T-tests R code
T-test for two independent samples
We will test the ability of a drug to kill bacteria. The response is the count of bacteria
colony forming units (CFU) x106 per milliliter of blood.
Specifically, we will use a two-sample t-test to test the null hypothesis:

mean bacteria count in the drug group = mean bacteria count in placebo group
The standard version of the two-sample t-test assumes the following.


the samples are normally distributed
the two samples have equal variance
The standard version of the two-sample t-test will usually give valid results if the data do
not completely meet these assumptions. However, if the samples are strongly nonnormal, or have greatly different variance, we should consider alternative analyses, such
as log transforming the data to get a normal distribution, or using non-parametric
Wilcoxon rank tests described later.
From our experiment, we collect the following data.
drug.bacteria.counts =c(0.7, 0.9,1.0, 1.2, 2.2, 2.5, 3.6, 3.9, 4.2, 4.5, 4.5, 5.6, 5.9, 6.1)
placebo.bacteria.counts=c(2.0,2.3, 2.3, 3.9, 4.1, 4.4, 5.3, 5.7, 6.3, 6.4, 7.2, 7.7, 8.0)
We'll use boxplots and qqnorm plots to see if the counts deviate greatly from normality,
or if there are outliers.
boxplot(drug.bacteria.counts )
boxplot(placebo.bacteria.counts )
qqnorm(drug.bacteria.counts )
qqline(drug.bacteria.counts)
qqnorm(placebo.bacteria.counts )
qqline(placebo.bacteria.counts )
The boxplots and qqnorm plots do not show significant non-normality or outliers. What
about the assumption of equal variance? We'll calculate the standard deviation of each of
the treatment groups using the sd() function.
sd(drug.bacteria.counts )
sd(placebo.bacteria.counts )
> sd(drug.bacteria.counts )
[1] 1.923024
> sd(placebo.bacteria.counts )
[1] 2.067452
The standard deviations of the two treatment groups are similar.
In this case, the boxplots and qqnorm plots do not show significant non-normality or
outliers, and the standard deviations of the two treatment groups are similar. So we'll use
the standard version of the t-test assuming equal variance.
We can do a t-test directly with these values, using the t.test() function.
t.test(drug.bacteria.counts, placebo.bacteria.counts, var.equal=TRUE)
Two Sample t-test
data: drug.bacteria.counts and placebo.bacteria.counts
t = -2.2182, df = 25, p-value = 0.03586
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-3.2847862 -0.1218072
sample estimates:
mean of x mean of y
3.342857 5.046154
Conclusion: We reject the null hypothesis. Bacterial count was significantly less in the
drug group than in the placebo group (p = 0.03586).
The 95% confidence interval for the true value of the difference of the means ranges from
-3.2847862 to -0.1218072. Because the 95% confidence interval does not include zero,
we reject the null hypothesis that the difference between the two groups is zero.
T-test using biomarker data set.
setwd("C:/Users/Walker/Desktop/UCSD Biom 285/Data")
biomarker.data= read.csv("biomarkers.csv", header=TRUE)
biomarker.data
> biomarker.data
ID
Sex Age Disease Biomarker.1 Biomarker.2 Biomarker.3
1
1 Female 30
Case
138
137
79
2
2 Female 30 Control
141
143
93
3
3 Female 40
Case
134
148
58
4
4 Female 40 Control
150
153
87
5
5 Female 50
Case
153
147
53
6
6 Female 50 Control
168
163
62
7
7 Female 60
Case
161
167
34
8
9
10
11
12
13
14
15
16
8 Female
9
Male
10
Male
11
Male
12
Male
13
Male
14
Male
15
Male
16
Male
60
30
30
40
40
50
50
60
60
Control
Case
Control
Case
Control
Case
Control
Case
Control
180
135
160
152
169
163
182
179
178
178
138
164
155
164
165
183
179
184
53
88
109
76
93
61
73
49
61
biomarker.data= read.csv("biomarkers.csv", header=TRUE)
When you have your data in a data frame, you can use the following syntax.
The text "Biomarker.1 ~ Disease" is called a formula, and it tells R to perform a t-test
with Biomarker.1 as the quantitative response (y) variable, and Disease as the categorical
(x) variable.
t.test(Biomarker.1 ~ Disease, data=biomarker.data, var.equal=TRUE)
Two Sample t-test
data: Biomarker.1 by Disease
t = -1.8493, df = 14, p-value = 0.08563
alternative hypothesis: true difference in means is not equal to
0
95 percent confidence interval:
-30.506601
2.256601
sample estimates:
mean in group Case mean in group Control
151.875
166.000
Conclusion: We do not reject the null hypothesis. Biomarker.1 was not significantly
different between the Disease groups (p = 0.08563).
The 95% confidence interval for the true value of the difference of the means ranges from
-30.5 to 2.26. Because the 95% confidence interval includes zero, we do not reject the
null hypothesis that the difference between the two groups is zero.
Do bears lose weight between winter and spring?
Here is another example of a two-sample t-test. We measure the weight of one group of
bears in winter, and measure the weight of a different group of bears the following
spring.
bears.winter=c(300,470,550,650,750,760,800,985,1100,1200)
bears.spring=c(280,420,500,620,690,710,790,935,1050,1110)
t.test(bears.spring, bears.winter, var.equal=TRUE)
> t.test(bears.spring, bears.winter, var.equal=TRUE)
Two Sample t-test
data: bears.spring and bears.winter
t = -0.3732, df = 18, p-value = 0.7134
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-304.9741 212.9741
sample estimates:
mean of x mean of y
710.5
756.5
The p-value=0.7134 is not significant, so we do not reject the null hypothesis that the
mean weight of bears in winter is the same as the mean weight of bears in spring. The
95% confidence interval for the true value of the mean is -304.9741 to 212.9741.
Because the 95% confidence interval includes zero, we do not reject the null hypothesis
that the difference between the two groups is zero.
A better design for this study would be to measure the same bears in winter and in spring,
to remove the effect of variability among bears from the experiment. We'll do that shortly
using a paired t-test.
T-test for two independent samples: unequal variance
In some cases, we may know or expect that two groups have unequal variance. For
example, we may want to compare the mean yield of two processes, where we know that
one process is more variable than the other. It may be that a group receiving an active
treatment (drug) is more variable in its response than the control (placebo) group. In these
cases, we may use an alternative version of the t-test, called Welch's test, which does not
assume equal variance of the two treatment groups.
Furness and Bryant (1996) compared the metabolic rates of male and female breeding
northern fulmars (data described in Logan (2010) and Quinn (2002)).
setwd("C:/Users/Walker/Desktop/UCSD Biom 285/Data")
furness= read.csv("furness.csv", header=TRUE)
furness
1
2
3
4
5
6
SEX
Male
Female
Male
Male
Male
Female
METRATE
2950.0
1956.1
2308.7
2135.6
1945.6
1490.5
BODYMASS
875
635
765
780
790
635
7
8
9
10
11
12
13
14
Female
Female
Female
Male
Female
Male
Male
Male
1361.3
1086.5
1091.0
1195.5
727.7
843.3
525.8
605.7
668
640
645
788
635
855
860
1005
boxplot(METRATE~SEX, data=furness)
The boxplots indicate that the variances of males and females are not equal. So we use
the t-test assuming unequal variances.
t.test(METRATE~SEX, data=furness, var.equal=F)
Welch Two Sample t-test
data: METRATE by SEX
t = -0.7732, df = 10.468, p-value = 0.4565
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-1075.3208
518.8042
sample estimates:
mean in group Female
mean in group Male
1285.517
1563.775
The p-value=0.4565 is not significant, so we do not reject the null hypothesis.
Paired t-test for matched samples:
Here we'll test the hypothesis that bears lose weight during hibernation. First, we'll
assume that different bears are weighed in November and March, and test if their mean
weight is different using an ordinary t-test. Then, we'll assume that the same bears are
weighed in November and March, and test if their mean weight is different using a paired
t-test.
The big advantage of measuring the same bears (and using the paired t-test) is that,
because each bear serves as its own control, we control for the variability among bears.
This has the effect of removing the (unexplained) variability due to variation in weight
among the bears.
bears.winter=c(300,470,550,650,750,760,800,985,1100,1200)
bears.spring=c(280,420,500,620,690,710,790,935,1050,1110)
diff= bears.spring - bears.winter
stripchart(diff, method="stack")
t.test(bears.winter, bears.spring, var.equal=TRUE, paired=TRUE)
# In R, we use the argument "paired=TRUE" to do the paired t.test.
> t.test(bears.winter, bears.spring, var.equal=TRUE, paired=TRUE)
Paired t-test
data: bears.winter and bears.spring
t = 6.5492, df = 9, p-value = 0.0001053
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
30.11113 61.88887
sample estimates:
mean of the differences
46
The p-value=0.0001053 is significant, so we reject the null hypothesis that the change in
weight between winter and spring is zero.
One-sample t-test
A one-sample t-test tests the hypothesis that the mean of a sample is different from a
specified mean.
Suppose that we have recruited several students for a study to see if body mass index
(BMI) is changed by exercise. In previous studies, the mean BMI of students at baseline
(before the exercise) was 18.2. Is the mean BMI =18.2 for the new students at baseline?
bmi.new=c(17,17, 18,18,18, 19, 20,20, 21, 22)
mean(bmi.new)
sd(bmi.new)
> mean(bmi.new)
[1] 19
> sd(bmi.new)
[1] 1.699673
Our null hypothesis is that the bmi of the new students is 18.2. Here's the t-test.
t.test(bmi.new, mu=18.2)
> t.test(bmi.new, mu=18.2)
One Sample t-test
data: bmi.new
t = 1.4884, df = 9, p-value = 0.1708
alternative hypothesis: true mean is not equal to 18.2
95 percent confidence interval:
17.78413 20.21587
sample estimates:
mean of x
19
The p-value=0.17 is not significant, so we do not reject the null hypothesis that the bmi
of the new students is 18.2.
A little while later we recruit another group of students. Is their mean BMI=18.2?
bmi.2=c(18,18,18, 19, 19, 20,20, 21, 22, 22, 23)
mean(bmi.2)
sd(bmi.2)
> mean(bmi.2)
[1] 20
> sd(bmi.2)
[1] 1.788854
t.test(bmi.2, mu=18.2)
> t.test(bmi.2, mu=18.2)
One Sample t-test
data: bmi.2
t = 3.3373, df = 10, p-value = 0.007525
alternative hypothesis: true mean is not equal to 18.2
95 percent confidence interval:
18.79823 21.20177
sample estimates:
mean of x
20
The p-value=0. 007525 is significant, so we reject the null hypothesis that the bmi of
this group of students is 18.2. The 95% confidence interval for the true value of the mean
is 18.79823 to 21.20177.
By default, R assumes that mu=0 unless you specify otherwise.
The paired t-test is equivalent to a one sample t-test using the difference between before
and after values. Recall that by default, R assumes that mu=0 in a one-sample t-test
unless you specify otherwise.
# Compare the paired t-test to one sample t-test:
diff= bears.spring - bears.winter
t.test(diff)
> t.test(diff)
One Sample t-test
data: diff
t = -6.5492, df = 9, p-value = 0.0001053
alternative hypothesis: true mean is not equal to 0
95 percent confidence interval:
-61.88887 -30.11113
sample estimates:
mean of x
-46
Log transform
If you have outliers and/or non-normal distributions, you may be able to apply a
transform to the data to make the distribution more normal, and to reduce the influence of
the outliers.
Do colon cancer patients have elevated level of the mucin protein in their blood?
We measure the level of the protein mucin in the blood of patients with colon cancer and
in healthy controls.
colon.cancer=c(83, 89, 90, 93, 98)
controls=c(99, 100, 103, 104, 141)
boxplot(colon.cancer, controls)
Let's do a t-test.
t.test(colon.cancer, controls, var.equal=TRUE)
> t.test(colon.cancer, controls, var.equal=TRUE)
Two Sample t-test
data: colon.cancer and controls
t = -2.258, df = 8, p-value = 0.05389
alternative hypothesis: true difference in means is not
equal to 0
95 percent confidence interval:
-37.9994752
0.3994752
sample estimates:
mean of x mean of y
90.6
109.4
Even though the boxplot shows a clear separation of the two groups, the p-value for the ttest is not significant (p=0.054), due to the outlier (and the small sample size).
We could try a log transform of the data to reduce the effect of the outlier.
> t.test(log(colon.cancer), log(controls), var.equal=TRUE)
Two Sample t-test
data: log(colon.cancer) and log(controls)
t = -2.5166, df = 8, p-value = 0.036
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-0.34620596 -0.01512052
sample estimates:
mean of x mean of y
4.504971 4.685634
In this case, the log transform yields a p-value of p=0.036, so we reject the null
hypothesis and conclude that colon cancer patients have mucin levels different from those
of healthy controls.