hypothesis in t test

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What is a t- test?

A t -test (also known as Student's t -test) is a tool for evaluating the means of one or two populations using hypothesis testing. A t-test may be used to evaluate whether a single group differs from a known value ( a one-sample t-test ), whether two groups differ from each other ( an independent two-sample t-test ), or whether there is a significant difference in paired measurements ( a paired, or dependent samples t-test ).

How are t -tests used?

First, you define the hypothesis you are going to test and specify an acceptable risk of drawing a faulty conclusion. For example, when comparing two populations, you might hypothesize that their means are the same, and you decide on an acceptable probability of concluding that a difference exists when that is not true. Next, you calculate a test statistic from your data and compare it to a theoretical value from a  t- distribution. Depending on the outcome, you either reject or fail to reject your null hypothesis. 

What if I have more than two groups?

You cannot use a t -test. Use a multiple comparison method. Examples are analysis of variance ( ANOVA ) , Tukey-Kramer pairwise comparison, Dunnett's comparison to a control, and analysis of means (ANOM).  

t -Test assumptions

While t -tests are relatively robust to deviations from assumptions, t -tests do assume that:

• The data are continuous.
• The sample data have been randomly sampled from a population.
• There is homogeneity of variance (i.e., the variability of the data in each group is similar).
• The distribution is approximately normal.

For two-sample t -tests, we must have independent samples. If the samples are not independent, then a paired t -test may be appropriate.

Types of t -tests

There are three t -tests to compare means: a one-sample t -test, a two-sample t -test and a paired t -test. The table below summarizes the characteristics of each and provides guidance on how to choose the correct test. Visit the individual pages for each type of t -test for examples along with details on assumptions and calculations.

The table above shows only the t -tests for population means. Another common t -test is for correlation coefficients .  You use this t -test to decide if the correlation coefficient is significantly different from zero. 

One-tailed vs. two-tailed tests

When you define the hypothesis, you also define whether you have a one-tailed or a two-tailed test. You should make this decision before collecting your data or doing any calculations. You make this decision for all three of the t -tests for means.

To explain, let’s use the one-sample t -test. Suppose we have a random sample of protein bars, and the label for the bars advertises 20 grams of protein per bar. The null hypothesis is that the unknown population mean is 20. Suppose we simply want to know if the data shows we have a different population mean. In this situation, our hypotheses are:

$ \mathrm H_o: \mu = 20 $

$ \mathrm H_a: \mu \neq 20 $

Here, we have a two-tailed test. We will use the data to see if the sample average differs sufficiently from 20 – either higher or lower – to conclude that the unknown population mean is different from 20.

Suppose instead that we want to know whether the advertising on the label is correct. Does the data support the idea that the unknown population mean is at least 20? Or not? In this situation, our hypotheses are:

$ \mathrm H_o: \mu >= 20 $

$ \mathrm H_a: \mu < 20 $

Here, we have a one-tailed test. We will use the data to see if the sample average is sufficiently less than 20 to reject the hypothesis that the unknown population mean is 20 or higher.

See the "tails for hypotheses tests" section on the t -distribution page for images that illustrate the concepts for one-tailed and two-tailed tests.

How to perform a t -test

For all of the t -tests involving means, you perform the same steps in analysis:

• Define your null ($ \mathrm H_o $) and alternative ($ \mathrm H_a $) hypotheses before collecting your data.
• Decide on the alpha value (or α value). This involves determining the risk you are willing to take of drawing the wrong conclusion. For example, suppose you set α=0.05 when comparing two independent groups. Here, you have decided on a 5% risk of concluding the unknown population means are different when they are not.
• Check the data for errors.
• Check the assumptions for the test.
• Perform the test and draw your conclusion. All t -tests for means involve calculating a test statistic. You compare the test statistic to a theoretical value from the t- distribution . The theoretical value involves both the α value and the degrees of freedom for your data. For more detail, visit the pages for one-sample t -test , two-sample t -test and paired t -test .

Independent t-test for two samples

Introduction.

The independent t-test, also called the two sample t-test, independent-samples t-test or student's t-test, is an inferential statistical test that determines whether there is a statistically significant difference between the means in two unrelated groups.

Null and alternative hypotheses for the independent t-test

The null hypothesis for the independent t-test is that the population means from the two unrelated groups are equal:

H 0 : u 1 = u 2

In most cases, we are looking to see if we can show that we can reject the null hypothesis and accept the alternative hypothesis, which is that the population means are not equal:

H A : u 1 ≠ u 2

To do this, we need to set a significance level (also called alpha) that allows us to either reject or accept the alternative hypothesis. Most commonly, this value is set at 0.05.

What do you need to run an independent t-test?

In order to run an independent t-test, you need the following:

• One independent, categorical variable that has two levels/groups.
• One continuous dependent variable.

Unrelated groups

Unrelated groups, also called unpaired groups or independent groups, are groups in which the cases (e.g., participants) in each group are different. Often we are investigating differences in individuals, which means that when comparing two groups, an individual in one group cannot also be a member of the other group and vice versa. An example would be gender - an individual would have to be classified as either male or female – not both.

Assumption of normality of the dependent variable

The independent t-test requires that the dependent variable is approximately normally distributed within each group.

Note: Technically, it is the residuals that need to be normally distributed, but for an independent t-test, both will give you the same result.

You can test for this using a number of different tests, but the Shapiro-Wilks test of normality or a graphical method, such as a Q-Q Plot, are very common. You can run these tests using SPSS Statistics, the procedure for which can be found in our Testing for Normality guide. However, the t-test is described as a robust test with respect to the assumption of normality. This means that some deviation away from normality does not have a large influence on Type I error rates. The exception to this is if the ratio of the smallest to largest group size is greater than 1.5 (largest compared to smallest).

What to do when you violate the normality assumption

If you find that either one or both of your group's data is not approximately normally distributed and groups sizes differ greatly, you have two options: (1) transform your data so that the data becomes normally distributed (to do this in SPSS Statistics see our guide on Transforming Data ), or (2) run the Mann-Whitney U test which is a non-parametric test that does not require the assumption of normality (to run this test in SPSS Statistics see our guide on the Mann-Whitney U Test ).

Assumption of homogeneity of variance

The independent t-test assumes the variances of the two groups you are measuring are equal in the population. If your variances are unequal, this can affect the Type I error rate. The assumption of homogeneity of variance can be tested using Levene's Test of Equality of Variances, which is produced in SPSS Statistics when running the independent t-test procedure. If you have run Levene's Test of Equality of Variances in SPSS Statistics, you will get a result similar to that below:

This test for homogeneity of variance provides an F -statistic and a significance value ( p -value). We are primarily concerned with the significance value – if it is greater than 0.05 (i.e., p > .05), our group variances can be treated as equal. However, if p < 0.05, we have unequal variances and we have violated the assumption of homogeneity of variances.

Overcoming a violation of the assumption of homogeneity of variance

If the Levene's Test for Equality of Variances is statistically significant, which indicates that the group variances are unequal in the population, you can correct for this violation by not using the pooled estimate for the error term for the t -statistic, but instead using an adjustment to the degrees of freedom using the Welch-Satterthwaite method. In all reality, you will probably never have heard of these adjustments because SPSS Statistics hides this information and simply labels the two options as "Equal variances assumed" and "Equal variances not assumed" without explicitly stating the underlying tests used. However, you can see the evidence of these tests as below:

From the result of Levene's Test for Equality of Variances, we can reject the null hypothesis that there is no difference in the variances between the groups and accept the alternative hypothesis that there is a statistically significant difference in the variances between groups. The effect of not being able to assume equal variances is evident in the final column of the above figure where we see a reduction in the value of the t -statistic and a large reduction in the degrees of freedom (df). This has the effect of increasing the p -value above the critical significance level of 0.05. In this case, we therefore do not accept the alternative hypothesis and accept that there are no statistically significant differences between means. This would not have been our conclusion had we not tested for homogeneity of variances.

Reporting the result of an independent t-test

When reporting the result of an independent t-test, you need to include the t -statistic value, the degrees of freedom (df) and the significance value of the test ( p -value). The format of the test result is: t (df) = t -statistic, p = significance value. Therefore, for the example above, you could report the result as t (7.001) = 2.233, p = 0.061.

Fully reporting your results

In order to provide enough information for readers to fully understand the results when you have run an independent t-test, you should include the result of normality tests, Levene's Equality of Variances test, the two group means and standard deviations, the actual t-test result and the direction of the difference (if any). In addition, you might also wish to include the difference between the groups along with a 95% confidence interval. For example:

Inspection of Q-Q Plots revealed that cholesterol concentration was normally distributed for both groups and that there was homogeneity of variance as assessed by Levene's Test for Equality of Variances. Therefore, an independent t-test was run on the data with a 95% confidence interval (CI) for the mean difference. It was found that after the two interventions, cholesterol concentrations in the dietary group (6.15 ± 0.52 mmol/L) were significantly higher than the exercise group (5.80 ± 0.38 mmol/L) ( t (38) = 2.470, p = 0.018) with a difference of 0.35 (95% CI, 0.06 to 0.64) mmol/L.

To know how to run an independent t-test in SPSS Statistics, see our SPSS Statistics Independent-Samples T-Test guide. Alternatively, you can carry out an independent-samples t-test using Excel, R and RStudio .

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8.2: Hypothesis Testing with t

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• Foster et al.
• University of Missouri-St. Louis, Rice University, & University of Houston, Downtown Campus via University of Missouri’s Affordable and Open Access Educational Resources Initiative

Hypothesis testing with the \(t\)-statistic works exactly the same way as \(z\)-tests did, following the four-step process of

• Stating the Hypothesis
• Finding the Critical Values
• Computing the Test Statistic
• Making the Decision.

We will work though an example: let’s say that you move to a new city and find a an auto shop to change your oil. Your old mechanic did the job in about 30 minutes (though you never paid close enough attention to know how much that varied), and you suspect that your new shop takes much longer. After 4 oil changes, you think you have enough evidence to demonstrate this.

Step 1: State the Hypotheses Our hypotheses for 1-sample t-tests are identical to those we used for \(z\)-tests. We still state the null and alternative hypotheses mathematically in terms of the population parameter and written out in readable English. For our example:

\(H_0\): There is no difference in the average time to change a car’s oil

\(H_0: μ = 30\)

\(H_A\): This shop takes longer to change oil than your old mechanic

\(H_A: μ > 30\)

Step 2: Find the Critical Values As noted above, our critical values still delineate the area in the tails under the curve corresponding to our chosen level of significance. Because we have no reason to change significance levels, we will use \(α\) = 0.05, and because we suspect a direction of effect, we have a one-tailed test. To find our critical values for \(t\), we need to add one more piece of information: the degrees of freedom. For this example:

\[df = N – 1 = 4 – 1 = 3 \nonumber \]

Going to our \(t\)-table, we find the column corresponding to our one-tailed significance level and find where it intersects with the row for 3 degrees of freedom. As shown in Figure \(\PageIndex{1}\): our critical value is \(t*\) = 2.353

We can then shade this region on our \(t\)-distribution to visualize our rejection region

Step 3: Compute the Test Statistic The four wait times you experienced for your oil changes are the new shop were 46 minutes, 58 minutes, 40 minutes, and 71 minutes. We will use these to calculate \(\overline{\mathrm{X}}\) and s by first filling in the sum of squares table in Table \(\PageIndex{1}\):

After filling in the first row to get \(\Sigma\)=215, we find that the mean is \(\overline{\mathrm{X}}\) = 53.75 (215 divided by sample size 4), which allows us to fill in the rest of the table to get our sum of squares \(SS\) = 564.74, which we then plug in to the formula for standard deviation from chapter 3:

\[s=\sqrt{\dfrac{\sum(X-\overline{X})^{2}}{N-1}}=\sqrt{\dfrac{S S}{d f}}=\sqrt{\dfrac{564.74}{3}}=13.72 \nonumber \]

Next, we take this value and plug it in to the formula for standard error:

\[s_{\overline{X}}=\dfrac{s}{\sqrt{n}}=\dfrac{13.72}{2}=6.86 \nonumber \]

And, finally, we put the standard error, sample mean, and null hypothesis value into the formula for our test statistic \(t\):

\[t=\dfrac{\overline{\mathrm{X}}-\mu}{s_{\overline{\mathrm{X}}}}=\dfrac{53.75-30}{6.86}=\dfrac{23.75}{6.68}=3.46 \nonumber \]

This may seem like a lot of steps, but it is really just taking our raw data to calculate one value at a time and carrying that value forward into the next equation: data  sample size/degrees of freedom  mean  sum of squares  standard deviation  standard error  test statistic. At each step, we simply match the symbols of what we just calculated to where they appear in the next formula to make sure we are plugging everything in correctly.

Step 4: Make the Decision Now that we have our critical value and test statistic, we can make our decision using the same criteria we used for a \(z\)-test. Our obtained \(t\)-statistic was \(t\) = 3.46 and our critical value was \(t* = 2.353: t > t*\), so we reject the null hypothesis and conclude:

Based on our four oil changes, the new mechanic takes longer on average (\(\overline{\mathrm{X}}\) = 53.75) to change oil than our old mechanic, \(t(3)\) = 3.46, \(p\) < .05.

Notice that we also include the degrees of freedom in parentheses next to \(t\). And because we found a significant result, we need to calculate an effect size, which is still Cohen’s \(d\), but now we use \(s\) in place of \(σ\):

\[d=\dfrac{\overline{X}-\mu}{s}=\dfrac{53.75-30.00}{13.72}=1.73 \nonumber \]

This is a large effect. It should also be noted that for some things, like the minutes in our current example, we can also interpret the magnitude of the difference we observed (23 minutes and 45 seconds) as an indicator of importance since time is a familiar metric.

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Understanding the t-test, using a t-test, which t-test to use.

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Read how this calculation can be used for hypothesis testing in statistics

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A t-test is an inferential statistic used to determine if there is a significant difference between the means of two groups and how they are related. T-tests are used when the data sets follow a normal distribution and have unknown variances, like the data set recorded from flipping a coin 100 times.

The t-test is a test used for hypothesis testing in statistics and uses the t-statistic, the t-distribution values, and the degrees of freedom to determine statistical significance.

Key Takeaways

• A t-test is an inferential statistic used to determine if there is a statistically significant difference between the means of two variables.
• The t-test is a test used for hypothesis testing in statistics.
• Calculating a t-test requires three fundamental data values including the difference between the mean values from each data set, the standard deviation of each group, and the number of data values.
• T-tests can be dependent or independent.

Investopedia / Sabrina Jiang

A t-test compares the average values of two data sets and determines if they came from the same population. In the above examples, a sample of students from class A and a sample of students from class B would not likely have the same mean and standard deviation. Similarly, samples taken from the placebo-fed control group and those taken from the drug prescribed group should have a slightly different mean and standard deviation.

Mathematically, the t-test takes a sample from each of the two sets and establishes the problem statement. It assumes a null hypothesis that the two means are equal.

Using the formulas, values are calculated and compared against the standard values. The assumed null hypothesis is accepted or rejected accordingly. If the null hypothesis qualifies to be rejected, it indicates that data readings are strong and are probably not due to chance.

The t-test is just one of many tests used for this purpose. Statisticians use additional tests other than the t-test to examine more variables and larger sample sizes. For a large sample size, statisticians use a  z-test . Other testing options include the chi-square test and the f-test.

Consider that a drug manufacturer tests a new medicine. Following standard procedure, the drug is given to one group of patients and a placebo to another group called the control group. The placebo is a substance with no therapeutic value and serves as a benchmark to measure how the other group, administered the actual drug, responds.

After the drug trial, the members of the placebo-fed control group reported an increase in average life expectancy of three years, while the members of the group who are prescribed the new drug reported an increase in average life expectancy of four years.

Initial observation indicates that the drug is working. However, it is also possible that the observation may be due to chance. A t-test can be used to determine if the results are correct and applicable to the entire population.

Four assumptions are made while using a t-test. The data collected must follow a continuous or ordinal scale, such as the scores for an IQ test, the data is collected from a randomly selected portion of the total population, the data will result in a normal distribution of a bell-shaped curve, and equal or homogenous variance exists when the standard variations are equal.

T-Test Formula

Calculating a t-test requires three fundamental data values. They include the difference between the mean values from each data set, or the mean difference, the standard deviation of each group, and the number of data values of each group.

This comparison helps to determine the effect of chance on the difference, and whether the difference is outside that chance range. The t-test questions whether the difference between the groups represents a true difference in the study or merely a random difference.

The t-test produces two values as its output: t-value and degrees of freedom . The t-value, or t-score, is a ratio of the difference between the mean of the two sample sets and the variation that exists within the sample sets.

The numerator value is the difference between the mean of the two sample sets. The denominator is the variation that exists within the sample sets and is a measurement of the dispersion or variability.

This calculated t-value is then compared against a value obtained from a critical value table called the T-distribution table. Higher values of the t-score indicate that a large difference exists between the two sample sets. The smaller the t-value, the more similarity exists between the two sample sets.

A large t-score, or t-value, indicates that the groups are different while a small t-score indicates that the groups are similar.

Degrees of freedom refer to the values in a study that has the freedom to vary and are essential for assessing the importance and the validity of the null hypothesis. Computation of these values usually depends upon the number of data records available in the sample set.

Paired Sample T-Test

The correlated t-test, or paired t-test, is a dependent type of test and is performed when the samples consist of matched pairs of similar units, or when there are cases of repeated measures. For example, there may be instances where the same patients are repeatedly tested before and after receiving a particular treatment. Each patient is being used as a control sample against themselves.

This method also applies to cases where the samples are related or have matching characteristics, like a comparative analysis involving children, parents, or siblings.

The formula for computing the t-value and degrees of freedom for a paired t-test is:

T = mean 1 − mean 2 s ( diff ) ( n ) where: mean 1  and  mean 2 = The average values of each of the sample sets s ( diff ) = The standard deviation of the differences of the paired data values n = The sample size (the number of paired differences) n − 1 = The degrees of freedom \begin{aligned}&T=\frac{\textit{mean}1 - \textit{mean}2}{\frac{s(\text{diff})}{\sqrt{(n)}}}\\&\textbf{where:}\\&\textit{mean}1\text{ and }\textit{mean}2=\text{The average values of each of the sample sets}\\&s(\text{diff})=\text{The standard deviation of the differences of the paired data values}\\&n=\text{The sample size (the number of paired differences)}\\&n-1=\text{The degrees of freedom}\end{aligned} ​ T = ( n ) ​ s ( diff ) ​ mean 1 − mean 2 ​ where: mean 1  and  mean 2 = The average values of each of the sample sets s ( diff ) = The standard deviation of the differences of the paired data values n = The sample size (the number of paired differences) n − 1 = The degrees of freedom ​

Equal Variance or Pooled T-Test

The equal variance t-test is an independent t-test and is used when the number of samples in each group is the same, or the variance of the two data sets is similar.

The formula used for calculating t-value and degrees of freedom for equal variance t-test is:

T-value = m e a n 1 − m e a n 2 ( n 1 − 1 ) × v a r 1 2 + ( n 2 − 1 ) × v a r 2 2 n 1 + n 2 − 2 × 1 n 1 + 1 n 2 where: m e a n 1  and  m e a n 2 = Average values of each of the sample sets v a r 1  and  v a r 2 = Variance of each of the sample sets n 1  and  n 2 = Number of records in each sample set \begin{aligned}&\text{T-value} = \frac{ mean1 - mean2 }{\frac {(n1 - 1) \times var1^2 + (n2 - 1) \times var2^2 }{ n1 +n2 - 2}\times \sqrt{ \frac{1}{n1} + \frac{1}{n2}} } \\&\textbf{where:}\\&mean1 \text{ and } mean2 = \text{Average values of each} \\&\text{of the sample sets}\\&var1 \text{ and } var2 = \text{Variance of each of the sample sets}\\&n1 \text{ and } n2 = \text{Number of records in each sample set} \end{aligned} ​ T-value = n 1 + n 2 − 2 ( n 1 − 1 ) × v a r 1 2 + ( n 2 − 1 ) × v a r 2 2 ​ × n 1 1 ​ + n 2 1 ​ ​ m e an 1 − m e an 2 ​ where: m e an 1  and  m e an 2 = Average values of each of the sample sets v a r 1  and  v a r 2 = Variance of each of the sample sets n 1  and  n 2 = Number of records in each sample set ​

Degrees of Freedom = n 1 + n 2 − 2 where: n 1  and  n 2 = Number of records in each sample set \begin{aligned} &\text{Degrees of Freedom} = n1 + n2 - 2 \\ &\textbf{where:}\\ &n1 \text{ and } n2 = \text{Number of records in each sample set} \\ \end{aligned} ​ Degrees of Freedom = n 1 + n 2 − 2 where: n 1  and  n 2 = Number of records in each sample set ​

Unequal Variance T-Test

The unequal variance t-test is an independent t-test and is used when the number of samples in each group is different, and the variance of the two data sets is also different. This test is also called Welch's t-test.

The formula used for calculating t-value and degrees of freedom for an unequal variance t-test is:

T-value = m e a n 1 − m e a n 2 ( v a r 1 n 1 + v a r 2 n 2 ) where: m e a n 1  and  m e a n 2 = Average values of each of the sample sets v a r 1  and  v a r 2 = Variance of each of the sample sets n 1  and  n 2 = Number of records in each sample set \begin{aligned}&\text{T-value}=\frac{mean1-mean2}{\sqrt{\bigg(\frac{var1}{n1}{+\frac{var2}{n2}\bigg)}}}\\&\textbf{where:}\\&mean1 \text{ and } mean2 = \text{Average values of each} \\&\text{of the sample sets} \\&var1 \text{ and } var2 = \text{Variance of each of the sample sets} \\&n1 \text{ and } n2 = \text{Number of records in each sample set} \end{aligned} ​ T-value = ( n 1 v a r 1 ​ + n 2 v a r 2 ​ ) ​ m e an 1 − m e an 2 ​ where: m e an 1  and  m e an 2 = Average values of each of the sample sets v a r 1  and  v a r 2 = Variance of each of the sample sets n 1  and  n 2 = Number of records in each sample set ​

Degrees of Freedom = ( v a r 1 2 n 1 + v a r 2 2 n 2 ) 2 ( v a r 1 2 n 1 ) 2 n 1 − 1 + ( v a r 2 2 n 2 ) 2 n 2 − 1 where: v a r 1  and  v a r 2 = Variance of each of the sample sets n 1  and  n 2 = Number of records in each sample set \begin{aligned} &\text{Degrees of Freedom} = \frac{ \left ( \frac{ var1^2 }{ n1 } + \frac{ var2^2 }{ n2 } \right )^2 }{ \frac{ \left ( \frac{ var1^2 }{ n1 } \right )^2 }{ n1 - 1 } + \frac{ \left ( \frac{ var2^2 }{ n2 } \right )^2 }{ n2 - 1}} \\ &\textbf{where:}\\ &var1 \text{ and } var2 = \text{Variance of each of the sample sets} \\ &n1 \text{ and } n2 = \text{Number of records in each sample set} \\ \end{aligned} ​ Degrees of Freedom = n 1 − 1 ( n 1 v a r 1 2 ​ ) 2 ​ + n 2 − 1 ( n 2 v a r 2 2 ​ ) 2 ​ ( n 1 v a r 1 2 ​ + n 2 v a r 2 2 ​ ) 2 ​ where: v a r 1  and  v a r 2 = Variance of each of the sample sets n 1  and  n 2 = Number of records in each sample set ​

The following flowchart can be used to determine which t-test to use based on the characteristics of the sample sets. The key items to consider include the similarity of the sample records, the number of data records in each sample set, and the variance of each sample set.

Image by Julie Bang © Investopedia 2019

Example of an Unequal Variance T-Test

Assume that the diagonal measurement of paintings received in an art gallery is taken. One group of samples includes 10 paintings, while the other includes 20 paintings. The data sets, with the corresponding mean and variance values, are as follows:

Though the mean of Set 2 is higher than that of Set 1, we cannot conclude that the population corresponding to Set 2 has a higher mean than the population corresponding to Set 1.

Is the difference from 19.4 to 21.6 due to chance alone, or do differences exist in the overall populations of all the paintings received in the art gallery? We establish the problem by assuming the null hypothesis that the mean is the same between the two sample sets and conduct a t-test to test if the hypothesis is plausible.

Since the number of data records is different (n1 = 10 and n2 = 20) and the variance is also different, the t-value and degrees of freedom are computed for the above data set using the formula mentioned in the Unequal Variance T-Test section.

The t-value is -2.24787. Since the minus sign can be ignored when comparing the two t-values, the computed value is 2.24787.

The degrees of freedom value is 24.38 and is reduced to 24, owing to the formula definition requiring rounding down of the value to the least possible integer value.

One can specify a level of probability (alpha level, level of significance,  p ) as a criterion for acceptance. In most cases, a 5% value can be assumed.

Using the degree of freedom value as 24 and a 5% level of significance, a look at the t-value distribution table gives a value of 2.064. Comparing this value against the computed value of 2.247 indicates that the calculated t-value is greater than the table value at a significance level of 5%. Therefore, it is safe to reject the null hypothesis that there is no difference between means. The population set has intrinsic differences, and they are not by chance.

How Is the T-Distribution Table Used?

The T-Distribution Table is available in one-tail and two-tails formats. The former is used for assessing cases that have a fixed value or range with a clear direction, either positive or negative. For instance, what is the probability of the output value remaining below -3, or getting more than seven when rolling a pair of dice? The latter is used for range-bound analysis, such as asking if the coordinates fall between -2 and +2.

What Is an Independent T-Test?

The samples of independent t-tests are selected independent of each other where the data sets in the two groups don’t refer to the same values. They may include a group of 100 randomly unrelated patients split into two groups of 50 patients each. One of the groups becomes the control group and is administered a placebo, while the other group receives a prescribed treatment. This constitutes two independent sample groups that are unpaired and unrelated to each other.

What Does a T-Test Explain and How Are They Used?

A t-test is a statistical test that is used to compare the means of two groups. It is often used in hypothesis testing to determine whether a process or treatment has an effect on the population of interest, or whether two groups are different from one another.

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Welcome to our t-test calculator! Here you can not only easily perform one-sample t-tests , but also two-sample t-tests , as well as paired t-tests .

Do you prefer to find the p-value from t-test, or would you rather find the t-test critical values? Well, this t-test calculator can do both! 😊

What does a t-test tell you? Take a look at the text below, where we explain what actually gets tested when various types of t-tests are performed. Also, we explain when to use t-tests (in particular, whether to use the z-test vs. t-test) and what assumptions your data should satisfy for the results of a t-test to be valid. If you've ever wanted to know how to do a t-test by hand, we provide the necessary t-test formula, as well as tell you how to determine the number of degrees of freedom in a t-test.

When to use a t-test?

A t-test is one of the most popular statistical tests for location , i.e., it deals with the population(s) mean value(s).

There are different types of t-tests that you can perform:

• A one-sample t-test;
• A two-sample t-test; and
• A paired t-test.

In the next section , we explain when to use which. Remember that a t-test can only be used for one or two groups . If you need to compare three (or more) means, use the analysis of variance ( ANOVA ) method.

The t-test is a parametric test, meaning that your data has to fulfill some assumptions :

• The data points are independent; AND
• The data, at least approximately, follow a normal distribution .

If your sample doesn't fit these assumptions, you can resort to nonparametric alternatives. Visit our Mann–Whitney U test calculator or the Wilcoxon rank-sum test calculator to learn more. Other possibilities include the Wilcoxon signed-rank test or the sign test.

Which t-test?

Your choice of t-test depends on whether you are studying one group or two groups:

One sample t-test

Choose the one-sample t-test to check if the mean of a population is equal to some pre-set hypothesized value .

The average volume of a drink sold in 0.33 l cans — is it really equal to 330 ml?

The average weight of people from a specific city — is it different from the national average?

Two-sample t-test

Choose the two-sample t-test to check if the difference between the means of two populations is equal to some pre-determined value when the two samples have been chosen independently of each other.

In particular, you can use this test to check whether the two groups are different from one another .

The average difference in weight gain in two groups of people: one group was on a high-carb diet and the other on a high-fat diet.

The average difference in the results of a math test from students at two different universities.

This test is sometimes referred to as an independent samples t-test , or an unpaired samples t-test .

Paired t-test

A paired t-test is used to investigate the change in the mean of a population before and after some experimental intervention , based on a paired sample, i.e., when each subject has been measured twice: before and after treatment.

In particular, you can use this test to check whether, on average, the treatment has had any effect on the population .

The change in student test performance before and after taking a course.

The change in blood pressure in patients before and after administering some drug.

How to do a t-test?

So, you've decided which t-test to perform. These next steps will tell you how to calculate the p-value from t-test or its critical values, and then which decision to make about the null hypothesis.

Decide on the alternative hypothesis :

Use a two-tailed t-test if you only care whether the population's mean (or, in the case of two populations, the difference between the populations' means) agrees or disagrees with the pre-set value.

Use a one-tailed t-test if you want to test whether this mean (or difference in means) is greater/less than the pre-set value.

Compute your T-score value :

Formulas for the test statistic in t-tests include the sample size , as well as its mean and standard deviation . The exact formula depends on the t-test type — check the sections dedicated to each particular test for more details.

Determine the degrees of freedom for the t-test:

The degrees of freedom are the number of observations in a sample that are free to vary as we estimate statistical parameters. In the simplest case, the number of degrees of freedom equals your sample size minus the number of parameters you need to estimate . Again, the exact formula depends on the t-test you want to perform — check the sections below for details.

The degrees of freedom are essential, as they determine the distribution followed by your T-score (under the null hypothesis). If there are d degrees of freedom, then the distribution of the test statistics is the t-Student distribution with d degrees of freedom . This distribution has a shape similar to N(0,1) (bell-shaped and symmetric) but has heavier tails . If the number of degrees of freedom is large (>30), which generically happens for large samples, the t-Student distribution is practically indistinguishable from N(0,1).

💡 The t-Student distribution owes its name to William Sealy Gosset, who, in 1908, published his paper on the t-test under the pseudonym "Student". Gosset worked at the famous Guinness Brewery in Dublin, Ireland, and devised the t-test as an economical way to monitor the quality of beer. Cheers! 🍺🍺🍺

p-value from t-test

Recall that the p-value is the probability (calculated under the assumption that the null hypothesis is true) that the test statistic will produce values at least as extreme as the T-score produced for your sample . As probabilities correspond to areas under the density function, p-value from t-test can be nicely illustrated with the help of the following pictures:

The following formulae say how to calculate p-value from t-test. By cdf t,d we denote the cumulative distribution function of the t-Student distribution with d degrees of freedom:

p-value from left-tailed t-test:

p-value = cdf t,d (t score )

p-value from right-tailed t-test:

p-value = 1 − cdf t,d (t score )

p-value from two-tailed t-test:

p-value = 2 × cdf t,d (−|t score |)

or, equivalently: p-value = 2 − 2 × cdf t,d (|t score |)

However, the cdf of the t-distribution is given by a somewhat complicated formula. To find the p-value by hand, you would need to resort to statistical tables, where approximate cdf values are collected, or to specialized statistical software. Fortunately, our t-test calculator determines the p-value from t-test for you in the blink of an eye!

t-test critical values

Recall, that in the critical values approach to hypothesis testing, you need to set a significance level, α, before computing the critical values , which in turn give rise to critical regions (a.k.a. rejection regions).

Formulas for critical values employ the quantile function of t-distribution, i.e., the inverse of the cdf :

Critical value for left-tailed t-test: cdf t,d -1 (α)

critical region:

(-∞, cdf t,d -1 (α)]

Critical value for right-tailed t-test: cdf t,d -1 (1-α)

[cdf t,d -1 (1-α), ∞)

Critical values for two-tailed t-test: ±cdf t,d -1 (1-α/2)

(-∞, -cdf t,d -1 (1-α/2)] ∪ [cdf t,d -1 (1-α/2), ∞)

To decide the fate of the null hypothesis, just check if your T-score lies within the critical region:

If your T-score belongs to the critical region , reject the null hypothesis and accept the alternative hypothesis.

If your T-score is outside the critical region , then you don't have enough evidence to reject the null hypothesis.

How to use our t-test calculator

Choose the type of t-test you wish to perform:

A one-sample t-test (to test the mean of a single group against a hypothesized mean);

A two-sample t-test (to compare the means for two groups); or

A paired t-test (to check how the mean from the same group changes after some intervention).

Two-tailed;

Left-tailed; or

Right-tailed.

This t-test calculator allows you to use either the p-value approach or the critical regions approach to hypothesis testing!

Enter your T-score and the number of degrees of freedom . If you don't know them, provide some data about your sample(s): sample size, mean, and standard deviation, and our t-test calculator will compute the T-score and degrees of freedom for you .

Once all the parameters are present, the p-value, or critical region, will immediately appear underneath the t-test calculator, along with an interpretation!

One-sample t-test

The null hypothesis is that the population mean is equal to some value μ 0 \mu_0 μ 0 ​ .

The alternative hypothesis is that the population mean is:

• different from μ 0 \mu_0 μ 0 ​ ;
• smaller than μ 0 \mu_0 μ 0 ​ ; or
• greater than μ 0 \mu_0 μ 0 ​ .

One-sample t-test formula :

• μ 0 \mu_0 μ 0 ​ — Mean postulated in the null hypothesis;
• n n n — Sample size;
• x ˉ \bar{x} x ˉ — Sample mean; and
• s s s — Sample standard deviation.

Number of degrees of freedom in t-test (one-sample) = n − 1 n-1 n − 1 .

The null hypothesis is that the actual difference between these groups' means, μ 1 \mu_1 μ 1 ​ , and μ 2 \mu_2 μ 2 ​ , is equal to some pre-set value, Δ \Delta Δ .

The alternative hypothesis is that the difference μ 1 − μ 2 \mu_1 - \mu_2 μ 1 ​ − μ 2 ​ is:

• Different from Δ \Delta Δ ;
• Smaller than Δ \Delta Δ ; or
• Greater than Δ \Delta Δ .

In particular, if this pre-determined difference is zero ( Δ = 0 \Delta = 0 Δ = 0 ):

The null hypothesis is that the population means are equal.

The alternate hypothesis is that the population means are:

• μ 1 \mu_1 μ 1 ​ and μ 2 \mu_2 μ 2 ​ are different from one another;
• μ 1 \mu_1 μ 1 ​ is smaller than μ 2 \mu_2 μ 2 ​ ; and
• μ 1 \mu_1 μ 1 ​ is greater than μ 2 \mu_2 μ 2 ​ .

Formally, to perform a t-test, we should additionally assume that the variances of the two populations are equal (this assumption is called the homogeneity of variance ).

There is a version of a t-test that can be applied without the assumption of homogeneity of variance: it is called a Welch's t-test . For your convenience, we describe both versions.

Two-sample t-test if variances are equal

Use this test if you know that the two populations' variances are the same (or very similar).

Two-sample t-test formula (with equal variances) :

where s p s_p s p ​ is the so-called pooled standard deviation , which we compute as:

• Δ \Delta Δ — Mean difference postulated in the null hypothesis;
• n 1 n_1 n 1 ​ — First sample size;
• x ˉ 1 \bar{x}_1 x ˉ 1 ​ — Mean for the first sample;
• s 1 s_1 s 1 ​ — Standard deviation in the first sample;
• n 2 n_2 n 2 ​ — Second sample size;
• x ˉ 2 \bar{x}_2 x ˉ 2 ​ — Mean for the second sample; and
• s 2 s_2 s 2 ​ — Standard deviation in the second sample.

Number of degrees of freedom in t-test (two samples, equal variances) = n 1 + n 2 − 2 n_1 + n_2 - 2 n 1 ​ + n 2 ​ − 2 .

Two-sample t-test if variances are unequal (Welch's t-test)

Use this test if the variances of your populations are different.

Two-sample Welch's t-test formula if variances are unequal:

• s 1 s_1 s 1 ​ — Standard deviation in the first sample;
• s 2 s_2 s 2 ​ — Standard deviation in the second sample.

The number of degrees of freedom in a Welch's t-test (two-sample t-test with unequal variances) is very difficult to count. We can approximate it with the help of the following Satterthwaite formula :

Alternatively, you can take the smaller of n 1 − 1 n_1 - 1 n 1 ​ − 1 and n 2 − 1 n_2 - 1 n 2 ​ − 1 as a conservative estimate for the number of degrees of freedom.

🔎 The Satterthwaite formula for the degrees of freedom can be rewritten as a scaled weighted harmonic mean of the degrees of freedom of the respective samples: n 1 − 1 n_1 - 1 n 1 ​ − 1 and n 2 − 1 n_2 - 1 n 2 ​ − 1 , and the weights are proportional to the standard deviations of the corresponding samples.

As we commonly perform a paired t-test when we have data about the same subjects measured twice (before and after some treatment), let us adopt the convention of referring to the samples as the pre-group and post-group.

The null hypothesis is that the true difference between the means of pre- and post-populations is equal to some pre-set value, Δ \Delta Δ .

The alternative hypothesis is that the actual difference between these means is:

Typically, this pre-determined difference is zero. We can then reformulate the hypotheses as follows:

The null hypothesis is that the pre- and post-means are the same, i.e., the treatment has no impact on the population .

The alternative hypothesis:

• The pre- and post-means are different from one another (treatment has some effect);
• The pre-mean is smaller than the post-mean (treatment increases the result); or
• The pre-mean is greater than the post-mean (treatment decreases the result).

Paired t-test formula

In fact, a paired t-test is technically the same as a one-sample t-test! Let us see why it is so. Let x 1 , . . . , x n x_1, ... , x_n x 1 ​ , ... , x n ​ be the pre observations and y 1 , . . . , y n y_1, ... , y_n y 1 ​ , ... , y n ​ the respective post observations. That is, x i , y i x_i, y_i x i ​ , y i ​ are the before and after measurements of the i -th subject.

For each subject, compute the difference, d i : = x i − y i d_i := x_i - y_i d i ​ := x i ​ − y i ​ . All that happens next is just a one-sample t-test performed on the sample of differences d 1 , . . . , d n d_1, ... , d_n d 1 ​ , ... , d n ​ . Take a look at the formula for the T-score :

Δ \Delta Δ — Mean difference postulated in the null hypothesis;

n n n — Size of the sample of differences, i.e., the number of pairs;

x ˉ \bar{x} x ˉ — Mean of the sample of differences; and

s s s  — Standard deviation of the sample of differences.

Number of degrees of freedom in t-test (paired): n − 1 n - 1 n − 1

t-test vs Z-test

We use a Z-test when we want to test the population mean of a normally distributed dataset, which has a known population variance . If the number of degrees of freedom is large, then the t-Student distribution is very close to N(0,1).

Hence, if there are many data points (at least 30), you may swap a t-test for a Z-test, and the results will be almost identical. However, for small samples with unknown variance, remember to use the t-test because, in such cases, the t-Student distribution differs significantly from the N(0,1)!

🙋 Have you concluded you need to perform the z-test? Head straight to our z-test calculator !

What is a t-test?

A t-test is a widely used statistical test that analyzes the means of one or two groups of data. For instance, a t-test is performed on medical data to determine whether a new drug really helps.

What are different types of t-tests?

Different types of t-tests are:

• One-sample t-test;
• Two-sample t-test; and
• Paired t-test.

How to find the t value in a one sample t-test?

To find the t-value:

• Subtract the null hypothesis mean from the sample mean value.
• Divide the difference by the standard deviation of the sample.
• Multiply the resultant with the square root of the sample size.

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ol{padding-top:0px;}.css-4okk7a ul:not(:first-child),.css-4okk7a ol:not(:first-child){padding-top:4px;} Test setup

Choose test type

t-test for the population mean, μ, based on one independent sample . Null hypothesis H 0 : μ = μ 0  

Alternative hypothesis H 1

Test details

Significance level α

The probability that we reject a true H 0 (type I error).

Degrees of freedom

Calculated as sample size minus one.

Test results

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NCBI Bookshelf. A service of the National Library of Medicine, National Institutes of Health.

StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2024 Jan-.

StatPearls [Internet].

Raoul R. Wadhwa ; Raghavendra Marappa-Ganeshan .

Affiliations

Last Update: January 16, 2023 .

• Definition/Introduction

T-test was first described by William Sealy Gosset in 1908, when he published his article under the pseudonym 'student' while working for a brewery. [1]  In simple terms, a Student's t-test is a ratio that quantifies how significant the difference is between the 'means' of two groups while taking their variance or distribution into account.

• Issues of Concern

Selecting appropriate statistical tests is a critical step in conducting research. [2]  Therefore, there are three forms of Student’s t-test about which physicians, particularly physician-scientists, need to be aware: (1) one-sample t-test; (2) two-sample t-test; and (3) two-sample paired t-test. The one-sample t-test evaluates a single list of numbers to test the hypothesis that a statistic of that set is equal to a chosen value, for instance, to test the hypothesis that the mean of the set of numbers is equal to zero. As an example, consider the following question: what is the average serum sodium concentration in adults? Currently, 140 mEq/L serves as an approximate center of a reference range of 135 to 145 mEq/L; thus, the null hypothesis is that the average serum sodium concentration in adults is equal to 140 mEq/L.

If you believe these numbers are wrong (alternate hypothesis) or you want to test the original hypothesis, you could collect blood from a set of subjects, measure the sodium concentration in each sample, and then take the mean of this set. If the mean is 140.1 mEq/L, you probably do not have convincing evidence that the numbers mentioned above are faulty (since 140 and 140.1 are fairly close). Thus, you would fail to reject the null hypothesis. However, if your sample has a mean of 70 mEq/L, this could be preliminary evidence (assuming, of course, rigorous methodology) and could end up rejecting the null hypothesis. The decision-making process would be trickier if the mean of the sample were 134 or 150 mEq/L. The t-test can be used to reduce subjective influence when testing a null hypothesis. Before testing a hypothesis, researchers should choose the alpha and beta values of the test. Loosely, the alpha parameter determines the threshold for false-positive results (e.g., if the actual mean serum sodium concentration is 140 mEq/L, but the t-test rejects the original hypothesis in favor of your new hypothesis), and the beta parameter determines the threshold for false-negative results (e.g., if true mean serum sodium concentration is 200 mEq/L, but the t-test fails to reject the old hypothesis). Methods of selection of alpha and beta are outside the scope of this article.

While the one-sample t-test allows you to test the statistic of a single set of numbers against a specific numeric value, the two-sample t-test allows testing the values of a statistic between two groups. In this case, a research question could be: do children and adults have the same mean serum sodium concentration? Testing this hypothesis would require sampling two groups, a group of adults and a group of children, and comparing the mean serum sodium concentrations between these two groups in a manner analogous to the one-sample t-test described above. The paired t-test is used in scenarios where measurements from the two groups have a link to one another. In the example above concerning the mean serum sodium concentration of children and adults, the implicit assumption was that all the measurements would all be completed at one point in time in a set of children and a distinct set of adults. However, it would also be possible to measure serum sodium concentrations in a set of children, wait a few years until they are adults, then measure the serum sodium concentrations again. Here, each adult sodium concentration corresponds to exactly one child sodium concentration. A paired two-sample t-test can be used to capture the dependence of measurements between the two groups.

These variations of the student's t-test use observed or collected data to calculate a test statistic, which can then be used to calculate a p-value. Often misinterpreted, the p-value is equal to the probability of collecting data that is at least as extreme as the observed data in the study, assuming that the null hypothesis is true. [3]  This concept is best illustrated by examples, as in the questions that accompany this article. Often, a threshold value is set prior to the study (equal to the alpha mentioned above); if the resulting p-value is below the preset threshold, there is sufficient evidence to reject the null hypothesis.

In the above scenarios, before using any form of the t-test, one must ensure that the assumptions for the test have been met. This article does not list or explain these assumptions in detail. Please follow the guidance of a trained statistician when designing research studies and conducting data analysis.

• Clinical Significance

Given the rate of research progress, disease management (medical or surgical) continuously evolves. To follow the framework of evidence-based medicine, physicians must be able to read and critically evaluate primary literature. [4] [5]  The ability to do this successfully requires at least a basic foundation of knowledge in statistics, including common biases (e.g., nonresponse bias), standard study designs (e.g., randomized controlled trials), and common statistical pitfalls researchers face (e.g., statistically significant results that are not clinically significant). [6] [7] Understanding a student’s t-test is a start to clinicians gaining this necessary foundation of knowledge.

• Review Questions
• Access free multiple choice questions on this topic.
• Comment on this article.

Disclosure: Raoul Wadhwa declares no relevant financial relationships with ineligible companies.

Disclosure: Raghavendra Marappa-Ganeshan declares no relevant financial relationships with ineligible companies.

This book is distributed under the terms of the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International (CC BY-NC-ND 4.0) ( http://creativecommons.org/licenses/by-nc-nd/4.0/ ), which permits others to distribute the work, provided that the article is not altered or used commercially. You are not required to obtain permission to distribute this article, provided that you credit the author and journal.

• Cite this Page Wadhwa RR, Marappa-Ganeshan R. T Test. [Updated 2023 Jan 16]. In: StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2024 Jan-.

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• T Test Formula

T-Test Formula

The t-test is any statistical hypothesis test in which the test statistic follows a Student’s t-distribution under the null hypothesis. It can be used to determine if two sets of data are significantly different from each other, and is most commonly applied when the test statistic would follow a normal distribution if the value of a scaling term in the test statistic were known.

T-test uses means and standard deviations of two samples to make a comparison. The formula for T-test is given below:

\begin{array}{l}\qquad t=\frac{\bar{X}_{1}-\bar{X}_{2}}{s_{\bar{\Delta}}} \\ \text { where } \\ \qquad s_{\bar{\Delta}}=\sqrt{\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}} \\ \end{array}

Where, \(\begin{array}{l}\overline{x}\end{array} \) = Mean of first set of values \(\begin{array}{l}\overline{x}_{2}\end{array} \)  = Mean of second set of values \(\begin{array}{l}S_{1}\end{array} \)   = Standard deviation of first set of values \(\begin{array}{l}S_{2}\end{array} \)   = Standard deviation of second set of values \(\begin{array}{l}n_{1}\end{array} \)   = Total number of values in first set \(\begin{array}{l}n_{2}\end{array} \)   = Total number of values in second set.

The formula for standard deviation is given by:

Where, x = Values given \(\begin{array}{l}\overline{x}\end{array} \) = Mean n = Total number of values.

T-Test Solved Examples

Question 1: Find the t-test value for the following two sets of values: 7, 2, 9, 8 and 1, 2, 3, 4?

Formula for standard deviation:  \(\begin{array}{l}S=\sqrt{\frac{\sum\left(x-\overline{x}\right)^{2}}{n-1}}\end{array} \)

Number of terms in first set:  \(\begin{array}{l}n_{1}\end{array} \) = 4

Mean for first set of data: \(\begin{array}{l}\overline{x}_{1}\end{array} \) = 6.5

Construct the following table for standard deviation:

Standard deviation for the first set of data: S 1 = 3.11

Number of terms in second set: n 2 = 4

Standard deviation for first set of data: \(\begin{array}{l}S_{2}\end{array} \) = 1.29

Formula for t-test value:

t = 2.3764 = 2.36 (approx)

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Statistics Made Easy

Paired Samples t-test: Definition, Formula, and Example

A paired samples t-test is used to compare the means of two samples when each observation in one sample can be paired with an observation in the other sample.

This tutorial explains the following:

• The motivation for performing a paired samples t-test.
• The formula to perform a paired samples t-test.
• The assumptions that should be met to perform a paired samples t-test.
• An example of how to perform a paired samples t-test.

Paired Samples t-test: Motivation

A paired samples t-test is commonly used in two scenarios:

1. A measurement is taken on a subject before and after some treatment – e.g. the max vertical jump of college basketball players is measured before and after participating in a training program.

2. A measurement is taken under two different conditions  – e.g. the response time of a patient is measured on two different drugs.

In both cases we are interested in comparing the mean measurement between two groups in which each observation in one sample can be paired with an observation in the other sample.

Paired Samples t-test: Formula

A paired samples t-test always uses the following null hypothesis:

• H 0 : μ 1  = μ 2 (the two population means are equal)

The alternative hypothesis can be either two-tailed, left-tailed, or right-tailed:

• H 1 (two-tailed): μ 1  ≠ μ 2 (the two population means are not equal)
• H 1 (left-tailed): μ 1  < μ 2  (population 1 mean is less than population 2 mean)
• H 1 (right-tailed):  μ 1 > μ 2  (population 1 mean is greater than population 2 mean)

We use the following formula to calculate the test statistic t:

t = x diff  / (s diff /√n)

• x diff :  sample mean of the differences
• s:  sample standard deviation of the differences
• n:  sample size (i.e. number of pairs)

If the p-value that corresponds to the test statistic t with (n-1) degrees of freedom is less than your chosen significance level (common choices are 0.10, 0.05, and 0.01) then you can reject the null hypothesis.

Paired Samples t-test: Assumptions

For the results of a paired samples t-test to be valid, the following assumptions should be met:

• The participants should be selected randomly from the population.
• The differences between the pairs should be approximately normally distributed.
• There should be no extreme outliers in the differences.

Paired Samples t-test : Example

Suppose we want to know whether or not a certain training program is able to increase the max vertical jump (in inches) of college basketball players.

To test this, we may recruit a simple random sample of 20 college basketball players and measure each of their max vertical jumps. Then, we may have each player use the training program for one month and then measure their max vertical jump again at the end of the month.

To determine whether or not the training program actually had an effect on max vertical jump, we will perform a paired samples t-test at significance level α = 0.05 using the following steps:

Step 1: Calculate the summary data for the differences.

• x diff :  sample mean of the differences =  -0.95
• s:  sample standard deviation of the differences =  1.317
• n:  sample size (i.e. number of pairs) =  20

Step 2: Define the hypotheses.

We will perform the paired samples t-test with the following hypotheses:

• H 0 :  μ 1  = μ 2 (the two population means are equal)
• H 1 :  μ 1  ≠ μ 2 (the two population means are not equal)

Step 3: Calculate the test statistic  t .

t = x diff  / (s diff /√n)  = -0.95 / (1.317/ √ 20) =  -3.226

Step 4: Calculate the p-value of the test statistic  t .

According to the T Score to P Value Calculator , the p-value associated with t = -3.226 and degrees of freedom = n-1 = 20-1 = 19 is  0.00445 .

Step 5: Draw a conclusion.

Since this p-value is less than our significance level α = 0.05, we reject the null hypothesis. We have sufficient evidence to say that the mean max vertical jump of players is different before and after participating in the training program.

Note:  You can also perform this entire paired samples t-test by simply using the Paired Samples t-test Calculator .

Additional Resources

The following tutorials explain how to perform a paired samples t-test using different statistical programs:

How to Perform a Paired Samples t-Test in Excel How to Perform a Paired Samples t-test in SPSS How to Perform a Paired Samples t-test in Stata How to Perform a Paired Samples t-test on a TI-84 Calculator How to Perform a Paired Samples t-test in R How to Perform a Paired Samples t-Test in Python How to Perform a Paired Samples t-Test by Hand

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Hey there. My name is Zach Bobbitt. I have a Masters of Science degree in Applied Statistics and I’ve worked on machine learning algorithms for professional businesses in both healthcare and retail. I’m passionate about statistics, machine learning, and data visualization and I created Statology to be a resource for both students and teachers alike.  My goal with this site is to help you learn statistics through using simple terms, plenty of real-world examples, and helpful illustrations.

5 Replies to “Paired Samples t-test: Definition, Formula, and Example”

What will happen if the p-value is more than the significance level of 0.05? Does it mean it is not significant or effective?

Thank you for this. It helped me a lot.

Excellent it really helps me

Hey Zach , thanks for statology! I love this site. It helps me a lot to remember and learn new things. I have a question. What if rather then test if the difference between tests is zero or greater/lower than zero, I’d like to test if the difference is at least x? Eg: I’d like to check if the time to change tire was reduced by at least 5 seconds after a training. Not only if the time is different or less than before. Could you include this example, please? Which changes in calculation we should do? Thanks a lot. Marcelo

in paired t test are we supposed to deduct the before from the after ?

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T-Test in Statistics: Formula, Types and Steps

• Data Types in Statistics
• What are the types of statistics?
• Introduction of Statistics and its Types
• What are some of the important formulae used in statistics?
• Type I and Type II Errors
• Statistics Formulas
• How to Calculate the P-Value of an F-Statistic in R
• How to Conduct a Paired Samples T-Test in Python
• Types of Statistical Series
• Mode in Statistics | Definition, Formula and Examples
• Upper Tail Test of Population Proportion in R
• Differences Between two-sample, t-test and paired t-test
• Measures of Central Tendency in Statistics
• Who is the Father of Statistics?
• What is n in Statistics?
• How to Conduct a Two Sample T-Test in Python
• How to Conduct a One Sample T-Test in Python
• SciPy - Statistical Significance Tests
• How to Conduct a Sobel Test in R
• How to Perform an F-Test in Python
• Hypothesis Testing Formula
• Two-Tailed Test of Population Proportion in R
• F-Test in Statistics
• Student's t-distribution in Statistics
• Python - Student’s t Distribution in Statistics
• TCS SDE Sheet: Interview Questions and Answers
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T-Test is a method used in statistics to determine if there is a significant difference between the means of two groups and how they are related. In T-Test statistics, the sample data is a subset of the two groups that we use to draw conclusions about the groups as a whole. 

T-Test in Statistics

For example, if we want to know the average weight of mangoes grown on a farm, the population would consist of all the mangoes that grew on the farm. However, it would be time-consuming to weigh each mango. Instead, we could take a sample of mangoes from trees at different locations on the farm and use their weights to make inferences about the average weight of all the mangoes grown on the farm.

T-Test Definition

T-Test is a statistics method to determine significance changes between means of two groups. It helps us to determine whether the data sets belong to the same group or not. This comparison is often called a T-test.

T-Test Formula

There is no specific formula for T-Test, as it is divided into various types such as One Samples T-Test, Independent Samples T-test, etc. which are used as per the need. The formula used in each type is defined under the specific headings. The formula allows us to calculate a T-value which helps to make a comparison between the data sets.

Value of T-Test

The value obtained by substituting required values to the t-test formula is called the t-value. A larger T-value implies that the sets belong to a different population, while a smaller T-value implies that they belong to the same population. The formula is comprised of the values of mean, standard deviation and variance of the data sets under consideration.

How to Calculate T Value in T-Test

To calculate T-value in T-Test, we can use the following steps:

Step 1: To perform a T-test, two hypotheses namely the null hypothesis and the alternative hypothesis are defined which have different meanings for different types of T-tests.

Step 2: And, a value for the level of significance is defined which signifies the probability of making a Type I error, which implies the rejection of the null hypothesis while it is actually true. Commonly used values of level of significance are 0.05 (5%) and 0.01 (1%).

Step 3: A higher significance level, such as α = 0.05, provides a higher tolerance for Type I errors, meaning that it is more likely to reject the null hypothesis even when it is true.

Step 4: On the other hand, a lower significance level, such as α = 0.01, reduces the risk of Type I errors but it may increase the chances of accepting the null hypothesis when it is actually false, resulting in a Type II error.

Types of T-Test

Below are the three types of T-Test mentioned below.

• One Sample T-test
• Independent Samples T-test
• Paired Samples T-test

Let’s discuss these types in detail as follows:

One Sample T-Test

As the name implies, this test is used when we have one data set for a sample and we need to determine whether this data set belongs to a particular population or not. The mean value for the population data must be known in this case. The formula to determine T-value, in this case, is as follows:

t = (x̄ – μ) / (σ / √n) Where, t is the t-value, x̄ is the Sample mean, μ is the Population mean, σ is the Sample standard deviation, and n is the Sample size.

Steps to Calculate T Value One Sample T-Test 

To perform the One Sample T-test, the steps listed below are generally followed:

Step 1: State a null hypothesis and an alternative hypothesis. The null hypothesis assumes that the sample mean and the known population mean (μ) are equal, while the other assumes that the sample mean is different from the population mean.

Step 2: Define values for the level of significance (α) and the degree of freedom (df). The degree of freedom equals (n – 1) for this case.

Step 3: Calculate the t-value using the formula stated above by putting all the known values of the sample mean (x̄), sample standard deviation (σ), the population mean (μ), and the sample size (n).

Step 4: Determine the associated p-value with the t-value using a t-distribution table.

Step 5: Compare the p-value to the level of significance. If the p-value is less than the level of significance, reject the null hypothesis and conclude that the sample mean is significantly different from the population mean. Otherwise, conclude that there is no significant difference between the sample mean and the population mean.

Independent Samples T-Test

As the name suggests, an Independent samples T-test is used when we need to compare the statistical means of two independent samples or groups. It helps us determine whether there is a significant difference between the means of the two groups. If there is a significant difference, it suggests that the groups likely have different population means; otherwise, they have the same population means.

For example, when an investigation aims to determine if there is a significant difference in the mean scores between athletes who follow a specific training camp (Team A) and those who do not (Team B), an independent samples t-test can be conducted.

This test is performed using either of two assumptions made about variances of the samples, one assumes equal variances for the sample and the other assumes unequal variances for the samples.

Unequal Variances T-Test

Under this test, variances of two groups considered are assumed to be equal. This is appropriate when we are uncertain about the variances of the two groups considered. The formula to calculate T-value, in this case, is as follows:

t = (x̄ 1 – x̄ 2 ) / √((σ 1 2 /n 1 ) + (σ 2 2 /n 2 )) Where, x̄ 1 is the sample mean of Group 1, x̄ 2 is the sample mean of Group 2, σ 1 is the sample standard deviation of Group 1, σ 2 is the sample standard deviation of Group 2, n 1 is the sample size of Group 1, and n 2 sample size of Group 2.

Equal Variance T-Test

Under this test, variances of two groups considered are assumed to be equal. This is appropriate when we have some assurance about variances of data considered to be equal. The formula to calculate T-value, in this case, is similar to the above formula with a slight change that σ 1 = σ 2 = σ.

t = (x̄ 1 – x̄ 2 ) / √(σ 2 (1/n 1 + 1/n 2 )) Where, x̄ 1 is the sample mean of Group 1, x̄ 2 is the sample mean of Group 2, σ is the standard deviation of both groups, n 1 is the sample size of Group 1, and n 2 sample size of Group 2.

T Test for Independent Samples

The steps listed below are generally followed to perform this test:

Step 1: State a null hypothesis and an alternate hypothesis. The null hypothesis assumes that the means of the two groups are equal (x̄ 1 = x̄ 2 ), while the other assumes that the means of the two groups are significantly different (x̄ 1 ≠ x̄ 2 ).

Step 2: Define the values for the level of significance (α) and the degrees of freedom (df). The degree of freedom equals (n 1 + n 2 – 2) in this case.

Step 3: Calculate the t-value from the formula defined above after obtaining the required data related to each group.

Step 4: Find the critical t-value from a t-distribution table with the corresponding degrees of freedom and level of significance.

Step 5: If the calculated t-value is greater than the critical t-value, then reject the null hypothesis. This indicates that there is a significant difference between the means of the two groups. Otherwise, the null hypothesis is not rejected. And, this suggests that there is no significant difference between the means of the two groups.

Paired Samples T-Test

The Paired samples t-test is used when we want to compare the means of two related groups or samples. For example, we may use this test to compare the average scores of the players of an athletics team before and after a training program. To calculate the t-value in this case, the following formula is used,

t = (x̄ d – μ d ) / (σ d / √n) Where: t is the t-value, x̄ d is the sample mean of the differences between the paired observations, μ d isthe population mean difference, σ d is the sample standard deviation of the differences, n is the number of paired observations.

Steps for Paired Samples T-Test

Following are the steps to perform this type of T-test:

Step 1: State the null hypothesis which assumes that there is no significant difference between the statistical means of the paired observations (μ d = 0) while the alternative hypothesis assumes that there is a significant difference between the statistical means of the paired observations (μ d ≠ 0).

Step 2: Match each observation in one group with a corresponding observation in the other group.

Step 3: Calculate the differences between each paired observation and then, calculate the mean of the differences (x̄ d ), and the sample standard deviation of the differences (σ d ). Furthermore, calculate the t-value from the formula.

Step 4: Obtain the critical t-value from a t-distribution table corresponding to the chosen level of significance (α) and degree of freedom (df). The degree of freedom (df) equals (n – 1) in this case.

Step 5: If the calculated t-value is greater than the critical t-value, then reject the null hypothesis. This indicates a significant difference in the sample before and after the intervention. Otherwise, it can be concluded that there is no significant difference in the sample before and after the intervention.

T-Distribution Table

A T-Distribution table is used to obtain a critical t-value that is used as a reference to the calculated t-value for obtaining further results. Critical t-value depends on values of the level of significance and degrees of freedom. A concise form of the table for critical t-values is as follows for your reference:

Solved Problems of T-Test Formula

Problem 1: Determine whether the average weight of a sample of 20 mangoes is significantly different from the population’s average weight of 70 grams. The sample mean weight is 70.55 grams, and the sample standard deviation is 2.82 grams. Use one sample T-test.

To perform a T-test, first of all, we define two hypotheses: Null hypothesis: The sample mean weight of mangoes is equal to the known population mean. (i.e., 70 grams). Alternative hypothesis: The sample mean weight of mangoes is not equal to the known mean value. Then, determine the degrees of freedom (d f ): d f = n – 1 = 20 – 1 = 19 and define the level of significance(α) as 0.05 for this case. Next, determine the t-value from the formula, t = (70.55 – 70) / (2.82 / √20) ⇒ t ≈ 1.041 From the t-distribution table, we find 1.041 < 2.093. (i.e. p-value for α = 0.05). So, the null hypothesis is true. Thus, we conclude that the sample does not vary significantly from the population.

Problem 2: Determine if there is a significant difference in the average scores between the two teams. The following data is given:  Team A: Score: 65, 68, 70, 63, 67 Team B: Score: 62, 66, 69, 64, 68

According to the question, we come to know that we need to perform an Independent Samples T-test. Set up the null hypothesis and alternative hypothesis: Null hypothesis: The means of the two groups are equal (μ A = μ B ). Alternative hypothesis: The means of the two groups are not equal (μ A ≠ μ B ). Next, we calculate the sample means (x̄ A and x̄ B ) and sample standard deviations (σ A and σ B ): Team A: Sample size (n A ) = 5 Sample mean (x̄ A ) = (65 + 68 + 70 + 63 + 67) / 5 = 66.6 Sample standard deviation (σ A ) ≈ 2.607 Team B: Sample size (n B ) = 5 Sample mean (x̄ B ) = (62 + 66 + 69 + 64 + 68) / 5 = 65.8 Sample standard deviation (σ B ) ≈ 2.588 Now, we calculate the t-value using the formula: t = (x̄ A – x̄ B ) / √((σ A 2 / n A ) + (σ B 2 / n B )) ⇒ t = (66.6 – 65.8) / √{(2.607 2 /5) + (2.588 2 /5)} ⇒ t ≈ 0.296 Then, determine the degrees of freedom (df):  df = n A + n B – 2 = 5 + 5 – 2 = 8  and set the level of significance as 0.05.  From the table, we get the critical t-value as 2.306. As the calculated t-value is less than the critical t-value, we conclude that the null hypothesis is not rejected, which suggests that there is no significant difference between the average scores of the two teams.

Problem 3: You need to assess the effectiveness of a new teaching scheme by comparing the test scores of the same group of students before and after the implementation of the scheme. The following data is given: Before scores: 76, 88, 65, 56, 76 After scores: 85, 95, 75, 60, 81 Determine if there is a significant difference in the average test scores before and after the implementation of the scheme.

Here, we need to perform a Paired Samples T-test, as we need to compare data of the same sample. Set up the null hypothesis and alternative hypothesis: Null hypothesis: The population mean difference between the before and after scores are zero (μ d = 0). Alternative hypothesis: The population mean difference between the before and after scores is not zero (μ d ≠ 0). Next, calculate the differences between the paired observations: Difference (d) = After score – Before score d 1 = 85 – 76 = 9 d 2 = 95 – 88 = 7 d 3 = 75 – 65 = 10 d 4 = 60 – 56 = 4 d 5 = 81 – 76 = 5 Now, calculate the sample mean (x̄d) and sample standard deviation (σ d ) of the differences: Sample size (n) = 5 Sample mean (x̄ d ) = (d 1 + d 2 + d 3 + d 4 + d 5 ) / 5 = (9 + 7 + 10 + 4 + 5) / 5 = 7 Sample standard deviation (σ d ) ≈ 2.828 Then, calculate the t-value using the formula: t = (x̄ d – μ d ) / (σ d / √n)  ⇒ t = (7 – 0) / (2.828 / √5) ⇒ t ≈ 5.535 Next, calculate the value of degrees of freedom (df):  df = n – 1 = 5 – 1 = 4.  And, define the level of significance(α) as 0.05.  Now, from the t-distribution table, we find that the critical t-value is 2.776. As the calculated t-value is greater than the critical t-value ( 5.535 > 2.776), thus, the null hypothesis is rejected. And we conclude that there is a significant difference in the average test scores before and after the implementation of the scheme.

T-Test in Statistics – FAQs

What is a t-test in statistics.

T-Test is the test in statistics to derive some conclusions for a population which is based upon some sample data using values of means and variances. 

When is a T-Test used?

The test is basically used to determine whether there is any significant difference in the statistical means of two samples of the data considered. The purpose to determine this can be to check if a sample data set belongs to the population data set, or if there is an effect of any variation on the data values before or after any specific treatment/intervention.

What are the Different Types of T-Tests?

There are three types of T-tests that are used as per the situation, listed as follows: One-sample T-test: It is used when we need to compare the mean of a single sample to a known (or assumed) population mean value. Independent T-test: It is used when we need to compare the means of two independent groups. Paired T-test: It is used to compare the means of two related or paired groups.

What does the T-Value obtain from the T-Test Formula Indicate?

The t-value indicates the magnitude of the difference between the means relative to the variability within the groups. A larger t-value suggests a greater difference between the means.

Are there any Assumptions related to Sample Data in Performing a T-Test on it?

The t-test assumes that the data within each group are normally distributed, the variances of the two groups are equal (in the case of an independent t-test), the observations are independent, and the data points represent their respective populations.

What are the Limitations of the T-Test?

The t-test assumes that the data meet the assumptions of normality, independence, and equal variances (in the case of an independent t-test). If these assumptions are not true, it can lead to inaccurate or misleading results. Also, the test is sensitive to outliers, and may not give accurate results for small sample sizes.

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Z-test vs T-test: the differences and when to use each

What is hypothesis testing, what is a z-test, examples of a z-test, what is a t-test, examples of a t-test, how to know when to use z-test vs t-test, difference between z-test and t-test: a comparative table.

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Testing is how you determine effectiveness. Whether you work as a data scientist , statistician, or software developer, to ensure quality, you must measure performance. Without tests, you could deploy flawed code, features, or data points.

With that in mind, the use cases of testing are endless. Machine learning models need statistical tests. Data analysis involves statistical tests to validate assumptions. Optimization of any kind requires evaluation. You even need to test the strength of your hypothesis before you begin an inquiry.

Let's explore two inferential statistics: the Z-test vs the T-test. That way you can understand their differences, their unique purposes, and when to use a Z-test vs T-test.

To start, imagine you have a good idea. At the moment of inception, you have no data to back up your idea. It is an unformed thought. But the idea is an excellent starting point that can launch a full investigation. We consider this starting point a hypothesis.

But what if your hypothesis is off-base? You don’t want to dive into a full-scale search if it is a pointless chase with no reward. That is a waste of resources. You need to determine if you have a workable hypothesis.

Enter hypothesis testing. It is a statistical act used to assess the viability of a hypothesis. The method discovers whether there is sufficient data to support your idea. If there is next to no significance, you do not have a very plausible hypothesis.

To confirm the validity of a hypothesis, you compare it against the status quo (also known as the null hypothesis). Your idea is something new, opposite from normal conditions (also known as the alternative hypothesis). It is zero sum: only one hypothesis between the null and alternate hypothesis can be true in a given set of parameters.

In such a comparison test, you can now determine validity. You can compare and contrast conditions to find meaningful conclusions. Whichever conditions become statistically apparent determines which hypothesis is plausible.

A Z-test is a test statistic. It works with two groups: the sample mean and the population mean. It will test whether these two groups are different.

With a Z-test, you know the population standard deviation. That is to ensure statistical accuracy as you compare one group (the sample mean) vs the second group (the population mean). In other words, you can minimize external confounding factors with a normal distribution. In addition, a defining characteristic of a Z-test is that it works with large sample sizes (typically more than 30, so we achieve normal distribution as defined by the central limit theorem). These are two crucial criteria for using a Z-test.

Within hypothesis testing, your null hypothesis states there is no difference between the two groups your Z-test will compare. Your alternative hypothesis will state there is a difference that your Z-test will expose.

How to perform a Z-test

A Z-test occurs in the following standard format:

• Formulate your hypothesis: First, define the parameters of your alternative and null hypothesis.
• Choose a critical value: Second, determine what you consider a viable difference between your two groups. This threshold determines when you can say the null hypothesis should be rejected. Common levels are 0.01 (1%) or 0.05 (5%) , values found to best balance Type I and Type II errors .
• Collect samples: Obtain the needed data. The data must be large enough and random.
• Calculate for a Z-score: Input your data into the standard Z-test statistics formula, shown below, where Z = standard score, x = observed value, mu = mean of the sample, sigma = standard deviation of the sample .
• Compare: If the statistical test is greater than the critical value, you have achieved statistical significance. The sample mean is so different so you can reject the null hypothesis. Your alternative hypothesis (something other than the status quo) is at work, and that's worth investigating.

There are different variations of a Z-test. Let's explore examples of one-sample and two-sample Z-tests.

One-sample Z-test

A one-sample Z-test looks for either an increase or a decrease. There is one sample group involved, taken from a population. We want to see if there is a difference between those two means.

For example, consider a school principal who believes their students' IQ is higher than the state average. The state average is 100 points (population mean), give or take 20 (the population standard deviation). To prove this hypothesis, the principal takes 50 students (the sample size) and finds their IQ scores. To their delight, they earn an average of 110.

But does the difference offer any statistical value? The principal then plugs the numbers into a Z-test. Any Z-score greater than the critical value would state there is sufficient significance. The claim that the students have an above-average IQ is valid.

Two-sample Z-test

A two-sample test involves comparing the average of two sample groups against the population means. It is to determine a difference between two independent samples.

For example, our principal wants to compare their students' IQ scores to the school across the street. They believe their students' average IQ is higher. They don’t need to know the exact numerical increase or decrease. All they want is proof that their student's average scores are higher than the other group.

To confirm the validity of this hypothesis, the principal will search for statistical significance. They can take a 50-student sample size from their school and a 50-student sample size from the rival school. Now in possession of both sample group's average IQ (and the sample standard deviation), they hope to find a number value that is not equal. And they need them to be unequal by a significant amount.

If the test statistic comes in less than the critical value, the differences are negligible. There is not enough evidence to say the hypothesis is worth exploring, the null hypothesis is maintained. He would not have enough proof that the IQ levels between the two schools are different.

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A T-test performs the same crucial function as a Z-test: determine if there is a difference between the means of two groups. If there is a significant difference, you have achieved statistical validity for your hypothesis.

However, a T-test involves a different set of factors. Most importantly, a T-test applies when you do not know the sample variance of your values. You must generalize the normal distribution (or T-distribution). Plus, there is an expectation that you do not possess all the data in a given scenario.

These conditions better match reality, as it is often hard to collect data from entire populations or always obtain a standard normal distribution. That is why T-tests are more widely applicable than Z-tests, though they operate with less precision.

How to perform a T-test

A T-test occurs in the following standard format:

• Formulate your hypothesis: First, define the parameters of your null and alternative hypothesis.
• Choose a critical value: Like a Z-test, determine what you consider a viable difference between your two groups.
• Collect data: Obtain the needed data. One of the key differences is degrees of freedom in the samples of a T-test, so try to define the typical values and range of values in each group.
• Calculate your T-score: Input your data into the T-test formula you chose. Here is a one-sample formula:
• Compare: If the statistical test is greater than the critical value, you have achieved statistical significance. The sample mean is so far from the population mean that you likely have a useful hypothesis.

There are several different kinds of T-tests as well. Let's go through the standard one-sample and two-sample T-tests.

One-sample T-test

A one-sample T-test looks for an increase or decrease compared to a population mean.

For example, your company just went through sales training. Now, the manager wants to know if the training helped improve sales.

Previous company sales data shows an average of $100 on each transaction from all workers. The null hypothesis would be no change. The alternative hypothesis (which you hope is significant), is that there is an improvement.

To test if there is significance, you take the sales average of 20 salesmen. That is the only available data, and you have no other data from nationwide stores. The average of that sample of salesmen in the past month is $130. We will also assume that the standard deviation is approximately equal .

With this set of factors, you can calculate your T-score with a T-test. You compare the sample result to the critical value. In addition, you assess it against the number of degrees of freedom. Since we know with smaller sample data sizes there is greater uncertainty, we allow more room for our data to vary.

After comparing, we may find a lot of significance. That means the data possesses enough strength to support our hypothesis that sales training likely impacted sales. Of course, this is an estimate, as we only assessed one factor with a small group. Sales could have risen for numerous other reasons. But with our set of assumptions, our hypothesis is valid.

Two-sample T-test

A two-sample T-test occurs the same as a two-sample Z-test and compares if two groups are equal when compared to a defined population parameter.

For example, consider English and non-native speakers. We want to see the effect of maternal language on test scores inside a country. To do that, we will offer both groups a reading test and compare those scores to the average.

Of course, finding the mean of an entire population of language speakers is impossible to procure. Still, we can make some assumptions and compare them with a smaller size. We take 15 English speakers and 15 non-native speakers and collect their results. We can decide on a critical score value on the reading test as well. If the average score on the test is not crucially different or outside the population standard deviation, our assumption failed. There is no significant difference between the groups, so the impact of maternal language is not worth investigating.

Both a Z-test and a T-test validate a hypothesis. Both are parametric tests that rely on assumptions. The key difference between Z-test and T-test is in their assumptions (e.g. population variance).

Key differences about the data used result in different applications. You want to use the appropriate tool, otherwise you won’t draw valid conclusions from your data.

So when should you use a Z-test vs a T-test? Here are some factors to consider:

• Sample size: If the available sample size is small, opt for a T-test. Small sample sizes are more variable, so the greater spread of distribution and estimation of error involved with T-tests is ideal.
• Knowledge of the population standard deviation: Z-tests are more precise and often simpler to execute. So if you know the standard deviation, use a Z-test.
• Test purpose: If you are assessing the validity of a mean, a T-test is the best choice. If you are working with a hypothesized population proportion, go for a Z-test.
• Assumption of normality: A Z-test assumes a normal distribution. This does not apply to all real-world scenarios. If you hope to validate a hypothesis that is not well-defined, opt for a T-test instead.
• Type of data: You can only work within the constraints of the available data. The more information the better, but that is often not possible given testing and collecting conditions. If you have limited data describing means between groups, opt for a T-test. If you have large data sets comparing means between populations, you can use a Z-test.

Knowing the key differences with each statistical test makes selecting the right tool far easier. Here is a table that can help you compare:

Statistical testing lets you determine the validity of a hypothesis. You discover validity by determining if there is a significant difference between your hypothesis and the status quo. If there is, you have a possible idea worth exploring.

That process has numerous applications in the field of computer science and data analysis . You might want to determine the performance of an app with an A/B test. Or you might need to test if an application fits within the defined limits and compare performance metrics. Z-tests and T-tests can depict whether there is significant evidence in each of these scenarios. With that information, you can take the appropriate measures to fix bugs or optimize processes.

Z-test and T-test are helpful tools, especially for hypothesis testing. For data engineers of the future, knowledge of statistical testing will only help your work and overall career trajectory.

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Computer Science > Cryptography and Security

Title: detecting compromised iot devices using autoencoders with sequential hypothesis testing.

Abstract: IoT devices fundamentally lack built-in security mechanisms to protect themselves from security attacks. Existing works on improving IoT security mostly focus on detecting anomalous behaviors of IoT devices. However, these existing anomaly detection schemes may trigger an overwhelmingly large number of false alerts, rendering them unusable in detecting compromised IoT devices. In this paper we develop an effective and efficient framework, named CUMAD, to detect compromised IoT devices. Instead of directly relying on individual anomalous events, CUMAD aims to accumulate sufficient evidence in detecting compromised IoT devices, by integrating an autoencoder-based anomaly detection subsystem with a sequential probability ratio test (SPRT)-based sequential hypothesis testing subsystem. CUMAD can effectively reduce the number of false alerts in detecting compromised IoT devices, and moreover, it can detect compromised IoT devices quickly. Our evaluation studies based on the public-domain N-BaIoT dataset show that CUMAD can on average reduce the false positive rate from about 3.57% using only the autoencoder-based anomaly detection scheme to about 0.5%; in addition, CUMAD can detect compromised IoT devices quickly, with less than 5 observations on average.

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