INDEPENDENT SAMPLES t TEST
Syntax
The syntax for the t.test( ) function is outlined in the Single Sample t Test tutorial.
The t Test With Two Independent Groups
As part of his senior research project in the Fall semester of 2001, Scott
Keats looked for a possible relationship between marijuana smoking and a deficit
in performance on a task measuring short term memory--the digit span task from
the Wechsler Adult Intelligence Scale. Two groups of ten subjects were tested.
One group, the "nonsmokers," claimed not to smoke marijuana. A second group, the
"smokers," claimed to smoke marijuana regularly. The data set is small and
easily entered by hand...
> nonsmokers = c(18,22,21,17,20,17,23,20,22,21)
> smokers = c(16,20,14,21,20,18,13,15,17,21)
The values of the response variable indicate number of items completely
successfully on the digit span task. An examination of the distributions would
be in order at this point, as the t-test assumes sampling from normal parent
populations...
> plot(density(nonsmokers)) # output not shown
> plot(density(smokers)) # output not shown
The smoothed distributions appear reasonably mound-shaped, although with samples
this small, it's hard to say for sure what the parent distribution looks like.
Another way to examine the data graphically is with side-by-side boxplots...
> boxplot(nonsmokers,smokers,ylab="Scores on Digit Span Task",
+ names=c("nonsmokers","smokers"),
+ main="Digit Span Performance by\n Smoking Status")
Note: The "\n" inside the main title causes it to be printed on two lines. I'm not
entirely sure this will work in Windows, but it should.
Boxplots are becoming the standard graphical method of displaying this sort of data, although I have an issue with the fact that boxplots are median oriented graphics, while the t-test is comparing means. We could also do a bar graph with error bars (which requires some finagling), and I will discuss this is a future tutorial, but the bar graph would display less information about the data. Meanwhile, the boxplots show the nonsmokers to have done better on the digit span task than the smokers. It also shows an absence of outliers, which is good news.
Of course, the standard numerical summaries could also be done...
> mean(nonsmokers)
[1] 20.1
> sd(nonsmokers)
[1] 2.131770
> mean(smokers)
[1] 17.5
> sd(smokers)
[1] 2.953341
If you want standard errors, there is no built-in function (we will write one
later), but they are easily enough calculated...
> sqrt(var(nonsmokers)/length(nonsmokers))
[1] 0.674125
> sqrt(var(smokers)/length(smokers))
[1] 0.9339284
There is little left to do but the statistical test.
The t-test can be done either by entering the names of the two groups (or
two numerical vectors) into the t.test( ) function, or by using a formula
interface if the data are in a data frame (below)...
> t.test(nonsmokers,smokers)
Welch Two Sample t-test
data: nonsmokers and smokers
t = 2.2573, df = 16.376, p-value = 0.03798
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
0.1628205 5.0371795
sample estimates:
mean of x mean of y
20.1 17.5
By default the two-tailed test is done, and the Welch correction for
nonhomogeneity of variance is applied. Both of these options can easily enough
be changed...
> t.test(nonsmokers,smokers,alternative="greater",var.equal=T)
Two Sample t-test
data: nonsmokers and smokers
t = 2.2573, df = 18, p-value = 0.01833
alternative hypothesis: true difference in means is greater than 0
95 percent confidence interval:
0.6026879 Inf
sample estimates:
mean of x mean of y
20.1 17.5
The output also includes a 95% CI (change this using "conf.level=") for the
difference between the means and the sample means. Note that the t.test( )
function always subtracts first group minus second group. Since the "nonsmokers"
were entered first into the function, and our hypothesis was that they would
score higher on the digit span test, this dictated the alternative hypothesis
to be "difference between means is greater than zero." The null hypothesized
difference can also be changed by setting the "mu=" option.
The Formula Interface
I hope you haven't erased those vectors, because now we are going to make a
data frame from them...
> scores = c(nonsmokers,smokers)
> groups = c("nonsmokers","smokers") # Make sure these are in the right order!
> groups = rep(groups,c(10,10))
> mj.data = data.frame(groups,scores)
> mj.data
groups scores
1 nonsmokers 18
2 nonsmokers 22
3 nonsmokers 21
4 nonsmokers 17
5 nonsmokers 20
6 nonsmokers 17
7 nonsmokers 23
8 nonsmokers 20
9 nonsmokers 22
10 nonsmokers 21
11 smokers 16
12 smokers 20
13 smokers 14
14 smokers 21
15 smokers 20
16 smokers 18
17 smokers 13
18 smokers 15
19 smokers 17
20 smokers 21
> rm(scores,groups) # Why is this necessary?
> attach(mj.data)
Now for some summary statistics...
> by(scores,groups,mean) # or tapply(scores,groups,mean)
groups: nonsmokers
[1] 20.1
----------------------------------------------------------
groups: smokers
[1] 17.5
> by(scores,groups,sd)
groups: nonsmokers
[1] 2.131770
----------------------------------------------------------
groups: smokers
[1] 2.953341
The boxplot function also accepts a formula interface...
> boxplot(scores ~ groups) # output not shown
Read the formula as "scores by groups". Notice you get group labels on the
x-axis this way, too. Finally, the t-test...
> t.test(scores ~ groups)
Welch Two Sample t-test
data: scores by groups
t = 2.2573, df = 16.376, p-value = 0.03798
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
0.1628205 5.0371795
sample estimates:
mean in group nonsmokers mean in group smokers
20.1 17.5
If you don't care for attaching data frames (and I usually try to avoid it), the
t.test( ) function also has a "data=" option...
> detach(mj.data)
> t.test(scores ~ groups, data=mj.data) ### output not shown
Note: I am told there is an easier way to create a data frame from individual
vectors representing group-by-group data, but I haven't yet been told what it
is. If I find out, I'll pass it on!
Textbook Problems
Problems in textbooks often leave out the raw data and just present summary
statistics. There is no provision for entering the summary stats into the
t.test( ) function, so such problems would have to be calculated at the
command prompt...
Two groups of ten subjects each were given the digit span subtest from
the Wechsler Adult Intelligence Scale. One group consisted of regular
smokers of marijuana, while the other group consisted of nonsmokers.
Below are summary statistics for number of items completed correctly
on the digit span task. Is there a significant difference between the
means of the two groups?
smokers nonsmokers
----------------------
mean 17.5 20.1
sd 2.95 2.13
> mean.diff = 17.5 - 20.1
> df = 10 + 10 - 2
> pooled.var = (2.95^2 * 9 + 2.13^2 * 9) / df
> se.diff = sqrt(pooled.var/10 + pooled.var/10)
> t.obt = mean.diff / se.diff
> t.obt
[1] -2.259640
> p.value = 2*pt(t.obt,df=df) # two-tailed
> p.value
[1] 0.03648139
A custom function could be written if these calculations had to be done
repeatedly, but there is a certain educational value to be had from doing them
by hand!
Power
The power.t.test( ) function has the following syntax (from the help
page for this function)...
power.t.test(n = NULL, delta = NULL, sd = 1, sig.level = 0.05, power = NULL,
type = c("two.sample", "one.sample", "paired"),
alternative = c("two.sided", "one.sided"),
strict = FALSE)
Exactly one of the options on the first line should be set to "NULL", and R will
then calculate it from the remaining values. Suppose we wanted a power of 85%
for the Scott Keats experiment, and were anticipating a mean difference of 2.5
and a pooled standard deviation of 2.6. How many subjects should be used per
group if a two-tailed test is planned?
> power.t.test(delta=2.5, sd=2.6, sig.level=.05, power=.85,
+ type="two.sample", alternative="two.sided")
Two-sample t test power calculation
n = 20.43001
delta = 2.5
sd = 2.6
sig.level = 0.05
power = 0.85
alternative = two.sided
NOTE: n is number in *each* group
So there you go!
Homogeneity of Variance
The standard, textbook, pooled-variance t-test assumes homogeneity of variance. The easiest way to deal with nonhomogeneity of variance is to allow R to do what it does by default anyway--run the t-test using the Welch correction for nonhomogeneity. Further discussion of nonhomogeneity of variance will occur in the ANOVA tutorial.
Alternatives to the t Test
If the normality assumption of the t-test is violated, and the sample sizes
are two small to heal that via an appeal to the central limit theorem, then a
nonparametric alternative test should be sought. The standard alternative to the
independent samples t-test is what we old timers were taught to call the
Mann-Whitney U test. For some reason, this name is now out of fashion, and the
test goes by the name Wilcoxin-Mann-Whitney test, or Wilcoxin rank sum test,
or some variant thereof. The syntax is very similar, and it will work with or
without the formula interface...
> wilcox.test(nonsmokers,smokers)
Wilcoxon rank sum test with continuity correction
data: nonsmokers and smokers
W = 76.5, p-value = 0.04715
alternative hypothesis: true location shift is not equal to 0
Warning message:
In wilcox.test.default(nonsmokers, smokers) :
cannot compute exact p-value with ties
>
> ### and now the formula interface...
> wilcox.test(scores ~ groups, data=mj.data)
Wilcoxon rank sum test with continuity correction
data: scores by groups
W = 76.5, p-value = 0.04715
alternative hypothesis: true location shift is not equal to 0
Warning message:
In wilcox.test.default(x = c(18, 22, 21, 17, 20, 17, 23, 20, 22, :
cannot compute exact p-value with ties
The test assumes continuous variables without ties (and neither is the case
here). And no, I don't know why the warning message about that takes two
different forms.