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Introduction
After having written an article on the Student’s ttest for two samples (independent and paired samples), I believe it is time to explain in details how to perform one sample ttests by hand and in R.
One sample ttest is an important part of inferential statistics (probably one of the first statistical test that students learn). Remind that, unlike descriptive statistics, inferential statistics is a branch of statistics aiming at drawing conclusions about one or two populations, based on a subset (or two) of that population (called samples). In other words, we first collect a random set of observations from a population, and then some measurements are calculated in order to generalize to the population the information found through the sample.
In this context, the one sample ttest is used to determine whether the mean of a measurement variable is different from a specified value (a belief or a theoretical expectation for example). It works as follows: if the mean of the sample is too distant from the specified value (the value under the null hypothesis), it is considered that the mean of the population is different from what is expected. On the contrary, if the mean of the sample is close to the specified value, we cannot reject the hypothesis that the population mean is equal to what is expected.
Like the Student’s ttest for two samples, there are also different versions of the one sample ttest. Luckily, there are only two different versions for this test (the Student’s ttest for two samples has 5 versions!). The difference between the two versions of the one sample ttest lies in the fact that one version is used when the variance of the population (not the variance of the sample!) is known, the other version being used when the variance of the population is unknown.
In this article, I will first detail step by step how to perform both versions of the one sample ttest by hand. The analyses will be done on a small set of observations for the sake of illustration and easiness. I will then show how to perform this test in R with the exact same data in order to verify the results found by hand. Reminders about the reasoning behind hypothesis testing, interpretations of the pvalue and the results, and assumptions of this test will also be presented.
Note that the aim of this article is to show how to compute the one sample ttest by hand and in R, so we refrain from testing the assumptions and we assume all assumptions are met for this exercise. For completeness, we still mention the assumptions and how to test them. Interested readers are invited to have a look at the end of the article for more information about these assumptions.
Null and alternative hypothesis
Before diving into the computations of the one sample ttest by hand, let’s recap the null and alternative hypotheses of this test:
 \(H_0\): \(\mu = \mu_0\)
 \(H_1\): \(\mu \ne \mu_0\)
where \(\mu\) is the population mean and \(\mu_0\) is the known or hypothesized value of the mean in the population.
This is in the general case where we simply want to determine whether the population mean is different (in terms of the dependent variable) compared to the hypothesized value. In this sense, we have no prior belief about the population mean being larger or smaller than the hypothesized value. This type of test is referred as a twosided or bilateral test.
If we have some prior beliefs about the population mean being larger or smaller than the hypothesized value, the one sample ttest also allows to test the following hypotheses:
 \(H_0\): \(\mu = \mu_0\)
 \(H_1\): \(\mu > \mu_0\)
or
 \(H_0\): \(\mu = \mu_0\)
 \(H_1\): \(\mu < \mu_0\)
In the first case, we want to test if the population mean is significantly larger than the hypothesized value, while in the latter case, we want to test if the population mean is significantly smaller than the hypothesized value. This type of test is referred as a onesided or unilateral test.
Hypothesis testing
In statistics, many statistical tests is in the form of hypothesis tests. Hypothesis tests are used to determine whether a certain belief can be deemed as true (plausible) or not, based on the data at hand (i.e., the sample(s)). Most hypothesis tests boil down to the following 4 steps:^{1}
 State the null and alternative hypothesis.
 Compute the test statistic, denoted tstat. Formulas to compute the test statistic differ among the different versions of the one sample ttest but they have the same structure. See scenarios 1 and 2 below to see the different formulas.
 Find the critical value given the theoretical statistical distribution of the test, the parameters of the distribution and the significance level \(\alpha\). For the two versions of the one sample ttest, it is either the normal or the Student’s t distribution (t denoting the Student distribution and z denoting the normal distribution).
 Conclude by comparing the tstat (found in step 2.) with the critical value (found in step. 3). If the tstat lies in the rejection region (determined thanks to the critical value and the direction of the test), we reject the null hypothesis, otherwise we do not reject the null hypothesis. These two alternatives (reject or do not reject the null hypothesis) are the only two possible solutions, we never “accept” an hypothesis. It is also a good practice to always interpret the decision in the terms of the initial question.
Two versions of the one sample ttest
There are two versions of the one sample ttest, depending on whether the variance of the population (not the variance of the sample!) is known or unknown. This criteria is rather straightforward, we either know the variance of the population or we do not. The variance of the population cannot be computed because if you can compute the variance of a population, it means you have the data for the whole population, then there is no need to do a hypothesis test anymore…
So the variance of the population is either given in the statement (use them in that case), or there is no information about the variance and in that case, it is assumed that the variance is unknown. In practice, the variance of the population is most of the time unknown. However, we still illustrate how to do both versions of this test by hand and in R in the next sections following the 4 steps of hypothesis testing.
How to compute the one sample ttest by hand?
Note that the data are artificial and do not represent any real variable. Furthermore, remind that the assumptions may or may not be met. The point of the article is to detail how to compute the different versions of the test by hand and in R, so all assumptions are assumed to be met. Moreover, assume that the significance level \(\alpha = 5\)% for all tests.
If you are interested in applying these tests by hand without having to do the computations yourself, here is a Shiny app which does it for you. You just need to enter the data and choose the appropriate version of the test thanks to the sidebar menu. There is also a graphical representation that helps you to visualize the test statistic and the rejection region. I hope you will find it useful!
Scenario 1: variance of the population is known
For the first scenario, suppose the data below. Moreover, suppose that the population variance \(\sigma^2 = 1\) and that we would like to test whether the population mean is different from 0.
value 

0.9 
0.8 
1.3 
0.3 
1.7 
So we have:
 5 observations: \(n = 5\)
 mean of the sample: \(\bar{x} = 0.56\)
 variance of the population: \(\sigma^2 = 1\)
 \(\mu_0 = 0\)
Following the 4 steps of hypothesis testing we have:
 \(H_0: \mu = 0\) and \(H_1: \mu \ne 0\). (\(\ne\) because we want to test whether the population mean is different from 0, we do not impose a direction in the test.)
 Test statistic: \[z_{obs} = \frac{\bar{x} – \mu_0}{\frac{\sigma}{\sqrt{n}}} = \frac{0.560}{0.447} = 1.252\]
 Critical value: \(\pm z_{\alpha / 2} = \pm z_{0.025} = \pm 1.96\) (see a guide on how to read statistical tables if you struggle to find the critical value)
 Conclusion: The rejection regions are thus from \(\infty\) to 1.96 and from 1.96 to \(+\infty\). The test statistic is outside the rejection regions so we do not reject the null hypothesis \(H_0\). In terms of the initial question: At the 5% significance level, we do not reject the hypothesis that the population mean is equal to 0, or there is no sufficient evidence in the data to conclude that the population mean is different from 0.
Scenario 2: variance of the population is unknown
For the second scenario, suppose the data below. Moreover, suppose that the variance in the population is unknown and that we would like to test whether the population mean is larger than 5.
value 

7.9 
5.8 
6.3 
7.3 
6.7 
So we have:
 5 observations: \(n = 5\)
 mean of the sample: \(\bar{x} = 6.8\)
 standard deviation of the sample: \(s = 0.825\)
 \(\mu_0 = 5\)
Following the 4 steps of hypothesis testing we have:
 \(H_0: \mu = 5\) and \(H_1: \mu > 5\). (> because we want to test whether the population mean is larger than 5.)
 Test statistic: \[t_{obs} = \frac{\bar{x} – \mu_0}{\frac{s}{\sqrt{n}}} = \frac{6.85}{0.369} = 4.881\]
 Critical value: \(t_{\alpha, n – 1} = t_{0.05, 4} = 2.132\) (see a guide on how to read statistical tables if you struggle to find the critical value)
 Conclusion: The rejection region is thus from 2.132 to \(+\infty\). The test statistic lies within the rejection region so we reject the null hypothesis \(H_0\). In terms of the initial question: At the 5% significance level, we conclude that the population mean is larger than 5.
This concludes how to perform the two versions of the one sample ttest by hand. In the next sections, we detail how to perform the exact same tests in R.
How to compute the one sample ttest in R?
A good practice before doing ttests in R is to visualize the data thanks to a boxplot (or a density plot, or eventually both). A boxplot gives a first indication on the location of the sample, and thus, a first indication on whether the null hypothesis is likely to be rejected or not. However, even if a boxplot or a density plot is great in showing the distribution of a sample, only a sound statistical test will confirm our first impression.
After a visualization of the data, we replicate in R the results found by hand. We will see that for the version of the ttest with a known population variance, there is no default function built in R (at least to my knowledge, do not hesitate to let me know if I’m mistaken). In this case, a function is written to replicate the results by hand.
Note that we use the same data, the same assumptions and the same question for both scenarios to facilitate the comparison between the tests performed by hand and in R.
Scenario 1: variance of the population is known
For the first scenario, suppose the data below. Moreover, suppose that the population variance \(\sigma^2 = 1\) and that we would like to test whether the population mean is different from 0.
dat1 < data.frame(
value = c(0.9, 0.8, 1.3, 0.3, 1.7)
)
dat1
## value
## 1 0.9
## 2 0.8
## 3 1.3
## 4 0.3
## 5 1.7
library(ggplot2)
ggplot(dat1) +
aes(y = value) +
geom_boxplot() +
theme_minimal()
Note that you can use the {esquisse}
RStudio addin if you want to draw a boxplot with the package {ggplot2}
without writing the code yourself. If you prefer the default graphics, use the boxplot()
function:
boxplot(dat1$value)
The boxplot shows that the distribution of the sample is not distant from 0 (the hypothesized value), so we tend to believe that we will not be able to reject the null hypothesis that the population mean is equal to 0. However, only a formal statistical test will confirm this belief.
Since there is no function in R to perform a ttest with a known population variance, here is one with arguments accepting the sample (x
), the variance of the population (V
), the mean under the null hypothesis (m0
, default is 0
), the significance level (alpha
, default is 0.05
) and the alternative (alternative
, one of "two.sided"
(default), "less"
or "greater"
):
t.test2 < function(x, V, m0 = 0, alpha = 0.05, alternative = "two.sided") {
M < mean(x)
n < length(x)
sigma < sqrt(V)
S < sqrt(V / n)
statistic < (M  m0) / S
p < if (alternative == "two.sided") {
2 * pnorm(abs(statistic), lower.tail = FALSE)
} else if (alternative == "less") {
pnorm(statistic, lower.tail = TRUE)
} else {
pnorm(statistic, lower.tail = FALSE)
}
LCL < (M  S * qnorm(1  alpha / 2))
UCL < (M + S * qnorm(1  alpha / 2))
value < list(mean = M, m0 = m0, sigma = sigma, statistic = statistic, p.value = p, LCL = LCL, UCL = UCL, alternative = alternative)
# print(sprintf("Pvalue = %g",p))
# print(sprintf("Lower %.2f%% Confidence Limit = %g",
# alpha, LCL))
# print(sprintf("Upper %.2f%% Confidence Limit = %g",
# alpha, UCL))
return(value)
}
test < t.test2(dat1$value,
V = 1
)
test
## $mean
## [1] 0.56
##
## $m0
## [1] 0
##
## $sigma
## [1] 1
##
## $statistic
## [1] 1.252198
##
## $p.value
## [1] 0.2104977
##
## $LCL
## [1] 0.3165225
##
## $UCL
## [1] 1.436523
##
## $alternative
## [1] "two.sided"
The output above recaps all the information needed to perform the test: the test statistic, the pvalue, the alternative used, the sample mean, the hypothesized value and the population variance (compare these results found in R with the results found by hand).
The pvalue can be extracted as usual:
test$p.value
## [1] 0.2104977
The pvalue is 0.21 so at the 5% significance level we do not reject the null hypothesis. There is no sufficient evidence in the data to reject the hypothesis that the population mean is equal to 0. This result confirms what we found by hand.
If you are unfamiliar with the concept of pvalue, I invite you to read my note on pvalue and significance level \(\alpha\).
To sum up what have been said in that article about pvalue and significance level \(\alpha\):
 If the pvalue is smaller than the predetermined significance level \(\alpha\) (usually 5%) so if pvalue < 0.05, we reject the null hypothesis
 If the pvalue is greater than or equal to the predetermined significance level \(\alpha\) (usually 5%) so if pvalue \(\ge\) 0.05, we do not reject the null hypothesis
This applies to all statistical tests without exception. Of course, the null and alternative hypotheses change depending on the test.
Scenario 2: variance of the population is unknown
For the second scenario, suppose the data below. Moreover, suppose that the variance in the population is unknown and that we would like to test whether the population mean is larger than 5.
dat2 < data.frame(
value = c(7.9, 5.8, 6.3, 7.3, 6.7)
)
dat2
## value
## 1 7.9
## 2 5.8
## 3 6.3
## 4 7.3
## 5 6.7
ggplot(dat2) +
aes(y = value) +
geom_boxplot() +
theme_minimal()
Unlike the previous scenario, the box is quite distant from the hypothesized value of 5. From this boxplot, we can expect the test to reject the null hypothesis of the population mean being equal to 5. Nonetheless, only a formal statistical test will confirm this expectation.
There is a function in R, and it is simply the t.test()
function. This version of the test is actually the “standard” ttest for one sample. Note that in our case the alternative hypothesis is \(H_1: \mu > 5\) so we need to add the arguments mu = 5
and alternative = "greater"
to the function because the default arguments are mu = 0
and the twosided test:
test < t.test(dat2$value,
mu = 5,
alternative = "greater"
)
test
##
## One Sample ttest
##
## data: dat2$value
## t = 4.8809, df = 4, pvalue = 0.004078
## alternative hypothesis: true mean is greater than 5
## 95 percent confidence interval:
## 6.013814 Inf
## sample estimates:
## mean of x
## 6.8
The output above recaps all the information needed to perform the test: the name of the test, the test statistic, the degrees of freedom, the pvalue, the alternative used, the hypothesized value and the sample mean (compare these results found in R with the results found by hand).
The pvalue can be extracted as usual:
test$p.value
## [1] 0.004077555
The pvalue is 0.004 so at the 5% significance level we reject the null hypothesis.
Unlike the first scenario, the pvalue in this scenario is below 5% so we reject the null hypothesis. At the 5% significance level, we can conclude that the population mean is significantly larger than 5. This result confirms what we found by hand.
Confidence interval
Note that the confidence interval can be extracted with $conf.int
:
test$conf.int
## [1] 6.013814 Inf
## attr(,"conf.level")
## [1] 0.95
You can see that the 95% confidence interval for the population mean is \([6.01; \infty]\), meaning that, at the significance level \(\alpha = 5\)%, we reject the null hypothesis as long as the hypothesized value \(\mu_0\) is below 6.01, otherwise the null hypothesis cannot be rejected.
Assumptions
As for many statistical tests, there are some assumptions that need to be met in order to be able to interpret the results. When one or several assumptions are not met, although it is technically possible to perform these tests, it would be incorrect to interpret the results. Below are the assumptions of the one sample ttest and how to test them:
 The data, collected from a representative and randomly selected portion of the total population, should be independent of one another.
 The dependent variable (i.e., the measured variable) must be continuous.
 Observations should follow a normal distribution. The normality assumption can be tested visually thanks to a histogram and a QQplot, and/or formally via a normality test such as the ShapiroWilk or KolmogorovSmirnov test (see more information about the normality assumption and how to test it here). Some transformations, such as, among others, taking the logarithm, the square root or the BoxCox transformation can be applied on the observations to transform you data to better fit the normal distribution.
Thanks for reading. I hope this article helped you to understand how the different versions of the one sample ttest work and how to perform them by hand and in R. If you are interested, here is a Shiny app to perform these tests by hand easily (you just need to enter your data and select the appropriate version of the test thanks to the sidebar menu). Moreover, read this article if you would like to know how to compute the Student’s ttest but this time, for two samples, in order to compare two dependent or independent groups.
As always, if you have a question or a suggestion related to the topic covered in this article, please add it as a comment so other readers can benefit from the discussion. If you find a mistake or bug, you can inform me by raising an issue on GitHub. For all other requests, you can contact me here.
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It is a least the case regarding parametric hypothesis tests. A parametric test means that it is based on a theoretical statistical distribution, which depends on some defined parameters. In the case of the one sample ttest, it is based on the Student’s t distribution with a single parameter, the degrees of freedom (\(df = n – 1\) where \(n\) is the sample size), or the normal distribution.↩
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