statistics for the social sciences
DESCRIPTION
Statistics for the Social Sciences. Psychology 340 Spring 2005. Using t-tests. Outline. Review t-tests One sample, related samples, independent samples. 1 sample. Two scores per subject. The related-samples t-test can be used when:. Statistical analysis follows design. 2 samples. - PowerPoint PPT PresentationTRANSCRIPT
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Statistics for the Social Sciences
Psychology 340Spring 2005
Using t-tests
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Statistics for the Social Sciences
Outline
• Review t-tests– One sample, related samples, independent samples
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Statistics for the Social Sciences
Statistical analysis follows design
• The related-samples t-test can be used when:
– 1 sample
t =D−μD
sD
– Two scores per subject
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Statistics for the Social Sciences
Statistical analysis follows design
• The related-samples t-test can be used when:
– 1 sample– Two scores per subject
t =D−μD
sD
– 2 samples
– Scores are related
- OR -
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Statistics for the Social Sciences
Performing your statistical test
€
t =X − μ
X
sX
€
zX
=X − μ
X
σX
€
sX
=s
n
Test statistic
Diff. Expected by chance
€
σX
=σ
n
One sample z One sample t
€
df = n −1
Related samples t
t =D−μD
sD
sD =sDnD
df =nD −1
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Statistics for the Social Sciences
Effect Sizes & Power for t Test for Dependent Means
d =μ1 −μ2
σD
estimated d =D−0sD
Remember we don’t know these
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Statistics for the Social Sciences
Approximate Sample Size Needed for 80% Power (.05 significance level)
• Using Power and effect sizes to determine how many participants you need
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Statistics for the Social Sciences
Independent samples
• What are we doing when we test the hypotheses?– Consider a new variation of our memory experiment example
Memory treatment
Memory patients Memory
Test
• the memory treatment sample are the same as those in the population of memory patients.• they aren’t the same as those in the population of memory patients
H0
:HA:
Memory placebo
MemoryTest
Compare these two means
XA
XB
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Statistics for the Social Sciences
Statistical analysis follows design
• The independent samples t-test can be used when:
– 2 samples
– Samples are independent
€
t =(X A − X B ) − (μA − μB )
sX A −X B
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Statistics for the Social Sciences
Performing your statistical test
Estimate of the standard error based on the variability of both samples
€
test statistic =observed difference
difference expected by chance
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Statistics for the Social Sciences
Performing your statistical test
€
test statistic =observed difference
difference expected by chance
Test statistic
€
t =(X A − X B ) − (μA − μB )
sX A −X B
€
t =X − μ
X
sX
One-sample tIndependent-samples t
Sample means
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Statistics for the Social Sciences
Performing your statistical test
€
test statistic =observed difference
difference expected by chance
Test statistic
€
t =(X A − X B ) − (μA − μB )
sX A −X B
€
t =X − μ
X
sX
One-sample tIndependent-samples t
Population means• from the hypotheses
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Statistics for the Social Sciences
Performing your statistical test
€
test statistic =observed difference
difference expected by chance
Test statistic
€
t =(X A − X B ) − (μA − μB )
sX A −X B
€
t =X − μ
X
sX
One-sample tIndependent-samples t
Population means• from the hypotheses
H0
:Memory performance by the treatment group is equal to memory performance by the no treatment group.So:
(μA −μB) =0
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Statistics for the Social Sciences
Performing your statistical test
€
test statistic =observed difference
difference expected by chance
Test statistic
€
t =X − μ
X
sX
One-sample t
€
t =(X A − X B ) − (μA − μB )
sX A −X B
Estimated standard error(difference expected by chance)
estimate is based on one
sample
We have two samples, so the estimate is based on two
samples
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Statistics for the Social Sciences
Performing your statistical test
€
sX A −X B
=sp
2
nA
+sp
2
nB
“pooled variance”
We combine the variance from the two
samples
Number of
subjects in group
A
Number of
subjects in group
B
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Statistics for the Social Sciences
s2 =SSn−1
variance
Performing your statistical test
€
sX A −X B
=sp
2
nA
+sp
2
nB
“pooled variance”
We combine the variance from the two
samples
Recall “weighted means,”
need to use “weighted
variances” here
€
sp2 =
SSA + SSB
dfA + dfB
sp2 =
sA2dfA( ) + sB
2dfB( )dfA +dfB
dfA =(nA −1)dfB =(nB −1)
Variance (s2) * degrees of freedom (df)
s2 (n −1) =SS
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Statistics for the Social Sciences
Performing your statistical test
€
sX A −X B
=sp
2
nA
+sp
2
nB
€
df = nA + nB − 2
€
t =(X A − X B ) − (μA − μB )
sX A −X B
€
=(nA −1) + (nB −1)
€
sp2 =
SSA + SSB
dfA + dfB
Independent-samples t• Compute your estimated standard error
sp2 =
sA2dfA( ) + sB
2dfB( )dfA +dfB
• Compute your t-statistic
• Compute your degrees of freedom
dfA =(nA −1)dfB =(nB −1)
This is the one you use to look up your tcrit
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Statistics for the Social Sciences
Performing your statistical test
PersonExp. group
Control
group1
23
4
45
5540
60
43
4935
51
Need to compute the mean and variability for each sample
Dr. Mnemonic develops a new treatment for patients with a memory disorder. He isn’t certain what impact, if any, it will have. To test it he randomly assigns 8 patients to one of two samples. He then gives one sample the new treatment but not the other. Following the treatment period he gives both groups a memory test. The data are presented below. Use = 0.05.
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Statistics for the Social Sciences
Performing your statistical test
PersonExp. group
Control
group1
23
4
45
5540
60
43
4935
51
Need to compute the mean and variability for each sampleControl group
= 50
(45-50)2 + (55-50)2 + (40-50)2 + (60-50)2
= 250
SS =A
Dr. Mnemonic develops a new treatment for patients with a memory disorder. He isn’t certain what impact, if any, it will have. To test it he randomly assigns 8 patients to one of two samples. He then gives one sample the new treatment but not the other. Following the treatment period he gives both groups a memory test. The data are presented below. Use = 0.05.
XA =45 + 55 + 40 + 60
4
XA =50SSA =250
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Statistics for the Social Sciences
Performing your statistical test
Exp. group
(43-44.5)2 + (49- 44.5)2 + (35- 44.5)2 + (51- 44.5)2
= 155
SS =B
PersonExp. group
Control
group1
23
4
45
5540
60
43
4935
51
Need to compute the mean and variability for each sample
Dr. Mnemonic develops a new treatment for patients with a memory disorder. He isn’t certain what impact, if any, it will have. To test it he randomly assigns 8 patients to one of two samples. He then gives one sample the new treatment but not the other. Following the treatment period he gives both groups a memory test. The data are presented below. Use = 0.05.
XB =43+ 49 + 35 + 51
4
XA =50SSA =250
XB =44.5SSB =155
= 44.5
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Statistics for the Social Sciences
Performing your statistical test
€
sX A −X B
=sp
2
nA
+sp
2
nB
€
t =(X A − X B ) − (μA − μB )
sX A −X B
€
sp2 =
SSA + SSB
dfA + dfB
€
=250 +155
3+ 3= 67.5€
=67.5
4+
67.5
4= 5.81€
=(50 − 44.5) − (0)
5.81
€
dfA = (nA −1)
€
dfB = (nB −1)
PersonExp. group
Control
group1
23
4
45
5540
60
43
4935
51XA =50
SSA =250XB =44.5
SSB =155
Dr. Mnemonic develops a new treatment for patients with a memory disorder. He isn’t certain what impact, if any, it will have. To test it he randomly assigns 8 patients to one of two samples. He then gives one sample the new treatment but not the other. Following the treatment period he gives both groups a memory test. The data are presented below. Use = 0.05.
= 0.95
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Statistics for the Social Sciences
Performing your statistical test
Tobs= 0.95Tcrit= ±2.447
€
sX A −X B
= 5.81
€
sp2 = 67.5
Proportion in one tail0.10 0.05 0.025 0.01 0.005
Proportion in two tailsdf 0.20 0.10 0.05 0.02 0.01: : : : : :5 1,476 2.015 2.571 3.365 4.0326 1.440 1.943 2.447 3.143 3.707: : : : : :
€
df = nA + nB − 2 = 6
= 0.05Two-tailed
PersonExp. group
Control
group1
23
4
45
5540
60
43
4935
51
Dr. Mnemonic develops a new treatment for patients with a memory disorder. He isn’t certain what impact, if any, it will have. To test it he randomly assigns 8 patients to one of two samples. He then gives one sample the new treatment but not the other. Following the treatment period he gives both groups a memory test. The data are presented below. Use = 0.05.
€
t =(X A − X B ) − (μA − μB )
sX A −X B
€
=(50 − 44.5) − (0)
5.81
€
dfA = (nA −1)
€
dfB = (nB −1)
XA =50SSA =250
XB =44.5SSB =155
= 0.95
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Statistics for the Social Sciences
Performing your statistical test
Tobs= 0.95= 0.05Two-tailedTcrit= ±2.447
PersonExp. group
Control
group1
23
4
45
5540
60
43
4935
51
€
sX A −X B
= 5.81
€
sp2 = 67.5
€
df = nA + nB − 2 = 6
Dr. Mnemonic develops a new treatment for patients with a memory disorder. He isn’t certain what impact, if any, it will have. To test it he randomly assigns 8 patients to one of two samples. He then gives one sample the new treatment but not the other. Following the treatment period he gives both groups a memory test. The data are presented below. Use = 0.05.
€
t =(X A − X B ) − (μA − μB )
sX A −X B
€
=(50 − 44.5) − (0)
5.81
€
dfA = (nA −1)
€
dfB = (nB −1)
XA =50SSA =250
XB =44.5SSB =155
+2.45 = tcrit
- Fail to Reject H0
tobs=0.95
= 0.95
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Statistics for the Social Sciences
Performing your statistical test
Tobs= 0.95= 0.05Two-tailedTcrit= ±2.447
Tobs Tcrit
Compare<
Fail to reject the H0
PersonExp. group
Control
group1
23
4
45
5540
60
43
4935
51
€
sX A −X B
= 5.81
€
sp2 = 67.5
€
df = nA + nB − 2 = 6
Dr. Mnemonic develops a new treatment for patients with a memory disorder. He isn’t certain what impact, if any, it will have. To test it he randomly assigns 8 patients to one of two samples. He then gives one sample the new treatment but not the other. Following the treatment period he gives both groups a memory test. The data are presented below. Use = 0.05.
€
t =(X A − X B ) − (μA − μB )
sX A −X B
€
=(50 − 44.5) − (0)
5.81
€
dfA = (nA −1)
€
dfB = (nB −1)
XA =50SSA =250
XB =44.5SSB =155
= 0.95
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Statistics for the Social Sciences
Assumptions
• Each of the population distributions follows a normal curve
• The two populations have the same variance– We’ll return to this next time
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Statistics for the Social Sciences
Effect Size for the t Test for Independent Means
• Estimated effect size after a completed study
Estimated d =X1 −X2
sPooled
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Statistics for the Social Sciences
Power for the t Test for Independent Means (.05 significance level)
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Statistics for the Social Sciences
Approximate Sample Size Needed for 80% Power (.05 significance level)
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Statistics for the Social Sciences
Statistical Tests Summary
Design Statistical test (Estimated) Standard error
tcrit =(μA −μB)−(XA −XB)
sXA−XBsXA −XB
=sP2
nA+sP2
nB€
sD
=sD
nD
tcrit =D−μD
sD
sX =sn
tcrit =X−μX
sX
zX =X−μX
σ X
σ X =σ
nOne sample, σ knownOne sample, σ unknown
Two related samples, σ unknown
Two independent samples, σ unknown