sd +1 sd +1.96 sd +2.58 sd

50%
There is a 50%
chance that if
you randomly
selected a
person that their
IQ score would
be below a 100
or a 0 SD
m = 100
+1 s
=115
-1 s
=85
+1.96 s
=130
+2.58 s
=145
-1.96 s
=70
-2.58 s
=55
-2.58 sd -1.96 sd -1 sd +1 sd +1.96 sd +2.58 sd

There is a 50%
chance that if
you randomly
selected a
person that their
IQ score would
be above a 100
or a 0 SD
50%
m = 100
+1 s
=115
-1 s
=85
+1.96 s
=130
+2.58 s
=145
-1.96 s
=70
-2.58 s
=55
-2.58 sd -1.96 sd -1 sd +1 sd +1.96 sd +2.58 sd

There is a 16.5%
chance that if
you randomly
selected a
person that their
IQ score would
be above a 115
or a +1 SD
16.5%
m = 100
+1 s
=115
-1 s
=85
+1.96 s
=130
+2.58 s
=145
-1.96 s
=70
-2.58 s
=55
-2.58 sd -1.96 sd -1 sd +1 sd +1.96 sd +2.58 sd

m = 100
+1 s
=115
-1 s
=85
-1 sd +1 sd
There is a 2.5%
chance that if
you randomly
selected a
person that their
IQ score would
be above a 130
or a +2.58 SD
+1.96 s
=130
+1.96 sd
-1.96 s
=70
-1.96 sd
-2.58 s
=55
-2.58 sd
2.5%
+2.58 s
=145
+2.58 sd

m = 100
+1 s
=115
-1 s
=85
-1 sd +1 sd
+1.96 s
=130
+1.96 sd
-1.96 s
=70
-1.96 sd
-2.58 s
=55
-2.58 sd
There is a 0.5%
chance that if
you randomly
selected a
person that their
IQ score would
be above a 145
or a +2.58 SD
.5%
+2.58 s
=145
+2.58 sd

What you have just seen illustrated is the concept of probability density
or the probability that a score or observation would be selected above,
m = 100
+1 s
=115
-1 s
=85
-1 sd +1 sd
+1.96 s
=130
+1.96 sd
-1.96 s
=70
-1.96 sd
-2.58 s
=55
-2.58 sd
+2.58 s
=145
+2.58 sd
below or between two points on a distribution.

m = 100
+1 s
=115
-1 s
=85
-1 sd +1 sd
+1.96 s
=130
+1.96 sd
-1.96 s
=70
-1.96 sd
-2.58 s
=55
-2.58 sd
+2.58 s
=145
+2.58 sd

m = 100
+1 s
=115
-1 s
=85
-1 sd +1 sd
+1.96 s
=130
+1.96 sd
-1.96 s
=70
-1.96 sd
-2.58 s
=55
-2.58 sd
+2.58 s
=145
+2.58 sd
Here is the
sample we
randomly
selected
풙 = 70

m = 100
+1 s
=115
-1 s
=85
-1 sd +1 sd
+1.96 s
=130
+1.96 sd
-1.96 s
=70
-1.96 sd
-2.58 s
=55
-2.58 sd
+2.58 s
=145
+2.58 sd
The sample
mean is 30 units
away from the
population mean
(100 – 70 = 30).
풙 = 70

m = 100
+1 s
=115
-1 s
=85
-1 sd +1 sd
+1.96 s
=130
+1.96 sd
-1.96 s
=70
-1.96 sd
-2.58 s
=55
-2.58 sd
+2.58 s
=145
+2.58 sd
The sample
mean is 30 units
away from the
population mean
(100 – 70 = 30).
풙 = 70
30

m = 100
+1 s
=115
-1 s
=85
-1 sd +1 sd
+1.96 s
=130
+1.96 sd
-1.96 s
=70
-1.96 sd
-2.58 s
=55
-2.58 sd
+2.58 s
=145
+2.58 sd
Is that far away
enough to be
called statistically
significantly
different than
the population?
풙 = 70

m = 100
+1 s
=115
-1 s
=85
-1 sd +1 sd
+1.96 s
=130
+1.96 sd
-1.96 s
=70
-1.96 sd
-2.58 s
=55
-2.58 sd
+2.58 s
=145
+2.58 sd
How far is too far
away?
풙 = 70

m = 100
+1 s
=115
-1 s
=85
-1 sd +1 sd
+1.96 s
=130
+1.96 sd
-1.96 s
=70
-1.96 sd
-2.58 s
=55
-2.58 sd
+2.58 s
=145
+2.58 sd
Fortunately,
statisticians have
come up with a
couple of
distances that are
considered too far
away to be a part
of the population.
풙 = 70

m = 100
+1 s
=115
-1 s
=85
-1 sd +1 sd
+1.96 s
=130
+1.96 sd
-1.96 s
=70
-1.96 sd
-2.58 s
=55
-2.58 sd
+2.58 s
=145
+2.58 sd
These distances
are measured in
z-scores
풙 = 70

m = 100
+1 s
=115
-1 s
=85
-1 sd +1 sd
+1.96 s
=130
+1.96 sd
-1.96 s
=70
-1.96 sd
-2.58 s
=55
-2.58 sd
+2.58 s
=145
+2.58 sd
풙 = 70
Which are what
these are:
z scores

Let’s say statisticians determined that if the sample mean you
collected is below a -1.96 z-score or above a +1.96 z-score that that’s
just too far away from the mean to be a part of the population.
m = 100
+1 s
=115
-1 s
=85
-1 sd +1 sd
+1.96 s
=130
+1.96 sd
-1.96 s
=70
-1.96 sd
-2.58 s
=55
-2.58 sd
+2.58 s
=145
+2.58 sd
풙 = 70

We know from previous slides that only 2.5% of the
scores are below a -1.96
m = 100
+1 s
=115
-1 s
=85
-1 sd +1 sd
+1.96 s
=130
+1.96 sd
-1.96 s
=70
-1.96 sd
-2.58 s
=55
-2.58 sd
+2.58 s
=145
+2.58 sd
풙 = 70

We know from previous slides that only 2.5% of the
scores are below a -1.96
m = 100
+1 s
=115
-1 s
=85
-1 sd +1 sd
+1.96 s
=130
+1.96 sd
-1.96 s
=70
-1.96 sd
-2.58 s
=55
-2.58 sd
+2.58 s
=145
+2.58 sd
풙 = 70
2.5%

Since anything at this point or below is considered to be too
rare to be a part of this population, we would conclude that
the population and the sample are statistically significantly
different from one another.
m = 100
+1 s
=115
-1 s
=85
-1 sd +1 sd
+1.96 s
=130
+1.96 sd
-1.96 s
=70
-1.96 sd
-2.58 s
=55
-2.58 sd
+2.58 s
=145
+2.58 sd
풙 = 70
2.5%

m = 100
+1 s
=115
-1 s
=85
-1 sd +1 sd
+1.96 s
=130
+1.96 sd
-1.96 s
=70
-1.96 sd
-2.58 s
=55
-2.58 sd
+2.58 s
=145
+2.58 sd
풙 = 70
And that’s our answer!
2.5%

What if the sample mean had been 105?

Since our decision rule is to determine that the sample
mean is statistically significantly different than the
population mean if the sample mean lies outside of the
top or bottom 2.5% of all scores,

m = 100
+1 s
=115
-1 s
=85
-1 sd +1 sd
+1.96 s
=130
+1.96 sd
-1.96 s
=70
-1.96 sd
-2.58 s
=55
-2.58 sd
+2.58 s
=145
+2.58 sd
2.5% 2.5%

m = 100
+1 s
=115
-1 s
=85
-1 sd +1 sd
+1.96 s
=130
+1.96 sd
-1.96 s
=70
-1.96 sd
-2.58 s
=55
-2.58 sd
+2.58 s
=145
+2.58 sd
풙 = 105
2.5% 2.5%

. . . and the sample mean (105) does not lie in these
outer regions,
m = 100
+1 s
=115
-1 s
=85
-1 sd +1 sd
+1.96 s
=130
+1.96 sd
-1.96 s
=70
-1.96 sd
-2.58 s
=55
-2.58 sd
+2.58 s
=145
+2.58 sd
풙 = 105
2.5% 2.5%

Therefore, we would say that this is not a rare event
and the probability that the sample is significantly
similar to the population is high.
m = 100
+1 s
=115
-1 s
=85
-1 sd +1 sd
+1.96 s
=130
+1.96 sd
-1.96 s
=70
-1.96 sd
-2.58 s
=55
-2.58 sd
+2.58 s
=145
+2.58 sd
풙 = 105
2.5% 2.5%

By the way, how do we figure out the z-score for an IQ
score of 105.

We use the following formula to compute z-scores
across the normal distribution:

풙 - m
SD

풙 - m
Here’s our SD
sample
mean: 70

70 - m
Here’s our SD
sample
mean: 70

70 - m
SD Here’s our
Population
mean: 100

70 - 100
SD
Here’s our
Population
mean: 100

70 - 100
SD
Here’s our
Standard
Deviation:
15

70 - 100
15
Here’s our
Standard
Deviation:
15

70 - 100
15

30
15

2.0

A z score of 2 is located right here
m = 100
+1 s
=115
-1 s
=85
-1 sd +1 sd
+1.96 s
=130
+1.96 sd
-1.96 s
=70
-1.96 sd
-2.58 s
=55
-2.58 sd
+2.58 s
=145
+2.58 sd
풙 = 70
2.5% 2.5%

In some instances we may not know the population
standard deviation s (in this case 15).

Without the standard deviation of the population we
cannot determine the z-scores or the probability that a
sample mean is too far away to be apart of the
population.

population.
m = 100
+1 s
=115
-1 s
=85
-1 sd +1 sd
+1.96 s
=130
+1.96 sd
-1.96 s
=70
-1.96 sd
-2.58 s
=55
-2.58 sd
+2.58 s
=145
+2.58 sd

population.
m = 100
+1 s
=115
-1 s
=85
-1 sd +1 sd
+1.96 s
=130
+1.96 sd
-1.96 s
=70
-1.96 sd
-2.58 s
=55
-2.58 sd
+2.58 s
=145
+2.58 sd
50%

population.
m = 100
+1 s
=115
-1 s
=85
-1 sd +1 sd
+1.96 s
=130
+1.96 sd
-1.96 s
=70
-1.96 sd
-2.58 s
=55
-2.58 sd
+2.58 s
=145
+2.58 sd
16.5%

population.
m = 100
+1 s
=115
-1 s
=85
-1 sd +1 sd
+1.96 s
=130
+1.96 sd
-1.96 s
=70
-1.96 sd
-2.58 s
=55
-2.58 sd
2.5%
+2.58 s
=145
+2.58 sd

population.
m = 100
+1 s
=115
-1 s
=85
-1 sd +1 sd
+1.96 s
=130
+1.96 sd
-1.96 s
=70
-1.96 sd
-2.58 s
=55
-2.58 sd
+2.58 s
=145
+2.58 sd
Etc.

Therefore these values below cannot be computed:
m = 100
+1 s
=115
-1 s
=85
-1 sd +1 sd
+1.96 s
=130
+1.96 sd
-1.96 s
=70
-1.96 sd
-2.58 s
=55
-2.58 sd
+2.58 s
=145
+2.58 sd

When we only know the population mean we
use the Single Sample t-test.

Actually whenever we are dealing with a
population and a sample, we generally use a
single-sample t-test.

In the last example we relied on the population
mean and standard deviation to determine if the
sample mean was too far away from the
population mean to be considered a part of the
population.

The single sample t-test relies on a concept
called the estimated standard error

The single sample t-test relies on a concept
called the estimated standard error to compute
something like a z-score to determine the
probability distance between the population
and the sample means.

We call it estimated because as you will see it is not
really feasible to compute.

Standard error draws on two concepts:

1. sampling distributions
2. t-distributions

Let’s begin with sampling distributions.

What you are about to see is purely theoretical, but it
provides the justification for the formula we will use to
run a single sample t-test.

What you are about to see is purely theoretical, but it
provides the justification for the formula we will use to
run a single sample t-test.
x̄ – μ
SEmean

x̄ – μ
SEmean
the mean of a
sample

x̄ – μ
SEmean
the mean of a
sample
the mean of a
population

x̄ – μ
SEmean
the mean of a
sample
the mean of a
population
the estimated
standard error

x̄ – μ
SEmean
In our example,
this is 70

70̄ – μ
SEmean
In our example,
this is 70

70̄ – μ
SEmean
And this is
100

70̄ – 100
SEmean
And this is
100

The numerator here
is easy to compute
70̄ – 100
SEmean

The numerator here
is easy to compute
-30
SEmean

-30
SEmean
This value will help
us know the distance
between 70 and 100
in t-values

-30
SEmean
If the estimated
standard error is
large, like 30, then
the t value would be:
-30/30 = -1

A -1.0 t-value is like a z-score as shown below:

A -1.0 t-value is like a z-score as shown below:
m = 100
+1 s
=115
-1 s
=85
-1 sd +1 sd
+1.96 s
=130
+1.96 sd
-1.96 s
=70
-1.96 sd
-2.58 s
=55
-2.58 sd
+2.58 s
=145
+2.58 sd

But since we don’t know the population standard
deviation, we have to use the standard deviation of the
sample (not the population as we did before) to
determine the distance in standard error units
(or t values).

The focus of our theoretical justification is to explain
our rationale for using information from the sample to
compute the standard error or standard error of the
mean.

The focus of our theoretical justification is to explain
our rationale for using information from the sample to
compute the standard error or standard error of the
mean.
x̄ – μ
SEmean

Here comes the theory behind standard error:

Imagine we took a sample of 20 IQ scores from the
population.

population. This sample of 20 would have its own IQ
mean, standard deviation and distribution.

mean, standard deviation and distribution.
x ̄= 70
SD = 10

mean, standard deviation and distribution. And then
let’s say we took another sample of 20 with its mean
and distribution,
x ̄= 70

and distribution,
x ̄= 70 x ̄= 100

and distribution, and another
x ̄= 70 x ̄= 100

and distribution, and another
x ̄= 70 x ̄= 100 x ̄= 120

and distribution, and another, and another
x ̄= 70 x ̄= 100 x ̄= 120 x ̄= 140

and distribution, and another, and another, and so on…
x ̄= 70 x ̄= 100 x ̄= 120 x ̄= 140

Let’s say, theoretically, that we do this one hundred
times.

times.
We now have 100 samples of 20 person IQ scores:

times.
We now have 100 samples of 20 person IQ scores: We
take the mean of each of those samples:
x ̄= 115 x ̄= 100
x ̄= 100 x ̄= 120 x ̄= 90
x ̄= 110 x ̄= 102 x ̄= 120 x ̄= 90
x ̄= 114 x ̄= 100 x ̄= 120 x ̄= 90 x ̄= 95

And we create a new distribution called the sampling
distribution of the means
x ̄= 115 x ̄= 100
x ̄= 100 x ̄= 120 x ̄= 90
x ̄= 110 x ̄= 102 x ̄= 120 x ̄= 90
x ̄= 114 x ̄= 100 x ̄= 120 x ̄= 90 x ̄= 95

And
x ̄= 115 x ̄= 105
x ̄= 100 x ̄= 120 x ̄= 90
x ̄= 110 x ̄= 100 x ̄= 100 x ̄= 90
x ̄= 115 x ̄= 70 x ̄= 120 x ̄= 90 x ̄= 95

And
x ̄= 115 x ̄= 105
x ̄= 100 x ̄= 120 x ̄= 90
x ̄= 110 x ̄= 100 x ̄= 100 x ̄= 90
x ̄= 115 x ̄= 70 x ̄= 120 x ̄= 90 x ̄= 95
70 75 70 85 90 95 100 105 110 115 120 125

And
x ̄= 115 x ̄= 105
x ̄= 100 x ̄= 120 x ̄= 90
etc. …
x ̄= 110 x ̄= 100 x ̄= 100 x ̄= 90
x ̄= 115 x ̄= 70 x ̄= 120 x ̄= 90 x ̄= 95
70 75 70 85 90 95 100 105 110 115 120 125

And then we do something interesting. We take the
standard deviation of this sampling distribution.

If these sample means are close to one another then
the standard deviation will be small.

If these sample means are close to one another then
the standard deviation will be small.
70 75 70 85 90 95 100 105 110 115 120 125

If these sample means are far apart from one another
then the standard deviation will be large.
70 75 70 85 90 95 100 105 110 115 120 125

This standard deviation of the sampling distribution of
the means has another name:

the means has another name: the standard error.

the means has another name: the standard error.
x̄ – μ
SEmean
the estimated
standard error

Standard error is a unit of measurement that makes it
possible to determine if a raw score difference is really
significant or not.

significant or not.
Think of it this way. If you get a 92 on a 100 point test
and the general population gets on average a 90, is
there really a significant difference between you and
the population at large? If you retook the test over and
over again would you likely outperform or
underperform their average of 90?

significant or not.
Think of it this way. If you get a 92 on a 100 point test
and the general population gets on average a 90, is
there really a significant difference between you and
the population at large? If you retook the test over and
over again would you likely outperform or
underperform their average of 90?
Standard error helps us understand the likelihood that
those results would replicate the same way over and
over again . . . or not.

So let’s say we calculate the standard error to be 0.2.
Using the formula below we will determine how many
standard error units you are apart from each other.

x̄ – μ
SEmean
t =

92 – 90
0.2
your score
t =
standard error
population average

2
0.2
t =

2
0.2
t =
This is the raw
score difference

t = 10.0
And this is the
difference in
standard error
units

t = 10.0
And this is the
difference in standard
error units, otherwise
known as a t statistic
or t value

So, while your test score is two raw scores above the
average for the population, you are 10.0 standard error
units higher than the population score.

While 2.0 raw scores do not seem like a lot, 10.0
standard error units constitute a big difference!

While 2.0 raw scores do not seem like a lot, 10.0
standard error units constitute a big difference!
This means that this result is most likely to replicate
and did not happen by chance.

But what if the standard error were much bigger, say,
4.0?

4.0?
x̄ – μ
SEmean
t =

4.0?
92 – 90
4.0
your score
t =
standard error
population average

4.0?
2
4.0
t =

4.0?
t = 0.5

Once again, your raw score difference is still 2.0 but you
are only 0.5 standard error units apart. That distance is
most likely too small to be statistically significantly
different if replicated over a hundred times.

Once again, your raw score difference is still 2.0 but you
are only 0.5 standard error units apart. That distance is
most likely too small to be statistically significantly
different if replicated over a hundred times.
We will show you how to determine when the number
of standard error units is significantly different or not.

One more example:
Let’s say the population average on the test is 80 and
you still received a 92. But the standard error is 36.0.

One more example:
Let’s do the math again.

One more example:
x̄ – μ
SEmean
t =

One more example:
92 – 80
36.0
your score
t =
standard error
population average

One more example:
12
36.0
t =

One more example:
t = 0.3

In this case, while your raw score is 12 points higher (a
large amount), you are only 0.3 standard error units
higher.

In this case, while your raw score is 12 points higher (a
large amount), you are only 0.3 standard error units
higher. The standard error is so large (36.0) that if you
were to take the test 1000 times with no growth in
between it is most likely that your scores would vary
greatly (92 on one day, 77 on another day, 87 on
another day and so on and so forth.)

In summary, the single sample test t value is the
number of standard error units that separate the
sample mean from the population mean:

In summary, the single sample test t value is the
number of standard error units that separate the
sample mean from the population mean:
x̄ – μ
SEmean
t =
Let’s see this play out with our original example.

Let’s say our of 20 has an average IQ score of 70.

x̄ – μ
SEmean
t =

70 – μ
SEmean
t =

We already know that the population mean is 100.
70 – μ
SEmean
t =

70 – 100
SEmean
t =

Let’s say the Standard Error of the Sampling
Distribution means is 5
70 – 100
SEmean
t =

Let’s say the Standard Error of the Sampling
Distribution means is 5
70 – 100
5
t =

The t value would be:
70 – 100
5
t =

-30
5
t =

So all of this begs the question: How do I know if a t
value of -6 is rare or common?

• If it is rare we can accept the null hypothesis and say
that there is a difference between the sample and
the population.

the population.
• If it is common we can reject the null hypothesis and
say there is not a significant difference between
veggie eating IQ scores and the general population
IQ scores (which in this unique case is what we want)

value of 5 is rare or common?
the population.
• If it is common we can reject the null hypothesis and
say there is not a significant difference between
veggie eating IQ scores and the general population
IQ scores (which in this unique case is what we want)
Here is what we do: We compare this value (6) with the
critical t value.

What is the critical t value?
The critical t value is a point in the distribution that
arbitrarily separates the common from the rare
occurrences.

occurrences.
rare
occurrence

occurrences.
rare
occurrence
rare
occurrence

occurrences.
rare
occurrence
common
occurrence
rare
occurrence

occurrences.
rare
occurrence
common
occurrence
rare
occurrence
This represents the rare/common possibilities used to
determine if the sample mean is similar the population
mean).

If this were a normal distribution the red line would
have a z critical value of + or – 1.96
rare
occurrence
common
occurrence
rare
occurrence

If this were a normal distribution the red line would
have a z critical value of + or – 1.96 (which is essentially
a t critical value but for a normal distribution.)
rare
occurrence
common
occurrence
rare
occurrence

A t or z value of + or – 1.96 means that 95% of the
scores are in the center of the distribution and 5% are
to the left and right of it.

common
occurrence
rare
occurrence
rare
occurrence

rare
occurrence
common
occurrence
rare
occurrence
-1.96 +1.96

So if the t value computed from this equation:

x̄ – μ
SEmean
t =

x̄ – μ
SEmean
t =
… is between -1.96 and +1.96 we would say that that
result is not rare and we would fail to reject the null
hypothesis.

x̄ – μ
SEmean
t =
… is between -1.96 and +1.96 we would say that that
result is not rare and we would fail to reject the null
hypothesis.
rare
occurrence
common
occurrence
rare
occurrence
-1.96 +1.96

However, if the t value is smaller than -1.96 or larger
than +1.96 we would say that that result is rare and we
would reject the null hypothesis.

However, if the t value is smaller than -1.96 or larger
than +1.96 we would say that that result is rare and we
would reject the null hypothesis.
rare
occurrence
common
occurrence
rare
occurrence
-1.96 +1.96

As a review, let’s say the distribution is normal. When
the distribution is normal and we want to locate the z
critical for a two tailed test at a level of significance of
.05 (which means that if we took 100 samples we are
willing to be wrong 5 times and still reject the null
hypothesis), the z critical would be -+1.96.

As a review, let’s say the distribution is normal. When
the distribution is normal and we want to locate the z
critical for a two tailed test at a level of significance of
.05 (which means that if we took 100 samples we are
willing to be wrong 5 times and still reject the null
hypothesis), the z critical would be -+1.96.
- 1.96 + 1.96

Because we generally do not have the resources to take
100 IQ samples of those who eat veggies, we have to
estimate the t value from one sample.

x ̄= 115 x ̄= 105
x ̄= 100 x ̄= 120 x ̄= 90
x ̄= 110 x ̄= 100 x ̄= 100 x ̄= 90
x ̄= 115 x ̄= 70 x ̄= 120 x ̄= 90 x ̄= 95

x ̄= 115 x ̄= 105
x ̄= 100 x ̄= 120 x ̄= 90
x ̄= 110
x ̄= 110 x ̄= 100 x ̄= 100 x ̄= 90
x ̄= 115 x ̄= 70 x ̄= 120 x ̄= 90 x ̄= 95

x ̄= 110

So let’s imagine we selected a sample
of 20 veggie eaters with an average IQ
score of 70.
x ̄= 70

x ̄= 70
score of 70.
Because we did not take hundreds of samples of 20
veggie eaters each, average each sample’s IQ scores,
and form a sampling distribution from which we could
compute the standard error and then the t value, we
have to figure out another way to compute an estimate
of the standard error.

x ̄= 70
score of 70.
Because we did not take hundreds of samples of 20
veggie eaters each, average each sample’s IQ scores,
and form a sampling distribution from which we could
compute the standard error and then the t value, we
have to figure out another way to compute an estimate
of the standard error. There is another way.

Since it is not practical to collect a hundreds of samples
of 20 from the population, compute their mean score
and calculate the standard error, we must estimate it
using the following equation:
S
n
SEmean =

Since it is not practical to collect a hundreds of samples
of 20 from the population, compute their mean score
and calculate the standard error, we must estimate it
using the following equation:
S
n
SEmean =
Standard Deviation
of the sample

We estimate the standard error using the following
equation:
S
n
SEmean =
Standard Deviation
of the sample

We estimate the standard error using the following
equation:
SEmean =
Standard Deviation
of the sample
Square root of the
sample size
S
n
We won’t go into the derivation of
this formula, but just know that this acts as a good
substitute in the place of taking hundreds of samples
and computing the standard deviation to get the actual
standard error.

So remember this point: The equation below is an
estimate of the standard error, not the actual standard
error, because the actual standard error is not feasible
to compute.

to compute.
S
n
SEmean =

to compute.
S
n
SEmean =
However, just know that when researchers have taken
hundreds of samples and computed their means and
then taken the standard deviation of all of those means
they come out pretty close to one another.

Now comes another critical point dealing with t-distributions.

Because we are working with a sample
that is generally small, we do not use the same normal
distribution to determine the critical z or t (-+1.96).

Here is what the z distribution looks like for samples
generally larger than 30:

Here is what the z distribution looks like for samples
generally larger than 30:
95% of the scores
- 1.96 + 1.96
Sample Size 30+

As the sample size decreases the critical values increase
- making it harder to get significance.

Notice how the t distribution gets shorter and wider
when the sample size is smaller. Notice also how the t
critical values increase as well.
95% of the scores
Sample Size 20
- 2.09 + 2.09

95% of the scores
Sample Size 10
- 2.26 + 2.26

95% of the scores
Sample Size 5
- 2.78 + 2.78

To determine the critical t we need two values:

• the degrees of freedom (sample size minus one [20-1 = 19])

• the significance level (.05 or .025 for two tailed test)

• the significance level (.05 or .025 for two tailed test)
Using these two
values we can
locate the
critical t

So, our t critical value that separates the common from
the rare occurrences in this case is + or – 2.09

So, our t critical value that separates the common from
the rare occurrences in this case is + or – 2.09
95% of the scores
Sample Size 20
- 2.09 + 2.09

What was our calculated t value again?

x̄ – μ
SEmean
t =

70 – μ
SEmean
t =

70 – 100
SEmean
t =
To calculate the estimated standard error of the mean
distribution we use the following equation:

70 – 100
SEmean
t =
To calculate the estimated standard error of the mean
distribution we use the following equation:
S
n
SEmean =

70 – 100
SEmean
t =
The Standard Deviation for this sample is 26.82 and the
sample size, of course, is 20
S
n
SEmean =

70 – 100
SEmean
t =
Let’s plug in those values:
S
n
SEmean =

70 – 100
SEmean
t =
26.82
20
SEmean =

70 – 100
SEmean
26.82
4.47
t =
SEmean =

70 – 100
SEmean
6.0
t =
SEmean =

70 – 100
6.0
6.0
t =
SEmean =

70 – 100
6.0
t =

Now do the math:
70 – 100
6.0
t =

70 – 100
6.0
t =

-30
6.0
t =

With a t value of -5 we are ready to compare it to the
critical t:

critical t: 2.093

critical t: 2.093 or 2.093 below the mean or -2.093.

Since a t value of +5.0 is below the cutoff point
(critical t) of -2.09, we would reject the null hypothesis

Since a t value of +5.0 is below the cutoff point
(critical t) of -2.09, we would reject the null hypothesis
95% of the scores
- 2.09 + 2.09

Here is how we would state our results:

Here is how we would state our results:
The randomly selected sample of twenty IQ scores with
a sample mean of 70 is statistically significantly
different then the population of IQ scores.

Now, what if the standard error had been much larger,
say, 15.0 instead of 6.0?

70 – 100
6.0
t =

70 – 100
6.0
t =
70 – 100
15.0
t =

70 – 100
6.0
t =
30
15.0
t =

70 – 100
t = t = 2.0
6.0

A t value of -2.0 is not below the cutoff point (critical t)
of -2.09 and would be considered a common rather
than a rare outcome. We therefore would fail to reject
the null hypothesis

A t value of -2.0 is not below the cutoff point (critical t)
of -2.09 and would be considered a common rather
than a rare outcome. We therefore would fail to reject
the null hypothesis
95% of the scores
- 2.09 - 2.0 + 2.09

So in summary, the single sample t-test helps us
determine the probability that the difference between
a sample and a population did or did not occur by
chance.

So in summary, the single sample t-test helps us
determine the probability that the difference between
a sample and a population did or did not occur by
chance.
It utilizes the concepts of standard error, common and
rare occurrences, and t-distributions to justify its use.

What single samples t test (2)?

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