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Statistics By Jim

Making statistics intuitive

How t-Tests Work: t-Values, t-Distributions, and Probabilities

By Jim Frost 12 Comments

T-tests are statistical hypothesis tests that you use to analyze one or two sample means. Depending on the t-test that you use, you can compare a sample mean to a hypothesized value, the means of two independent samples, or the difference between paired samples. In this post, I show you how t-tests use t-values and t-distributions to calculate probabilities and test hypotheses.

As usual, I’ll provide clear explanations of t-values and t-distributions using concepts and graphs rather than formulas! If you need a primer on the basics, read my hypothesis testing overview .

What Are t-Values?

The term “t-test” refers to the fact that these hypothesis tests use t-values to evaluate your sample data. T-values are a type of test statistic. Hypothesis tests use the test statistic that is calculated from your sample to compare your sample to the null hypothesis. If the test statistic is extreme enough, this indicates that your data are so incompatible with the null hypothesis that you can reject the null. Learn more about Test Statistics .

Don’t worry. I find these technical definitions of statistical terms are easier to explain with graphs, and we’ll get to that!

When you analyze your data with any t-test, the procedure reduces your entire sample to a single value, the t-value. These calculations factor in your sample size and the variation in your data. Then, the t-test compares your sample means(s) to the null hypothesis condition in the following manner:

• If the sample data equals the null hypothesis precisely, the t-test produces a t-value of 0.
• As the sample data become progressively dissimilar from the null hypothesis, the absolute value of the t-value increases.

Read the companion post where I explain how t-tests calculate t-values .

The tricky thing about t-values is that they are a unitless statistic, which makes them difficult to interpret on their own. Imagine that we performed a t-test, and it produced a t-value of 2. What does this t-value mean exactly? We know that the sample mean doesn’t equal the null hypothesis value because this t-value doesn’t equal zero. However, we don’t know how exceptional our value is if the null hypothesis is correct.

To be able to interpret individual t-values, we have to place them in a larger context. T-distributions provide this broader context so we can determine the unusualness of an individual t-value.

What Are t-Distributions?

A single t-test produces a single t-value. Now, imagine the following process. First, let’s assume that the null hypothesis is true for the population. Now, suppose we repeat our study many times by drawing many random samples of the same size from this population. Next, we perform t-tests on all of the samples and plot the distribution of the t-values. This distribution is known as a sampling distribution, which is a type of probability distribution.

Related posts : Sampling Distributions and Understanding Probability Distributions

If we follow this procedure, we produce a graph that displays the distribution of t-values that we obtain from a population where the null hypothesis is true. We use sampling distributions to calculate probabilities for how unusual our sample statistic is if the null hypothesis is true.

Luckily, we don’t need to go through the hassle of collecting numerous random samples to create this graph! Statisticians understand the properties of t-distributions so we can estimate the sampling distribution using the t-distribution and our sample size.

The degrees of freedom (DF) for the statistical design define the t-distribution for a particular study. The DF are closely related to the sample size. For t-tests, there is a different t-distribution for each sample size.

Related posts : Degrees of Freedom in Statistics and T Distribution: Definition and Uses .

Use the t-Distribution to Compare Your Sample Results to the Null Hypothesis

T-distributions assume that the null hypothesis is correct for the population from which you draw your random samples. To evaluate how compatible your sample data are with the null hypothesis, place your study’s t-value in the t-distribution and determine how unusual it is.

The sampling distribution below displays a t-distribution with 20 degrees of freedom, which equates to a sample size of 21 for a 1-sample t-test. The t-distribution centers on zero because it assumes that the null hypothesis is true. When the null is true, your study is most likely to obtain a t-value near zero and less liable to produce t-values further from zero in either direction.

On the graph, I’ve displayed the t-value of 2 from our hypothetical study to see how our sample data compares to the null hypothesis. Under the assumption that the null is true, the t-distribution indicates that our t-value is not the most likely value. However, there still appears to be a realistic chance of observing t-values from -2 to +2.

We know that our t-value of 2 is rare when the null hypothesis is true. How rare is it exactly? Our final goal is to evaluate whether our sample t-value is so rare that it justifies rejecting the null hypothesis for the entire population based on our sample data. To proceed, we need to quantify the probability of observing our t-value.

Related post : What are Critical Values?

t-Tests Use t-Values and t-Distributions to Calculate Probabilities

Hypothesis tests work by taking the observed test statistic from a sample and using the sampling distribution to calculate the probability of obtaining that test statistic if the null hypothesis is correct. In the context of how t-tests work, you assess the likelihood of a t-value using the t-distribution. If a t-value is sufficiently improbable when the null hypothesis is true, you can reject the null hypothesis.

I have two crucial points to explain before we calculate the probability linked to our t-value of 2.

Because I’m showing the results of a two-tailed test, we’ll use the t-values of +2 and -2. Two-tailed tests allow you to assess whether the sample mean is greater than or less than the target value in a 1-sample t-test. A one-tailed hypothesis test can only determine statistical significance for one or the other.

Additionally, it is possible to calculate a probability only for a range of t-values. On a probability distribution plot, probabilities are represented by the shaded area under a distribution curve. Without a range of values, there is no area under the curve and, hence, no probability.

Related posts : One-Tailed and Two-Tailed Tests Explained and T-Distribution Table of Critical Values

t-Test Results for Our Hypothetical Study

Considering these points, the graph below finds the probability associated with t-values less than -2 and greater than +2 using the area under the curve. This graph is specific to our t-test design (1-sample t-test with N = 21).

The probability distribution plot indicates that each of the two shaded regions has a probability of 0.02963—for a total of 0.05926. This graph shows that t-values fall within these areas almost 6% of the time when the null hypothesis is true.

There is a chance that you’ve heard of this type of probability before—it’s the P value! While the likelihood of t-values falling within these regions seems small, it’s not quite unlikely enough to justify rejecting the null under the standard significance level of 0.05.

Learn how to interpret the P value correctly and avoid a common mistake!

Related posts : How to Find the P value: Process and Calculations and Types of Errors in Hypothesis Testing

t-Distributions and Sample Size

The sample size for a t-test determines the degrees of freedom (DF) for that test, which specifies the t-distribution. The overall effect is that as the sample size decreases, the tails of the t-distribution become thicker. Thicker tails indicate that t-values are more likely to be far from zero even when the null hypothesis is correct. The changing shapes are how t-distributions factor in the greater uncertainty when you have a smaller sample.

You can see this effect in the probability distribution plot below that displays t-distributions for 5 and 30 DF.

Sample means from smaller samples tend to be less precise. In other words, with a smaller sample, it’s less surprising to have an extreme t-value, which affects the probabilities and p-values. A t-value of 2 has a P value of 10.2% and 5.4% for 5 and 30 DF, respectively. Use larger samples!

Click here for step-by-step instructions for how to do t-tests in Excel !

If you like this approach and want to learn about other hypothesis tests, read my posts about:

• How the F-test Works in ANOVA .
• How Chi-Squared Tests of Independence Work

To see an alternative to traditional hypothesis testing that does not use probability distributions and test statistics, learn about bootstrapping in statistics !

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Reader Interactions

May 25, 2021 at 10:42 pm

what statistical tools, is recommended for measuring the level of satisfaction

May 26, 2021 at 3:55 pm

Hi McKienze,

The correct analysis depends on the nature of the data you have and what you want to learn. You don’t provide enough information to be able to answer the question. However, read my hypothesis testing overview to learn about the options.

August 23, 2020 at 1:33 am

Hi Jim, I want to ask about standardizing data before the t test.. For example I have USD prices of a big Mac across the world and this varies by quite a bit. Doing the t-test here would be misleading since some countries would have a higher mean… Should the approach be standardizing all the usd values? Or perhaps even local values?

August 24, 2020 at 12:37 am

Yes, that makes complete sense. I don’t know what method is best. If you can find a common scale to use for all prices, I’d do that. You’re basically using a data transformation before analysis, which is totally acceptable when you have a good reason.

April 3, 2020 at 4:47 am

Hey Jim. Your blog is one of the only few ones where everything is explained in a simple and well structured manner, in a way that both an absolute beginner and a geek can equally benefit from your writing. Both this article as well as your article on one tailed and two tailed hypothesis tests have been super helpful. Thank you for this post

March 6, 2020 at 11:04 am

Thank you, Jim, for sharing your knowledge with us.

I have a 2 part question. I am testing the difference in walking distance within a busy environment compared with a simple environment. I am also testing walking time within the 2 environments. I am using the same individuals for both scenarios. I was planning to do a paired ttest for distance difference between busy and simple environments and a 2nd paired ttest for time difference between the environments.

My question(s) for you is: 1. Do you feel that a paired ttest is the best choice for these? 2. Do you feel that, because there are 2 tests, I should do a bonferroni correction or do you believe that because the data is completely different (distance as opposed to time), it is okay not to do a multiple comparison test?

August 13, 2019 at 12:43 pm

thank you very eye opening on the use of two or one tailed test

April 19, 2019 at 3:49 pm

Hi Mr. Frost,

Thanks for the breakdown. I have a question … if I wanted to run a test to show that the medical professionals could use more training with data set consisting of questions which in your opinion would be my best route?

January 14, 2019 at 2:22 pm

Hello Jim, I find this statement in this excellent write up contradicting : 1)This graph shows that t-values fall within these areas almost 6% of the time when the null hypothesis is true I mean if this is true the t-value =0 hypothesis is rejected.

January 14, 2019 at 2:51 pm

I can see how that statement sounds contradictory, but I can assure that it is quite accurate. It’s often forgotten but the underlying assumption for the calculations surrounding hypothesis testing, significance levels, and p-values is that the null hypothesis is true.

So, the probabilities shown in the graph that you refer to are based on the assumption that the null hypothesis is true. Further, t-values for this study design have a 6% chance of falling in those critical areas assuming the null is true (a false positive).

Significance levels are defined as the maximum acceptable probability of a false positive. Usually, we set that as 5%. In the example, there’s a large probability of a false positive (6%), so we fail to reject the null hypothesis. In other words, we fail to reject the null because false positives will happen too frequently–where the significance level defines the cutoff point for too frequently.

Keep in mind that when you have statistically significant results, you’re really saying that the results you obtained are improbable enough assuming that the null is true that you can reject the notion that the null is true. But, the math and probabilities are all based on the assumption that the null is true because you need to determine how unlikely your results are under the null hypothesis.

Even the p-value is defined in terms of assuming the null hypothesis is true. You can read about that in my post about interpreting p-values correctly .

I hope this clarifies things!

November 9, 2018 at 2:36 am

Jim …I was involved in in a free SAT/ACT tutoring program that I need to analyze for effectiveness .

I have pre test scores of a number of students and the post test scores after they were tutored (treatment ).

Glenn dowell

November 9, 2018 at 9:05 am

It sounds like you need to perform a paired t-test assuming.

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• Knowledge Base

An Introduction to t Tests | Definitions, Formula and Examples

Published on January 31, 2020 by Rebecca Bevans . Revised on June 22, 2023.

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 actually has an effect on the population of interest, or whether two groups are different from one another.

• The null hypothesis ( H 0 ) is that the true difference between these group means is zero.
• The alternate hypothesis ( H a ) is that the true difference is different from zero.

Table of contents

When to use a t test, what type of t test should i use, performing a t test, interpreting test results, presenting the results of a t test, other interesting articles, frequently asked questions about t tests.

A t test can only be used when comparing the means of two groups (a.k.a. pairwise comparison). If you want to compare more than two groups, or if you want to do multiple pairwise comparisons, use an   ANOVA test  or a post-hoc test.

The t test is a parametric test of difference, meaning that it makes the same assumptions about your data as other parametric tests. The t test assumes your data:

• are independent
• are (approximately) normally distributed
• have a similar amount of variance within each group being compared (a.k.a. homogeneity of variance)

If your data do not fit these assumptions, you can try a nonparametric alternative to the t test, such as the Wilcoxon Signed-Rank test for data with unequal variances .

Prevent plagiarism. Run a free check.

When choosing a t test, you will need to consider two things: whether the groups being compared come from a single population or two different populations, and whether you want to test the difference in a specific direction.

One-sample, two-sample, or paired t test?

• If the groups come from a single population (e.g., measuring before and after an experimental treatment), perform a paired t test . This is a within-subjects design .
• If the groups come from two different populations (e.g., two different species, or people from two separate cities), perform a two-sample t test (a.k.a. independent t test ). This is a between-subjects design .
• If there is one group being compared against a standard value (e.g., comparing the acidity of a liquid to a neutral pH of 7), perform a one-sample t test .

One-tailed or two-tailed t test?

• If you only care whether the two populations are different from one another, perform a two-tailed t test .
• If you want to know whether one population mean is greater than or less than the other, perform a one-tailed t test.
• Your observations come from two separate populations (separate species), so you perform a two-sample t test.
• You don’t care about the direction of the difference, only whether there is a difference, so you choose to use a two-tailed t test.

The t test estimates the true difference between two group means using the ratio of the difference in group means over the pooled standard error of both groups. You can calculate it manually using a formula, or use statistical analysis software.

T test formula

The formula for the two-sample t test (a.k.a. the Student’s t-test) is shown below.

In this formula, t is the t value, x 1 and x 2 are the means of the two groups being compared, s 2 is the pooled standard error of the two groups, and n 1 and n 2 are the number of observations in each of the groups.

A larger t value shows that the difference between group means is greater than the pooled standard error, indicating a more significant difference between the groups.

You can compare your calculated t value against the values in a critical value chart (e.g., Student’s t table) to determine whether your t value is greater than what would be expected by chance. If so, you can reject the null hypothesis and conclude that the two groups are in fact different.

T test function in statistical software

Most statistical software (R, SPSS, etc.) includes a t test function. This built-in function will take your raw data and calculate the t value. It will then compare it to the critical value, and calculate a p -value . This way you can quickly see whether your groups are statistically different.

In your comparison of flower petal lengths, you decide to perform your t test using R. The code looks like this:

Download the data set to practice by yourself.

Sample data set

If you perform the t test for your flower hypothesis in R, you will receive the following output:

The output provides:

• An explanation of what is being compared, called data in the output table.
• The t value : -33.719. Note that it’s negative; this is fine! In most cases, we only care about the absolute value of the difference, or the distance from 0. It doesn’t matter which direction.
• The degrees of freedom : 30.196. Degrees of freedom is related to your sample size, and shows how many ‘free’ data points are available in your test for making comparisons. The greater the degrees of freedom, the better your statistical test will work.
• The p value : 2.2e-16 (i.e. 2.2 with 15 zeros in front). This describes the probability that you would see a t value as large as this one by chance.
• A statement of the alternative hypothesis ( H a ). In this test, the H a is that the difference is not 0.
• The 95% confidence interval . This is the range of numbers within which the true difference in means will be 95% of the time. This can be changed from 95% if you want a larger or smaller interval, but 95% is very commonly used.
• The mean petal length for each group.

When reporting your t test results, the most important values to include are the t value , the p value , and the degrees of freedom for the test. These will communicate to your audience whether the difference between the two groups is statistically significant (a.k.a. that it is unlikely to have happened by chance).

You can also include the summary statistics for the groups being compared, namely the mean and standard deviation . In R, the code for calculating the mean and the standard deviation from the data looks like this:

flower.data %>% group_by(Species) %>% summarize(mean_length = mean(Petal.Length), sd_length = sd(Petal.Length))

In our example, you would report the results like this:

If you want to know more about statistics , methodology , or research bias , make sure to check out some of our other articles with explanations and examples.

• Chi square test of independence
• Statistical power
• Descriptive statistics
• Degrees of freedom
• Pearson correlation
• Null hypothesis

Methodology

• Double-blind study
• Case-control study
• Research ethics
• Data collection
• Hypothesis testing
• Structured interviews

Research bias

• Hawthorne effect
• Unconscious bias
• Recall bias
• Halo effect
• Self-serving bias
• Information bias

A t-test is a statistical test that compares the means of two samples . It is used in hypothesis testing , with a null hypothesis that the difference in group means is zero and an alternate hypothesis that the difference in group means is different from zero.

A t-test measures the difference in group means divided by the pooled standard error of the two group means.

In this way, it calculates a number (the t-value) illustrating the magnitude of the difference between the two group means being compared, and estimates the likelihood that this difference exists purely by chance (p-value).

Your choice of t-test depends on whether you are studying one group or two groups, and whether you care about the direction of the difference in group means.

If you are studying one group, use a paired t-test to compare the group mean over time or after an intervention, or use a one-sample t-test to compare the group mean to a standard value. If you are studying two groups, use a two-sample t-test .

If you want to know only whether a difference exists, use a two-tailed test . If you want to know if one group mean is greater or less than the other, use a left-tailed or right-tailed one-tailed test .

A one-sample t-test is used to compare a single population to a standard value (for example, to determine whether the average lifespan of a specific town is different from the country average).

A paired t-test is used to compare a single population before and after some experimental intervention or at two different points in time (for example, measuring student performance on a test before and after being taught the material).

A t-test should not be used to measure differences among more than two groups, because the error structure for a t-test will underestimate the actual error when many groups are being compared.

If you want to compare the means of several groups at once, it’s best to use another statistical test such as ANOVA or a post-hoc test.

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Bevans, R. (2023, June 22). An Introduction to t Tests | Definitions, Formula and Examples. Scribbr. Retrieved June 9, 2024, from https://www.scribbr.com/statistics/t-test/

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Understanding t-Tests: t-values and t-distributions

Topics: Hypothesis Testing , Data Analysis

T-tests are handy hypothesis tests in statistics when you want to compare means. You can compare a sample mean to a hypothesized or target value using a one-sample t-test. You can compare the means of two groups with a two-sample t-test. If you have two groups with paired observations (e.g., before and after measurements), use the paired t-test.

How do t-tests work? How do t-values fit in? In this series of posts, I’ll answer these questions by focusing on concepts and graphs rather than equations and numbers. After all, a key reason to use statistical software like Minitab  is so you don’t get bogged down in the calculations and can instead focus on understanding your results.

In this post, I will explain t-values, t-distributions, and how t-tests use them to calculate probabilities and assess hypotheses.

What Are t-Values?

T-tests are called t-tests because the test results are all based on t-values. T-values are an example of what statisticians call test statistics. A test statistic is a standardized value that is calculated from sample data during a hypothesis test. The procedure that calculates the test statistic compares your data to what is expected under the null hypothesis .

Each type of t-test uses a specific procedure to boil all of your sample data down to one value, the t-value. The calculations behind t-values compare your sample mean(s) to the null hypothesis and incorporates both the sample size and the variability in the data. A t-value of 0 indicates that the sample results exactly equal the null hypothesis. As the difference between the sample data and the null hypothesis increases, the absolute value of the t-value increases.

Assume that we perform a t-test and it calculates a t-value of 2 for our sample data. What does that even mean? I might as well have told you that our data equal 2 fizbins! We don’t know if that’s common or rare when the null hypothesis is true.

By itself, a t-value of 2 doesn’t really tell us anything. T-values are not in the units of the original data, or anything else we’d be familiar with. We need a larger context in which we can place individual t-values before we can interpret them. This is where t-distributions come in.

What Are t-Distributions?

When you perform a t-test for a single study, you obtain a single t-value. However, if we drew multiple random samples of the same size from the same population and performed the same t-test, we would obtain many t-values and we could plot a distribution of all of them. This type of distribution is known as a sampling distribution .

Fortunately, the properties of t-distributions are well understood in statistics, so we can plot them without having to collect many samples! A specific t-distribution is defined by its degrees of freedom (DF) , a value closely related to sample size. Therefore, different t-distributions exist for every sample size.  You can graph t-distributions u sing Minitab’s probability distribution plots .

T-distributions assume that you draw repeated random samples from a population where the null hypothesis is true. You place the t-value from your study in the t-distribution to determine how consistent your results are with the null hypothesis.

The graph above shows a t-distribution that has 20 degrees of freedom, which corresponds to a sample size of 21 in a one-sample t-test. It is a symmetric, bell-shaped distribution that is similar to the normal distribution, but with thicker tails. This graph plots the probability density function (PDF), which describes the likelihood of each t-value.

The peak of the graph is right at zero, which indicates that obtaining a sample value close to the null hypothesis is the most likely. That makes sense because t-distributions assume that the null hypothesis is true. T-values become less likely as you get further away from zero in either direction. In other words, when the null hypothesis is true, you are less likely to obtain a sample that is very different from the null hypothesis.

Our t-value of 2 indicates a positive difference between our sample data and the null hypothesis. The graph shows that there is a reasonable probability of obtaining a t-value from -2 to +2 when the null hypothesis is true. Our t-value of 2 is an unusual value, but we don’t know exactly how unusual. Our ultimate goal is to determine whether our t-value is unusual enough to warrant rejecting the null hypothesis. To do that, we'll need to calculate the probability.

Ready for a demo of Minitab Statistical Software? Just ask! 

Using t-Values and t-Distributions to Calculate Probabilities

The foundation behind any hypothesis test is being able to take the test statistic from a specific sample and place it within the context of a known probability distribution. For t-tests, if you take a t-value and place it in the context of the correct t-distribution, you can calculate the probabilities associated with that t-value.

A probability allows us to determine how common or rare our t-value is under the assumption that the null hypothesis is true. If the probability is low enough, we can conclude that the effect observed in our sample is inconsistent with the null hypothesis. The evidence in the sample data is strong enough to reject the null hypothesis for the entire population.

Before we calculate the probability associated with our t-value of 2, there are two important details to address.

First, we’ll actually use the t-values of +2 and -2 because we’ll perform a two-tailed test. A two-tailed test is one that can test for differences in both directions. For example, a two-tailed 2-sample t-test can determine whether the difference between group 1 and group 2 is statistically significant in either the positive or negative direction. A one-tailed test can only assess one of those directions.

Second, we can only calculate a non-zero probability for a range of t-values. As you’ll see in the graph below, a range of t-values corresponds to a proportion of the total area under the distribution curve, which is the probability. The probability for any specific point value is zero because it does not produce an area under the curve.

With these points in mind, we’ll shade the area of the curve that has t-values greater than 2 and t-values less than -2.

The graph displays the probability for observing a difference from the null hypothesis that is at least as extreme as the difference present in our sample data while assuming that the null hypothesis is actually true. Each of the shaded regions has a probability of 0.02963, which sums to a total probability of 0.05926. When the null hypothesis is true, the t-value falls within these regions nearly 6% of the time.

This probability has a name that you might have heard of—it’s called the p-value!  While the probability of our t-value falling within these regions is fairly low, it’s not low enough to reject the null hypothesis using the common significance level of 0.05.

Learn how to correctly interpret the p-value.

t-Distributions and Sample Size

As mentioned above, t-distributions are defined by the DF, which are closely associated with sample size. As the DF increases, the probability density in the tails decreases and the distribution becomes more tightly clustered around the central value. The graph below depicts t-distributions with 5 and 30 degrees of freedom.

The t-distribution with fewer degrees of freedom has thicker tails. This occurs because the t-distribution is designed to reflect the added uncertainty associated with analyzing small samples. In other words, if you have a small sample, the probability that the sample statistic will be further away from the null hypothesis is greater even when the null hypothesis is true.

Small samples are more likely to be unusual. This affects the probability associated with any given t-value. For 5 and 30 degrees of freedom, a t-value of 2 in a two-tailed test has p-values of 10.2% and 5.4%, respectively. Large samples are better!

I’ve showed how t-values and t-distributions work together to produce probabilities. To see how each type of t-test works and actually calculates the t-values, read the other post in this series, Understanding t-Tests: 1-sample, 2-sample, and Paired t-Tests .

If you'd like to learn how the ANOVA F-test works, read my post, Understanding Analysis of Variance (ANOVA) and the F-test .

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The Ultimate Guide to T Tests

Get all of your t test questions answered here

The ultimate guide to t tests

The t test is one of the simplest statistical techniques that is used to evaluate whether there is a statistical difference between the means from up to two different samples. The t test is especially useful when you have a small number of sample observations (under 30 or so), and you want to make conclusions about the larger population.

The characteristics of the data dictate the appropriate type of t test to run. All t tests are used as standalone analyses for very simple experiments and research questions as well as to perform individual tests within more complicated statistical models such as linear regression. In this guide, we’ll lay out everything you need to know about t tests, including providing a simple workflow to determine what t test is appropriate for your particular data or if you’d be better suited using a different model.

What is a t test?

A t test is a statistical technique used to quantify the difference between the mean (average value) of a variable from up to two samples (datasets). The variable must be numeric. Some examples are height, gross income, and amount of weight lost on a particular diet.

A t test tells you if the difference you observe is “surprising” based on the expected difference. They use t-distributions to evaluate the expected variability. When you have a reasonable-sized sample (over 30 or so observations), the t test can still be used, but other tests that use the normal distribution (the z test) can be used in its place.

Sometimes t tests are called “Student’s” t tests, which is simply a reference to their unusual history.

It got its name because a brewer from the Guinness Brewery, William Gosset , published about the method under the pseudonym "Student". He wanted to get information out of very small sample sizes (often 3-5) because it took so much effort to brew each keg for his samples.

When should I use a t test?

A t test is appropriate to use when you’ve collected a small, random sample from some statistical “population” and want to compare the mean from your sample to another value. The value for comparison could be a fixed value (e.g., 10) or the mean of a second sample.

For example, if your variable of interest is the average height of sixth graders in your region, then you might measure the height of 25 or 30 randomly-selected sixth graders. A t test could be used to answer questions such as, “Is the average height greater than four feet?”

How does a t test work?

Based on your experiment, t tests make enough assumptions about your experiment to calculate an expected variability, and then they use that to determine if the observed data is statistically significant. To do this, t tests rely on an assumed “null hypothesis.” With the above example, the null hypothesis is that the average height is less than or equal to four feet.

Say that we measure the height of 5 randomly selected sixth graders and the average height is five feet. Does that mean that the “true” average height of all sixth graders is greater than four feet or did we randomly happen to measure taller than average students?

To evaluate this, we need a distribution that shows every possible average value resulting from a sample of five individuals in a population where the true mean is four. That may seem impossible to do, which is why there are particular assumptions that need to be made to perform a t test.

With those assumptions, then all that’s needed to determine the “sampling distribution of the mean” is the sample size (5 students in this case) and standard deviation of the data (let’s say it’s 1 foot).

That’s enough to create a graphic of the distribution of the mean, which is:

Notice the vertical line at x = 5, which was our sample mean. We (use software to) calculate the area to the right of the vertical line, which gives us the P value (0.09 in this case). Note that because our research question was asking if the average student is greater than four feet, the distribution is centered at four. Since we’re only interested in knowing if the average is greater than four feet, we use a one-tailed test in this case.

Using the standard confidence level of 0.05 with this example, we don’t have evidence that the true average height of sixth graders is taller than 4 feet.

What are the assumptions for t tests?

• One variable of interest : This is not correlation or regression, where you are interested in the relationship between multiple variables. With a t test, you can have different samples, but they are all measuring the same variable (e.g., height).
• Numeric data: You are dealing with a list of measurements that can be averaged. This means you aren’t just counting occurrences in various categories (e.g., eye color or political affiliation).
• Two groups or less: If you have more than two samples of data, a t test is the wrong technique. You most likely need to try ANOVA.
• Random sample : You need a random sample from your statistical “population of interest” in order to draw valid conclusions about the larger population. If your population is so small that you can measure everything, then you have a “census” and don’t need statistics. This is because you don’t need to estimate the truth, since you have measured the truth without variability.
• Normally Distributed : The smaller your sample size, the more important it is that your data come from a normal, Gaussian distribution bell curve. If you have reason to believe that your data are not normally distributed, consider nonparametric t test alternatives . This isn’t necessary for larger samples (usually 25 or 30 unless the data is heavily skewed). The reason is that the Central Limit Theorem applies in this case, which says that even if the distribution of your data is not normal, the distribution of the mean of your data is, so you can use a z-test rather than a t test.

How do I know which t test to use?

There are many types of t tests to choose from, but you don’t necessarily have to understand every detail behind each option.

You just need to be able to answer a few questions, which will lead you to pick the right t test. To that end, we put together this workflow for you to figure out which test is appropriate for your data.

Do you have one or two samples?

Are you comparing the means of two different samples, or comparing the mean from one sample to a fixed value? An example research question is, “Is the average height of my sample of sixth grade students greater than four feet?”

If you only have one sample of data, you can click here to skip to a one-sample t test example, otherwise your next step is to ask: 

Are observations in the two samples matched up or related in some way?

This could be as before-and-after measurements of the same exact subjects, or perhaps your study split up “pairs” of subjects (who are technically different but share certain characteristics of interest) into the two samples. The same variable is measured in both cases.

If so, you are looking at some kind of paired samples t test . The linked section will help you dial in exactly which one in that family is best for you, either difference (most common) or ratio.

If you aren’t sure paired is right, ask yourself another question:

Are you comparing different observations in each of the two samples?

If the answer is yes, then you have an unpaired or independent samples t test. The two samples should measure the same variable (e.g., height), but are samples from two distinct groups (e.g., team A and team B). 

The goal is to compare the means to see if the groups are significantly different. For example, “Is the average height of team A greater than team B?” Unlike paired, the only relationship between the groups in this case is that we measured the same variable for both. There are two versions of unpaired samples t tests (pooled and unpooled) depending on whether you assume the same variance for each sample.

Have you run the same experiment multiple times on the same subject/observational unit?

If so, then you have a nested t test (unless you have more than two sample groups). This is a trickier concept to understand. One example is if you are measuring how well Fertilizer A works against Fertilizer B. Let’s say you have 12 pots to grow plants in (6 pots for each fertilizer), and you grow 3 plants in each pot.

In this case you have 6 observational units for each fertilizer, with 3 subsamples from each pot. You would want to analyze this with a nested t test . The “nested” factor in this case is the pots. It’s important to note that we aren’t interested in estimating the variability within each pot, we just want to take it into account.

You might be tempted to run an unpaired samples t test here, but that assumes you have 6*3 = 18 replicates for each fertilizer. However, the three replicates within each pot are related, and an unpaired samples t test wouldn’t take that into account.

What if none of these sound like my experiment?

If you’re not seeing your research question above, note that t tests are very basic statistical tools. Many experiments require more sophisticated techniques to evaluate differences. If the variable of interest is a proportion (e.g., 10 of 100 manufactured products were defective), then you’d use z-tests. If you take before and after measurements and have more than one treatment (e.g., control vs a treatment diet), then you need ANOVA.

How do I perform a t test using software?

If you’re wondering how to do a t test, the easiest way is with statistical software such as Prism or an online t test calculator .

If you’re using software, then all you need to know is which t test is appropriate ( use the workflow here ) and understand how to interpret the output. To do that, you’ll also need to:

• Determine whether your test is one or two-tailed
• Choose the level of significance

Is my test one or two-tailed? 

Whether or not you have a one- or two-tailed test depends on your research hypothesis. Choosing the appropriately tailed test is very important and requires integrity from the researcher. This is because you have more “power” with one-tailed tests, meaning that you can detect a statistically significant difference more easily. Unless you have written out your research hypothesis as one directional before you run your experiment, you should use a two-tailed test.

Two-tailed tests

Two-tailed tests are the most common, and they are applicable when your research question is simply asking, “is there a difference?”

One-tailed tests

Contrast that with one-tailed tests, where the research questions are directional, meaning that either the question is, “is it greater than ” or the question is, “is it less than ”. These tests can only detect a difference in one direction.

Choosing the level of significance

All t tests estimate whether a mean of a population is different than some other value, and with all estimates come some variability, or what statisticians call “error.” Before analyzing your data, you want to choose a level of significance, usually denoted by the Greek letter alpha, 𝛼. The scientific standard is setting alpha to be 0.05.

An alpha of 0.05 results in 95% confidence intervals, and determines the cutoff for when P values are considered statistically significant.

One sample t test

If you only have one sample of a list of numbers, you are doing a one-sample t test. All you are interested in doing is comparing the mean from this group with some known value to test if there is evidence, that it is significantly different from that standard. Use our free one-sample t test calculator for this.

A one sample t test example research question is, “Is the average fifth grader taller than four feet?”

It is the simplest version of a t test, and has all sorts of applications within hypothesis testing. Sometimes the “known value” is called the “null value”. While the null value in t tests is often 0, it could be any value. The name comes from being the value which exactly represents the null hypothesis, where no significant difference exists. 

Any time you know the exact number you are trying to compare your sample of data against, this could work well. And of course: it can be either one or two-tailed.

One sample t test formula

Statistical software handles this for you, but if you want the details, the formula for a one sample t test is:

• M: Calculated mean of your sample
• μ: Hypothetical mean you are testing against
• s: The standard deviation of your sample
• n: The number of observations in your sample.

In a one-sample t test, calculating degrees of freedom is simple: one less than the number of objects in your dataset (you’ll see it written as n-1 ).

Example of a one sample t test

For our example within Prism, we have a dataset of 12 values from an experiment labeled “% of control”. Perhaps these are heights of a sample of plants that have been treated with a new fertilizer. A value of 100 represents the industry-standard control height. Likewise, 123 represents a plant with a height 123% that of the control (that is, 23% larger).

We’ll perform a two-tailed, one-sample t test to see if plants are shorter or taller on average with the fertilizer. We will use a significance threshold of 0.05. Here is the output:

You can see in the output that the actual sample mean was 111. Is that different enough from the industry standard (100) to conclude that there is a statistical difference?

The quick answer is yes, there’s strong evidence that the height of the plants with the fertilizer is greater than the industry standard (p=0.015). The nice thing about using software is that it handles some of the trickier steps for you. In this case, it calculates your test statistic (t=2.88), determines the appropriate degrees of freedom (11), and outputs a P value.

More informative than the P value is the confidence interval of the difference, which is 2.49 to 18.7. The confidence interval tells us that, based on our data, we are confident that the true difference between our sample and the baseline value of 100 is somewhere between 2.49 and 18.7. As long as the difference is statistically significant, the interval will not contain zero.

You can follow these tips for interpreting your own one-sample test.

Graphing a one-sample t test

For some techniques (like regression), graphing the data is a very helpful part of the analysis. For t tests, making a chart of your data is still useful to spot any strange patterns or outliers, but the small sample size means you may already be familiar with any strange things in your data.

Here we have a simple plot of the data points, perhaps with a mark for the average. We’ve made this as an example, but the truth is that graphing is usually more visually telling for two-sample t tests than for just one sample.

Two sample t tests

There are several kinds of two sample t tests, with the two main categories being paired and unpaired (independent) samples.

Paired samples t test

In a paired samples t test, also called dependent samples t test, there are two samples of data, and each observation in one sample is “paired” with an observation in the second sample. The most common example is when measurements are taken on each subject before and after a treatment. A paired t test example research question is, “Is there a statistical difference between the average red blood cell counts before and after a treatment?”

Having two samples that are closely related simplifies the analysis. Statistical software, such as this paired t test calculator , will simply take a difference between the two values, and then compare that difference to 0.

In some (rare) situations, taking a difference between the pairs violates the assumptions of a t test, because the average difference changes based on the size of the before value (e.g., there’s a larger difference between before and after when there were more to start with). In this case, instead of using a difference test, use a ratio of the before and after values, which is referred to as ratio t tests .

Paired t test formula

The formula for paired samples t test is:

• Md: Mean difference between the samples
• sd: The standard deviation of the differences
• n: The number of differences

Degrees of freedom are the same as before. If you’re studying for an exam, you can remember that the degrees of freedom are still n-1 (not n-2) because we are converting the data into a single column of differences rather than considering the two groups independently.

Also note that the null value here is simply 0. There is no real reason to include “minus 0” in an equation other than to illustrate that we are still doing a hypothesis test. After you take the difference between the two means, you are comparing that difference to 0.

For our example data, we have five test subjects and have taken two measurements from each: before (“control”) and after a treatment (“treated”). If we set alpha = 0.05 and perform a two-tailed test, we observe a statistically significant difference between the treated and control group (p=0.0160, t=4.01, df = 4). We are 95% confident that the true mean difference between the treated and control group is between 0.449 and 2.47.

Graphing a paired t test

The significant result of the P value suggests evidence that the treatment had some effect, and we can also look at this graphically. The lines that connect the observations can help us spot a pattern, if it exists. In this case the lines show that all observations increased after treatment. While not all graphics are this straightforward, here it is very consistent with the outcome of the t test. 

Prism’s estimation plot is even more helpful because it shows both the data (like above) and the confidence interval for the difference between means. You can easily see the evidence of significance since the confidence interval on the right does not contain zero.

Here are some more graphing tips for paired t tests .

Unpaired samples t test

Unpaired samples t test, also called independent samples t test, is appropriate when you have two sample groups that aren’t correlated with one another. A pharma example is testing a treatment group against a control group of different subjects. Compare that with a paired sample, which might be recording the same subjects before and after a treatment.

With unpaired t tests, in addition to choosing your level of significance and a one or two tailed test, you need to determine whether or not to assume that the variances between the groups are the same or not. If you assume equal variances, then you can “pool” the calculation of the standard error between the two samples. Otherwise, the standard choice is Welch’s t test which corrects for unequal variances. This choice affects the calculation of the test statistic and the power of the test, which is the test’s sensitivity to detect statistical significance. 

It’s best to choose whether or not you’ll use a pooled or unpooled (Welch’s) standard error before running your experiment, because the standard statistical test is notoriously problematic. See more details about unequal variances here .

As long as you’re using statistical software, such as this two-sample t test calculator , it’s just as easy to calculate a test statistic whether or not you assume that the variances of your two samples are the same. If you’re doing it by hand, however, the calculations get more complicated with unequal variances.

Unpaired (independent) samples t test formula

The general two-sample t test formula is:

• M1 and M2: Two means you are comparing, one from each dataset
• SE : The combined standard error of the two samples (calculated using pooled or unpooled standard error)

The denominator (standard error) calculation can be complicated, as can the degrees of freedom. If the groups are not balanced (the same number of observations in each), you will need to account for both when determining n for the test as a whole.

As an example for this family, we conduct a paired samples t test assuming equal variances (pooled). Based on our research hypothesis, we’ll conduct a two-tailed test, and use alpha=0.05 for our level of significance. Our samples were unbalanced, with two samples of 6 and 5 observations respectively. 

The P value (p=0.261, t = 1.20, df = 9) is higher than our threshold of 0.05. We have not found sufficient evidence to suggest a significant difference. You can see the confidence interval of the difference of the means is -9.58 to 31.2.

Note that the F-test result shows that the variances of the two groups are not significantly different from each other.

Graphing an unpaired samples t test

For an unpaired samples t test, graphing the data can quickly help you get a handle on the two groups and how similar or different they are. Like the paired example, this helps confirm the evidence (or lack thereof) that is found by doing the t test itself.

Below you can see that the observed mean for females is higher than that for males. But because of the variability in the data, we can’t tell if the means are actually different or if the difference is just by chance. 

Nonparametric alternatives for t tests

If your data comes from a normal distribution (or something close enough to a normal distribution), then a t test is valid. If that assumption is violated, you can use nonparametric alternatives.

T tests evaluate whether the mean is different from another value, whereas nonparametric alternatives compare either the median or the rank. Medians are well-known to be much more robust to outliers than the mean.

The downside to nonparametric tests is that they don’t have as much statistical power, meaning a larger difference is required in order to determine that it’s statistically significant.

Wilcoxon signed-rank test

The Wilcoxon signed-rank test is the nonparametric cousin to the one-sample t test. This compares a sample median to a hypothetical median value. It is sometimes erroneously even called the Wilcoxon t test (even though it calculates a “W” statistic).

And if you have two related samples, you should use the Wilcoxon matched pairs test instead. The two versions of Wilcoxon are different, and the matched pairs version is specifically for comparing the median difference for paired samples. 

Mann-Whitney and Kolmogorov-Smirnov tests

For unpaired (independent) samples, there are multiple options for nonparametric testing. Mann-Whitney is more popular and compares the mean ranks (the ordering of values from smallest to largest) of the two samples. Mann-Whitney is often misrepresented as a comparison of medians, but that’s not always the case. Kolmogorov-Smirnov tests if the overall distributions differ between the two samples. 

More t test FAQs

What is the formula for a t test.

The exact formula depends on which type of t test you are running, although there is a basic structure that all t tests have in common. All t test statistics will have the form:

• t : The t test statistic you calculate for your test
• Mean1 and Mean2: Two means you are comparing, at least 1 from your own dataset
• Standard Error of the Mean : The standard error of the mean , also called the standard deviation of the mean, which takes into account the variance and size of your dataset

The exact formula for any t test can be slightly different, particularly the calculation of the standard error. Not only does it matter whether one or two samples are being compared, the relationship between the samples can make a difference too.

What is a t-distribution?

A t-distribution is similar to a normal distribution. It’s a bell-shaped curve, but compared to a normal it has fatter tails, which means that it’s more common to observe extremes. T-distributions are identified by the number of degrees of freedom. The higher the number, the closer the t-distribution gets to a normal distribution. After about 30 degrees of freedom, a t and a standard normal are practically the same.

What are degrees of freedom?

Degrees of freedom are a measure of how large your dataset is. They aren’t exactly the number of observations, because they also take into account the number of parameters (e.g., mean, variance) that you have estimated.

What is the difference between paired vs unpaired t tests?

Both paired and unpaired t tests involve two sample groups of data. With a paired t test, the values in each group are related (usually they are before and after values measured on the same test subject). In contrast, with unpaired t tests, the observed values aren’t related between groups. An unpaired, or independent t test, example is comparing the average height of children at school A vs school B. 

When do I use a z-test versus a t test?

Z-tests, which compare data using a normal distribution rather than a t-distribution, are primarily used for two situations. The first is when you’re evaluating proportions (number of failures on an assembly line). The second is when your sample size is large enough (usually around 30) that you can use a normal approximation to evaluate the means.

When should I use ANOVA instead of a t test?

Use ANOVA if you have more than two group means to compare.

What are the differences between t test vs chi square?

Chi square tests are used to evaluate contingency tables , which record a count of the number of subjects that fall into particular categories (e.g., truck, SUV, car). t tests compare the mean(s) of a variable of interest (e.g., height, weight).

What are P values?

P values are the probability that you would get data as or more extreme than the observed data given that the null hypothesis is true. It’s a mouthful, and there are a lot of issues to be aware of with P values.

What are t test critical values?

Critical values are a classical form (they aren’t used directly with modern computing) of determining if a statistical test is significant or not. Historically you could calculate your test statistic from your data, and then use a t-table to look up the cutoff value (critical value) that represented a “significant” result. You would then compare your observed statistic against the critical value.

How do I calculate degrees of freedom for my t test?

In most practical usage, degrees of freedom are the number of observations you have minus the number of parameters you are trying to estimate. The calculation isn’t always straightforward and is approximated for some t tests.

Statistical software calculates degrees of freedom automatically as part of the analysis, so understanding them in more detail isn’t needed beyond assuaging any curiosity.

Perform your own t test

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What Is a T-Test?

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

• T-Test FAQs
• Fundamental Analysis

T-Test: What It Is With Multiple Formulas and When To Use Them

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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8.2.3.1 - one sample mean t test, formulas, five step hypothesis testing procedure.

Data must be quantitative. In order to use the t distribution to approximate the sampling distribution either the sample size must be large (\(\ge\ 30\)) or the population must be known to be normally distributed. The possible combinations of null and alternative hypotheses are:

where \( \mu_{0} \) is the hypothesized population mean.

For the test of one group mean we will be using a \(t\) test statistic:

\(t=\dfrac{\overline{x}-\mu_0}{\frac{s}{\sqrt{n}}}\)

\(\overline{x}\) = sample mean \(\mu_{0}\) = hypothesized population mean \(s\) = sample standard deviation \(n\) = sample size

Note that structure of this formula is similar to the general formula for a test statistic:

\(\dfrac{sample\;statistic-null\;value}{standard\;error}\)

When testing hypotheses about a mean or mean difference, a \(t\) distribution is used to find the \(p\)-value. These \(t\) distributions are indexed by a quantity called degrees of freedom, calculated as \(df = n – 1\) for the situation involving a test of one mean or test of mean difference. The \(p\)-value can be found using Minitab.

If \(p \leq \alpha\) reject the null hypothesis.

If \(p>\alpha\) fail to reject the null hypothesis.

Based on your decision in Step 4, write a conclusion in terms of the original research question.

The new few pages will walk you through examples before giving you the opportunity to do two on your own.

t-test Calculator

Table of contents

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

Hypothesis Testing with the Student t-Distribution

Contents Toggle Main Menu 1 Hypothesis Testing with the Student $t$-distribution 2 Worked Example 1 2.1 Video Example 3 Paired Samples 4 Worked Example 2 5 Worked Example 3 5.1 Video Example 6 Unpaired Samples 6.1 Unknown but Approximately Equal Variances 7 Worked Example 4 7.1 Video Example 7.2 Unknown and unequal variances 8 Workbooks 9 Test Yourself 10 External Resources 11 See Also

Hypothesis Testing with the Student $t$-distribution

We test hypotheses with the Student $t$-distribution when we have a sample of the population and the population standard deviation is not known.

We estimate the population standard deviation using \[s = \sqrt{\frac{1}{n-1}\sum\limits_{i=1}^n(x_i-\bar{x})^2}\] so the test statistic becomes \[t= \frac{\bar{x}-\mu}{\sqrt{\frac{s^2}{n}}}\] where $n$ is the number of samples, $\bar{x}$ is the sample mean, $\mu$ is the population mean and $x_1, x_2, \ldots, x_n$ are the $n$ observations obtained in the sample.

Because we are dividing by an estimate, the test statistic does not have a standard normal distribution under the null hypothesis, instead it has a Student $t$-distribution with $n-1$ degrees of freedom.

Worked Example 1

Worked example.

The population mean of the heights of five-year old boys is 100cm. A teacher measures the height of her twenty five students, obtaining a mean height of $105$cm and standard deviation $18$. This means $\bar{x}=105, \mu=100, s=18, n=25$. Perform a Student $t$-test with a $5$% significance level to calculate whether the true mean is actually greater than $100$cm.

We need to set up a hypothesis test. \begin{align} &H_0 : \mu = 100\\ &H_1 : \mu > 100 \end{align}

First calculate the $t$-statistic, \begin{align} t&=\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n} } } \\ &=\frac{105 - 100}{\frac{18}{5} } \\ &= 1.3889 \text{.} \end{align} Now looking at the $t$-distribution table gives $t_{\alpha, v} = t_{0.05, 24} = 1.7109$.

We have $t = 1.3889 < 1.7109 = t_{\alpha, v}$ so we accept the null hypothesis. The mean height of the population of five-year old boys is $100$ at a $5$% significance level.

Video Example

In this video, Dr Lee Fawcett conducts a hypothesis test for the mean of a distribution whose population variance is unknown, using a one-sample t-test.

Paired Samples

Paired data is when each value in one sample corresponds to a specific value in the other sample.

If we have paired data, both samples must be of the same size. Call this sample size $n$. Let $x$ be the first sample and $y$ be the second sample. Then $x_1$ is paired with $y_1$, $x_2$ is paired with $y_2$ etc. so for $i = 1, 2, \ldots, n$, every $ x_i$ is paired with $y_i$.

Now calculate the difference $d_i$ between the pairs, where $d_i=y_i-x_i$. From this the mean difference $\bar{d}$ can be calculated where \[\bar{d} = \frac{1}{n} \sum\limits_{i=1}^{n} d_i \text{.}\] Now the test statistic is given by \[t = \frac{\bar{d}}{\frac{s}{\sqrt{n}}}\text{.}\]

An alternative formula for the test statistic is \[t= \frac{\sum\limits_{i=1}^n d}{\sqrt{\frac{n\left(\sum\limits_{i=1}^n d^2\right) - \left(\sum\limits_{i=1}^n d\right)^2}{n-1}}} \text{.}\] This is the same, but sometimes makes calculation easier.

Worked Example 2

A sports governing body wants to test whether a supplement used by professional athletes ought to be banned for increasing levels of testosterone in the body.

The levels of testosterone in picograms/millilitre of ten athletes were tested before and after taking the supplement, with the results summarised in the following table.

Perform a hypothesis test using a $1$% significance level to decide whether or not the supplement should be banned.

Worked Example 3

A disgruntled member of public sits at a train station for over two hours recording the time at which trains arrive. He records the following data.

The company responsible claim that on average, trains arrive $3$ minutes late which they believe is acceptable. Perform a Student $t$-test at a $1$% significance level to see if the actual mean is greater than the company claim.

First, write down the null and alternative hypotheses.

From the alternative hypothesis, we see that we need to perform a one-tailed test.

As the test has been performed for us, we can move onto calculating the test statistic. As we are using the Student $t$-distribution, we first need to calculate the sample mean difference $\bar{d}$ and the sample standard deviation $s$:

\[\bar{x} = \dfrac{-1+5+3+0+2+13+6+4+2+1+5+8}{12} = 4\text{,}\] \[s = \sqrt{\sum_{i=1}^{12}\frac{(d_i - \bar{d})^2}{n-1} }.\]

Now calculate the differences: \begin{align} &(d_1-\bar{d})^2 = (-1 - 4)^2 = 25, \\ &(d_2-\bar{d})^2 = (5-4)^2 =1, \\ &(d_3-\bar{d})^2 = (3-4)^2=1,\\ &(d_4-\bar{d})^2 = (0-4)^2=16,\\ &(d_5-\bar{d})^2 = (2-4)^2 =4,\\ &(d_6-\bar{d})^2 = (13-4)^2=81,\\ &(d_7-\bar{d})^2 = (6-4)^2 =4,\\ &(d_8-\bar{d})^2 = (4-4)^2 =0,\\ &(d_9-\bar{d})^2 = (2-4)^2 =4, \\ &(d_{10} - \bar{d})^2 =(1-4)^2=9, \\ &(d_{11} - \bar{d})^2 = (5-4)^2 =1, \\ &(d_{12} - \bar{d})^2 = (8-4)^2 =16. \end{align} Substitute these into the formula for $s$,

\[s=\sqrt{\dfrac{25+1+1+16+4+81+4+0+4+9+1+16}{11} } = \sqrt{\dfrac{162}{11} } = \sqrt{14.7273} = 3.8376.\]

Now calculate the test statistic \[t= \dfrac{\bar{x} - \mu}{\frac{s}{\sqrt{n} } } = \dfrac{4-3}{\frac{3.8376}{\sqrt{12} } } = \dfrac{\sqrt{12} }{3.8376} = 0.9027.\]

Check the one-tailed tables with $11 = (12-1)$ degrees of freedom at the $1$% significance level. We see that the critical value is $2.7181$ so the critical region is $t>2.7181$. As $0.9027<2.7181$, the null hypothesis is accepted.

In this video, Daniel Organisciak performs a hypothesis test on a paired set of data with variance unknown.

Unpaired Samples

If the data is unpaired, we have to compare the variances, even though we don't actually know the true value of the variance. If the sample variances are approximately equal we combine them when computing the test statistic, whereas if they are completely different, the test statistic is computed by considering each variance separately.

Unknown but Approximately Equal Variances

When the variances are equal we need to use pooled data. When we test hypotheses with two means, we will look at the difference $\mu_1 - \mu_2$. The null hypothesis will be of the form

where $a$ is a constant. Often $a=0$ is used to compare if the two means are the same. Given two continuous random variables $X_1$ and $X_2$ with means $\mu_1$ and $\mu_2$ and variances $\frac{\sigma_1^2}{n_1}$ and $\frac{\sigma_2^2}{n_2}$ respectively, \[\mathrm{E} [\bar{X_1} - \bar{X_2} ] = \mathrm{E} [\bar{X_1}] - \mathrm{E} [\bar{X_2}] = \mu_1 - \mu_2\] and \[\mathrm{Var}[\bar{X_1} - \bar{X_2}] = \mathrm{Var}[\bar{X_1}] - \mathrm{Var}[\bar{X_2}]=\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}.\]

As we don't know the value of $\sigma$ we need to estimate it. This is done by pooling (combining) the sample variances. We introduce $S_p$ to mean the pooled standard deviation where \[S_p^2 = \frac{(n_1-1)S_1^2 + (n_2-1)S_2^2}{n_1+n_2-2} .\] Note this is a weighted average of the two variances. We then obtain the $t$-statistic using the formula \[t = \frac{(\bar{X_1}-\bar{X_2})-(\mu_1 - \mu_2)}{S_p \sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}\] with $n_1+n_2-2$ degrees of freedom.

Worked Example 4

Two energy drinks both claim that their average caffeine content is $30$mg/$100$ml. The following data was recorded from random cans of each drink.

Perform a hypothesis test at a $1$% significance level to check if they have the same mean caffeine content.

We are not given the population standard deviation so we cannot perform a test using the normal distribution. This means we have to perform a Student $t$-test.

Mean of $X_1$ is \[\bar{X_1}=\frac{29.37+31.24+29.98+29.04+28.93+31.34+30.16}{7} = 30.0086.\] Mean of $X_2$ is \[\bar{X_2}=\frac{29.79+31.89+33.84+32.37+31.40}{5}=31.6980.\] Firstly, we notice our data isn't paired. One observation of $X_1$ does not have a correspondence with any observation of $X_2$.

We next need to calculate the two standard deviations:

\begin{align} s_1 &= \sqrt{\frac{1}{n_1-1}\sum\limits_{i=1}^n (x_i-\bar{x})^2} \\\\ &=\sqrt{\frac{1}{6} \times (0.4078+1.5164+0.0008+0.9381+1.1633+1.7727+0.0229 } \\\\ &=0.9851 \text{,} \\\\ s_2 &= \sqrt{\frac{1}{n_1-1}\sum\limits_{i=1}^n (x_i-\bar{x})^2} \\\\ &=\sqrt{\frac{1}{4} \times (3.6406+0.0369+2.0794+0.3272+0.0888)} \\\\ &=1.2422 . \end{align} These standard deviations are of a similar size. The corresponding variances are $0.9703$ and $1.5431$ which are also similar so we assume that the population variances are the same. This means we need to calculate the pooled standard deviation. The pooled standard deviation $s_p$ is given by \begin{align} s_p&= \frac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2} \\ &= \frac{6 \times 0.9851^2 + 4 \times 1.2422^2}{6+4-2} \\ &=1.1995 . \end{align}

The test statistic $t$ is given by \begin{align} t &= \frac{(\bar{X_1}-\bar{X_2})-(\mu_1 - \mu_2)}{S_p \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }\\ &=\frac{30.0086 - 31.6980}{1.1195 \times \sqrt{\frac{1}{5} + \frac{1}{7} } } \\ &=-2.4052. \end{align}

In this video, Holly Ainsworth conducts a test of the hypothesis that two samples come from distributions with the same population mean, using a two-sample t-test.

Unknown and unequal variances

What if the variances aren't equal? This time the $t$-statistic is calculated using the formula \[t = \frac{(\bar{X_1}-\bar{X_2})-(\mu_1-\mu_2)}{ \sqrt{\frac{S_1^2}{n_1}+\frac{S_2^2}{n_2}}}\] where $\bar{X_1}$ and $\bar{X_2}$ are the sample means, $S_1^2$ and $S_2^2$ are the sample variances and $n_1$ and $n_2$ are the sample sizes. The number of degrees of freedom $v$ is given by \[v=\frac{\left(\frac{S_1^2}{n_1}+\frac{S_2^2}{n_2} \right)^2}{\frac{1}{n_1+1}\left(\frac{S_1^2}{n_1}\right)^2+\frac{1}{n_2+1}\left(\frac{S_2^2}{n_2}\right)^2}-2.\]

These workbooks produced by HELM are good revision aids, containing key points for revision and many worked examples.

• Tests concerning a single sample
• Tests concerning two samples

Test Yourself

Test yourself: Numbas quiz on $t$-tests

Test yourself: Numbas test on two sample $t$-tests

External Resources

• Unpaired t-tests at
• Paired t-test at

Selecting a Hypothesis Test

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9.4: Student's t-Distribution

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T-Distributions

Hypothesis testing with small populations and sample sizes.

Back in the early 1900’s a chemist at a brewery in Ireland discovered that when he was working with very small samples, the distributions of the mean differed significantly from the normal distribution. He noticed that as his sample sizes changed, the shape of the distribution changed as well. He published his results under the pseudonym ‘Student’ and this concept and the distributions for small sample sizes are now known as “ Student’s t−distributions. ”

T−distributions are a family of distributions that, like the normal distribution, are symmetrical and bell-shaped and centered on a mean. However, the distribution shape changes as the sample size changes. Therefore, there is a specific shape or distribution for every sample of a given size (see figure below; each distribution has a different value of k, the number of degrees of freedom , which is 1 less than the size of the sample).

We use the Student's t−distribution in hypothesis testing the same way that we use the normal distribution. Each row in the t distribution table (see link below) represents a different t−distribution and each distribution is associated with a unique number of degrees of freedom (the number of observations minus one). The column headings in the table represent the portion of the area in the tails of the distribution – we use the numbers in the table just as we used the z−scores.

As the number of observations gets larger, the t−distribution approaches the shape of the normal distribution. In general, once the sample size is large enough - usually about 30 - we would use the normal distribution or the z−table instead. Note that usually in practice, if the standard deviation is known then the normal distribution is used regardless of the sample size.

In calculating the t−test statistic, we use the formula:

t is the test statistic and has n−1 degrees of freedom.

x̄ is the sample mean

μ 0 is the population mean under the null hypothesis.

s is the sample standard deviation

n is the sample size

s / n 0.5 is the estimated standard error

Significance Testing

The high school athletic director is asked if football players are doing as well academically as the other student athletes. We know from a previous study that the average GPA for the student athletes is 3.10. After an initiative to help improve the GPA of student athletes, the athletic director samples 20 football players and finds that the average GPA of the sample is 3.18 with a sample standard deviation of 0.54. Is there a significant improvement? Use a .05 significance level.

First, we establish our null and alternative hypotheses.

H0:μHa:μ=3.10≠3.10

Next, we use our alpha level of .05 and the t−distribution table to find our critical values. For a two-tailed test with 19 degrees of freedom and a .05 level of significance, our critical values are equal to ±2.093.

In calculating the test statistic, we use the formula:

This means that the observed sample mean 3.18 of football players is .66 standard errors above the hypothesized value of 3.10. Because the value of the test statistic is less than the critical value of 2.093, we fail to reject the null hypothesis.

Therefore, we can conclude that the difference between the sample mean and the hypothesized value is not sufficient to attribute it to anything other than sampling error. Thus, the athletic director can conclude that the mean academic performance of football players does not differ from the mean performance of other student athletes.

Calculating a Test Statistic

The masses of newly produced bus tokens are estimated to have a mean of 3.16 grams. A random sample of 11 tokens was removed from the production line and the mean weight of the tokens was calculated as 3.21 grams with a standard deviation of 0.067. What is the value of the test statistic for a test to determine how the mean differs from the estimated mean?

If the value of t from the sample fits right into the middle of the distribution of t constructed by assuming the null hypothesis is true, the null hypothesis is true. On the other hand, if the value of t from the sample is way out in the tail of the t−distribution, then there is evidence to reject the null hypothesis. Now that the distribution of t is known when the null hypothesis is true, the location of this value on the distribution. The most common method used to determine this is to find a p−value (observed significance level). The p−value is a probability that is computed with the assumption that the null hypothesis is true.

The p−value for a two-sided test is the area under the t−distribution with df=11−1=10 that lies above t=2.48 and below t=−2.48. This p−value can be calculated by using technology.

Technology Note: Using the tcdf command to calculate probabilities associated with the t distribution

Press 2ND [DIST] Use ↓ to select 5.tcdf (lower bound, upper bound, degrees of freedom). This will be the total area under both tails. To calculate the area under one tail divide by 2.

There is only a .016 chance of getting an absolute value of t as large as or even larger than the one from this sample. The small p−value tells us that the sample is inconsistent with the null hypothesis. The population mean differs from the estimated mean of 3.16.

When the p−value is close to zero, there is strong evidence against the null hypothesis. When the p−value is large, the result from the sample is consistent with the estimated or hypothesized mean and there is no evidence against the null hypothesis.

A visual picture of the P−value can be obtained by using the graphing calculator.

The spread of any t distribution is greater than that of a standard normal distribution. This is due to the fact that that in the denominator of the formula σ has been replaced with s. Since s is a random quantity changing with various samples, the variability in t is greater, resulting in a larger spread.

Notice in the first distribution graph the spread of the first (inner curve) is small but in the second one the both distributions are basically overlapping, so are roughly normal. This is due to the increase in the degrees of freedom.

Here are the t−distributions for df=1 and for df=12 as graphed on the graphing calculator

You are now on the Y= screen.

Y=tpdf(X,1) [Graph]

Repeat the steps to plot more than one t−distribution on the same screen.

Notice the difference in the two distributions.

The one with 12 degrees of freedom approximates a normal curve.

The t−distribution can be used with any statistic having a bell-shaped distribution. The Central Limit Theorem states the sampling distribution of a statistic will be close to normal with a large enough sample size. As a rough estimate, the Central Limit Theorem predicts a roughly normal distribution under the following conditions:

• The population distribution is normal.
• The sampling distribution is symmetric and the sample size is ≤15.
• The sampling distribution is moderately skewed and the sample size is 16≤n≤30.
• The sample size is greater than 30, without outliers.

The t−distribution also has some unique properties. These properties are:

• The mean of the distribution equals zero.
• The population standard deviation is unknown.
• The variance is equal to the degrees of freedom divided by the degrees of freedom minus 2. This means that the degrees of freedom must be greater than two to avoid the expression being undefined.
• The variance is always greater than one, although it approaches 1 as the degrees of freedom increase. This is due to the fact that as the degrees of freedom increase, the distribution is becoming more of a normal distribution.
• Although the Student t−distribution is bell-shaped, the smaller sample sizes produce a flatter curve. The distribution is not as mounded as a normal distribution and the tails are thicker. As the sample size increases and approaches 30, the distribution approaches a normal distribution.
• The population is unimodal and symmetric.

Using Technology

Duracell manufactures batteries that the CEO claims will last 300 hours under normal use. A researcher randomly selected 15 batteries from the production line and tested these batteries. The tested batteries had a mean life span of 290 hours with a standard deviation of 50 hours. If the CEO’s claim were true, what is the probability that 15 randomly selected batteries would have a life span of no more than 290 hours?

Using the graphing calculator or a table of values, the cumulative probability is 0.226, which means that if the true life span of a battery were 300 hours, there is a 22.6% chance that the life span of the 15 tested batteries would be less than or equal to 290 days. This is not a high enough level of confidence to reject the null hypothesis and count the discrepancy as significant.

Y=tcdf(−1E99,−.7745993,14)=[0.226]

You have just taken ownership of a pizza shop. The previous owner told you that you would save money if you bought the mozzarella cheese in a 4.5 pound slab. Each time you purchase a slab of cheese, you weigh it to ensure that you are receiving 72 ounces of cheese. The results of 7 random measurements are 70, 69, 73, 68, 71, 69 and 71 ounces. Are these differences due to chance or is the distributor giving you less cheese than you deserve?

State the hypotheses.

For H0 the mean weight of cheese μ=72; and for Ha:μ≠72.

Calculate the test statistic.

Begin by determining the mean of the sample and the sample standard deviation. This can be done using the graphing calculator. x¯=70.143 and s=1.676.

Find and interpret the p-value.

The test statistic computed in part b) was -2.9315. Using technology, the p value is .0262. If the mean weight of cheese is 72 ounces, the probability that the weight of 7 random measurements would give a value of t greater than 2.9315 or less than -2.9315 is about 0.0262.

Would the null hypothesis be rejected at the 10% level? The 5% level? The 1% level?

Because the p−value of 0.0262 is less than both .10 and .05, the null hypothesis would be rejected at these levels. However, the p−value is greater than .01 so the null hypothesis would not be rejected if this level of confidence was required.

• Explain how you will do one run of this simulation.
• Produce four values of t using this simulation.
• The dean from UCLA is concerned that the students’ grade point averages have changed dramatically in recent years. The graduating seniors’ mean GPA over the last five years is 2.75. The dean randomly samples 30 seniors from the last graduating class and finds that their mean GP is 2.85 with a sample standard deviation of 0.65. Suppose that the dean samples only 30 students. Would a t−distribution now be the appropriate sampling distribution for the mean? Why or why not?
• Using the appropriate t−distribution, test the same null hypothesis with a sample of 30.
• With a sample size of 30, do you need to have a larger or smaller difference between then hypothesized population mean and the sample mean to obtain statistical significance than with a sample size of 256? Explain your answer.
• Is there a way to determine where the t−statistic lies on a distribution?
• If a way does exist, what is the meaning of its placement?
• A department store claims that a customer spends an average of $25 per visit. A random sample of 36 customers is drawn and the sample mean is $18 with a standard deviation of 3. Test the claim with a 1% level of significance.
• A person claims that people spend an average of 10 hours a week watching a particular TV show. It is found, in a random sample of 50 people, that the mean time watching this show was 11 hours with a standard deviation of 1.2. Test the claim at the 1% level of significance.
• A football coach claims that the average number of penalties per game is at least 19. A random sample of 45 games is drawn. The sample mean is 17 penalties with a standard deviation of 4. Test the claim at the .02 level.
• An insurance agent claims that the average number of accidents per day is at most 7. A random sample of 60 days finds the mean number of accidents is 9 with a standard deviation of 4. Test the claim at the 5% level of significance.
• H 0 :μ=50,x̄=60,s=90,n=100
• H 0 :μ=100,x̄=98,s=15,n=40
• A company claims that the average distance to work for its employees is 3.7 miles. A random sample of 81 employees found an average commute of 4.1 miles to work with a standard deviation of 1.7 miles. The company administrators believe this reflects a chance error. What do you think?
• Test the hypothesis that the average weight of packages shipped by a certain mail order company in December was no more than 6.0 pounds. A simple random sample of 144 packages that were shipped by the company in December was selected for inspection. It was found that the average weight of the 144 parcels was 5.7 pounds with a standard deviation of 2.1 pounds.
• For each of the past two years, the average verbal SAT score of students entering a particular university was 612. A simple random sample of 150 students is taken from this year’s freshman class. The average SAT verbal score for these students is 580 points, with a standard deviation of 75 points. Does this data indicate a decline in the verbal scores of entering students?
• A battery company claims that is battery can run a mechanical device for more than 50 minutes. 100 batteries are tested and have an average life of 40.8 minutes with a standard deviation of 4.8 minutes. Can we accept the company’s claim?
• H 0 :μ=μ 0 ,H a :μ>μ0,n=28,t=2.00
• H 0 :μ=μ 0 ,H a :μ>μ0,n=28,t=−2.00
• H 0 :μ=μ 0 ,H a :μ≠μ0,n=64,t=2.00
• H 0 :μ=μ 0 ,H a :μ≠μ0,n=64,t=−2.00
• H a :μ>100,n=21,α=0.5,t=2.30
• H a :μ>100,n=21,α=0.1,t=2.30
• H a :μ≠100,n=21,α=0.5,t=2.30
• H a :μ≠100,n=21,α=0.1,t=2.30
• Use the following data to test the hypothesis that the mean of the population is no more than 450. Use the data set: 440 490 600 540 540 600 240 440 360 600 490 400 490 540 440 490

Review (Answers)

To view the Review answers, open this PDF file and look for section 8.5.

Additional Resources

Video: Student's t-Distribution

Practice: Student's t-Distribution

Module 9: Hypothesis Testing With One Sample

Distribution needed for hypothesis testing, learning outcomes.

• Conduct and interpret hypothesis tests for a single population mean, population standard deviation known
• Conduct and interpret hypothesis tests for a single population mean, population standard deviation unknown

Earlier in the course, we discussed sampling distributions.  Particular distributions are associated with hypothesis testing. Perform tests of a population mean using a normal distribution or a Student’s t- distribution . (Remember, use a Student’s t -distribution when the population standard deviation is unknown and the distribution of the sample mean is approximately normal.) We perform tests of a population proportion using a normal distribution (usually n is large or the sample size is large).

If you are testing a  single population mean , the distribution for the test is for means :

[latex]\displaystyle\overline{{X}}\text{~}{N}{\left(\mu_{{X}}\text{ , }\frac{{\sigma_{{X}}}}{\sqrt{{n}}}\right)}{\quad\text{or}\quad}{t}_{{{d}{f}}}[/latex]

The population parameter is [latex]\mu[/latex]. The estimated value (point estimate) for [latex]\mu[/latex] is [latex]\displaystyle\overline{{x}}[/latex], the sample mean.

If you are testing a  single population proportion , the distribution for the test is for proportions or percentages:

[latex]\displaystyle{P}^{\prime}\text{~}{N}{\left({p}\text{ , }\sqrt{{\frac{{{p}{q}}}{{n}}}}\right)}[/latex]

The population parameter is [latex]p[/latex]. The estimated value (point estimate) for [latex]p[/latex] is p′ . [latex]\displaystyle{p}\prime=\frac{{x}}{{n}}[/latex] where [latex]x[/latex] is the number of successes and [latex]n[/latex] is the sample size.

Assumptions

When you perform a  hypothesis test of a single population mean μ using a Student’s t -distribution (often called a t-test), there are fundamental assumptions that need to be met in order for the test to work properly. Your data should be a simple random sample that comes from a population that is approximately normally distributed . You use the sample standard deviation to approximate the population standard deviation. (Note that if the sample size is sufficiently large, a t-test will work even if the population is not approximately normally distributed).

When you perform a  hypothesis test of a single population mean μ using a normal distribution (often called a z -test), you take a simple random sample from the population. The population you are testing is normally distributed or your sample size is sufficiently large. You know the value of the population standard deviation which, in reality, is rarely known.

When you perform a  hypothesis test of a single population proportion p , you take a simple random sample from the population. You must meet the conditions for a binomial distribution which are as follows: there are a certain number n of independent trials, the outcomes of any trial are success or failure, and each trial has the same probability of a success p . The shape of the binomial distribution needs to be similar to the shape of the normal distribution. To ensure this, the quantities np  and nq must both be greater than five ( np > 5 and nq > 5). Then the binomial distribution of a sample (estimated) proportion can be approximated by the normal distribution with μ = p and [latex]\displaystyle\sigma=\sqrt{{\frac{{{p}{q}}}{{n}}}}[/latex] . Remember that q = 1 – p .

Concept Review

In order for a hypothesis test’s results to be generalized to a population, certain requirements must be satisfied.

When testing for a single population mean:

• A Student’s t -test should be used if the data come from a simple, random sample and the population is approximately normally distributed, or the sample size is large, with an unknown standard deviation.
• The normal test will work if the data come from a simple, random sample and the population is approximately normally distributed, or the sample size is large, with a known standard deviation.

When testing a single population proportion use a normal test for a single population proportion if the data comes from a simple, random sample, fill the requirements for a binomial distribution, and the mean number of success and the mean number of failures satisfy the conditions:  np > 5 and nq > n where n is the sample size, p is the probability of a success, and q is the probability of a failure.

Formula Review

If there is no given preconceived  α , then use α = 0.05.

Types of Hypothesis Tests

• Single population mean, known population variance (or standard deviation): Normal test .
• Single population mean, unknown population variance (or standard deviation): Student’s t -test .
• Single population proportion: Normal test .
• For a single population mean , we may use a normal distribution with the following mean and standard deviation. Means: [latex]\displaystyle\mu=\mu_{{\overline{{x}}}}{\quad\text{and}\quad}\sigma_{{\overline{{x}}}}=\frac{{\sigma_{{x}}}}{\sqrt{{n}}}[/latex]
• A single population proportion , we may use a normal distribution with the following mean and standard deviation. Proportions: [latex]\displaystyle\mu={p}{\quad\text{and}\quad}\sigma=\sqrt{{\frac{{{p}{q}}}{{n}}}}[/latex].
• Distribution Needed for Hypothesis Testing. Provided by : OpenStax. Located at : . License : CC BY: Attribution
• Introductory Statistics . Authored by : Barbara Illowski, Susan Dean. Provided by : Open Stax. Located at : http://cnx.org/contents/[email protected] . License : CC BY: Attribution . License Terms : Download for free at http://cnx.org/contents/[email protected]

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9.4: A Single Population Mean using the Student t-Distribution

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All hypotheses tests have the same basic steps:

• The alternative hypothesis, \(H_{a}\), never has a symbol that contains an equal sign.
• The alternative hypothesis, \(H_{a}\), tells you if the test is left, right, or two-tailed. It is the key to conducting the appropriate test.
• In a hypothesis test problem, you may see words such as "the level of significance is 1%." The "1%" is the preconceived or preset \(\alpha\). The statistician setting up the hypothesis test selects the value of α to use before collecting the sample data. If no level of significance is given, a common standard to use is \(\alpha = 0.05\).
• When you calculate the \(p\)-value and draw the picture, the \(p\)-value is the area in the left tail, the right tail, or split evenly between the two tails. For this reason, we call the hypothesis test left, right, or two tailed.
• Never, ever, Accept the Null Hypothesis.
• Thinking about the meaning of the \(p\)-value: A data analyst (and anyone else) should have more confidence that he made the correct decision to reject the null hypothesis with a smaller \(p\)-value (for example, 0.001 as opposed to 0.04) even if using the 0.05 level for alpha. Similarly, for a large p -value such as 0.4, as opposed to a \(p\)-value of 0.056 (\(\alpha = 0.05\) is less than either number), a data analyst should have more confidence that she made the correct decision in not rejecting the null hypothesis. This makes the data analyst use judgment rather than mindlessly applying rules.
• Determine the conclusion : What does the decision mean in terms of the problem given?

Using the t-Distribution

Just like with confidence intervals, for some hypothesis tests of a mean, we need to use \(z\), the Standard Normal Distribution and for other tests of a mean, we need to the use \(t\)-distribution instead.

Similarly to confidence intervals, use the Student's t distribution if the population standard deviation is not known .

Full Hypothesis Test Examples

Example \(\pageindex{1}\).

Statistics students believe that the mean score on the first statistics test is 65. A statistics instructor thinks the mean score is higher than 65. He samples ten statistics students and obtains the scores 65 65 70 67 66 63 63 68 72 71. He performs a hypothesis test using a 5% level of significance. The data are assumed to be from a normal distribution.

\(P\)-value Solution

Determine the hypothesis :

A 5% level of significance means that \(\alpha = 0.05\). This is a test of a single population mean .

\(H_{0}: \mu = 65\)

\(H_{a}: \mu > 65\) (claim)

Since the instructor thinks the average score is higher, use a "\(>\)" in the alternative hypothesis. The "\(>\)" means the test is right-tailed.

Calculate the evidence :

There is no population standard deviation given in the problem statement. You are only given \(n = 10\) sample data values. Notice also that the data come from a normal distribution. This means that the distribution for the test is a student's \(t\).

Calculate the test statistic using the formula for a \(t\) test.

\[t=\frac{\bar{x}-\mu}{\frac{s}{\sqrt{n}}}\nonumber\]

\(\mu=65\) comes from \(H_{0}\) and not the data. Enter the data into Excel, and use the Excel formula \(=\text{AVERAGE}()\) to find \(\bar{x}=67\) and the formula \(=\text{STDEV.S}()\) to find \(s=3.197\).

\[t=\frac{67-65}{\frac{3.197}{\sqrt{10}}}=\frac{2}{\frac{3.197}{3.162}}=\frac{2}{1.011}=1.979\nonumber\]

Now calculate the \(p\)-value based on the test statistic found.

This is a right-tailed test, using the \(t\)-distribution, so use the Excel formula \(=\text{T.DIST.RT}(t,df)\).

In this problem, \(df = n-1 = 10 - 1 = 9\), and we found \(t\), which is the test statistic, to be \(t=1.979\).

Use the Excel formula \(=\text{T.DIST.RT}(1.979,9)=0.0396\).

Interpretation of the p -value: If the null hypothesis is true, then there is a 0.0396 probability (3.96%) that the sample mean is 65 or more.

Make a decision :

\(\alpha\) is the minimum area that could be considered to make our result significant.

Compare \(\alpha\) and the \(p-\text{value}\)

• If \(p\)-value is less than the \(\alpha\) then we will Reject \(H_{0}\).
• If \(\alpha\) is less than the \(p\)-value then we will Fail to Reject \(H_{0}\).

Compare \(\alpha = 0.05\) and \(p\text{-value} = 0.0396\).

Since \(p\)-value \(<\alpha\), reject \(H_{0}\)

This means you reject \(\mu = 65\). In other words, you believe the average test score is more than 65.

Conclusion : At a 5% level of significance, the sample data show sufficient evidence that the mean (average) test score is more than 65, just as the math instructor thinks.

Critical Value Solution

Determine the hypothesis (Same as the \(P\)-value solution) :

Calculate the critical value. Since this is right-tailed, using the \(t\)-distribution, use the Excel formula \(=\text{T.INV}(1-\alpha,df)\).

In this problem, \(df = n-1 = 10 - 1 = 9\), and \(\alpha=0.05\).

Use Excel formula \(=\text{T.INV}(1-0.05,9)=1.8331\).

Now calculate the test statistic using the formula for a \(t\) test.

Graph the critical value and the test statistic along the number line of the Standard Normal Distribution graph.

Since this is right-tailed, everything greater than the critical value, \(\text{CV}=1.8331\) will be the rejection region.

Since the test statistic, \(t=1.979\) is greater than the critical value, \(\text{CV}=1.8331\), the decision will be to Reject the Null Hypothesis.

Conclusion (Same as the \(P\)-value solution) : At a 5% level of significance, the sample data show sufficient evidence that the mean (average) test score is more than 65, just as the math instructor thinks.

Exercise \(\PageIndex{1}\)

It is believed that a stock price for a particular company will grow at a rate of $5 per week with a standard deviation of $1. An investor believes the stock won’t grow at that rate. The changes in stock price is recorded for ten weeks and are as follows: $4, $3, $2, $3, $1, $7, $2, $1, $1, $2. Perform a critical value hypothesis test using an 8% level of significance. Identify the Type I and Type II errors.

An 8% level of significance means that \(\alpha = 0.08\). This is a test of a single population mean .

\(H_{0}: \mu = 5\)

\(H_{a}: \mu \neq 5\) (claim)

The "\(\neq\)" in the alternative hypothesis means the test is two-tailed.

There is no population standard deviation given in the problem statement. You are only given \(n = 10\) sample data values. This means that the distribution for the test is a student's \(t\).

Calculate the critical value. Since this is two-tailed, using the \(t\)-distribution, use the Excel formula \(=\text{T.INV.2T}(\alpha,df)\).

In this problem, \(df = n-1 = 10 - 1 = 9\), and \(\alpha=0.08\).

Use Excel formula \(=\text{T.INV.2T}(0.08,9)=1.9726\).

Since this is a two-tailed test, we will have two critical values, at \(1.9726\) and at \(-1.9726\).

\(\mu=5\) comes from \(H_{0}\) and not the data. Enter the data into Excel, and use the Excel formula \(=\text{AVERAGE}()\) to find \(\bar{x}=2.6\) and the formula \(=\text{STDEV.S}()\) to find \(s=1.838\).

\[t=\frac{2.6-5}{\frac{1.838}{\sqrt{10}}}=\frac{-2.4}{\frac{1.838}{3.162}}=\frac{-2.4}{0.581}=-4.129\nonumber\]

Since this is two-tailed, everything less than the negative critical value, \(\text{CV}=-1.9726\), and everything greater than the positive critical value, \(\text{CV}=1.9726\) will be the rejection region.

Since the test statistic, \(t=-4.129\) is less than the negative critical value, \(\text{CV}=-1.9726\), the decision will be to Reject the Null Hypothesis.

Conclusion :

At an 8% level of significance, the sample data show sufficient evidence that the mean (average) stock growth will not be $5 per week.

Type I and Type II Errors :

Type I Error: To conclude that the stock price is growing slower than $5 a week when, in fact, the stock price is growing at $5 a week (reject the null hypothesis when the null hypothesis is true).

Type II Error: To conclude that the stock price is growing at a rate of $5 a week when, in fact, the stock price is growing slower than $5 a week (do not reject the null hypothesis when the null hypothesis is false).

Example \(\PageIndex{2}\)

The National Institute of Standards and Technology provides exact data on conductivity properties of materials. Following are conductivity measurements for 11 randomly selected pieces of a particular type of glass.

1.11; 1.07; 1.11; 1.07; 1.12; 1.08; 0.98; 0.98 1.02; 0.95; 0.95

Is there convincing evidence that the average conductivity of this type of glass is less than one? Use the \(p\)-value method at a significance level of 0.02. Assume the population is normal.

A 0.02 level of significance means that \(\alpha = 0.02\). This is a test of a single population mean .

\(H_{0}: \mu \geq 1\)

\(H_{a}: \mu <1\) (claim)

The "\(<\)" in the alternative hypothesis means the test is left-tailed.

There is no population standard deviation given in the problem statement. You are only given \(n = 11\) sample data values. Notice also that the data come from a normal distribution. This means that the distribution for the test is a student's \(t\).

\(\mu=1\) comes from \(H_{0}\) and not the data. Enter the data into Excel, and use the Excel formula \(=\text{AVERAGE}()\) to find \(\bar{x}=1.04\) and the formula \(=\text{STDEV.S}()\) to find \(s=0.0669\).

\[t=\frac{1.04-1}{\frac{0.0669}{\sqrt{11}}}=\frac{0.04}{\frac{0.0669}{3.317}}=\frac{0.04}{0.0199}=2.014\nonumber\]

This is a right-tailed test, using the \(t\)-distribution, so use the Excel formula \(=\text{T.DIST}(t,df,\text{true})\).

In this problem, \(df = n-1 = 11 - 1 = 10\), and we found \(t\), which is the test statistic, to be \(t=2.014\).

Use the Excel formula \(=\text{T.DIST}(2.014,10,\text{true})=0.9642\).

Interpretation of the p -value: If the null hypothesis is true, then there is a 0.9642 probability that the sample mean is 1 or more.

Compare \(\alpha = 0.02\) and \(p\text{-value} = 0.9642\).

Since \(p\)-value \(>\alpha\), do not reject \(H_{0}\)

This means you reject \(\mu \geq 1\).

At a 2% level of significance, the sample data do not show sufficient evidence that the mean (average) conductivity of this type of glass is less than 1.

The hypothesis test itself has an established process. This can be summarized as follows:

• Determine \(H_{0}\) and \(H_{a}\). Remember, they are contradictory.
• Determine the random variable.
• Determine the distribution for the test.
• Draw a graph, calculate the test statistic, and use the test statistic to calculate the \(p\text{-value}\). (A z -score and a t -score are examples of test statistics.)
• Compare the preconceived α with the p -value, make a decision (reject or do not reject H 0 ), and write a clear conclusion using English sentences.

Notice that in performing the hypothesis test, you use \(\alpha\) and not \(\beta\). \(\beta\) is needed to help determine the sample size of the data that is used in calculating the \(p\text{-value}\). Remember that the quantity \(1 – \beta\) is called the Power of the Test . A high power is desirable. If the power is too low, statisticians typically increase the sample size while keeping α the same.If the power is low, the null hypothesis might not be rejected when it should be.

• Data from Amit Schitai. Director of Instructional Technology and Distance Learning. LBCC.
• Data from Bloomberg Businessweek . Available online at www.businessweek.com/news/2011- 09-15/nyc-smoking-rate-falls-to-record-low-of-14-bloomberg-says.html.
• Data from energy.gov. Available online at http://energy.gov (accessed June 27. 2013).
• Data from Gallup®. Available online at www.gallup.com (accessed June 27, 2013).
• Data from Growing by Degrees by Allen and Seaman.
• Data from La Leche League International. Available online at www.lalecheleague.org/Law/BAFeb01.html.
• Data from the American Automobile Association. Available online at www.aaa.com (accessed June 27, 2013).
• Data from the American Library Association. Available online at www.ala.org (accessed June 27, 2013).
• Data from the Bureau of Labor Statistics. Available online at http://www.bls.gov/oes/current/oes291111.htm .
• Data from the Centers for Disease Control and Prevention. Available online at www.cdc.gov (accessed June 27, 2013)
• Data from the U.S. Census Bureau, available online at quickfacts.census.gov/qfd/states/00000.html (accessed June 27, 2013).
• Data from the United States Census Bureau. Available online at www.census.gov/hhes/socdemo/language/.
• Data from Toastmasters International. Available online at http://toastmasters.org/artisan/deta...eID=429&Page=1 .
• Data from Weather Underground. Available online at www.wunderground.com (accessed June 27, 2013).
• Federal Bureau of Investigations. “Uniform Crime Reports and Index of Crime in Daviess in the State of Kentucky enforced by Daviess County from 1985 to 2005.” Available online at http://www.disastercenter.com/kentucky/crime/3868.htm (accessed June 27, 2013).
• “Foothill-De Anza Community College District.” De Anza College, Winter 2006. Available online at research.fhda.edu/factbook/DA...t_da_2006w.pdf.
• Johansen, C., J. Boice, Jr., J. McLaughlin, J. Olsen. “Cellular Telephones and Cancer—a Nationwide Cohort Study in Denmark.” Institute of Cancer Epidemiology and the Danish Cancer Society, 93(3):203-7. Available online at http://www.ncbi.nlm.nih.gov/pubmed/11158188 (accessed June 27, 2013).
• Rape, Abuse & Incest National Network. “How often does sexual assault occur?” RAINN, 2009. Available online at www.rainn.org/get-information...sexual-assault (accessed June 27, 2013).

Contributors and Attributions

Barbara Illowsky and Susan Dean (De Anza College) with many other contributing authors. Content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/[email protected] .

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How do the t-distribution and standard normal distribution differ, and why is t-distribution used more?

For statistical inference (e.g., hypothesis testing or computing confidence intervals), why do we use the t-distribution instead of the standard normal distribution? My class started with the standard normal distribution and shifted to the t-distribution, and I am not fully sure why. Is it because t-distributions can a) deal with small sample sizes (because it gives more emphasis to the tails) or b) be more robust to a non-normally distributed sample?

• hypothesis-testing
• mathematical-statistics
• confidence-interval
• t-distribution

• $\begingroup$ Possibly related: stats.stackexchange.com/questions/285649/… $\endgroup$ –  Henry Aug 22, 2018 at 23:09
• $\begingroup$ Searcher like stats.stackexchange.com/search?q=t-distribution+normal and stats.stackexchange.com/search?q=t-test+normal will include a number of relevant posts (and a lot of other hits so you may need to add further keywords to reduce the clutter). $\endgroup$ –  Glen_b Aug 23, 2018 at 2:18

2 Answers 2

The normal distribution (which is almost certainly returning in later chapters of your course) is much easier to motivate than the t distribution for students new to the material. The reason why you are learning about the t distribution is more or less for your first reason: the t distribution takes a single parameter—sample size minus one—and more correctly accounts for uncertainty due to (small) sample size than the normal distribution when making inferences about a sample mean of normally-distributed data, assuming that the true variance is unknown.

With increasing sample size, both t and standard normal distributions are both approximately as robust with respect to deviations from normality (as sample size increases the t distribution converges to the standard normal distribution). Nonparametric tests (which I start teaching about half way through my intro stats course) are generally much more robust to non-normality than either t or normal distributions.

Finally, you are likely going to learn tests and confidence intervals for many different distributions by the end of your course ( F , $\chi^{2}$, rank distributions—at least in their table p -values, for example).

• 3 $\begingroup$ Thank you so much for this awesome response. I now get that t-distributions can better account for small sample sizes. However, if the sample size is large (> 30), it doesn't matter whether we use a t or standard normal distribution, right? $\endgroup$ –  Jane Sully Aug 22, 2018 at 19:26
• 2 $\begingroup$ they become very similar when the degrees of freedom rise. $\endgroup$ –  Bernhard Aug 22, 2018 at 19:37
• 2 $\begingroup$ @JaneSully Sure, but, for inference about means of normal data, it is never wrong to use the t distribution. $\endgroup$ –  Alexis Aug 22, 2018 at 21:13
• 1 $\begingroup$ (Also, when/if you like an answer enough to say that it has answered your question, you can "accept" it by clicking on the check mark to the top left of the question. :). $\endgroup$ –  Alexis Aug 22, 2018 at 21:24
• 2 $\begingroup$ I disagree with this statement: "the t distribution takes a single parameter—sample size minus one—and more correctly accounts for uncertainty due to (small) sample size than the normal distribution when making inferences about a sample mean of normally-distributed data." E.g. see this lecture: onlinecourses.science.psu.edu/stat414/node/173 There's no need for t-distribution on Gaussian data when standard deviation is known. The key here is whether you do or do not know the variance, not the n-1 adjustment $\endgroup$ –  Aksakal Aug 23, 2018 at 3:49

The reason t-distribution is used in inference instead of normal is due to the fact that the theoretical distribution of some estimators is normal (Gaussian) only when the standard deviation is known, and when it is unknown the theoretical distribution is Student t.

We rarely know the standard deviation. Usually, we estimate from the sample, so for many estimators it is theoretically more solid to use Student t distribution and not normal.

Some estimators are consistent, i.e. in layman terms, they get better when the sample size increases. Student t becomes normal when sample size is large.

Example: sample mean

Consider a mean $\mu$ of the sample $x_1,x_2,\dots,x_n$. We can estimate it using a usual average estimator: $\bar x=\frac 1 n\sum_{i=1}^nx_i$, which you may call a sample mean.

If we want to make inference statements about the mean, such as whether a true mean $\mu<0$, we can use the sample mean $\bar x$ but we need to know what is its distribution. It turns out that if we knew the standard deviation $\sigma$ of $x_i$, then the sample mean would be distributed around the true mean according to Gaussian: $\bar x\sim\mathcal N(\mu,\sigma^2/n)$, for large enough $n$

The problem's that we rarely know $\sigma$, but we can estimate its value from the sample $\hat\sigma$ using one of the estimators. In this case the distribution of the sample mean is no longer Gaussian , but closer to Student t distribution.

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P-Value And Statistical Significance: What It Is & Why It Matters

Saul Mcleod, PhD

Editor-in-Chief for Simply Psychology

BSc (Hons) Psychology, MRes, PhD, University of Manchester

Saul Mcleod, PhD., is a qualified psychology teacher with over 18 years of experience in further and higher education. He has been published in peer-reviewed journals, including the Journal of Clinical Psychology.

Learn about our Editorial Process

Olivia Guy-Evans, MSc

Associate Editor for Simply Psychology

BSc (Hons) Psychology, MSc Psychology of Education

Olivia Guy-Evans is a writer and associate editor for Simply Psychology. She has previously worked in healthcare and educational sectors.

On This Page:

The p-value in statistics quantifies the evidence against a null hypothesis. A low p-value suggests data is inconsistent with the null, potentially favoring an alternative hypothesis. Common significance thresholds are 0.05 or 0.01.

Hypothesis testing

When you perform a statistical test, a p-value helps you determine the significance of your results in relation to the null hypothesis.

The null hypothesis (H0) states no relationship exists between the two variables being studied (one variable does not affect the other). It states the results are due to chance and are not significant in supporting the idea being investigated. Thus, the null hypothesis assumes that whatever you try to prove did not happen.

The alternative hypothesis (Ha or H1) is the one you would believe if the null hypothesis is concluded to be untrue.

The alternative hypothesis states that the independent variable affected the dependent variable, and the results are significant in supporting the theory being investigated (i.e., the results are not due to random chance).

What a p-value tells you

A p-value, or probability value, is a number describing how likely it is that your data would have occurred by random chance (i.e., that the null hypothesis is true).

The level of statistical significance is often expressed as a p-value between 0 and 1.

The smaller the p -value, the less likely the results occurred by random chance, and the stronger the evidence that you should reject the null hypothesis.

Remember, a p-value doesn’t tell you if the null hypothesis is true or false. It just tells you how likely you’d see the data you observed (or more extreme data) if the null hypothesis was true. It’s a piece of evidence, not a definitive proof.

Example: Test Statistic and p-Value

Suppose you’re conducting a study to determine whether a new drug has an effect on pain relief compared to a placebo. If the new drug has no impact, your test statistic will be close to the one predicted by the null hypothesis (no difference between the drug and placebo groups), and the resulting p-value will be close to 1. It may not be precisely 1 because real-world variations may exist. Conversely, if the new drug indeed reduces pain significantly, your test statistic will diverge further from what’s expected under the null hypothesis, and the p-value will decrease. The p-value will never reach zero because there’s always a slim possibility, though highly improbable, that the observed results occurred by random chance.

P-value interpretation

The significance level (alpha) is a set probability threshold (often 0.05), while the p-value is the probability you calculate based on your study or analysis.

A p-value less than or equal to your significance level (typically ≤ 0.05) is statistically significant.

A p-value less than or equal to a predetermined significance level (often 0.05 or 0.01) indicates a statistically significant result, meaning the observed data provide strong evidence against the null hypothesis.

This suggests the effect under study likely represents a real relationship rather than just random chance.

For instance, if you set α = 0.05, you would reject the null hypothesis if your p -value ≤ 0.05. 

It indicates strong evidence against the null hypothesis, as there is less than a 5% probability the null is correct (and the results are random).

Therefore, we reject the null hypothesis and accept the alternative hypothesis.

Example: Statistical Significance

Upon analyzing the pain relief effects of the new drug compared to the placebo, the computed p-value is less than 0.01, which falls well below the predetermined alpha value of 0.05. Consequently, you conclude that there is a statistically significant difference in pain relief between the new drug and the placebo.

What does a p-value of 0.001 mean?

A p-value of 0.001 is highly statistically significant beyond the commonly used 0.05 threshold. It indicates strong evidence of a real effect or difference, rather than just random variation.

Specifically, a p-value of 0.001 means there is only a 0.1% chance of obtaining a result at least as extreme as the one observed, assuming the null hypothesis is correct.

Such a small p-value provides strong evidence against the null hypothesis, leading to rejecting the null in favor of the alternative hypothesis.

A p-value more than the significance level (typically p > 0.05) is not statistically significant and indicates strong evidence for the null hypothesis.

This means we retain the null hypothesis and reject the alternative hypothesis. You should note that you cannot accept the null hypothesis; we can only reject it or fail to reject it.

Note : when the p-value is above your threshold of significance,  it does not mean that there is a 95% probability that the alternative hypothesis is true.

One-Tailed Test

Two-Tailed Test

How do you calculate the p-value ?

Most statistical software packages like R, SPSS, and others automatically calculate your p-value. This is the easiest and most common way.

Online resources and tables are available to estimate the p-value based on your test statistic and degrees of freedom.

These tables help you understand how often you would expect to see your test statistic under the null hypothesis.

Understanding the Statistical Test:

Different statistical tests are designed to answer specific research questions or hypotheses. Each test has its own underlying assumptions and characteristics.

For example, you might use a t-test to compare means, a chi-squared test for categorical data, or a correlation test to measure the strength of a relationship between variables.

Be aware that the number of independent variables you include in your analysis can influence the magnitude of the test statistic needed to produce the same p-value.

This factor is particularly important to consider when comparing results across different analyses.

Example: Choosing a Statistical Test

If you’re comparing the effectiveness of just two different drugs in pain relief, a two-sample t-test is a suitable choice for comparing these two groups. However, when you’re examining the impact of three or more drugs, it’s more appropriate to employ an Analysis of Variance ( ANOVA) . Utilizing multiple pairwise comparisons in such cases can lead to artificially low p-values and an overestimation of the significance of differences between the drug groups.

How to report

A statistically significant result cannot prove that a research hypothesis is correct (which implies 100% certainty).

Instead, we may state our results “provide support for” or “give evidence for” our research hypothesis (as there is still a slight probability that the results occurred by chance and the null hypothesis was correct – e.g., less than 5%).

Example: Reporting the results

In our comparison of the pain relief effects of the new drug and the placebo, we observed that participants in the drug group experienced a significant reduction in pain ( M = 3.5; SD = 0.8) compared to those in the placebo group ( M = 5.2; SD  = 0.7), resulting in an average difference of 1.7 points on the pain scale (t(98) = -9.36; p < 0.001).

The 6th edition of the APA style manual (American Psychological Association, 2010) states the following on the topic of reporting p-values:

“When reporting p values, report exact p values (e.g., p = .031) to two or three decimal places. However, report p values less than .001 as p < .001.

The tradition of reporting p values in the form p < .10, p < .05, p < .01, and so forth, was appropriate in a time when only limited tables of critical values were available.” (p. 114)

• Do not use 0 before the decimal point for the statistical value p as it cannot equal 1. In other words, write p = .001 instead of p = 0.001.
• Please pay attention to issues of italics ( p is always italicized) and spacing (either side of the = sign).
• p = .000 (as outputted by some statistical packages such as SPSS) is impossible and should be written as p < .001.
• The opposite of significant is “nonsignificant,” not “insignificant.”

Why is the p -value not enough?

A lower p-value  is sometimes interpreted as meaning there is a stronger relationship between two variables.

However, statistical significance means that it is unlikely that the null hypothesis is true (less than 5%).

To understand the strength of the difference between the two groups (control vs. experimental) a researcher needs to calculate the effect size .

When do you reject the null hypothesis?

In statistical hypothesis testing, you reject the null hypothesis when the p-value is less than or equal to the significance level (α) you set before conducting your test. The significance level is the probability of rejecting the null hypothesis when it is true. Commonly used significance levels are 0.01, 0.05, and 0.10.

Remember, rejecting the null hypothesis doesn’t prove the alternative hypothesis; it just suggests that the alternative hypothesis may be plausible given the observed data.

The p -value is conditional upon the null hypothesis being true but is unrelated to the truth or falsity of the alternative hypothesis.

What does p-value of 0.05 mean?

If your p-value is less than or equal to 0.05 (the significance level), you would conclude that your result is statistically significant. This means the evidence is strong enough to reject the null hypothesis in favor of the alternative hypothesis.

Are all p-values below 0.05 considered statistically significant?

No, not all p-values below 0.05 are considered statistically significant. The threshold of 0.05 is commonly used, but it’s just a convention. Statistical significance depends on factors like the study design, sample size, and the magnitude of the observed effect.

A p-value below 0.05 means there is evidence against the null hypothesis, suggesting a real effect. However, it’s essential to consider the context and other factors when interpreting results.

Researchers also look at effect size and confidence intervals to determine the practical significance and reliability of findings.

How does sample size affect the interpretation of p-values?

Sample size can impact the interpretation of p-values. A larger sample size provides more reliable and precise estimates of the population, leading to narrower confidence intervals.

With a larger sample, even small differences between groups or effects can become statistically significant, yielding lower p-values. In contrast, smaller sample sizes may not have enough statistical power to detect smaller effects, resulting in higher p-values.

Therefore, a larger sample size increases the chances of finding statistically significant results when there is a genuine effect, making the findings more trustworthy and robust.

Can a non-significant p-value indicate that there is no effect or difference in the data?

No, a non-significant p-value does not necessarily indicate that there is no effect or difference in the data. It means that the observed data do not provide strong enough evidence to reject the null hypothesis.

There could still be a real effect or difference, but it might be smaller or more variable than the study was able to detect.

Other factors like sample size, study design, and measurement precision can influence the p-value. It’s important to consider the entire body of evidence and not rely solely on p-values when interpreting research findings.

Can P values be exactly zero?

While a p-value can be extremely small, it cannot technically be absolute zero. When a p-value is reported as p = 0.000, the actual p-value is too small for the software to display. This is often interpreted as strong evidence against the null hypothesis. For p values less than 0.001, report as p < .001

Further Information

• P-values and significance tests (Kahn Academy)
• Hypothesis testing and p-values (Kahn Academy)
• Wasserstein, R. L., Schirm, A. L., & Lazar, N. A. (2019). Moving to a world beyond “ p “< 0.05”.
• Criticism of using the “ p “< 0.05”.
• Publication manual of the American Psychological Association
• Statistics for Psychology Book Download

Bland, J. M., & Altman, D. G. (1994). One and two sided tests of significance: Authors’ reply.  BMJ: British Medical Journal ,  309 (6958), 874.

Goodman, S. N., & Royall, R. (1988). Evidence and scientific research.  American Journal of Public Health ,  78 (12), 1568-1574.

Goodman, S. (2008, July). A dirty dozen: twelve p-value misconceptions . In  Seminars in hematology  (Vol. 45, No. 3, pp. 135-140). WB Saunders.

Lang, J. M., Rothman, K. J., & Cann, C. I. (1998). That confounded P-value.  Epidemiology (Cambridge, Mass.) ,  9 (1), 7-8.

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