INDEPENDENT SAMPLES t TEST Syntax The syntax for the t.test() function is given
here from the help page in R.
## Default S3 method: t.test(x, y = NULL, alternative = c("two.sided", "less", "greater"), mu = 0, paired = FALSE, var.equal = FALSE, conf.level = 0.95, ...) ## S3 method for class 'formula': t.test(formula, data, subset, na.action, ...)"S3" refers to the S language (version 3), which is often the same as the methods and syntax used by R. In the case of the t.test() function, there are two alternative syntaxes, the default, and the "formula" syntax. Both syntaxes are relevant to the two-sample t-tests. The default syntax requires two data vectors, "x" and "y", to be specified. To get the independent measures t-test, leave the option "paired=" set to FALSE. The "alternative=" option is set by default to "two.sided" but can be set to any of the three values shown above. The default null hypothesis is "mu = 0", which in this case should be read as "mu1-mu2=0". This is usually what we want, but doesn't have to be, and should be changed by the user if this is not the null hypothesis. The "var.equal=" option determines whether or not the variances are pooled to estimate the population variance. The default is no (FALSE). The rest can be ignored for the moment. 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 with c() or scan(). I'll leave off the command prompts so you can copy
and paste these two lines if you want to take the easy way out. (Actually, it's
so I can do it, but I didn't want to admit it.)
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 completed 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. > par(mfrow=c(1,2)) # set graphics window to plot side by side > plot(density(nonsmokers), main="nonsmokers") > plot(density(smokers), main="smokers")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. There may be a wee bit of bimodality. Another way to examine the data graphically is with side-by-side boxplots.
(NOTE: Close the graphics window before executing the following command, so
that the parameters of the graphics device will be reset.)
> boxplot(nonsmokers, smokers, ylab="Scores on Digit Span Task", # plot and label y-axis + names=c("nonsmokers","smokers"), # group names on x-axis + main="Digit Span Performance by\n Smoking Status") # main titleNote: 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 (comparison of groups), although I have an issue with the fact that
boxplots are median-oriented graphics, while the t-test is comparing means.
It's easy enough to plot the means inside the boxes, and I'll show you how in
another tutorial. 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 obtained.
> mean(nonsmokers) [1] 20.1 > sd(nonsmokers) [1] 2.131770 > mean(smokers) [1] 17.5 > sd(smokers) [1] 2.953341If 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)) # assumes no missing values [1] 0.674125 > sqrt(var(smokers)/length(smokers)) # ditto [1] 0.9339284There 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 numeric 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 17.5By 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 17.5The output also includes a 95% CI (change this using "conf.level=") for the difference between the means, and it includes 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, times=c(10,10)) # 10 nonsmokers, 10 smokers > 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 a good idea? > attach(mj.data)Now for some descriptive 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] 2.953341The boxplot function also accepts a formula interface. > boxplot(scores ~ groups) # output not shownRead the formula as "scores by groups". Notice you get group labels on the x-axis this way, too. The boxplots will be plotted in the order in which R sees the levels of the groups, which is usually (but not always!) in alphabetical order. If you want some other order, the factor can be releveled, and I'll discuss this in a future tutorial. 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 17.5Important note: The difference between the means is calculated by subtracting one group mean from the other, of course, and for the two-tailed (two.sided) test, it doesn't really matter what order the means are subtracted in. For directional tests, it does. If you do levels(mj.data$groups), R will let you in on what order that is: first listed group minus second listed group. In this case, that would be nonsmokers' mean minus smokers' mean. Set your "alternative=" option accordingly. 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 when used
with the formula interface (but not otherwise).
> detach(mj.data) > t.test(scores ~ groups, data=mj.data) ### output not shown 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(abs(t.obt), df=df, lower.tail=F) # two-tailed > p.value [1] 0.03648139A custom function could be written if these calculations had to be done repeatedly, but (STUDENTS!) 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, # n=NULL by default + 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* groupSo there you go: to play it safe, 21 subjects per group should be used. 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. I'm not entirely happy that nonhomogeneity is assumed by default and then "swept under the rug" by applying the Welch correction. If the variances are different, that could be an interesting effect in itself. In any event, I don't think it should just be ignored. In my opinion, the pooled-variance t-test should be the default. But then, they didn't ask me. 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) # if these vectors are still in your workspace 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 tiesThe test assumes continuous variables without ties (neither of which is the case here). revised 2016 January 30 |