Karl L. Wuensch
Department of Psychology
East Carolina University

• We wish to investigate the relationship between a dichotomous variable and a normally distributed variable.
• We have two groups of scores.
• The null is that the population means are identical.
• The alternative is that they are not.

• Independent Samples – there is no reason to expect that the scores in the one group are correlated with the scores in the other group.
• Correlated Sample – there is a good reason to expect such a correlation
– aka related samples, matched pairs, repeated measures, within-subjects, randomized blocks, and split plot.

4
5
7
3
4
1 25
36
27
38
40
46
• M
1
= 4
• M
2
= 35.4
• r = .84

• Subjects are rating the social desirability of two persons.
• One is physically attractive, the other not.
• I have arranged the scores to produce a positive correlation, as would be expected if we had a within-subjects design
• That is, each subject rated both persons.
• The observed correlation is r = .92

Attractive Unattractive Difference
5 1 4
6 3 3
7
8
8
5
5
7
2
3
1
M = 6.8
SD = 1.304
4.2
2.280
2.6
1.140

• H
0
:

1
=

2
H
1
:

1
≠ 
2
• For each case, find a difference score.
• If 
1
=

2
, then
 diff
= 0.
• Now we have converted the problem to a one sample problem, testing the null that the mean difference score is zero.

t

M
1 s
Diff

/
M
2 n

2 .
1 .
14 /
6
5

2 .
6
.
510

5 .
100
• df = n – 1 = 5 – 1 = 4 p = .007
• Jurors rated the physically attractive defendant as significantly more socially desirable ( M = 6.8, SD = 1.30) than the physically unattractive defendant ( M = 4.2,
SD = 2.28), t (4) = 5.10, p = .007.

• One should also report a confidence interval for the difference between means.
• CI = ( M
1
– M
2
)

CV ( SE
• CV is critical value for t
)
• On 4 df , for 95% confidence, CV = 2.776.
• SE is the standard error of the difference between means, s
Diff
/ n

.
51
• 2.6 
2.776(.51) = [1.18, 4.02]

proc means mean stddev t prt; run
;
Variable Label Mean Std Dev t Value Pr > |t|
Group1 Group1 6.8000000
Group2 Group2 4.2000000
1.3038405
2.2803509
11.66
4.12
Diff Diff 2.6000000 1.1401754 5.10 proc means CLM; var diff; run ;
Analysis Variable : Diff
Lower 95%
CL for Mean
Upper 95%
CL for Mean
1.1842852
4.0157148
0.0003
0.0146
0.0070

• The design was independent samples.
• I artificially created the correlation by altering the order of the scores
• for pedagogical purposes.
• Now we do the appropriate analysis.

Attractive Unattractive
5 1
6 3
7
8
8
5
5
7
M = 6.8
SD = 1.304
4.2
2.280

1
2 s
M
1

M
2
 s
1
2  s
2
2 n

1 .
304 2 
2 .
280 2
5

1 .
175

t t


M
1

M
2
6 .
s
M
1

M
2
80

4 .
20
1 .
175

2 .
213

p
• COMPUTE P=2*CDF.T(-2.213,8).
EXECUTE.
• df = N – 2 = 8 (assuming homogeneity of variance) ; p = .0578
• The effect of the defendant’s physical attractiveness on rated happiness fell short of statistical significance.
• Note: The independent samples analysis had less power that the correlated samples analysis.

CI
( 6 .
8



M
1

M
2
4 .
2 )


 t critical
2 .
306 ( 1
 s
M 1

M 2
.
175 )

2 .
6

2 .
71
 
0 .
11

5 .
31


Group N
1 5
2
Diff (1-2)
5
Group
1
2
Diff (1-2)
Diff (1-2)
Mean Std Dev Std Err Minimum Maximum
6.8000
1.3038
0.5831
5.0000
8.0000
4.2000
2.2804
1.0198
1.0000
7.0000
2.6000
1.8574
1.1747
Method Mean
6.8000
Pooled
4.2000
2.6000
Satterthwaite 2.6000
95% CL Mean
5.1811
8.4189
1.3686
-0.1089
-0.2352
7.0314
5.3089
5.4352
Method
Pooled
Variances
Equal
Satterthwaite Unequal
DF
8
6.3628
t Value
2.21
2.21
Pr > |t|
0.0578
0.0663

• Regardless of the design, s
M
1

M
2
 s
2
M
1
 s
2
M
2

2
 r
 s
M
1
 s
M
2
• With independent samples, r = 0.
• With correlated samples, r > 0, which will reduce the standard error.
• Reducing the standard error will give you a larger absolute value of t .

• It is theoretically possible for correlated t to lower power because the degrees of freedom are half those in the independent samples design.
• But practically speaking, the reduction of the standard error will more than compensate for the loss of df .

n
1
 n
2
s
M
1

M
2



SS
1 n
1

 n
2
SS
2

2




1 n
1

1 n
2


• SS = s 2 ( n - 1), df = n
1
+ n
2
– 2
• Must assume homogeneity of variance
• Donald Zimmerman says don’t do it if sample sizes differ more than a little.

n
1
 n
2
s
M
1

M
2
 s
1
2 n
1
 s
2
2 n
2
• df will be between ( N -2) and the smaller of
( n
1
-1) and ( n
2
-1).
• See our text for formula for computing the exact df .
• No assumption of homogeneity of variance.

• Most popular is Cohen’s d .
d

M
1

M
2 s pooled s pooled
 
( p j s
2 j
) p j
 n j
N
• Use my programs to put a CI on the estimate.

s pooled


1 .
857

( p j s 2 j
)

.
5 ( 1 .
304 ) 2 
.
5 ( 2 .
280 ) 2 d

M
1

M
2 s pooled

2 .
6
1 .
857

1 .
40
• Whether the design is correlated samples or independent samples does not affect how d is computed.

• I have SAS and SPSS macros that will put a confidence interval about d . Here are the values were the design independent samples.

• Compute d exactly as you would were the data from independent samples.
• The denominator should be the pooled standard deviation, not the standard deviation of the difference scores.
• For confidence intervals, see my document . Here it does matter whether the design is correlated samples or independent samples.

 
M
1

M
2 s control
• Useful when the control group SD is a better estimate of the population SD than is the treatment group SD .
• For example, when the treatment exists only in the lab, not in the field.

• This is the simple correlation between groups (numerically codes) and scores.
• the ratio of the two sample sizes can have a drastic effect on the value of the pointbiserial r , but does not affect d .
• From pooled t , r pb
 t 2 t
2
 df

• For a two-sample t ,

2 is simply the squared point-biserial r .
• It estimates the proportion of variance in the scores that is explained by group membership.
• It tends to overestimate the population value.
• Omega-square, 
2 , is less biased.

• The estimated parameter is the probability that a score randomly drawn from the one population will be larger than one randomly drawn from the other population.
• Neat idea, but it never caught on.
• See my document on this statistic.

• Here the null is that the difference between two population means is trivially small.
• For example, -.1 < 
< .1
• Simply construct a CI for 
.
• If the CI is totally contained within the null interval, assert the null.

• Suppose that the mean amount of cholesterol lowering achieved with drug A was 40 with a variance of 100 and for drug
B the mean was 42 with a variance of 400. n
A
= 11, n
B
= 9.
• The differences in means is trivial, that for variances is large. Is it significant?
• The null is 
2 
A

B
2

• Simply divide the larger variance by the smaller, obtaining an F of 400/100 = 4.0
• df = (9-1), (11-1) = 8,10.
• in SAS, p = 2 *( 1 -PROBF( 4 , 8 , 10 ));
• p = .044.

• The t tests comparing means are moderately robust to their assumptions
(normality and homogeneity of variance), especially with equal or near equal sample sizes.
• The F test of variances is not robust to its normality assumption.
• There are more robust alternatives.

• Transform each score to either |Y ij
(Y ij
– M j
– M j
| or
) 2 . j is the group number, i is the subject number.
• Use independent samples t to compare the transformed scores.
• Alternatives include comparing scores to the group medians or to their group trimmed means.

• Used when the samples are correlated.
t

( F

1 )
2 n

F ( 1
 r
2
)
2
• F = larger variance divided by smaller variance.
• df = n - 2

• When comparing independent means, first test the null of equal variances.
• If that test is not significant, use the pooled variances test of means.
• If that test is significant, use the separate variances test.
• This procedure is BS (bad statistics)

• The test of variances will have very little power when sample size is small, and thus will not detect even rather large deviations from homogeneity of variance. It is with small sample sizes that pooled t is likely least robust to the homogeneity of variance assumption.

• The test of variances will have very much power when sample size is large, and thus will detect as significant even very small differences in variance, differences that are of no concern given the pooled t test’s great robustness when sample sizes are large.

• Heterogeneity of variance is often accompanied by non-normal distributions, and some tests of variances are often not robust to their normality assumption.

• Box (1953) was an early critic of testing variances prior to conducting a test of means. He wrote “to make the preliminary test on variances is rather like putting to sea in a rowing boat to find out whether conditions are sufficiently calm for an ocean liner to leave port.”

• Using the Howell data (participants were students in Vermont), compare boys’ GPA with girls’ GPA.
• Please check out the computer output .

Among Vermont schoolchildren, girls’
GPA ( M = 2.82, SD = .83, N = 33) was significantly higher than boys’ GPA ( M =
2.24, SD = .81, N = 55), t (65.9) = 3.24, p =
.002, d = .72. A 95% confidence interval for the difference between girls’ and boys’ mean
GPA runs from .23 to .95 in raw score units and from .27 to 1.16 in standardized units.

• For more details on summary statements, please read my document Two Mean
Inference .
• Statistics is my cup of tea – Student’s t , that is, aka a pint of stout.