Title: ttest for Correlated Samples
1t-test for Correlated Samples
- A researcher believes that relaxation training
will reduce the severity of asthma attacks in
asthmatic subjects. A sample of five subjects is
chosen for the study. The researcher records the
severity of their symptoms by measuring how many
doses of medication are needed for asthma
attacks. The subjects then receive relaxation
training. The week after, the researcher once
again measures the number of doses required by
each patient. The data are provided below. Is
the researcher correct? Use normal decision
rules? - This is a Before and After design.
2D 5 3 0 4 4
D2
25 9 0 16 16
?D2 66
?D 16
Note, D is the difference between each patients
score before and after training.
3- H0 ?D 0 (No change in symptoms)
- H1 ?D gt 0 (There is a change)
Calculate the sums of squares of the difference
scores.
N number of subjects per group
66 - (16)2 5
14.8
4Calculate the standard error of the difference
between the means.
0.86
Now we can calculate t.
t 3.2 - 0 3.72 0.86
Calculate df
df N - 1 Where N number of
subjects per group
5 - 1 4
t crit 4 2.132
5- Since t obs 3.72 gt t crit4 2.132 we
reject the H0. There is a change in symptoms
following relaxation training (t obs 3.72, p lt
0.05).
6t-test for Correlated Samples
- A researcher wishes to compare a new method of
teaching reading to slow learners to the
current standard method. Eight pairs of slow
learners with similar IQs are found, and one
member of each pair is randomly assigned to the
standard teaching method, while the other is
assigned to the new method. Reading ability is
then measured with a standard reading test. Is
one method better than the other? Use normal
decision rules. - This is a matched group design because subjects
are paired based on their IQ scores.
7Pair New Standard
Difference Method
Method 1 77 72 5 2 74 68 6 3
82 76 6 4 73 68 5 5 87 84 3
6 69 68 1 7 66 61 5 8
80 76 4
D 4.375
?D 35
?D2 173
8Calculate sums of squares.
173 (35)2 19.875 8
9Calculate standard error
Calculate t
4.375 7.343 0.5957
Calculate df
df N - 1 Where N number of pairs
8 1 7
t crit 2.365
Since tobs 7.343 gt t crit7 2.365, we
reject the H0. The new method leads to
significantly higher reading scores (t7 7, p lt
0.05)
10More Examples
- E.g. An investigator wants to determine whether
whether a banned performance enhancing drug
increases endurance when injected into an athlete
just before a competitive event. Ten volunteer
athletes are chosen and randomly assigned into
two groups of 5. The first group is given the
banned substance whereas the second group is
injected with a harmless red fluid. The subjects
then run on a treadmill until exhaustion. Total
time on treadmill is measured. Is there a
difference between the two group? Use normal
decision rules. - Independent t-test with equal N.
11N 5 N 5 X1 7
X2 6 ?X1 35 ?X2 30
?X12 259 ?X22 190
12H0 ?1 - ?2 0
H1 ?1 - ?2 0
259 - (35)2 5
190 - (30)2 5
14
10
13 1.10
t (X1 - X2) - (?1 - ?2) SX1 - X2
7 - 6 1.10
0.91
df N1 N2 - 2 5 5 - 2 8
t crit 8 2.306
Since t obs 0.91 lt t crit 2.306, do not
reject H0. There is no difference between the
groups (t 8 0.91, p gt 0.05)
14More Examples
- A researcher wishes to investigate the claim that
vitamin C reduces the frequency of the common
cold. To eliminate the variability due to
different family environments, pairs of children
from the same family are randomly assigned to
either a group that receives vitamin C or a group
that receives fake vitamin C. The researcher
then measures number of days sick during the
school year. Is there a difference in frequency
of sickness between the two groups? Use normal
decision rules. - Matched group design.
15 Days Missed Due to Illness Pair No.
Vitamin C Fake Vitamin C 1
2 3 2 5 4 3 7 9 4 0 3 5 3 5 6
7 7 7 4 6 8 5 8 9 1 2 10 3 5
D -1 1 -2 -3 -2 0 -2 -3 -1 -2
D -1.5 ?D -15 ?D2 37
16H0 ? D 0
H1 ? D lt 0
SSD ?D2 - (?D)2 N
37 - (-15)2 14.5 10
0.401
-1.5 - 0 -3.74 0.401
df N - 1 10 - 1 9
t crit 9 1.833
Since t obs -3.74 gt t crit -1.833, reject
H0. Children who take vitamin C have
significantly fewer illnesses (t9 -3.74, p lt
0.05)
17More Examples
- A researcher believes that women show more
empathy than men. Five men and four women are
interviewed. The number of times in which they
show empathy is recorded. Do the results support
the researchers hypothesis? Use normal decision
rules. - Independent t-test for unequal N.
18(No Transcript)
19H0 ?1 - ?2 0
H1 ?1 - ?2 lt 0
262 - (32)2 4
100 - (20)2 5
20
6
20 1.29
21 4 - 8 -3.10 1.29
df N1 N2 - 2 5 4 - 2 7
t crit 7 -1.895
Since t obs -3.10 gt t crit -1.895, reject
H0. Women show significantly more empathy than
men (t 7 -3.10, p lt 0.5)
22More Examples
- A researcher is interested in attitudes towards
public schools. A group of ten subjects is given
a test (pretest) regarding their attitudes and
then shown a movie about public schools.
Following the movie, the group is tested again
(posttest). Are attitudes as measured by the
tests were more or less favorable after seeing
the movie than they were before? Use normal
decision rules. - This is a before and after design.
23D 5 ?D 50 ?D2 322
24H0 ?D 0
H1 ?D 0
SSD ?D2 - (?D)2 N
322 - (50)2 72 10
0.894
5 - 0 5.59 0.894
25- df N - 1 10 - 1 9
- t crit 9 2.262
- Since t obs 5.59 gt t crit 2.262, we reject
H0. Attitudes after the movie are different from
attitudes before the movie (t 9 5.59, p lt 0.05).