Download Independent sample t-test

Hypothesis Tests: Two
Independent Samples
Violent Videos Again
• Bushman (1998) Violent videos and
aggressive behavior
 Doing the study Bushman’s way--almost
• Bushman had two independent groups
 Violent video versus educational video
 We want to compare mean number of
aggressive associations between groups
The Data
Condition
Mean St. Dev. Variance
n
Violent video
7.10
4.40
19.36
100
NonViolent
video
5.65
3.20
10.24
100
Analysis
• These are Bushman’s data, though he
had more dimensions.
• We still have means of 7.10 and 5.65,
but both of these are sample means.
• We want to test differences between
sample means.
 Not between a sample and a population
mean
Cont.
Analysis--cont.
• How are sample means distributed if H0
is true?
• Need sampling distribution of differences
between means
 Same idea as before, except statistic is
(X1 - X2)
Sampling Distribution of Mean
Differences
• Mean of sampling distribution = m1 - m2
• Standard deviation of sampling
distribution (standard error of mean
differences) =
sX X
1
2
1
2
2
2
s s


n1 n2
Cont.
Sampling Distribution--cont.
• Distribution approaches normal as n
increases.
• Later we will modify this to “pool”
variances.
Analysis--cont.
• Same basic formula as before, but with
accommodation to 2 groups.
X X
X X
t

s s
s

n n
1
2
X1X 2
1
2
2
2
1
2
1
2
• Note parallels with earlier t
Our Data
t
X1  X 2
7.10  5.65

19.36 10.24
s12 s22


100
100
n1 n2
1.45 1.45


 2.66
.296 .544
Degrees of Freedom
• Each group has 100 subjects.
• Each group has n - 1 = 100 - 1 = 99 df
• Total df = n1 - 1 + n2 - 1 = n1 + n2 - 2
100 + 100 - 2 = 198 df
• t.025(198) = +1.97 (approx.)
Conclusions
• Since 2.66 > 1.97, reject H0.
• Conclude that those who watch violent
videos produce more aggressive
associates (M = 7.10) than those who
watch nonviolent videos (M = 5.65),
t(198) = 2.66, p < .05 (one-tailed).
IQ Pill
Independent Samples
Mean
St. Dev.
Before
After
101
98
110
90
98
104
100
105
100
103
IQ Pill
Independent Samples
Mean
St. Dev.
Before
After
101
98
110
90
98
104
100
105
100
103
99.4
7.2
102.4
2.3
Computing t
X X
X X
t

s s
s

n n
1
2
X1X 2
99.4  102.4
t
51.8 5.3

5
5
1
2
2
2
1
2
1
2
3
t
 .888
11.42
Conclusions
Since -.88 < 1.97, fail to reject H0.
Conclude that those who took the IQ
pill did not score significantly higher
(M = 102.4) on an IQ test than
those who did not take the IQ pill (M
= 99.4), t(8) = -.88, p > .10.
t value for related samples
D  m 3
3
t


 1.236
s
5.43 2.428
n
5
D
Larger t value for same data
Assumptions
• Two major assumptions
 Both groups are sampled from populations
with the same variance
• “homogeneity of variance”
 Both groups are sampled from normal
populations
• Assumption of normality
 Frequently violated with little harm.
Pooling Variances
• If we assume both population variances
are equal, then average of sample
variances would be better estimate.
(n1  1) s  n2  1s
s 
n1  n2  2
2
p
2
1
2
2
99(19.36)  9910.24 29304


 14.8
99  99  2
198
Cont.
Pooling Variances--cont.
• Substitute sp2 in place of separate
variances in formula for t.
• Will not change result if sample sizes
equal
• Do not pool if one variance more than 4
times the other.
 Discuss
Heterogeneous Variances
• Refers to case of unequal population
variances.
• We don’t pool the sample variances.
• We adjust df and look t up in tables for
adjusted df.
• Minimum df = smaller n - 1.
 Most software calculates optimal df.
Effect Size for Two Groups
• Extension of what we already know.
• We can often simply express the effect as the
difference between means (=1.45)
• We can scale the difference by the size of the
standard deviation.
 Gives d
 Which standard deviation?
Cont.
Effect Size, cont.
• Use either standard deviation or their
pooled average.
 We will pool because neither has any claim
to priority.
 Pooled st. dev. = √14.8 = 3.85
Cont.
Effect Size, cont.
X viol  X nonviol 7.10  5.65
d

 0.38
s
3.85
• This difference is approximately .38
standard deviations.
• This is a medium effect.
• Elaborate
Confidence Limits


CI.95  X 1  X 2  t.025 s X 1  X 2
 (7.1  5.65)  1.97  .296
 1.45  1.97  .544  1.45  1.07
 0.38  m  2.53
Cont.
Confidence Limits--cont.
• p = .95 that interval formed in this way
includes the true value of m1 - m2.
• Not probability that m1 - m2 falls in the
interval, but probability that interval
includes m1 - m2.
 Comment that reference is to “interval
formed in this way,” not “this interval.”