Two-sample tests
Q: Is there a significant difference between
the population means of two groups?
• Are the data paired or not?
– If paired, work with differences
• If not paired – two independent samples
– Are the variances approximately equal?
– If approx equal variances use two-sample ttest with pooled variance estimate
– If not equal, use two-sample t-test with
unequal variances
Elastic Band Stretching
ambient
225
247
249
253
245
259
242
255
286
diff
19
8
4
1
6
10
6
-3
6
Scatterplot of heatedvs ambient
290
280
heated
heated
244
255
253
254
251
269
248
252
292
270
260
250
240
220
230
240
250
260
ambient
270
280
290
Paired T for heated - ambient
N
Mean
StDev
SE Mean
heated
9
257.556
14.604
4.868
ambient
9
251.222
16.285
5.428
Difference
9
6.33333
6.10328
2.03443
95% CI for mean difference: (1.64194, 11.02473)
T-Test of mean difference = 0 (vs not = 0): TValue = 3.11 P-Value = 0.014
One-Sample T: difference
… is equivalent to:
Test of mu = 0 vs not = 0
Variable
N
Mean
StDev
SE Mean
difference
9
6.33333
6.10328
2.03443
95% CI
T
(1.64194, 11.02473)
P
3.11
0.014
Why is pairing useful?
• If measurements on the same individual or
unit, them pairing accounts for the
individual variation which gives association
between the measurements
– Differencing removes the individual differences
– Leads to a simple one-sample t-test.
– Two-sample t-test is wrong – groups not
independent
Soil Contamination
• Measure of growth inhibitor, phthalide on
reclaimed (11 samples) and farm land (9
samples)
• Is there any evidence of a difference in
phthalide levels between reclaimed and
farm land?
– Null Hypothesis: no difference between
population levels
– Alternative Hypothesis: ?
1. Step: Check if variances are equal
Boxplot of farm, reclaimed
H0: Variances in both
groups are equal
40
38
Data
H1: Variances are not equal
42
36
34
32
30
farm
reclaimed
F-Test (normal distribution)
Test statistic = 0.83, p-value = 0.802
Levene's Test (any continuous distribution)
Test statistic = 0.07, p-value = 0.788
H0 is not rejected
=> One can apply a two-sample t-test with equal variances
2. Step: Two-sample t-test with equal variances
N
farm
reclaimed
9
11
Mean
38.33
35.09
StDev
2.92
3.21
SE Mean
0.97
0.97
Difference = mu (farm) - mu (reclaimed)
Estimate for difference: 3.24242
95% CI for difference: (0.33269, 6.15216)
T-Test of difference = 0 (vs not =):
T-Value = 2.34 P-Value = 0.031 DF = 18
Both use Pooled StDev = 3.0814
=> There is a significant difference in phthalide
levels between farm and reclaimed land
Shellfish Data: Cadmium
1. Step: Check if variances are equal
Test for Equal Variances for Cadmium
SpeciesGroup
F-Test
Test Statistic
P-Value
M
Levene's Test
Test Statistic
P-Value
O
0.06
SpeciesGroup
0.33
0.000
0.08
0.10
0.12
0.14
0.16
95% Bonferroni Confidence Intervals for StDevs
0.18
M
O
0.0
0.2
0.4
Cadmium
0.6
0.8
24.41
0.000
2. Step: Two-sample t-test with unequal variances
Two-sample T for Cadmium
SpeciesGroup
M
O
N
89
79
Mean
0.1656
0.385
StDev
0.0870
0.151
SE Mean
0.0092
0.017
Difference = mu (M) - mu (O)
Estimate for difference: -0.219192
95% CI for difference: (-0.257466, -0.180918)
T-Test of difference = 0 (vs not =):
T-Value = -11.34 P-Value = 0.000 DF = 121
Effect of log-transformation
equal variances!
Test for Equal Variances for log cadmium
SpeciesGroup
F-Test
Test Statistic
P-Value
M
Lev ene's Test
Test Statistic
P-Value
O
0.40
SpeciesGroup
1.10
0.660
0.45
0.50
0.55
95% Bonferroni Confidence Intervals for StDevs
0.60
M
O
-3.5
-3.0
-2.5
-2.0
-1.5
log cadmium
-1.0
-0.5
0.0
0.29
0.591
Non-parametric Tests
• T-test make certain assumptions about the data
• For comparing means normality is not so
important – central limit theorem says that means
tend to be normally distributed for “large samples”
• Skewness, multiple modes, outliers may result in
poor estimates of means and standard deviations
• Non-parametric tests simply use the rank order of
the data and make fewer assumptions
• Give tests based on medians
• But still make some assumptions!!!
Correlation
• Gives measure of linear association
between two variables
• Values between –1 and +1
• 0
no linear association
• Correlation does not imply a causal
relationship
Can be a useful summary, but grossly
overused. Plots are often more informative
or regression modelling more useful.
Mussel mass data
Correlations: edible, height, width, length, shell
edible
height
width
height
0.881
width
0.910
0.922
length
0.879
0.946
0.941
shell
0.932
0.881
0.934
length
0.899
Large correlations between all variables
• could use formal hypothesis test that correlation is
non-zero, but not much point!
• Pictures tell more – but not everything ….!
Matrix Plot of edible, height, width, length, shell
80
120
160
160
240
320
40
edible
20
0
160
120
height
80
60
width
40
20
320
240
length
160
400
200
shell
0
0
20
40
20
40
60
0
200
400