Twosample tests
11 Twosample tests
Twosample tests match one sample against another. Their implementation
in
R is similar to a onesample test but there are differences to
be aware of.
11.1 Twosample tests of proportion
As before, we use the command
prop.test to handle these
problems. We just need to learn when to use it and how.
Example: Two surveys
A survey is taken two times over the course of two weeks. The
pollsters wish to see if there is a difference in the results as
there has been a new advertising campaign run. Here is the data

Week 1 
Week 2 
Favorable 
45 
56 
Unfavorable 
35 
47 
The standard hypothesis test is
H_{0}:
p_{1} =
p_{2} against the
alternative (twosided)
H_{1}:
p_{1} ¹ p_{2}. The function
prop.test is used to being called as
prop.test(x,n) where
x is the number favorable and
n is the total. Here it is no different, but since
there are two
x's it looks slightly different. Here is how
> prop.test(c(45,56),c(45+35,56+47))
2sample test for equality of proportions with continuity correction
data: c(45, 56) out of c(45 + 35, 56 + 47)
Xsquared = 0.0108, df = 1, pvalue = 0.9172
alternative hypothesis: two.sided
95 percent confidence interval:
0.1374478 0.1750692
sample estimates:
prop 1 prop 2
0.5625000 0.5436893
We let
R do the work in finding the
n, but otherwise
this is straightforward. The conclusion is similar to ones before,
and we observe that the
pvalue is 0.9172 so we accept the null
hypothesis that
p_{1} =
p_{2}.
11.2 Twosample ttests
The onesample
ttest was based on the statistic
and was used when the data was approximately normal and
s was
unknown.
The twosample
ttest is based on the statistic
t = 
( 

_{1}  

_{2})  (µ_{1} 
µ_{2}) 




.

and the assumptions that the
X_{i} are normally or approximately
normally distributed.
We observe that the denominator is much different that the
onesample test and that gives us some things to discuss. Basically,
it simplifies if we can further assume the two samples have the same
(unknown) standard deviation.
11.3 Equal variances
When the two samples are assumed to have equal variances, then the
data can be
pooled to find an estimate for the
variance. By default,
R assumes unequal variances. If the
variances are assumed equal, then you need to
specify
var.equal=TRUE when using
t.test.
Example: Recovery time for new drug
Suppose the recovery time for patients taking a new drug is
measured (in days). A placebo group is also used to avoid the
placebo effect. The data are as follows
with drug: 15 10 13 7 9 8 21 9 14 8
placebo: 15 14 12 8 14 7 16 10 15 12
After a sidebyside boxplot (
boxplot(x,y), but not
shown), it is determined that the assumptions of equal variances
and normality are valid. A onesided
test for equivalence of means using the
ttest is found. This
tests the null hypothesis of equal variances against the
onesided alternative that the drug group has a smaller
mean. (µ
_{1}  µ
_{2} < 0). Here are the results
> x = c(15, 10, 13, 7, 9, 8, 21, 9, 14, 8)
> y = c(15, 14, 12, 8, 14, 7, 16, 10, 15, 12)
> t.test(x,y,alt="less",var.equal=TRUE)
Two Sample ttest
data: x and y
t = 0.5331, df = 18, pvalue = 0.3002
alternative hypothesis: true difference in means is less than 0
95 percent confidence interval:
NA 2.027436
sample estimates:
mean of x mean of y
11.4 12.3
We accept the null hypothesis based on this test.
11.4 Unequal variances
If the variances are unequal, the denominator in the
tstatistic
is harder to compute mathematically. But not with
R. The only
difference is that you don't have to specify
var.equal=TRUE (so it is actually easier with
R).
If we continue the same example we would get the following
> t.test(x,y,alt="less")
Welch Two Sample ttest
data: x and y
t = 0.5331, df = 16.245, pvalue = 0.3006
alternative hypothesis: true difference in means is less than 0
95 percent confidence interval:
NA 2.044664
sample estimates:
mean of x mean of y
11.4 12.3
Notice the results are slightly different, but in this example the
conclusions are the same  accept the null hypothesis. When we
assume equal variances, then the sampling distribution of the test
statistic has a
t distribution with fewer degrees of
freedom. Hence less area is in the tails and so the
pvalues are
smaller (although just in this example).
11.5 Matched samples
Matched or paired
ttests use a different statistical
model. Rather than assume the two samples are independent normal
samples albeit perhaps with different means and standard
deviations, the matchedsamples test assumes that the two samples
share common traits.
The basic model is that
Y_{i} =
X_{i} +
e_{i} where
e_{i} is the randomness. We want to test if the
e_{i} are mean 0 against the alternative that they are not
mean 0. In order to do so, one subtracts the
X's from the
Y's
and then performs a regular onesample
ttest.
Actually,
R does all that work. You only need to specify
paired=TRUE when calling the
t.test function.
Example: Dilemma of two graders
In order to promote fairness in grading, each application was
graded twice by different graders. Based on the grades, can we
see if there is a difference between the two graders? The data
is
Grader 1: 3 0 5 2 5 5 5 4 4 5
Grader 2: 2 1 4 1 4 3 3 2 3 5
Clearly there are differences. Are they described by random
fluctuations (mean
e_{i} is 0), or is there a bias of one
grader over another? (mean
e ¹ 0).
A matched sample test will give us some insight. First we should
check the assumption of normality with normal plots say. (The
data is discrete due to necessary rounding, but the general
shape is seen to be normal.) Then we can apply the
ttest as
follows
> x = c(3, 0, 5, 2, 5, 5, 5, 4, 4, 5)
> y = c(2, 1, 4, 1, 4, 3, 3, 2, 3, 5)
> t.test(x,y,paired=TRUE)
Paired ttest
data: x and y
t = 3.3541, df = 9, pvalue = 0.008468
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
0.3255550 1.6744450
sample estimates:
mean of the differences
Which would lead us to reject the null hypothesis.
Notice, the data are not independent of each other as grader 1
and grader 2 each grade the same papers. We expect that if
grader 1 finds a paper good, that grader 2 will also and vice
versa. This is exactly what nonindependent means. A
ttest
without the
paired=TRUE yields
> t.test(x,y)
Welch Two Sample ttest
data: x and y
t = 1.478, df = 16.999, pvalue = 0.1577
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
0.4274951 2.4274951
sample estimates:
mean of x mean of y
3.8 2.8
which would lead to a different conclusion.
11.6 Resistant twosample tests
Again the resistant twosample test can be done with the
wilcox.test function. It's usage is similar to its usage
with a single sample test.
Example: Taxi out times
Let's compare taxi out
times at Newark airport for American and Northwest
Airlines. This data is in the dataset
ewr
, but we need to
work a little to get it. Here's one way using the command
subset:
> data(ewr) # read in data set
> attach(ewr) # unattach later
> tmp=subset(ewr, inorout == "out",select=c("AA","NW"))
> x=tmp[['AA']] # alternately AA[inorout=='out']
> y=tmp[['NW']]
> boxplot(x,y) # not shown
A boxplot shows that the distributions are skewed. So a test for the
medians is used.
> wilcox.test(x,y)
Wilcoxon rank sum test with continuity correction
data: x and y
W = 460.5, pvalue = 1.736e05
alternative hypothesis: true mu is not equal to 0
Warning message:
Cannot compute exact pvalue with ties in: wilcox.test(x,y)
One gets from
wilcox.test strong evidence to
reject the null hypothesis and accept the alternative that the medians
are not equal.
11.7 Problems

11.1
 Load the Simple dataset homework. This measures
study habits of students from private and public high schools. Make
a sidebyside boxplot. Use the appropriate test to test for
equality of centers.
 11.2
 Load the Simple data set corn. Twelve plots of
land are divided into two and then one half of each is planted with
a new corn seed, the other with the standard. Do a twosample
ttest on the data. Do the assumptions seems to be met. Comment
why the matched sample test is more appropriate, and then
perform the test. Did the two agree anyways?
 11.3
 Load the Simple dataset blood. Do a significance
test for equivalent centers. Which one did you use and why? What was
the pvalue?
 11.4
 Do a test of equality of medians on the Simple
cabinets data set. Why might this be more appropriate
than a test for equality of the mean or is it?
Copyright © John Verzani, 20012. All rights reserved.