T-tests
Part 1
PS1006 Lecture 2
Sam Cromie
What has
this got to
do with
statistics?
Overview
•
•
•
•
Review: hypothesis testing
The need for the t-test
The logic of the t-test
Application to:
– Single sample designs
– Two group designs
• Within group designs (Related samples)
• Between group designs (Independent samples)
• Assumptions
• Advantages and disadvantages
Generic Hypothesis testing steps
1. State hypotheses – Null and alternative
2. State alpha value
3. Calculate the statistic (z score, t score,
etc.)
4. Look the statistic up on the appropriate
probability table
5. Accept or reject null hypothesis
Generic form of a statistic
Data – Hypothesis
Error
What you got – what you expected (null)
The unreliability of your data
Z = Individual score – Population mean
Population standard deviation
Hypothesis testing with an
individual data point…
1. State null hypothesis
2. State α value
3. Convert the score to
a z score
4. Look z score up on z
score tables
z
x

In this case we have…
Data of
interest
Error
Sample
Population
Individual
score, n=1
Mean
known
Standard
deviation
known
Hypothesis testing with one sample
(n>1) …
• 100 participants saw video containing violence
• Then they free associated to 26 homonyms with
aggressive & non-aggressive forms - e.g.,
pound, mug,
• Mean number of aggressive free associates =
7.10
• Suppose we know that without an aggressive
video the mean ()=5.65 and the standard
deviation () = 4.5
• Is 7.10 significantly larger than 5.65?
In this case we have…
Data of
interest
Error
Sample
Population
Mean,
n=100
Mean
known
Standard
deviation
known
Hypothesis testing with one sample
(n>1) …
• Use the sample mean instead of x in the z score formula
• Use standard error of sample instead of the population standard
deviation
z
x

becomes
where
X 
z
x
X

n
n = the number of scores in the sample
Standard error:
 X is thestandarddeviationof a samplingdistribution
but is known as a standarderrorfor thepurposesof
differentiation
• If we know  then  X can be calculated using
the formula
X 

n
Sampling distribution
• http://www.ruf.rice.edu/~lane/stat_sim/sam
pling_dist/index.html
• Will always be narrower than the parent
population
• The more samples that are taken the more
normal the distribution
• As sample size increases standard error
decreases
Back to video violence
• H0:  = 5.65
• H1:  5.65 (two-tailed)
• Calculate p for sample mean of 7.10 assuming
=5.65
• Use z from normal distribution as sampling
distribution can be assumed to be normal
• Calculate z
X 
7.1  5.65
1.45
X 
4.5
z
=
=
 3.22

=
X
.45
100
n
• If z > + 1.96, reject H0
• 3.22 > 1.96  the difference is significant
But mostly we do not know σ
• E.g. do penalty-takers show a preference
for right or left?
• 16 penalty takers; 60 penalties each; null
hypothesis = 50% or 30 each way
• Result mean of 39 penalties to the left; is
this significantly different?
• µ = 30, but how do we calculate the
standard error without the σ?
In this case we have…
Sample
Population
Data of
interest
Mean
n=100
Hypothesis
= 30
Error
Variance
?
Using s to estimate σ
• Can’t substitute s for  in a z score because
s likely to be too small
• So we need:
– a different type of score – a t-score
– a different type of distribution – Student’s t
distribution
T distribution
• First published in 1908
by William Sealy
Gosset,
• Worked at a Guinness
Brewery in Dublin on
best yielding varieties
of barley
• Prohibited from
publishing under his
own name
• so the paper was
written under the
pseudonym Student.
T-test in a nut-shell…
Allows us to calculate precisely
what small samples tell us
Uses three critical bits of
information – mean, standard
deviation and sample size
t test for one mean
• Calculated the same way as z except 
is replaced by s.
• For the video example we gave before,
s = 4.40
X 
z
X  =
X
X 
t
sX

n
X 

s
n
=
7.1  5.65
4 .5
100
1.45
=
 3.22
.45
7.1 5.65

4.40
100
1.45

 3.30
.44
Degrees of freedom
• t distribution is dependent on the sample
size and this must be taken into account
when calculating p
• Skewness of sampling distribution
decreases as n increases
• t will differ from z less as sample size
increases
• t based on df where df = n - 1
t table
Two-Tailed Significance Level
df
10
15
20
25
30
100
.1 0
1 .8 1 2
1 .7 5 3
1 .7 2 5
1 .7 0 8
1 .6 9 7
1 .6 6 0
.0 5
2 .2 2 8
2 .1 3 1
2 .0 8 6
2 .0 6 0
2 .0 4 2
1 .9 8 4
.0 2
2 .7 6 4
2 .6 0 2
2 .5 2 8
2 .4 8 5
2 .4 5 7
2 .3 6 4
.0 1
3 .1 6 9
2 .9 4 7
2 .8 4 5
2 .7 8 7
2 .7 5 0
2 .6 2 6
Statistical inference made
• With n = 100, t.02599 = 1.98
• Because t = 3.30 > 1.98, reject H0
• Conclude that viewing violent video leads
to more aggressive free associates than
normal
Factors affecting t
• Difference between sample &
population means
– As value increases so t increases
• Magnitude of sample variance
– As sample variance decreases t increases
• Sample size - as it increases
– The value of t required to be significant
decreases
– The distribution becomes more like a
normal distribution
Application of t-test to Within
group designs
t for repeated measures
scores
• Same participants give data on two
measures
• Someone high on one measure probably
high on other
• Calculate difference between first and
second score
• Base subsequent analysis on these
difference scores. Before and after data are
ignored
Example - Therapy for PTSD
• Therapy for victims of psychological trauma-Foa et al (1991)
– 9 Individuals received Supportive Counselling
– Measured post-traumatic stress disorder symptoms before and after
therapy
Mean
St. Dev.
Before
After
Diff.
21
24
21
26
32
27
21
25
18
23.84
4.20
15
15
17
20
17
20
8
19
10
15.67
4.24
6
9
4
6
15
7
13
6
8
8.22
3.60
In this case we have…
Sample
Data of
interest
Error
Population
Mean
Hypothesis:
Difference
=0
(n=9)
S of the
?
Difference
Results
• The Supportive Counselling group
decreased number of symptoms - was
difference significant?
• If no change, mean of differences
should be zero
• So, test the obtained mean of
difference scores against  = 0.
• We don’t know , so use s and solve
for t
Repeated measures t test
• D and sD = mean and standard deviation of
differences respectively
X 
t
sX
X 

s
n
7.1 5.65

4.40
100
1.45

 3.30
.44
D
t
sD
D
 s
D
n
8.22

3.6
9
8.22

 6.85
1.2
df = n - 1 = 9 - 1 = 8
Inference made
•
•
•
•
With 8 df, t.025 = +2.306
We calculated t = 6.85
Since 6.85 > 2.306, reject H0
Conclude that the mean number of
symptoms after therapy was less than
mean number before therapy.
• Infer that supportive counselling seems to
work
+ & - of Repeated measures
design
• Advantages
– Eliminate subject-to-subject variability
– Control for extraneous variables
– Need fewer subjects
• Disadvantages
– Order effects
– Carry-over effects
– Subjects no longer naïve
– Change may just be a function of time
t test is robust
• Test assumes that variances are the same
– Even if the variances are not the same, the
test still works pretty well
• Test assumes data are drawn from a
normally distributed population
– Even if the population is not normally
distributed, the test still works pretty well