Transcript Comparing Two Samples - University of Hong Kong

Comparing Two Samples: Part I
Measurements
(data)
Descriptive
statistics
Normality Check
Frequency histogram
(Skewness & Kurtosis)
Probability plot, K-S
test
YES
Data transformation
NO
Median, range,
Q1 and Q3
Mean, SD, SEM,
95% confidence
interval
Data transformation
Check the
Homogeneity
of Variance
YES
Parametric Tests
Student’s t tests for
2 samples; ANOVA
for  2 samples; post
hoc tests for
multiple comparison
of means
NO
Non-Parametric
Test(s)
For 2 samples: MannWhitney
For 2-paired samples:
Wilcoxon
For >2 samples:
Kruskal-Wallis
Sheirer-Ray-Hare
Procedures for comparing two samples
• Test normality of the data; if passed, then
• Compare the measurements both of location (central
value) and dispersion (spread or range) of the
thickness indices between the two sites
• If there is a difference only in the measure of location,
a parametric statistical test based upon the difference
between the two sample means
One sample t test: t = (X - )/SX
• The 2-tailed t test for significant difference between a
mean longevity of horses and a hypothesized population
mean of  = 22 yr
• Ho:  = 22 year
• Age at death (in year) of 25 horses:
Age (yr)
17.20
18.00
18.70
19.80
20.30
20.90
21.00
21.70
22.30
22.60
23.10
23.40
23.80
24.20
24.60
25.80
26.00
26.30
27.20
27.60
28.10
28.60
29.30
30.10
35.10
t = (24.23 - 22)/ (4.25/25)
= 2.624
 = n - 1 = 25 - 1 = 24
t 0.05 (2), 24 = 2.064
mean
SD
SE
24.23
4.25
0.85
SX = Standard error of mean = SD/n
Thus reject Ho
0.01<p<0.02
Confidence interval for mean
• A researcher often needs to know how close a sample
mean (x) is to the population mean ().
For example, the amount of dissolved/dispersed petroleum
hydrocarbons (DDPH) in the upper ocean.
It is impossible to sample the entire ocean so the population
mean at any time is unknown. If data of the population follow the
normality, the sample mean will be normally distributed.
• The standard deviation (or standard error ) of the mean can
be obtained by:
SX = S.E.M. = S/n
• In general, the population mean lies within one standard deviation
(S) of the sample mean with about 68% confidence. It is usual to
quote 95% confidence intervals by multiplying S.E.M. with the
(1) appropriate value of z, namely 1.96 (for n >30) or (2)
appropriate value of t, df = n -1 (for n < 30).
Confidence interval for mean
• e.g. 50 measurements of DDPH in the upper ocean were made.
The sample mean = 4.75 ppb and S = 3.99 ppb
Then SX = 3.99/ 50 = 0.5643
The 95% confidence intervals are obtained as
X  z 0.05 SX = 4.75  (1.96)(0.5643) = 4.75  1.105 ppb
i.e. 5.855 and 3.633 ppb
• e.g. Only 10 measurements of DDPH in the upper ocean were
made. The sample mean = 78.2 ppb and S = 38.6 ppb
Then SX = 38.6/ 10 = 12.206
The 95% confidence intervals are obtained as
X  t 0.05, n-1 SX = 78.2  (2.262)(12.206) = 78.2  27.61 ppb
Two-tailed and one-tailed tests
• If we do not know which of the means is greater than the other,
we reject extreme values in either direction (i.e. HA: a  b); this
procedure is known as a two-tailed test.
• However, if sufficient information is available for us to specify the
direction of the population means, we could frame the alternative
hypothesis as HA: a > b or a < b
• For example, comparison of total organic carbon (TOC)
concentrations in the sewage effluent before and after being
upgraded to an advance treatment. In this case, a reduced in
TOC would be expected after improvement of the sewage
treatment, and therefore, HA: before > after
• The corresponding critical value of z for 1-tailed tests is slightly
different from those for 2-tailed tests (see Table B3).
The difference between two sample means
with limited data
• If n<30, the above method gives an unreliable estimate of z
• This problem was solved by ‘Student’ who introduced the t-test
early in the 20th century
• Similar to z-test, but instead of referring to z, a value of t is
required (Table B3)
• df = 2n - 2 for n1 = n2
• For all degrees of freedom
below infinity, the curve
appears leptokurtic
compared with the normal
distribution, and this
property becomes
extreme at small degrees
of freedom.
Mean measured unit
Comparison of two samples
A
B
Measured unit
Two-sample t test
A
B
Error bar = 2 SD
Measured unit
Two-sample t test
A
B
Error bar = 2 SD
Measured unit
A
B
Error bar = 2 SD
Importance of Equal Variance
• If the measure of dispersion (i.e. homogeneity of
variance) also differs significantly between the
samples:
– data transformation procedure: if there was no
significant difference between the variances of
transformed data, then a parametric test could be
applied; otherwise you should consider
– non-parametric test or Welch’s approximate t’ (Zars
p. 128-129)
Assumptions for z and t tests
• For the t and z tests to be valid:
– the measurements should be at least
approximately normally distributed and
– with similar sample variances (i.e.,
homoscedastic or homogeneity of variance)
• Need to check Normality of both datasets
• Homoscedasticity can be checked by a F-test
The largest sample variance
F ratio =
The smallest variance
Check for homogeneity of variance
Ho: equal variance between the two samples
HA: unequal variances
• Given that Sa2 = 0.2898 (n = 15) and Sb2 = 0.1701 (n =12),
• Then, F ratio = 0.28929/0.1701 = 1.703
• Use the F table (Table B4, Zar 99, App.34),
Critical F 0.05(2), 14, 11 = 3.36 > 1.703
Accept Ho
• If Ho rejected, then transform data and redo F test
• If still failed, non-parametric or Welch’s approximate t’
An example
• Birds’ eggshells are thought
to be influenced by acid rain
which reduces the egg
thickness index (egg shell
mass/ surface area; mg cm-2)
• We investigate gulls’ eggs at
two nesting sites: a control
site and a site affected by
acid rain
• Taking a single egg at
random from a number of
nests at each site;
determining the egg
thickness index
Two sample Z test
• Ho: no significant difference in the the thickness
indices between the two sites
• i.e. the thickness indices belong to the same
population (or probability density function). Then the
population means for the two sites will be the same
and their difference will be zero.
• Based on the central limit theorem, a collection of
sample means will follow a normal distribution. It can
also be shown that the distribution of the difference of
two means (Xa - Xb) will also be normally distributed.
• Recall the distribution of z: if (Xa - Xb) can be divided
by an appropriate standard deviation, a value of z can
be calculated:
Two sample Z test
• The standard deviation of (Xa - Xb) =
[(1/n)(Sa2 + Sb2)]
• Then, Z = (Xa - Xb)/[(1/n)(Sa2 + Sb2)]
• This equation can be used provided the number of
measurements in each sample is reasonably large (n > 30),
the measurements are approximately normally distributed
and n in each sample is the same.
• Ho: a = b ; HA: a  b
• If the sample means are similar, the z value will be small
and Ho will be accepted. The critical values of z can be
obtained from the t table, Table B3, (df = ).
Two sample Z test
• The following data were obtained at two gull nesting sites.
50 eggs were taken at random from each site, with 1 egg
taken randomly from each nest.
Egg thickness index results (mg cm-2)
Site
a (control)
n
50
mean
33.2
S2
16.41
b
50
31.89
17.39
Ho: a = b ; HA: a  b
z = (Xa - Xb)/[(1/n)(Sa2 + Sb2)] = (33.2-31.89)/[(1/50)(16.41 + 17.39)]
z = 1.593
As critical z=0.05, df =  = 1.96, accept Ho.
Remember to always check the homogeneity of variance before running the t test.
Exercise
• The following data were obtained at two gull nesting sites.
50 eggs were taken at random from each site, with 1 egg
taken randomly from each nest.
Egg thickness index results (mg cm-2)
Site
a (control)
n
50
mean
33.2
S2
16.41
b
50
27.8
15.21
Test the Ho: a = b (HA: a  b) using the two sample z test
z = (Xa - Xb)/[(1/n)(Sa2 + Sb2)]
Remember to always check the homogeneity of variance before running the t test.
A t test with equal measurements in each sample (na = nb)
Example
• e.g. The chemical oxygen demand (COD) is measured at two
industrial effluent outfalls, a and b, as part of consent procedure. Test
the null hypothesis: Ho: a = b while HA: a  b (Given that the data
are normally distributed)
SS = sum of square = S2 × υ
a
3.48
2.99
3.32
4.17
3.78
4.00
3.20
4.40
3.85
4.52
3.09
3.62
b
3.89
3.19
2.80
4.31
3.42
3.41
3.55
2.40
2.99
3.08
3.31
4.52
n
12
12
mean
3.701
3.406
S2
0.257
0.366
sp2 = (SS1+ SS2) / (υ1+ υ2) = [(0.257 × 11) + (0.366 × 11)]/(11+11)
= 0.312
sX1 – X2 = √(sp2/n1 + sp2/n2) = √(0.312/12) × 2 = 0.228
t = (X1 – X2) / sX1 – X2 = (3.701 – 3.406) / 0.228 = 1.294
df = 2n - 2 = 22
t = 0.05, df = 22, 2-tailed = 2.074 > t observed = 1.294, p > 0.05
The calculated t-value is less than the critical t value.
Thus, accept Ho.
Remember to always check the homogeneity of variance before running the t test.
Example
• Growth of 30-weeks old non-transgenic and transgenic tilapia
was determined by measuring the body mass (wet weight).
Since transgenic fish cloned with growth hormone (GH) related
gene OPAFPcsGH are known to grow faster in other fish
species (Rahman et al. 2001), it is hypothesized that HA:
transgenic > non-transgenic while the null hypothesis is given as Ho:
transgenic  non-transgenic
Ho: transgenic  non-transgenic
HA: transgenic > non-transgenic
Given that mass (g) of tilapia are normally distributed.
transgenic non-transgenic
700
305
680
280
500
275
510
250
670
490
670
275
620
275
650
300
n
8
8
mean 625.0
306.25
S2
5798.2
6028.6
Example
sp2 = (SS1+ SS2) / (υ1+ υ2) = 5913.4
sX1 – X2 = √(sp2/n1 + sp2/n2) = 38.45
t = (X1 – X2) / sX1 – X2 = 8.29
df = 2n - 2 = 14
t = 0.05, df = 14, 1-tailed = 1.761 << 8.29 ; p < 0.001
The t-value is greater than the critical t value. Thus,
reject Ho.
Remember to always check the homogeneity of variance before running the t test.
Example
Given drug a Given drug b
• e.g. The data are human blood-clotting time (in
minutes) of individuals given one of two different
9.90
9.00
drugs. It is hypothesized that Ho: a = b while HA: a
11.10
 b (Given that the data are normally distributed)
8.80
8.40
7.90
8.70
9.10
9.60
9.60
8.70
10.40
9.50
sp2 = (SS1+ SS2) / (υ1+ υ2) = 0.519
sX1 – X2 = √(sp2/n1 + sp2/n2) = 0.401
n
6
7
mean
8.75
9.74
t = (X1 – X2) / sX1 – X2 = -2.470
S2
0.339
0.669
t = 0.05, df = 6 + 7 -2 = 11, 2-tailed = 2.201 < 2.470; 0.02<p<0.05
Thus, reject Ho but accept HA.
Power of the two-sample t test
• Power of two-sample t test is greatest when the number
of measurements in each sample is the same (n1 = n2)
• Power of two-sample t test for different numbers of
measurements in each sample (n1  n2) is smaller
• When n1  n2, effective n = 2n1n2 /(n1+n2)
– e.g. n1=6, n2=7,
effective n = 2(6 × 7)/ (6+7) = 6.46,
which is smaller than the average of 6 and 7 (6.5)
• Therefore, we should always use ‘balanced design’
where possible.
Measurements
(data)
For comparison of two
independent samples
Descriptive
statistics
Normality Check
Frequency histogram
(Skewness & Kurtosis)
Probability plot, K-S
test
YES
Data transformation
NO
Median, range,
Q1 and Q3
Mean, SD, SEM,
95% confidence
interval
Data transformation
F-test
Check the
Homogeneity
of Variance
YES
z test
t tests
Parametric Tests
Student’s t tests for
2 samples; ANOVA
for  2 samples; post
hoc tests for
multiple comparison
of means
NO
Non-Parametric
Test(s)
For 2 samples: MannWhitney
For 2-paired samples:
Wilcoxon
For >2 samples:
Kruskal-Wallis
Sheirer-Ray-Hare
MannWhitney
test
Non-parametric methods for comparison of two independent samples
Why do we need to transform the data into
normal distribution for parametric tests
when we can apply non-parametric tests?
Parametric tests are more powerful and more
sensitive than non-parametric ones.
Non-parametric tests for two samples the Mann-Whitney test
• Non-parametric tests are also called ‘distribution-free’ test because
they are independent of the underlying population distribution, the
only assumption being independence of observations and continuity
of the variable which is being measured. Most of these tests involve
a ranking procedure
• Mann-Whitney test is useful for two independent samples and can be
used as an alternative to the t-test, particularly where the
assumptions for t-test cannot be demonstrated
• It can also be applied to ordinal data
• It is usually assumed that the distribution of the measurements in the
two samples are of the same general form (shown by frequency
histogram or stem-and-leaf plot)
• This test can be undertaken with any sample size. But the method of
calculating the test statistic (U) depends upon the size of n.
Mann-Whitney test - basic principle
Case 1
35 24 23 21
19
17
15
9
4
3
1
Case 2
35 24
23
21
19
17
15
9
4
3
1
Case 3
35 24
23
21
19
17
15
9
4
3
1
Case 4
35 24
23
21
19
17
15
9
4
3
1
Group A
Group B
Unequal medians
Non-parametric tests for two samples the Mann-Whitney test
Example
• In Mann-Whitney test, the null hypothesis cannot be stated in terms
of population parameters, but is defined as the equality of the
medians of the populations from which the two samples are drawn.
• Example: Mann-Whiney test where n is small
Sample A
Sample B
9
10
12
15
6
11
7
13
18
• These measurements are combined and ranked in descending order:
B
B
B
A
18
15
13 12
B
B
A A A
11 10 9
7
6
Example
Example I: Mann-Whiney test (2-tailed test)
• These measurements are combined and ranked in descending order :
B
B
B
A
18
15
13 12
B
B
A A A
11 10 9
7
6
• Ho: equal medians
• Summation of ranks for A score
 R2 = 4 + 7 + 8 + 9 = 28
• Summation of ranks for B score
 R1 = 1 + 2 + 3 + 5 + 6 = 17
• U = n1n2 + [n1(n1 + 1)/2] - R1
• where the greatest measurement (smaller sum of ranks) in either of
the two groups is given as rank 1 (R1); Thus group B is R1 in this case.
• U = (5)(4) + [(5)(5 +1)/2] - 17 = 18
• U0.05(2), 5, 4 = U0.05(2), 4, 5 = 19 > 18 ; thus accept Ho; i.e. equal medians
Example
Sample A
Sample B
9
Sample A
Sample B
10
12
18
12
15
9
15
6
11
7
13
7
13
6
11
18
18 15 13 12 11 10 9 7 6
Ho: equal medians
U = n1n2 + [n1(n1 + 1)/2] -  R1
10
Sample A
Rank
Sample B
Rank
12
4
18
1
9
7
15
2
7
8
13
3
6
9
11
5
10
6
R2 = 28
R1 = 17
where the greatest measurement in either of the two groups is given as rank 1 (R1);
Thus group B is R1 in this case.
U = (5)(4) + [(5)(5 +1)/2] - 17 = 18
U0.05(2), 5, 4 = U0.05(2), 4, 5 = 19 > 18 ; thus accept Ho; i.e. equal medians
Example
Sample A
Sample B
9
Sample A
Sample B
10
11
18
11
15
9
15
6
12
7
13
7
13
6
12
18
18 15 13 12 11 10 9 7 6
10
Sample A
Rank
Sample B
Rank
11
5
18
1
9
7
15
2
7
8
13
3
6
9
12
4
10
6
R2 = 29
U = n1n2 + [n1(n1 + 1)/2] -  R1
U = (5)(4) + [(5)(5 +1)/2] - 16 = 19
U0.05(2), 5, 4 = U0.05(2), 4, 5 = 19; thus reject Ho at p = 0.05; i.e. unequal medians
R1 = 16
Example
Sample A
Sample B
9
Sample A
Sample B
11
10
18
10
15
9
15
6
12
7
13
7
13
6
12
18
18 15 13 12 11 10 9 7 6
11
Sample A
Rank
Sample B
Rank
11
6
18
1
9
7
15
2
7
8
13
3
6
9
12
4
11
5
R2 = 32
U = n1n2 + [n1(n1 + 1)/2] -  R1
U = (5)(4) + [(5)(5 +1)/2] - 15 = 20
U0.05(2), 5, 4 = U0.05(2), 4, 5 = 19 < 20; thus reject Ho at p = 0.02; i.e. unequal medians
R1 = 15
Example
Example II: Mann-Whiney test (2-tailed test)
• The following data provide representative soil moisture contents on
south and north-facing slopes under grassland in June. The MannWhitney test is used to test the null hypothesis that the population
medians of the two samples are the same.
• A variance ratio test will
show that the measurements
are heteroscedastic.
• Also, the measurements are
percentages, which would
not be expected to be
normally distributed.
Soil Moisture (% dry wt)
North-facing
Rank
73.0
1
72.2
2
71.3
3
70.1
5
69.4
6
69.1
7
69.1
8
68.0
9
67.6
10
67.2
11
66.5
12
66.1
13
65.4
14
63.6
15
South-facing
70.7
60.0
55.9
54.2
53.0
50.5
49.4
48.5
47.1
46.4
45.3
41.6
40.6
39.5
38.3
33.4
26.7
Rank
4
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
Example
• A variance ratio test will
show that the measurements
are heteroscedastic.
• Also, the measurements are
percentages, which would
not be expected to be
normally distributed.
• U = n1n2 + [n1(n1 + 1)/2] - R1
• U = (14)(17) + [(14)(14 + 1)/2]
- 116 = 227
• U0.05(2), 14, 17 = 169 < 227
Soil Moisture (% dry wt)
North-facing
Rank
73.0
1
72.2
2
71.3
3
70.1
5
69.4
6
69.1
7
69.1
8
68.0
9
67.6
10
67.2
11
66.5
12
66.1
13
65.4
14
63.6
15
• p < 0.001, thus reject Ho
• The medians of the two samples
are significantly different
n = 14
R1 = 116
South-facing
70.7
60.0
55.9
54.2
53.0
50.5
49.4
48.5
47.1
46.4
45.3
41.6
40.6
39.5
38.3
33.4
26.7
n = 17
Rank
4
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
R2 = 380
Non-parametric tests for two samples the Mann-Whitney test for n > 20
• Critical tables for U become unwieldy as n increases above 20 and
Mann & Whitney obtained z as a function of U and n as shown
below:
• z = (U - n1n2/2)/[n1n2(n1 + n2 + 1)/12]
• Suppose that the test value of U was found to be 148 with n1 = 16
and n2 = 29. Then
• z = (148 - (16  29/ 2)/[(16)(29)(16 + 29 + 1)/12] = -84/42.174 = -1.99
• For a 2-tailed test, the modulus (1.99) is taken.
• Referring Table B3, df = infinity, at p = 0.05, z = 1.96 < 1.99
• Thus, reject Ho
Important Notes
•
Comparisons between two samples can be made with reference
to the difference between the sample means or medians
•
Comparison between sample means are made by calculating
their difference and dividing by an appropriate sample standard
deviation
•
Where the sample size is large (e.g. n > 50) a two-sample z test
can be used. For small samples, a Student’s t-test can be used
•
The parametric t and z tests should be applied to independent
interval/ ratio measurements which are at least approximately
normally distributed
•
A simple test of homoscedasticity (F ratio test) should be applied
prior to application of t and z tests
•
For non-normally distributed or heteroscedastic data, the nonparametric Mann-Whitney test can be utilized