ttest for Correlated Samples

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t-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.

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D 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.
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• 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
4
Calculate 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
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• 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).

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t-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.

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Pair 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
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• H0 µD 0
• H1 µD 0

Calculate sums of squares.
173 (35)2 19.875 8
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Calculate 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)
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More 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.

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N 5 N 5 X1 7
X2 6 ?X1 35 ?X2 30
?X12 259 ?X22 190
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H0 ?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)
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More 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.

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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
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H0 ? 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)
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More 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.

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H0 ?1 - ?2 0
H1 ?1 - ?2 lt 0
262 - (32)2 4
100 - (20)2 5
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6
20
1.29
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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)
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More 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.

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D 5 ?D 50 ?D2 322
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H0 ?D 0
H1 ?D 0
SSD ?D2 - (?D)2 N
322 - (50)2 72 10
0.894
5 - 0 5.59 0.894
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• 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).