The t-test, the paired t-test,
and introduction to non-parametric tests
July 8, 2004
1. The t-test:
for comparing means (averages)
Comparing two means

Is the difference in means that we observe
between two groups more than we’d expect
to see based on chance alone?
Are the two means
different enough to
conclude that the observed
difference is greater than
would be expected by
chance?
When background noise
is high, it’s difficult to tell
if the two groups are
different.
When background noise is
low, it’s easier to
distinguish between
groups.
Comparing two means

What is the sampling distribution of the
difference in the means of two samples?

First we need to know: What is the distribution of
a difference between two normally distributed
random variables?
~N(12,25)
simulation of 500 averages of 30 from a normal distribution with mean 12 and standard deviation 5
(variance 25)
SE 
5
 .91
30
Most experiments will
yield a mean between
10 and 14 (±2 se)
~N(8,25)
simulation of 500 averages of 30 from a normal distribution with mean
8 and standard deviation 5 (variance 25)
SE 
5
 .91
30
Most experiments will
yield a mean between
6 and 10 (±2 se)
Distribution of the difference…
simulation of 500 differences between means from above distributions
Notice that most
experiments will yield a
difference value between
1 and 7 (wider than the
above sampling
distributions!)
SE(diff ) 
25 25

 1.29
30 30
Distribution of differences
if X and Y are independent and X ~ N(x, x2)
and Y ~ N(y, y2)
• recall that averages are normally distributed if n is large
enough, by the central limit theorem
then (X-Y) ~ N(y-x, x2+y2) and (X+Y) ~
N(y+x, x2+y2)

Therefore, if X and Y are the averages of n and m
subjects, respectively:
X n  Ym ~N (  x   y ,
x
n
2

 y2
m
)
Example

A particular IQ test is designed to have a range of
0 to 200 with a standard deviation of 10 when
given to U.S. adults. You suspect that female
doctors have higher IQ’s than male doctors. To
test this hypothesis, you take a random sample of
30 female doctors and 30 male doctors. The
women score an average of 152 and the men an
average of 149. What is your conclusion?
Recall steps of a hypothesis
test:
1. Define your hypotheses (null,
alternative)
 2. Specify your null distribution:
 3. Do an experiment
 4. Calculate the p-value of what you
observed
 5. Reject or fail to reject (~accept) the
null hypothesis

1.
Define your hypotheses
(null, alternative)
H0: ♀-doctor IQ = ♂-doctor IQ; (♀ - ♂ = 0)
Ha: ♀-doctor IQ ≠ ♂-doctor IQ; (♀- ♂ ≠ 0 )
[two-sided]
2. Specify your null distribution
Null hypothesis is that the
difference is zero.
F30  M 30
100 100
~N (0,

)
30
30
100 100
s.e.(diff ) 

 2.58
30
30
3. Do an experiment
Observed difference in our experiment = 3.0
IQ points
4. Calculate the p-value of
what you observed
3/2.58=1.16
Z = (FROM SAS):
data _null_;
x=(1-probnorm(1.16))*2;
put x;
run;
0.2460488061
Two sided test!
Both tails are
possible, so
must double the
area from one
tail.
(two-sided p-value)
5. Reject or fail to reject
(~accept) the null hypothesis
Not enough evidence to reject at the .05
significance level. (.24>.05)
Complication 1…

The harsh reality is, we hardly ever know the true
standard deviation a priori. If we knew that much,
we probably wouldn’t need to run an experiment! In
most cases, we must use the sample standard
deviation as a stand-in for the truth. However, by
estimating the population standard deviation we are
adding more uncertainty to our experiment. The null
distribution is slightly wider than a normal
curve…called a “t-distribution.”
Recall: sample variance and
standard deviation
N

The variance of a population: 2 =
( xi   ) 2
i 1
N
N
The variance of a sample: s2 =

( xi  x ) 2
i 1
n 1
N
The standard deviation of a sample: s=
(x  x)
i
i 1
n 1
2
Example: calculation of
sample standard deviation

systolic blood pressures: 104, 114, 120,
148, 130, 132, 143, 152, 133, 124
 Mean = 1300/10 = 130
 Sample standard deviation =
(104  130) 2  (114  130) 2  (120  130) 2  0
 (132  130) 2  (143  130) 2  (152  130) 2 
 (133  130) 2  (124  130) 2  2070
2070
 230
10  1
230  15
Estimated
standard error of the mean=
15
10
Complication 1…

The null distribution is slightly wider than a
normal curve…called a “t-distribution.”
The “t” probability density
function
Where:
is the degrees of freedom
(gamma) is the Gamma function
is the constant Pi (3.14...)
The “t” distributions

The t distribution depends on the degrees of
freedom.

Degrees of freedom here=number of observations
used to calculate the standard deviation (n) minus
the number of sample means (1 or 2) used in
calculation of the sample standard deviation
The “t” distributions

The t distribution is just a slightly flattened version
of the normal curve.
 The t distribution is actually a family of distributions
that comes closer and closer to the normal
probability distribution as degrees of freedom
increase.
 With n>30, the t distribution is approximately
normal.
The t-function in SAS is:
probt(t-statistic, df)
Degrees of freedom
Example

A one-sample test when the standard
deviation is unknown (one-sample t-test)
Example: One sample t-test

A British sleep researcher claims that the British sleep
an average of 6.0 hours a night. If you ask 30 Brits how
many hours they sleep per night and your sample
average is 6.9 hours with a sample standard deviation
of 3.0, do you think the researcher was mistaken in his
claim?
1. Specify hypothesis:
H0: average hours = 6.0
Ha: average hourse ≠ 6.0
[two-sided]
One sample t-test
2. Specify null distribution.
The null distribution here actually follows a “t-distribution
with 29 (=n-1) “degrees of freedom” (the higher the
number of degrees of freedom, the more the t-distribution
looks like a normal curve).
X 30 ~T29 (6.0,
3.0
30
 0.55)
One sample t-test
3. Observed data=6.9 hours with a sample standard of
3.0
One sample t-test
4. USE SAS TO CALCULATE p-value:
data _null_;
pval=1-probt(1.64, 29);
put pval;
run;
0.0559046876
For two-sided test, multiply by 2: p-value=.11
– This gives just a slightly higher answer than the Z-test (Z=1.64), which
yields a two-sided p-value of .10. Diminished certainty due to estimating
the standard deviation.
One sample t-test
5. .11>.05; do not reject null at a
significance level of .05
Example: two-sample t-test

In 1980, some researchers reported that “men have
more mathematical ability than women” as
evidenced by the 1979 SAT’s, where a sample of
30 random male adolescents had a mean score ± 1
standard deviation of 436±77 and 30 random
female adolescents scored lower: 416±81 (genders
were similar in educational backgrounds, socioeconomic status, and age). Do you agree with the
authors’ conclusions?
Two-sample t-test
1. Define your hypotheses (null, alternative)
H0: ♂-♀ math SAT = 0
Ha: ♂-♀ math SAT ≠ 0 [two-sided]
Two-sample t-test
2. Specify your null distribution:
F and M have approximately equal standard
deviations/variances, so make a “pooled”
estimate of variance.
s 2p

(n  1)sm2  (m  1)s 2f
nm2
(29)77 2  (29)812

 6245
58
6245 6245
M 30  F30 ~T58 (0,

)
30
30
6245 6245

 20.4
30
30
Two-sample t-test
3. Observed difference in our experiment = 20
points
Two-sample t-test
4. Calculate the p-value of what you observed
20  0
T58 
 .98
20.4
data _null_;
pval=(1-probt(.98, 58))*2;
put pval;
run;
0.3311563454
5. Do not reject null! No evidence that men are better
in math ;)
Example 2

Example: Rosental, R. and Jacobson, L.
(1966) Teachers’ expectancies:
Determinates of pupils’ I.Q. gains.
Psychological Reports, 19, 115-118.
The Experiment
(note: exact numbers have been altered)

Grade 3 at Oak School were given an IQ test at the
beginning of the academic year (n=90).
 Classroom teachers were given a list of names of
students in their classes who had supposedly
scored in the top 20 percent; these students were
identified as “academic bloomers” (n=18).
 BUT: the children on the teachers lists had
actually been randomly assigned to the list.
 At the end of the year, the same I.Q. test was readministered.
The results
Children who had been randomly assigned to the
“top-20 percent” list had mean I.Q. increase of
12.2 points (sd=2.0) vs. children in the control
group only had an increase of 8.2 points (sd=2.5)
Is this a statistically significant difference? Give a
confidence interval for this difference.
1. Hypotheses
H0: mean change (“gifted”) – mean change
(control) = 0
Ha: mean change (“gifted”) – mean change
(control) ≠ 0
2. Null distribution
Null distribution of difference of two means:
2
2
(
17
)
2
.
0

(
71
)
2
.
5
s 2p
 5.81
88
5.81 5.81
 "gifted"   control ~T88 (0,

)
18
72
5.81 5.81

 .64
18
72
3. Empirical data
Observed difference in our experiment =
12.2-8.2 = 4.0
4. P-value
t-curve with 88 df’s has slightly wider
cut-off’s for 95% area (t=1.99) than a
normal curve (Z=1.96)
t 88
12.2  8.2 4


 6.25
.64
.64
p-value <.0001
5. Reject null!

Conclusion: I.Q. scores can bias
expectancies in the teachers’ minds and
cause them to unintentionally treat “bright”
students differently from those seen as less
bright.
Confidence interval (more
information!!)
95% CI for the difference: 4.0±1.99(.64) =
(2.7 – 5.3)
t-curve with 88 df’s
has slightly wider cutoff’s for 95% area
(t=1.99) than a normal
curve (Z=1.96)
2. The paired T-test
The Paired T-test

Paired data: either the same person on different
occasions or pairs of people who are more similar
to each other than to individuals from other pairs
(husband-wife pairs, twin pairs, matched cases and
controls, etc.)
 For example, evaluates whether an observed
change in mean (before vs. after) represents a true
improvement (or decrease).
 Null hypothesis: difference (after-before)=0
Did the control group in the
previous experiment improve
at all during the year?
t71 
8.2
8.2

 28
2
.29
2.5
72
p-value <.0001
Summary
True standard
deviation is known
One sample (or
paired sample)
Two samples
One-sample Z-test
Two-sample Z-test
Two-sample t-test
Standard deviation
is estimated by the
sample
One-sample t-test
Equal
variances
are pooled
Unequal
variances
(unpooled)
Non-parametric tests

t-tests require your outcome variable to be
normally distributed (or close enough).
 Non-parametric tests are based on RANKS
instead of means and standard deviations
(=“population parameters”).
Example: non-parametric tests
10 dieters following Atkin’s diet vs. 10 dieters following
Jenny Craig
Hypothetical RESULTS:
Atkin’s group loses an average of 34.5 lbs.
J. Craig group loses an average of 18.5 lbs.
Conclusion: Atkin’s is better?
Example: non-parametric tests
BUT, take a closer look at the individual data…
Atkin’s, change in weight (lbs):
+4, +3, 0, -3, -4, -5, -11, -14, -15, -300
J. Craig, change in weight (lbs)
-8, -10, -12, -16, -18, -20, -21, -24, -26, -30
Jenny Craig
30
25
20
P
e
r
c 15
e
n
t
10
5
0
-30
-25
-20
-15
-10
-5
0
5
Weight Change
10
15
20
Atkin’s
30
25
20
P
e
r
c 15
e
n
t
10
5
0
-300
-280
-260
-240
-220
-200
-180
-160
-140
-120
-100
-80
Weight Change
-60
-40
-20
0
20
t-test doesn’t work…

Comparing the mean weight loss of the two
groups is not appropriate here.
 The distributions do not appear to be
normally distributed.
 Moreover, there is an extreme outlier (this
outlier influences the mean a great deal).
Statistical tests to compare
ranks:

Wilcoxon Mann-Whitney test is analogue of
two-sample t-test.
Wilcoxon Mann-Whitney test







RANK the values, 1 being the least weight loss
and 20 being the most weight loss.
Atkin’s
+4, +3, 0, -3, -4, -5, -11, -14, -15, -300
1, 2, 3, 4, 5, 6, 9, 11, 12, 20
J. Craig
-8, -10, -12, -16, -18, -20, -21, -24, -26, -30
7, 8, 10, 13, 14, 15, 16, 17, 18, 19
Wilcoxon Mann-Whitney test
Sum of Atkin’s ranks:
 1+ 2 + 3 + 4 + 5 + 6 + 9 + 11+ 12 + 20=73
 Sum of Jenny Craig’s ranks:
7 + 8 +10+ 13+ 14+ 15+16+ 17+ 18+19=137


Jenny Craig clearly ranked higher!
 P-value *(from computer) = .018