Chapter 15
Comparing Two
Populations:
Dependent samples
Dependent Samples

Dependent Samples - measurements
are somehow related to one another
either because:
– The measurements came from the same
subject (Within-subjects design), or
– Subjects were paired or matched
(Matched-pairs design)

May also be called a “Repeatedmeasures” designs
Why use dependent
samples?

Advantages
– Fewer subjects required with a withinsubjects design (each subject is
measured twice)
– Increased power through decreased
variability

Disadvantages
– “Carry-over” effects
Carry-over Effects
Let’s say we are interested in methods to
teach reading to pre-schoolers
 We

– introduce the whole-language method,
measure,
– then use the phonics method, and measure

We have a repeated-measures (withinsubjects design), but
Carry-over Effects
How can we remove the effects of
teaching reading with the wholelanguage method?
 There is a “carry-over” to the next
condition (phonics) condition, because
the preschoolers may have learned to
read
 Do we expect to forget how to read
once they learn to?

Carry-over Effects
Matched-pairs designs will eliminate
the carry-over effect problem
 Matching, though, must be done well,
without respect to the obtained scores,

– matching is done before the experiment

Along a dimension that is relevant to
the study
Dependent Samples t-test
With pairs of scores, either the result of
matching or taking multiple
measurements from the same subject
 Calculate the difference (D) between
the two scores (maintaining positive
and negative values)
 Calculate the mean difference,
standard error, and then t

Dependent Samples t-test
M D - D0
t=
sM D
where
sM D
sD
=
np
Dependent Samples t-test
MD is the mean of the differences
sD is the standard deviation of the
differences
sMD is the estimated standard error of the
sampling distribution of MD
Δ0 is the difference between the
population means (parameter)
Example of a Dependent
Samples t-test
We want to compare a new teaching
method with a traditional method on
math
 2 groups - 10 12th grades in each,
 Group “Old” receives old method,
Group “New” gets new method
 Subjects are matched on their IQ

Hypothesis test of
Old vs. New (teaching math)

1. State and Check Assumptions
– About the population


Normally distributed? - don’t know
Homogeneity of variance – we’ll check
– About the sample


Independent Random sample? – yes
Dependent samples
– About the sample

Interval level
Hypothesis test of
Old vs. New (teaching math)

2. Hypotheses
HO : μOld = μNew
(the effectiveness of the old and new method are the same)
μOld - μNew = 0
(the difference between the effectiveness of Old and New is 0)
HA : μ1 ≠ μ2
(the effectiveness of Old and New are not equal)
μ1 - μ2 ≠ 0
(there is a difference between the effectiveness the Old and New)
Hypothesis test of
Old vs. New (teaching math)

3. Choose test statistic
– 2 groups

dependent samples
Dependent-sample t-test
Hypothesis test of
Old vs. New (teaching math)

4. Set Significance Level
α = .05
Critical Value
Non-directional Hypothesis with
df = np - 1 = 10 - 1 = 9
From Table C, critical value of α /2 = .025
tcrit = 2.262, so we reject HO if
t ≤ - 2.262 or t ≥ 2.262
Hypothesis test of
Old vs. New (teaching math)

5. Compute Statistic
– We need:
n p , å D,å D , M D ,s
2
df ,sM
D
2
D
Scores on the WAT
(Wisconsin Achievement test)
Old
78
55
95
57
60
80
50
83
90
70
New
74
45
88
65
64
75
41
68
80
64
D
4
10
7
-8
-4
5
9
15
10
6
Need: Δ
MD
sD
sMD
Scores on the WAT
Old
78
55
95
57
60
80
50
83
90
70
New
74
45
88
65
64
75
41
68
80
64
D
4
10
7
-8
-4
5
9
15
10
6
ΣD = 54
ΣD2 = 712
MD = 5.4
SS(D) = 420.4
s2D = 46.7111
sD = 6.833
np = 10
sMD = sD/√np
= 6.833/ √10
= 2.161
Example
M D - D0
t=
sM D
5.4 - 0
t=
= 2.499
2.161
df = n p - 1 = 10 - 1 = 9
t crit (9) = 2.262
(non- directional HA)
t is greater than tcrit,
reject HO
Hypothesis test of
Old vs. New (teaching math)

6. Draw Conclusions
– because our t falls within the rejection
region, we reject the HO, and
– conclude that the old and new method
differ in their effectiveness in teaching
math, with the old method superior
Comparing Dependent and
Independent Samples t-tests

Suppose, instead of pairing subjects
based on IQ, we compared the mean
of the two groups using a 2independent samples t-test
M1 - M 2
t=
sM1 -M 2
Scores on the WAT

Old
78
55
95
57
2219.6
60
80
50
83
90
70

ΣXO = 718
ΣX2O = 53772
MO = 71.8
SS(XO) =
s2O = 246.622
sO = 15.70
nO = 10
New
74
45
88
65
1902.4
64
75
41
68
80
64
ΣXN = 664
ΣX2N = 45992
MN = 66.4
SS(XN) =
s2N = 211.377
sN = 14.54
nN = 10
Independent Samples t-test
sM O -M N
æ 1
1ö
SS(XO ) + SS(X N ) æ 1
1ö
= s ç + ÷ =
+ ÷
ç
df
è nO nN ø
è nO nN ø
sM O -M N
(2219.6 + 1902.4) æ 1 1 ö
=
çè + ÷ø
18
10 10
2
p
sM O -M N = 6.77
M O - M N 71.8 - 66.4
t=
=
= 0.7978
sM O -M N
6.77
Cannot reject HO
Comparison (cont.)
By matching subjects, a priori (prior to
the experiment)
 the variance is reduced substantially,
thereby
 Allowing us to find a difference that
was obscured

Ranking
When completing a Wilcoxon, ranking
must be done with the smallest
difference receiving the lowest rank (1)
 The biggest difference gets a rank of
np

Hypothesis Test with nonparametric alternative to t-test
Nicotine patches supposedly reduces
smoking
 A psychologist records the number of
cigarettes each of 9 patients smokes
per day without a patch and with a
nicotine patch.

OOPS!
Hypothesis Test

1. State and Check assumptions
– Random Sample - yes
– Dependent Sample (Within-subjects
design)

measures come from the same person
(wearing the patch, and not wearing the
patch)
– Number of cigarettes smokes normally
distributed? - not sure
– Homogeneity of Variance - nope
Hypothesis Test

2. State Hypotheses
HO : μBEFORE = μAFTER
HA : μBEFORE > μAFTER
Hypothesis Test

3. Choose Test
– 2 dependent samples
– H of V - no
Wilcoxon Tm (Wilcoxon signed-ranks test)
Hypothesis Test

4. Set Significance level
α = .05
with np > 8, the Wilcoxon Tm approaches a z
directional alternative at .05, using Table A
z = 1.65
if obtained z ≥ 1.65, Reject HO
Hypothesis Test

5. Compute Test Statistic
T    positiveRanks
z
T   T 
T 

T 

T

n p (n p  1)
4
n p (n p  1)(2n p  1)
24
Wilcoxon Test

Non-parametric alternative to
dependent sample t-test
– Calculate the differences between paired
scores (D)
– Calculate the absolute value of each (|D|)
– Rank |D|
– “Sign” the ranks using the sign from D
– Add all positive Ranks (T+)
Hypothesis Test

6. Draw conclusions
– Since the obtained z was greater that the
critical z of +1.65, HO is rejected, and
– Conclude that the distributions are
different with the before values greater
than the after values
Review
Procedure for calculating t
Independent samples
 Compute (for each group):

M1 , M 2
2
1
2
2
s ,s
n1 , n2
SS ( X 1 ), SS ( X 2 )
Calculating t - Independent
Samples

To find t, the estimated standard error
of the sampling distribution is
calculated:
1 1
sM 1  M 2  s   
 n1 n2 
SS ( X 1 )  SS ( X 2 )
2
sp 
n1  n2  2
2
p
Computing t

The t statistic ( a family of distributions
which varies with df) is computed
using:
M1  M 2
t
sM 1  M 2
Review
Procedure for finding t

Dependent Samples
– matched-pairs, correlated samples,
repeated-measures, etc. (2 samples)
– Compute
D (the difference between pairs of scores)
 np (the number of pairs of scores
 MD (the mean of the differences)

Dependent Samples t (cont.)
– Compute
sD (the standard deviation of the differences)
 sM
(the estimated standard error)
D

Estimated Standard Error
and t
sM =
D
sD
np
M D - D0
t=
sM
D
Conceptually, what is the t
statistic?
In both tests, the numerator is a
measure of the observed difference
between scores (either unpaired or
paired)
 The denominator, of both tests, is an
estimate of the standard deviation of
the sampling distribution of M1 - M2
(we call it “standard error”)

Error Terms
When a sample is taken from a
population, statistics computed from
the sample are estimates of
parameters, however,
 A certain amount of variability is
expected (no two sample means are
exactly alike)
 This variability from sample to sample
is called “error”

Error and t-tests
The amount of error obtained from a
sample is used to estimate the
“standard” error that might be
expected in the situation (given n, etc.)
 This error can produce large
(significant) differences between
sample statistics, just by chance,
because no two sample means are
exactly alike

t statistic as a ratio

Therefore,
obtained difference
t = ————————————————
difference expected by chance (“error”)
We will return to this type of conception again
Can’t tell the difference
between independent and
dependent samples?
If measurements are related to one
another - dependent samples
 You will not be told when to use
independent or dependent samples ttests
 You have to decide

Example
A psychologist believes that
environment is more important than
genetics in influencing intelligence.
 She locates 12 pairs of identical twins
that have been reared apart, one twin
in each pair in an enriched
environment and the other in an
impoverished one.
 Independent or Dependent Samples?

Another
A new experimental drug, ABZ, is
thought to have beneficial effects on
AIDS.
 20 AIDS patients are randomly
selected and assigned to one of two
conditions: ABZ for 90 days or Placebo
for 90 days.
 Independent or Dependent Samples?

A third example
A researcher suspects that increasing
the level of lighting during the winter
months will increase mood.
 The researcher selects 36 students,
tests their mood, then replaces the
light bulbs in the dorm from 75-W to
100-W for 1 month, then tests each
student again
 Independent or Dependent Sample?
