Outline Introduction Comparing Two Means Comparing Two Proportions
Power and Sample Size Calculation – NOT COVERED
Comparing Two Variances – NOT COVERED
Lesson 10
Chapter 9: Comparing Two Populations
Michael Akritas
Department of Statistics
The Pennsylvania State University
Michael Akritas
Lesson 10 Chapter 9: Comparing Two Populations
Power and Sample Size Calculation – NOT COVERED
Comparing Two Variances – NOT COVERED
1
2
3
4
Power and Sample Size Calculation – NOT COVERED
5
6
Comparing Two Variances – NOT COVERED
7
Michael Akritas
Lesson 10 Chapter 9: Comparing Two Populations
Power and Sample Size Calculation – NOT COVERED
Comparing Two Variances – NOT COVERED
The comparisons will be based on a simple r.s. from each of the two populations: and X
21
, . . . , X
2 n
2
X
11
, . . . , X
1 n
1 from population 2.
from population 1,
In the Bernoulli case only p i
(or T i
= n i b i
) is typically given.
With the exception of the section on paired data, we will make the further assumption that the two samples are independent.
The sample mean and sample variances are denoted by
X i
=
1 n i n i
X j = 1
X in i
, S i
2
= n i
1
− 1 n i
X j = 1
X ij
− X i
2
, i = 1 , 2 .
Michael Akritas
Lesson 10 Chapter 9: Comparing Two Populations
Power and Sample Size Calculation – NOT COVERED
Comparing Two Variances – NOT COVERED
The Two Approaches
One approach for comparing means is based on the contrast
X
1
− X
2
An alternative approach for comparing means uses the ranks of the observations.
We will distinguish the cases where σ
1
= σ
2 and σ
1
= σ
2
:
Michael Akritas
Lesson 10 Chapter 9: Comparing Two Populations
Power and Sample Size Calculation – NOT COVERED
Comparing Two Variances – NOT COVERED
1 If σ
2
1
= σ
2
2
, under normality and for any n
1
, n
2
,
X
1
− X
2
− ( µ
1
− µ
2
) r
S
2 p
1 n
1
+
1 n
2
∼ t n
1
+ n
2
− 2
.
2 where S 2 p
= [( n
1
− 1 ) S 2
1
+ ( n
2
− 1 ) S 2
2
] / ( n
1
+ n
2
− 2 ) .
( Smith-Satterthwaite ) Under normality and for any n
1
, n
2
,
X
1
− X
2
− ( µ
1
− µ
2
) q
S
2
1 n
1
+
S
2 n
2
2
·
∼ t
ν
, where ν =
S
2
1 n
1
( S
2
1
/ n
1 n
1
− 1
) 2
+
+
S
2
2 n
2
2
( S
2
2
/ n
2 n
2
− 1
) 2
.
Without normality, both hold approximately if
n
Michael Akritas
Lesson 10 Chapter 9: Comparing Two Populations
Power and Sample Size Calculation – NOT COVERED
Comparing Two Variances – NOT COVERED
T
CIs for µ
1
−
µ
2
Let σ
2
1
= σ
2
2
. Under normality a ( 1 − α ) 100 % CI is:
X
1
− X
2
± t
α/ 2 , n
1
+ n
2
− 2 s
S
2 p
1 n
1
1
+ n
2
.
Under normality an approximate ( 1 − α ) 100 % CI is:
X
1
− X
2
± t
α/ 2 ,ν s
S
2
1 n
1
+
S
2
2
, n
2 where ν is the Smith-Satterthwaite DF (rounded down!).
Without normality, both hold approximately if
n
Michael Akritas
Lesson 10 Chapter 9: Comparing Two Populations
Power and Sample Size Calculation – NOT COVERED
Comparing Two Variances – NOT COVERED
A rule of thumb, for deciding if σ
2
1 approximately equal, is:
' σ
2
2
, i.e. that σ
2
1
, σ
2
2 are
max { S 2
1
, S 2
2
} min { S 2
1
, S 2
2
}
<
5 if n
1
, n
2
3 if n
1
, n
2
2 if n
1
, n
2
'
'
7
15
' 30
Rule of thumb for checking if
σ
2
1
' σ
2
2
Michael Akritas
Lesson 10 Chapter 9: Comparing Two Populations
Power and Sample Size Calculation – NOT COVERED
Comparing Two Variances – NOT COVERED
T
Tests for
µ
1
− µ
2
We will test H
0
: µ
1
− µ
2
= ∆
0 vs
H a
: µ
1
− µ
2
> ∆
0
, or µ
1
− µ
2
< ∆
0
, or µ
1
− µ
2
= ∆
0
.
The test statistic is T
H
0
=
X
1
− X
2
− ∆
0
, where
σ b
X
1
− X
2
If σ
2
1
= σ
2
2
, σ b
X
1
− X
2
= s
S 2 p
1 n
1
1
+ n
2
.
Without assuming σ
2
1
= σ
2
2
, σ b
X
1
− X
2
= s
S 2
1 n
1
+
S 2
2 n
2
.
Michael Akritas
Lesson 10 Chapter 9: Comparing Two Populations
Power and Sample Size Calculation – NOT COVERED
Comparing Two Variances – NOT COVERED
Rejection Rules
µ
1
− µ
2
> ∆
0
µ
µ
1
1
H a
− µ
2
− µ
2
< ∆
= ∆
0
0
RR at level α
|
T
T
T
H
0
H
0
H
0
|
> t
<
>
− t
∗
α
∗ t
∗
α
α/ 2
, where
If σ
2
1
= σ
2
2
, t
∗
α
= t
α, n
1
+ n
2
− 2
, t
∗
α/ 2
= t
α/ 2 , n
1
+ n
2
− 2
.
Without assuming σ
2
1
= σ
2
2
, t
∗
α
= t
α,ν denotes the Smith-Satterthwaite DF.
, t
∗
α/ 2
= t
α/ 2 ,ν
, where ν
Without normality these test procedures require n
1
, n
2
≥ 30.
Michael Akritas
Lesson 10 Chapter 9: Comparing Two Populations
Power and Sample Size Calculation – NOT COVERED
Comparing Two Variances – NOT COVERED
p
-values
The p -values are computed in a way similar to that for the one-sample T -test:
P − value =
1 − G
ν
∗
( T
H
0
) for upper-tailed test
G
ν
∗
( T
H
0
) for lower-tailed test
2 1 − G
ν ∗
( | T
H
0
| ) for two-tailed test where ν
∗
= n
1
+ n
2
− 2 if σ
2
1
= σ
2
2
, and
ν
∗
= ν , the Smith-Satterthwaite DF, if σ
2
1
= σ
2
2 is not assumed.
Michael Akritas
Lesson 10 Chapter 9: Comparing Two Populations
Power and Sample Size Calculation – NOT COVERED
Comparing Two Variances – NOT COVERED
Example
A simple r.s. of n
1
= average strength X
1
32 specimens of cold-rolled steel give
= 29 .
8 ksi, and S
1
= 4. A simple r.s. of n
2
X
2
= 35 specimens of two-sided galvanized steel give
= 34 .
7 ksi, and S
2
= 6 .
74. Construct a 99% CI for µ
1
− µ
2
.
Solution.
Assuming σ
2
1
S
2 p
= 5 .
6029
2 and use t
= σ
2
2
, we calculate the pooled variance
0 .
005 , 65
= 2 .
653. This gives a 99% CI of
( − 8 .
54 , − 1 .
26 ) . [NOTE: By the rule of thumb σ
1
= σ
2
.]
Without the assumption of to be ν = 56 .
12, and use t
σ
2
1
= σ
0 .
005 , 56
2
2
, we find the Smith-Satt. DF
= 2 .
666. This gives a 99% CI of ( − 8 .
48 , − 1 .
32 ) .
Michael Akritas
Lesson 10 Chapter 9: Comparing Two Populations
Power and Sample Size Calculation – NOT COVERED
Comparing Two Variances – NOT COVERED
Example (Cold-rolled and two-sided galvanized steel)
Are the strengths of the two types of steel different? Test at
α = 0 .
01 and compute the p-value.
Solution.
Since zero is not included in the 99% CIs,
H
0
: µ
1
= µ
2 is rejected at α = 0 .
01 with both test statistics.
As illustration, without the assumption of σ
2
1
= σ
2
2
, the TS is
T
H
0
= X
1
− X
2
/ q
S
2
1 n
1
+
S
2
2 n
2
= − 3 .
65 .
Since | T
H
0
| = 3 .
65 > t
0 .
005 , 56
= 2 .
666, p -value is 2 1 − G
56
( 3 .
65 ) = 0 .
0006.
H
0 is rejected. The
Michael Akritas
Lesson 10 Chapter 9: Comparing Two Populations
Power and Sample Size Calculation – NOT COVERED
Comparing Two Variances – NOT COVERED
The R command ”t.test”
The default is to assume σ
1 p -value for H
0
= σ
2
, do 95% CI, and give the
: µ
1
− µ
2
= 0 vs H a
: µ
1
− µ
2
= 0: t.test(y ∼ x) # For values in y and sample index in x t.test(y[1:32],y[33:67]) # Specify the two samples separately or: x1=y[which(x==1)]; x2=y[which(x==2)]; t.test(x1,x2)
For the pooled variance T test, and 99% CI do: t.test(y ∼ x, var.equal = TRUE, conf.level = 0.99)
To test H
0
: µ
1
− µ
2
= 1 .
8 vs H a
: µ
1
− µ
2
< 1 .
8 do: t.test(y ∼ x,mu=1.8, alternative = ”less”)
Other options: alternative = ”greater” or the default
”two.sided”
Michael Akritas
Lesson 10 Chapter 9: Comparing Two Populations
Power and Sample Size Calculation – NOT COVERED
Comparing Two Variances – NOT COVERED
Z
CIs for
p
1
− p
2
When n i p i
≥ 8, n i
( 1 − p i
) ≥ 8 for both i = 1 , 2, p
1
− p
2
·
∼ N p
1
− p
2
, σ b
2 p
1
− b
2
, where
σ b
2 p
1
− p
2
= p
1
( 1 − p
1
) n
1
+ p
2
( 1 − p
2
)
.
n
2
Thus the (approximate) ( 1 − α ) 100 % CI for p
1
− p
2 is p
1
− p
2
± z
α/ 2 s p
1
( 1 − n
1 p
1
)
+ p
2
( 1 − p
2
)
.
n
2
Michael Akritas
Lesson 10 Chapter 9: Comparing Two Populations
Power and Sample Size Calculation – NOT COVERED
Comparing Two Variances – NOT COVERED
Z
Tests for
p
1
− p
2
We will test H
0
: p
1
− p
2
= ∆
0 vs
H a
: p
1
− p
2
> ∆
0
, or p
1
− p
2
< ∆
0
, or p
1
− p
2
= ∆
0
.
The test statistic for testing this null hypothesis is
Z
H
0
σ b p
1
− b
2
=
= p
1
− p
2
− ∆
0
, where
σ b p
1
− p
2 s b
1
( 1 − b
1
) n
1
+ p
2
( 1 − p
2
)
.
n
2
Michael Akritas
Lesson 10 Chapter 9: Comparing Two Populations
Power and Sample Size Calculation – NOT COVERED
Comparing Two Variances – NOT COVERED
For testing H
0
: p
1
= p
2
(so ∆
0
= 0), it is customary to use:
σ e p
1
− b
2
= s p b
( 1 − p b
)
1 n
1
1
+ n
2
.
where b
= n
1 p
1 n
1
+ n
2 p
2
.
+ n
2
Thus the TS for testing H
0
: p
1
= p
2 is:
Z
H
0
= p
1
− p
2
.
σ e p
1
− b
2
Michael Akritas
Lesson 10 Chapter 9: Comparing Two Populations
Power and Sample Size Calculation – NOT COVERED
Comparing Two Variances – NOT COVERED
Rejection Rules and
p
-values
The rules for rejecting H
0 in favor of H a are: p p p
1
1
1
− p
−
H a p
− p
2
2
2
>
<
∆
∆
= ∆
0
0
0
RR at level α
Z
Z
H
0
H
0
> z
α
< − z
α
| Z
H
0
| > z
α/ 2
The p -value is computed the same way as for the one-sample Z -test:
P − value =
1 − Φ( Z
H
0
) for upper-tailed test
Φ( Z
H
0
)
2 1 − Φ( | Z
H
0 for lower-tailed test
| ) for two-tailed test
Michael Akritas
Lesson 10 Chapter 9: Comparing Two Populations
Power and Sample Size Calculation – NOT COVERED
Comparing Two Variances – NOT COVERED
Example
Tractors are assembled in assembly lines L
1 and L
2
. A simple r.s. of
200 tractors from L
1 yield 16 tractor requiring adjustment. A simple r.s. of 400 tractors from L
2 yield 14 requiring extensive adjustments.
Do a 99% CI for p
1
− p
2
, the difference of the two proportions.
Solution.
z
α/ 2
= z
Here
.
005 p b
1
= 16 / 200 = 0 .
08, p b
2
=
= 2 .
575. Thus the 99% C.I. is
14 / 400 = 0 .
035, and
0 .
08 − .
035 ± 2 .
575 r
( .
08 )( .
92 )
200
+
( .
05 )( .
95 )
400
= ( − 0 .
012 , 0 .
102 ) .
Michael Akritas
Lesson 10 Chapter 9: Comparing Two Populations
Power and Sample Size Calculation – NOT COVERED
Comparing Two Variances – NOT COVERED
Example (Assembly lines L
1 and L
2 continued)
It is suspected that a higher proportion of tractors from L
1 require adjustments. Formulate a hypothesis testing problem, test at α = .
01, and calculate the p -value.
Solution.
We want to test H
0
: p
1
= p
2 vs H a
: p
1 pooled estimate of the common (under H
0
) p is b
= ( 16 + 14 ) / 600 = 0 .
05. The TS is
> p
2
. The
Z
H
0
=
.
08 − .
035 p
( .
05 )( .
95 )( 1 / 200 + 1 / 400 )
= 2 .
38 .
Since 2 .
38 > z
.
01
= 2 .
33, H
0 is rejected. The p -value is p -value = 1 − Φ( 2 .
38 ) = 1 − .
9913 = .
0087 .
Michael Akritas
Lesson 10 Chapter 9: Comparing Two Populations
Power and Sample Size Calculation – NOT COVERED
Comparing Two Variances – NOT COVERED
The R command ”prop.test”
Set the number of successes and the number of trials adj=c(16,14); tract=c(200,400)
# X
1
= 16, X
2
= 14, n
1
= 200, n
2
= 400.
prop.test(adj,tract) # is equivalent to prop.test(adj, tract, alternative = ”two.sided”, conf.level =
0.95, correct = TRUE)
It tests H
0
: p
1
− p
2
= 0 vs alternative = ”two.sided”, ”less”, or ”greater”. No option for testing, e.g.
H
0
: p
1
− p
2
= 0 .
1
Michael Akritas
Lesson 10 Chapter 9: Comparing Two Populations
Power and Sample Size Calculation – NOT COVERED
Comparing Two Variances – NOT COVERED
Probability of Type II Error,
β (∆
1
)
, when
µ
1
− µ
2
= ∆
1
H a
β (∆
1
)
µ
µ
1
1
− µ
− µ
2
2
>
<
∆
∆
0
0
Φ z
α
−
∆
1
σ b
X
− ∆
0
1
− X
2
Φ − z
α
−
∆
1
σ b
X
− ∆
0
1
− X
2
µ
1
− µ
2
= ∆
0
Φ z
α/ 2
−
∆
1
− ∆
0
σ b
X
1
− X
2
− Φ − z
α/ 2
−
∆
σ b
1
X
− ∆
1
− X
2 where σ b
X
1
− X
2 is either of its two versions. These hold under normality with known variances (in which case be replaced by σ
X
1
− X
), but can also be used when n
2
Michael Akritas
σ b
X
1
− X
2
should
Lesson 10 Chapter 9: Comparing Two Populations
0
Power and Sample Size Calculation – NOT COVERED
Comparing Two Variances – NOT COVERED
Formulas for Sample Size
We will give the formula for the common sample size n = n
1
= n
2
II error, β (∆) , at alternative µ
H
0
: µ
1
− µ
2 required to achieve a specified level, β , of type
− µ
2
= ∆ when testing
= ∆
0
.
1
1 For a one-sided alternative, the formula is
2 n = ( σ
2
1
+ σ
2
2
) z
α
+ z
β
∆ − ∆
0
2 For a two-sided test the simple size formula is as above except that α is replaced by α/ 2.
Michael Akritas
Lesson 10 Chapter 9: Comparing Two Populations
Power and Sample Size Calculation – NOT COVERED
Comparing Two Variances – NOT COVERED
Example
Consider the data from the comparison of cold-rolled steel and two-sided galvanized steel. a) What is the probability of type II error when µ
1
− µ
2
= 5 ? Use α = 0 for 95% power at µ
1
− µ
2
= 5 ?
.
01 . b) What sample size is needed
Solution.
a) The formula for the probability of Type II error, without the assumption that σ
1
= q
S
2
1
/ n
1
+ S
2
2
/ n
2
= σ
√
=
2
(so σ b
X
1
− X
2
1 .
798 ), with ∆
0
= 0 , ∆
1
= 5 yields
β ( 5 ) = Φ( − 1 .
15 ) − Φ( − 6 .
31 ) = 0 .
1251 .
b) 95% power means β = 0 .
05 . The formula for the common sample size needed yields n = 4
2
+ 6 .
74
2
[( 2 .
Michael Akritas
576 + 1 .
645
Lesson 10 Chapter 9: Comparing Two Populations
) / 5 ]
2
= 43 .
7778 , or n = 44 .
Power and Sample Size Calculation – NOT COVERED
Comparing Two Variances – NOT COVERED
Comparison of Proportions
The formulas for the probability of type II error remain the same but with the following substitutions:
Substitute µ
1
, µ
2
Substitute σ b
X
1
− X
2 by p
1
, p
2
, respectively.
by σ b p
1
− p
2
.
The formula for the sample size remains the same with the following substitution:
Substitute σ where p i
2
1
, σ
2
2 by e
1
( 1 − p
1
) , p
2
( 1 − p is a preliminary estimator of p i
,
2 i
) , respectively,
= 1 , 2.
Michael Akritas
Lesson 10 Chapter 9: Comparing Two Populations
Power and Sample Size Calculation – NOT COVERED
Comparing Two Variances – NOT COVERED
Example
Consider the data from the tractor assembly lines L
1 is the probability of type II error when p
1
− p
2 and L
2
. a) What
= 0 .
04 ? Use
α = 0 .
01 . b) What sample size is needed for 95% power at p
1
− p
2
= 0 .
04 ?
Solution.
a) According to the formula for β (∆
1
) for the alternative
H a
: p
1
> p
2
, we have β ( 0 .
04 ) = Φ( z
Φ( 2 .
326 − 0 .
04 / 0 .
0189 ) = 0 .
5819 .
0 .
01
− 0 .
04 / σ b p
1
− ˆ
2
) = b) The desired level of probability of type II error is 0.05. According to the formula for common sample size we have n = ( 0 .
08 × 0 .
92 + 0 .
035 × 0 .
965 )
2 .
326 + 1 .
645
0 .
04
2
= 1058 .
24 , or n = 1059 .
Michael Akritas
Lesson 10 Chapter 9: Comparing Two Populations
Power and Sample Size Calculation – NOT COVERED
Comparing Two Variances – NOT COVERED
R for Power and Sample Size: Two Proportions
library(pwr) # For new R sessions with package pwr installed
Set h: h=2*asin(sqrt( p
1
))-2*asin(sqrt( p
2
))
There are two commands: pwr.2p.test(h = , n = , sig.level =, power = )
# For n
1
= n
2
= n . See ?pwr.2p.test for documentation.
pwr.2p2n.test(h = , n1 = , n2 = , sig.level = , power = )
# For n
1
= n
2
. See ?pwr.2p2n.test for documentation.
alternative = ”two.sided” is the default; ”less” or ”greater” are other options.
See also ?power.prop.test for the documentation of this function. It does not require installation of a package but it works only for n
1
= n
2
= n .
Michael Akritas
Lesson 10 Chapter 9: Comparing Two Populations
Power and Sample Size Calculation – NOT COVERED
Comparing Two Variances – NOT COVERED
R for Power and Sample Size: Two Means
library(pwr) # For new R sessions with package pwr installed
There are two commands (both assume σ
1
= σ
2
): pwr.t.test(n = , d = , sig.level = , power = , type =
”two.sample”))
# For n
1
= n
2
= n . Type ?pwr.t.test for the documentation.
pwr.t2n.test(n1 = , n2= , d = , sig.level =, power = )
# For n
1
= n
2
. Type ?pwr.t2n.test for the documentation.
alternative = c(”two.sided”,”less”,”greater”); ”two.sided” is default.
The command: power.t.test(n = NULL, delta = NULL, sd = 1, sig.level = 0.05, power = NULL, type = ”two.sample”, alternative
= c(”two.sided”, ”one.sided”), strict = FALSE)
only for n
1
= n
2
= n . Type ?power.t.test for the documentation.
Power and Sample Size Calculation – NOT COVERED
Comparing Two Variances – NOT COVERED
Motivation
The Mann-Whitney-Wilcoxon rank-sum test (or rank-sum test for short), which will be described, can be used with both small and large sample sizes.
If the two populations are continuous, the null distribution of the TS is known even with very small sample sizes.
For discrete populations, the null distribution of the TS can be well approximated with much smaller sample sizes than those required by contrast-based procedure.
The rank-sum test has high power, especially if the two population distributions are heavy tailed, or skewed.
Michael Akritas
Lesson 10 Chapter 9: Comparing Two Populations
Power and Sample Size Calculation – NOT COVERED
Comparing Two Variances – NOT COVERED
The Null Hypothesis and the TS
The rank sum procedure tests H
F
0
: F
1
= F
2
.
Let R ij denote the (mid-)rank of observation X ij combined set of N = n
1
+ n
2 in the observations, and set
W
1
= n
1
X
R
1 j
, R
1
= j = 1
W
1
, W
2 n
1
= n
2
X
R
2 j
, R
2
= j = 1
W
2
.
n
2
Then, the Mann-Whitney-Wilcoxon TS is
R
1
− R
2
=
N n
1 n
2
W
1
− n
1
N + 1
2
, or simply W
1
Michael Akritas
Lesson 10 Chapter 9: Comparing Two Populations
Power and Sample Size Calculation – NOT COVERED
Comparing Two Variances – NOT COVERED
The Standardized Rank-Sum TS and RR
If there are no ties,
Z
H
0
=
W
1
− n
1
( N + 1 ) / 2 p n
1 n
2
( N + 1 ) / 12
If H F
0 holds, Z
H
0
·
∼ N ( 0 , 1 ) , for n
1
, n
2
> 8. The RR are:
µ
1
−
H a
µ
2
> 0
µ
1
− µ
2
< 0
µ
1
− µ
2
= 0
Rejection region at level α
Z
H
0
≥ z
α
Z
H
0
| Z
H
0
≤ − z
α
| ≥ z
α/ 2
Michael Akritas
Lesson 10 Chapter 9: Comparing Two Populations
Power and Sample Size Calculation – NOT COVERED
Comparing Two Variances – NOT COVERED
Example
Data on sputum histamine levels from 9 allergic and 13 non-allergic individuals are given in http://sites.stat.
psu.edu/˜mga/401/Data/HistaminData.txt
. Is there a difference between the two populations? Test at α = .
01.
Solution.
R
15
= 17,
Here
R
16
R
=
11
=
21, R
18,
17
R
=
12
7,
=
R
11,
18
=
R
13
20,
= 22, R
14
R
19
= 19,
= 16. Thus
W =
X
R
1 j
= 151 and Z
H
0
= j
151 − 9 ( 23 ) / 2 p
9 ( 13 )( 23 ) / 12
= 3 .
17 .
Since n
1
, n
2
> 8, p -value=2 difference is significant.
[ 1 − Φ( 3 .
17 )] = .
0016. Thus the
Michael Akritas
Lesson 10 Chapter 9: Comparing Two Populations
Power and Sample Size Calculation – NOT COVERED
Comparing Two Variances – NOT COVERED
Effect of Outliers
In the above example, the t test does not reject at level
0.01.
With data in data frame hi , t.test(hi$Level ∼ hi$Sample) gives p-value of 0.13, and t.test(hi$Level ∼ hi$Sample, var.equal=TRUE) gives p-value of 0.06.
In general, using a procedure when the underlying assumptions are violated will give misleading results.
Michael Akritas
Lesson 10 Chapter 9: Comparing Two Populations
Power and Sample Size Calculation – NOT COVERED
Comparing Two Variances – NOT COVERED
R Command for the Rank Sum Test
With data in data frame hi , wilcox.test(hi$Level ∼ hi$Sample) is short for wilcox.test(hi$Level ∼ hi$Sample, alternative=”two.sided”,mu = 0, paired = FALSE)
Can also get a CI, whose description is not in the book, by wilcox.test(hi$Level ∼ hi$Sample, conf.int = TRUE), which is short for wilcox.test(hi$Level ∼ hi$Sample, alternative=”two.sided”,conf.int = TRUE, conf.level = 0.95)
More advanced application: boxplot(Ozone ∼ Month, data = airquality) wilcox.test(Ozone ∼ Month, data = airquality, subset =
Month %in% c(5, 8))
Michael Akritas
Lesson 10 Chapter 9: Comparing Two Populations
Power and Sample Size Calculation – NOT COVERED
Comparing Two Variances – NOT COVERED
Levene’s Test
It is based on the idea that if the variances are equal, then the two samples
V
1 j
= | X
1 j
− e 1
| , j = 1 , . . . , n
1
, and V
2 j
= | X
2 j
− e 2
| , j = 1 , . . . , n
2
, where e i
, i = 1 , 2 is the median from i th sample, correspond to populations with equal means and variances.
Thus, equality of variances can be tested by testing the hypothesis
H
0
: µ
V
1
= µ
V
2 vs µ
V
1
= µ
V
2 using the two-sample t -test with pooled variance.
Michael Akritas
Lesson 10 Chapter 9: Comparing Two Populations
Power and Sample Size Calculation – NOT COVERED
Comparing Two Variances – NOT COVERED
Example
The following data represent the plasma ascorbic acid (vitamin
C) concentration measurements ( µ mol/l) of five randomly selected smokers and nonsmokers:
Nonsmokers 41 .
48 41 .
71 41 .
98 41 .
68 41 .
18 s
1
Smokers
= 0 .
297
40 .
42 40 .
68 40 .
51 40 .
73 40 .
91 s
2
= 0 .
192
Here the two medians are e 1
= 41 .
68 , e 2
= 40 .
68. Subtracting them from their corresponding sample values we have
V
1
V
2 values for Nonsmokers 0 .
20 0 .
03 0 .
30 0 .
00 0 .
50 values for Smokers 0 .
26 0 .
00 0 .
17 0 .
05 0 .
23
The two-sample test statistic with pooled variance yields a
p
-value of 0.558. Thus, there is not enough evidence to reject
the null hypothesis of equality of the two population variances.
Power and Sample Size Calculation – NOT COVERED
Comparing Two Variances – NOT COVERED
The
F
Test Under Normality
When the two samples have been drawn from normal populations, the exact distribution of S 2
1
/ S 2
2 an F distribution.
is a multiple of
Theorem
Let X
11 sample from a normal distribution with variance σ
2
2
, and let S and S
2
2
, . . . , X
1 n
1 be a random sample from a normal distribution with variance σ
2
1
, let X
21
, . . . , X
2 n
2 be another denote the two sample variances. Then the rv
2
1
F =
S 2
1
/σ
2
1
S 2
2
/σ 2
2 has an F distribution with ν
1 of freedom.
= n
1
2
2
Power and Sample Size Calculation – NOT COVERED
Comparing Two Variances – NOT COVERED
The theorem suggests that H
0 the F test statistic:
: σ
2
1
= σ
2
2 can be tested with
F
H
0
=
S
2
1
.
S
2
2
If the ratio differs sufficiently from 1, the null hypothesis is rejected. In particular the RRs for testing H
0
: σ
2
1
= σ
2
2 are
σ
σ
σ
2
1
2
1
2
1
H a
> σ
< σ
= σ
2
2
2
2
2
2
RR at level α
F
H
0
F
H
0
> F n
1
− 1 , n
2
− 1 ; α
< F n
1
− 1 , n
2
− 1 ; α either F
H
0
> F n
1
− 1 , n
2
− 1 ; α/ 2 or F
H
0
< F n
1
− 1 , n
2
− 1 ; α
Michael Akritas
Lesson 10 Chapter 9: Comparing Two Populations
Power and Sample Size Calculation – NOT COVERED
Comparing Two Variances – NOT COVERED
Example
Consider the data in the previous example, and assume the underlying populations are normal. The test statistic is
F
H
0
=
0 .
297
0 .
192
= 2 .
40 .
The value 2.4 corresponds to the 79th percentile of the F distribution with ν
1
= 4 and ν
2
= 4 degrees of freedom. Thus, the p -value is 2 ( 1 − 0 .
79 ) = 0 .
42.
Michael Akritas
Lesson 10 Chapter 9: Comparing Two Populations
Power and Sample Size Calculation – NOT COVERED
Comparing Two Variances – NOT COVERED
Introduction, Motivation
Paired data arise when each experimental unit receives each of the two treatments that being compared.
1
2
3
Compare the durability of two types of tires.
Compare two labs for the analysis of mercury content.
Two acne treatments, two cataract treatments, etc.
Paired data are of the form: ( X
11
, X
21
) , . . . , ( X
1 n
, X
2 n
) .
CIs and the TS are again based on X
1
− X
2
. But now they are not independent. Thus, previous formulas do not apply.
For example, σ
2
X
1
− X
2
= σ
2
X
1
+ σ
2
X
2
− 2Cov ( X
Similarly, the rank sum test is not valid now.
1
, X
2
) .
Michael Akritas
Lesson 10 Chapter 9: Comparing Two Populations
Power and Sample Size Calculation – NOT COVERED
Comparing Two Variances – NOT COVERED
While Cov ( X
1
, X
2
) can be estimated, it is easier to use
D
1
= X
11
− X
21
, . . . , D n
= X
1 n
− X
2 n
D
σ
1
2
X
1
, . . . ,
− X
2
D
= n
σ are independent, and
2
D can be estimated by
D
σ b
=
2
D
X
1
−
= S
2
D
/
X n
2
. Thus,
, where
S
2
D
=
1 n − 1
" n
X
D i
2
− i = 1
1 n n
(
X i = 1
D i
)
2
#
CIs and testing are based on the fact:
D
S
D
−
/
µ
D n
∼ t n − 1 if normality holds, or if n ≥ 30
Michael Akritas
Lesson 10 Chapter 9: Comparing Two Populations
Power and Sample Size Calculation – NOT COVERED
Comparing Two Variances – NOT COVERED
Example
A total of 12 water samples are analyzed for mercury content by labs A and B . The paired data yields D = X
1
− X
2
= − and S
D
= 0 .
02645. Does lab B give, on average, higher concentration results than lab A ? Test at α = 0 .
05.
0 .
0167
Solution.
Here H
0
: µ
1
− µ
2
= 0 , H a
: µ
1
− µ
2
< 0. Because n < 30, we must assume normality. Doing so we have:
T
H
0
=
S
D
D
/
√ n
=
− 0 .
0167
.
02645 /
√
12
= − 2 .
1865 .
Since T
H
0
< − t
.
05 , 11
= − 1 .
796, H
0 is rejected.
Michael Akritas
Lesson 10 Chapter 9: Comparing Two Populations
Power and Sample Size Calculation – NOT COVERED
Comparing Two Variances – NOT COVERED
It is important to be able to recognize paired data. Below are two examples:
1. A study was conducted to see whether two cars, A and B, having very different wheel bases and turning radii, took the same time to parallel park. 7 drivers were randomly obtained and the time required for each of them to parallel park each of the 2 cars was measured. The results are as follows:
Car 1 2 3
Driver
4 5 6 7
A 19.0
21.8
16.8
24.2
22.0
34.7
23.8
B 17.8
20.2
16.2
41.4
21.4
28.4
22.7
Michael Akritas
Lesson 10 Chapter 9: Comparing Two Populations
Power and Sample Size Calculation – NOT COVERED
Comparing Two Variances – NOT COVERED
2. Two brands of motorcycle tires are to be compared for durability. Eight motorcycles are selected at random and one tire from each brand is randomly assigned (front or back) on each motorcycle. The motorcycles are then run until the tires wear out. The data in http://stat.psu.edu/˜mga/401/
Data/motorcycleTiresLifetimes.txt
are in km.
Michael Akritas
Lesson 10 Chapter 9: Comparing Two Populations
Power and Sample Size Calculation – NOT COVERED
Comparing Two Variances – NOT COVERED
Two Proportions (NOT COVERED)
Here each pair ( X
1 j
( 0 , 1 ) or ( 0 , 0 ) .
, X
2 j
) can be either ( 1 , 1 ) or ( 1 , 0 ) or
As an example, if n voters are asked, both before and after a presidential speech, whether or not they support a certain policy, X
1 j
= 1 or 0 if the speech, and X
2 j after the speech.
= j th voter supports or not before the
1 or 0 if the same voter supports or not
Now
X
1
= p b
1
, X
2
= p b
2
, D = p b
1
− p b
2
.
As before, S
2
D
/ n estimates the variance of p
1
− p
2
.
However: a) we have different sample size requirements, and b) typically, the pairs ( X
1 j
, X
2 j
) are not given.
Michael Akritas
Lesson 10 Chapter 9: Comparing Two Populations
Power and Sample Size Calculation – NOT COVERED
Comparing Two Variances – NOT COVERED
Table Format of the Data
Before
1
1 Y
1
After
0
Y
2
0 Y
3
Y
4
Y
1 is the number of ( 1 , 1 ) pairs,
Y
2 is the number of ( 1 , 0 ) pairs,
Y
3 is the number of ( 0 , 1 ) pairs,
Y
4 is the number of ( 0 , 0 ) pairs,
Y
1
+ · · · + Y
4
= n
Michael Akritas
Lesson 10 Chapter 9: Comparing Two Populations
Power and Sample Size Calculation – NOT COVERED
Comparing Two Variances – NOT COVERED
The test statistic can be computed from:
Y
1
+ Y
2 n
= p b
1
,
Y
1
+ Y n
3
= p b
2
, D =
Y
2
− Y
3
, and n n − 1 n
S
2
D
=
1 n n
X
( D i
− D )
2 i = 1
= q
2
+ b
3
− ( b
2
− q
3
)
2
, where q
2
= Y
2
/ n , q
3
= Y
3
/ n .
Thus, the test statistic for H
0
: p
1
− p
2
= ∆
0 is:
T
H
0
=
( b
2
− q
3
) − ∆
0 p
( q
2
+ q
3
− ( b
2
− q
3
) 2 ) / ( n − 1 )
It is approximately normal provided n
10
+ n
01
≥ 16.
Michael Akritas
Lesson 10 Chapter 9: Comparing Two Populations
Power and Sample Size Calculation – NOT COVERED
Comparing Two Variances – NOT COVERED
McNemar’s Test
It can be used only for testing H
0
: p
1
− p
2
= 0:
MN =
( q b
2
− q b
3
)
2
( q b
2
+ q b
3
) / n
=
( Y
2
− Y
3
)
2
Y
2
+ Y
3
∼ χ
2
1
Note that MN = ( MN
2
)
2
, where
MN
2
=
( q
2
− q
3
) p
( q
2
+ b
3
) / n
=
Y
√
2
− Y
3
Y
2
+ Y
3 is a variation of the paired t statistic:
The different denominator is a consistent estimator of S
2
D under the null hypothesis.
Michael Akritas
Lesson 10 Chapter 9: Comparing Two Populations
Power and Sample Size Calculation – NOT COVERED
Comparing Two Variances – NOT COVERED
The Signed Rank Test – NOT COVERED
2
3
1 Rank the absolute differences | D
1
| , . . . , | D n
| from smallest to largest. Let R i
Assign to R i denote the rank of the sign of D i
| D i
| .
, forming thus signed ranks.
Let S
+ be the sum of the ranks R i with positive sign, i.e.
the sum of the positive signed ranks.
If
If
H
H
0
0 holds, µ
S
+ holds, and
= n n ( n + 1 )
>
4
10,
, σ
2
S
+
S
+
= n ( n + 1 )( 2 n + 1 )
·
∼ N ( µ
S
+
24
, σ
2
S
+
) .
.
The TS for testing H
0
: µ
D
= 0 is
Z
H
0
= S
+
− n ( n + 1 )
4
/ r n ( n + 1 )( 2 n + 1 )
24
,
The RRs are the usual RRs of a Z -test.
Power and Sample Size Calculation – NOT COVERED
Comparing Two Variances – NOT COVERED
Example (Mercury concentrations from Labs A and B )
The 12 differences, D i and the ranks of their absolute values are given in the table below. Test
H
0
: µ
1
− µ
2
= 0 , H a
: µ
1
− µ
2
< 0 at α = 0 .
05.
D i
R i
-0.0206
5
-0.0350
10
-0.0161
4
-0.0017
1
0.0064
2
-0.0219
6
D i
R i
-0.0250
8
-0.0279
9
-0.0232
7
-0.0655
Solution: Here S
+
= 2 + 11 = 13. Thus
12
0.0461
11
-0.0159
3
Z
H
0
=
13 − 39
√
162 .
5
= − 2 .
04 .
The p -value = Φ( − 2 .
04 ) = 0 .
02. The exact p -value, as obtained from Minitab, is 0.0193.
Michael Akritas
Lesson 10 Chapter 9: Comparing Two Populations
Power and Sample Size Calculation – NOT COVERED
Comparing Two Variances – NOT COVERED
R Commands for Paired Data
Rank sum test:
Only for testing: wilcox.test(x, y, alternative = c(”two.sided”,
”less”, ”greater”), mu = 0)
For testing and CI: wilcox.test(x, y, alternative = c(”two.sided”, ”less”, ”greater”), mu = 0, conf.int = TRUE, conf.level = 0.95) x,y above can be replaced by y ∼ Sample
Paired data t test for means: t.test(x, y, alternative = c(”two.sided”, ”less”, ”greater”), mu
= 0, paired = TRUE, conf.level = 0.95) x,y above can be replaced by y ∼ Sample
McNemar’s test for proportions: mcnemar.test(table)
See next page for example.
Michael Akritas
Lesson 10 Chapter 9: Comparing Two Populations
Power and Sample Size Calculation – NOT COVERED
Comparing Two Variances – NOT COVERED
Example (Example of McNemar’s test)
Data on approval of the President’s performance in office in two surveys, one month apart, for a random sample of 1600 voting-age Americans.
table=matrix(c(794, 86, 150, 570), nrow = 2, dimnames = list(c(”Approve”, ”Disapprove”), c(”Approve”, ”Disapprove”))) table mcnemar.test(table)
Michael Akritas
