Last updated: 16th Dec, 2023
In statistics, the ttest is often used in research when the researcher wants to know if there is a significant difference between the mean of sample and the population, or whether there is a significant difference between the means of two groups (unpaired / independent or paired). There are three types of ttests: the one sample ttest, two samples or independent samples ttest, and paired samples ttest. In this blog post, we will focus on the one sample ttest and explain with formula and examples. As data scientists, it is important for us to understand the concepts of ttest and how to use it in our data analysis. Check out our onesample ttest calculator tool for means.
What is Onesample Ttest?
Onesample Ttest is a statistical hypothesis testing technique in which the mean of a sample is tested against a hypothesized value, e.g., a population mean. The ttest is used to determine whether the difference between the sample mean and the hypothesized value, e.g., the population mean is statistically significant or not. This test is particularly useful when the population standard deviation is unknown and the sample size is small (typically less than 30). The distribution used is Tdistribution with certain degrees of freedom.
Steps in Conducting a OneSample TTest:
We will understand the steps in conducting onesample ttest with an example.
Assume that the national average score (population mean) for a high school mathematics exam is known to be 70 out of 100. A school wants to evaluate whether its new teaching approach has significantly changed the performance of its students in mathematics compared to the national average. The school selects a random sample of 20 students (sample size) who have been taught using the new teaching method. The average score of these 30 students is calculated to be 75 (sample mean).

State the Hypotheses:
 Null Hypothesis ($H_0$): There is no difference between the sample mean and the population mean. Considering the example, the mean score of the students (sample mean) is equal to the national average (population mean), i.e., $μ=70$.
 Alternative Hypothesis ($$H__{1$}$): There is a significant difference. The mean score of the students is not equal to the national average, i.e., $≠70$.

Calculate the TStatistic: Using the formula and the data from the sample.

Determine the Critical Value: This value is obtained from the tdistribution table based on the desired significance level (commonly 0.05 for a 95% confidence level) and the degrees of freedom ($n−1$).

Make a Decision: If the calculated tstatistic is greater than the critical value, reject the null hypothesis; otherwise, fail to reject the null hypothesis.

Interpret the Results: If the null hypothesis is rejected, it suggests that the teaching method significantly impacts the students’ scores compared to the national average.
Onesample Ttest Formula
The following is onesample ttest formula / equation of tstatistics:
T = (X̄ – μ) / S/√n
Where, X̄ is the sample mean, μ is the hypothesized population mean, S is the standard deviation of the sample and n is the number of sample observations.
When working with Ttest, Tdistribution is used in place of the normal distribution. The tdistribution is a family of curves that are symmetrical about the mean, and have increasing variability as the degrees of freedom increase. The ttest statistic (T) follows a tdistribution with n – 1 degrees of freedom, where n is the number of observations in the sample.
Onesample Ttest Example
In this section, we will learn about how to calculate tstatistics in onesample ttest.
Suppose a claim is made that the average number of days a person spends on vacation is more than or equal to 5 days (hypothesized population mean) based on a sample of 16 people whose mean came out to be 9 days. As a first step, we will formulate the null and alternate hypothesis.
In hypothesis testing, we start by formulating the null hypothesis ($H_{0}$) and the alternative hypothesis ($H_{a}$ or $H_{1}$). The null hypothesis represents a position of no change, no effect, or no difference—it is the hypothesis that the researcher tries to disprove. The alternative hypothesis represents a new theory or the proposition that there is an effect, a change, or a difference.
Based on the claim that the average number of days a person spends on vacation is more than or equal to 5 days, the hypotheses can be formulated as follows:
For a OneTailed Test:

Null Hypothesis ($H_{0}$): The average number of days a person spends on vacation is equal to 5 days. Mathematically, $H_{0}:μ=5$.

Alternative Hypothesis ($H_{a}$): The average number of days a person spends on vacation is more than 5 days. Mathematically, $H_{a}:μ>5$.
Here, we are specifically looking to see if there is evidence to support the claim that people spend more than 5 days on vacation on average, which is why the alternative hypothesis is set up as greater than 5 days.
For a TwoTailed Test:
If, however, you wanted to test the hypothesis that the average number of vacation days is not equal to 5 (either less than or more than 5), the hypotheses would be formulated differently:

Null Hypothesis ($H_{0}$): The average number of days a person spends on vacation is equal to 5 days. Mathematically, $H_{0}:μ=5$.

Alternative Hypothesis ($H_{a}$): The average number of days a person spends on vacation is not equal to 5 days. Mathematically, $H_{a}:μ is not equal to5$.
In this case, the alternative hypothesis is testing for any significant difference, regardless of direction (more or fewer vacation days than 5).
Onetailed or a Twotailed test? Which one to use? For the data we have, where the sample mean is 9 days, and we are testing against the claim that the average number of vacation days is more than or equal to 5, a onetailed test is most appropriate.
We will use onesample ttest to test this hypothesis. A onetailed test will be performed.
T = (X̄ – μ) / S/√n
Where, X̄ is the sample mean, μ is the hypothesized population mean, S is the standard deviation of the sample and n is the number of observations in the sample.
A sample size of 16 persons is taken. The mean number of days spent on vacation by the persons in sample is found to be 9 days with a sample standard deviation is found to be 3 days.
T = (X̄ – μ) / S/√n
= (9 – 5)/(3/ √16)
= 5.33
If the calculated tvalue is 5.33 and the critical tvalue for a onetailed test at the alpha level of 0.05 is 1.753, you can make the following conclusions about the null hypothesis:
Since the calculated tvalue (5.33) is greater than the critical tvalue (1.753), you have sufficient evidence to reject the null hypothesis at the 0.05 significance level. This means that there is a statistically significant difference between the sample mean and the hypothesized population mean, and the sample provides enough evidence to support the claim that the average number of days a person spends on vacation is more than 5 days. The following plot can help you visualize the rejection of null hypothesis:
Another way to test is to get the pvalue for getting the Tstatistics of 5.33. You can use this OneSample TTest calculator to get the Tstatistics, degrees of freedom. . For a Tstatistics of 5.33 and the types of tailtest (onetailed or twotailed test), you can arrive at the pvalue of 0.000042. This means that there is a probability of only 0.000042 to get this kind of sample given the null hypothesis holds good. As this value is less than 0.05, one can reject the null hypothesis given the evidence of current sample. Calculate the Tstatistics using the following calculator:
Calculating TStatistics, Critical Value, PValue using Python
Calculating the tstatistic, critical value, and pvalue is central to this test, providing evidence to support or refute hypotheses. Python, with its simplicity and the powerful scipy.stats library, offers a streamlined approach to these calculations. By learning Python code for these purposes, one can efficiently automate the iterative process of hypothesis testing, reduce the potential for manual errors, and focus more on interpreting results rather than getting bogged down in calculations.
The following python code helps you calculate standard error and tstatistics value as 0.75 and 5.33 respectively.
from scipy import stats
import numpy as np
# Given values
sample_mean = 9 # sample mean
population_mean = 5 # hypothesized population mean
sample_size = 16 # number of people
# Assuming standard deviation of the sample is known
std_dev = 1.5 # You'll need to provide this value
# Calculating the standard error
standard_error = sample_std_dev / np.sqrt(sample_size)
# Calculating the tstatistic
t_statistic = (sample_mean  population_mean) / standard_error
# Display results
print(f"Standard Error: {standard_error}")
print(f"Tstatistic: {t_statistic}")
# Degrees of freedom
df = 15
# Calculate the pvalue for the onetailed test
p_value = stats.t.sf(t_statistic, df)
print(f"Pvalue for onetailed test: {p_value}")
The following gets printed:
Standard Error: 0.75
Tstatistic: 5.333333333333333
Pvalue for onetailed test: 4.1794868572493856e05
The following Python code helps calculate critical value:
from scipy import stats
# Degrees of freedom
df = 15 # for a sample size of 16, df = n  1
# Significance level
alpha = 0.05
# For a onetailed test, we use the 'ppf' method to find the critical tvalue
critical_t_value = stats.t.ppf(1  alpha, df)
print(f"Critical Tvalue for onetailed test at alpha = 0.05: {critical_t_value}")
The critical tvalue comes out to be ~1.753. The tstatistics is much larger than this. This is why we can reject the null hypothesis.
Check out our onesample ttest calculator tool.
Tscore / Tstatistics for Estimating Population Mean
The population mean can be estimated as a function of the tscore using the following equation:
Population mean = Sample mean + T*(Standard error of the mean)
Where T is a statistic that has a Tdistribution with known properties. The standard error of the mean (SE) is an estimate of the standard deviation of the sampling distribution of the tstatistic. The Tstatistic can be used to calculate confidence intervals for population means given the sample size is small and the population standard deviation is unknown. When the population standard deviation is know, we use Zstatistics and Zdistribution instead of Tstatistics.
The value of standard error of the mean can be calculated as :
SE of the mean = S/√n
Where, S is the standard deviation of the sample and n is the number of observations in the sample.
Summary
The onesample ttest is a statistical test that can be used to determine whether there is a significant difference between the sample mean and the population mean. The ttest statistic (T) follows a tdistribution with n – 1 degrees of freedom, where n is the number of observations in the sample. Tstatistics can be used to estimate the population mean when the population standard deviation is unknown. The ttest can be used to calculate confidence intervals for population means when the sample size is small and the population standard deviation is unknown.
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Question: In the Onesample Ttest example wouldn’t the hypotheses as stated denote a twotailed test? Therefore the critical value would be 2.131
Null hypothesis, H0: There is no difference between the sample mean and the population mean. Thus H0 xbar = u
Alternate hypothesis, Ha: There is a significant difference between the sample mean and the population mean. Thus H0 xbar u
If the alternate hypothesis, Ha was stated differently such as: There is a significant positive difference between the sample mean and the population mean. Thus H0 xbar > u; denoting a right hand onetailed test then the critical value would be 1.75. [1] I will note that in either case the 5.33 value does exceed the critical values both the onetailed and twotailed.
Thanks, Dave
Source: [1] https://www.nipissingu.ca/sites/default/files/OnetailedTestorTwotailedTest.pdf
Hi Dave,
You are correct in pointing out that the hypotheses mentioned in the example denote a twotailed test, which tests for the possibility of the sample mean being significantly greater or less than the hypothesized population mean.
For a twotailed test with α = 0.05 and 15 degrees of freedom (n1), the critical tvalue is approximately 2.131. This value will reject the null hypothesis if the calculated tstatistic is either less than 2.131 or greater than 2.131. Since 5.33 is greater than 2.131, we can reject the null hypothesis.
Made the appropriate changes.
Thank you
Hello. May I ask, the problem states that that the average number of days on vacation is more than or equal to 16, so shouldn’t that mean that µ≥5 is the null hypothesis while the alternative hypothesis is µ<5?
Please answer speedily. God bless and thanks!
As the claim is made about average number of days spent on vacation is greater than or equal to 5 days, we are talking about establishing a new truth such as µ≥5. The null hypothesis would rather be µ<5. Read my post on hypothesis testing for more details (https://vitalflux.com/datasciencehowtoformulatehypothesisforhypothesistesting/)
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