t test problem solving

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

• Math Formulas
• T Test Formula

T-Test Formula

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

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

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

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

The formula for standard deviation is given by:

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

T-Test Solved Examples

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

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

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

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

Construct the following table for standard deviation:

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

Number of terms in second set: n 2 = 4

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

Formula for t-test value:

t = 2.3764 = 2.36 (approx)

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T Test (Students T Test) – Understanding the math and how it works

• October 1, 2020
• Selva Prabhakaran

T Test (Students T Test) is a statistical significance test that is used to compare the means of two groups and determine if the difference in means is statistically significant.

In this one, you’ll understand when to use the T-Test, the different types of T-Test, math behind it, how to determine which test to choose in what situation and why, how to read from the t-tables, example situations and how to apply it in R and Python.

Introduction

Story behind the t-test.

• What are the three types of T-Tests
• How to perform a T-Test?

T-Test Example Exercise

How to implement t-test in r and python.

• Related Posts

T Test is one of the foundational statistical tests. It is used to compare the means of two groups and determine if the difference is statistically significant. It is a very common test often used in data and statistical analysis.

So when to use the T-Test?

Let’s suppose, you have measurements from two different groups, say, you are measuring the marks scored by students from a class you attended special coaching versus those who did not. And you want to determine if the scores of those who attending the coaching is significantly higher than those who did not. You can use the T-Test to determine this.

However, depending on how the groups are chosen and how the measurements were made you will have to choose a different type of T-Test for different situations. You will see more such examples coming up.

By the end of this you will know:

• How the Student’s T Test got its name
• What is the T-Test
• Three Types of T-Tests
• How to pick the right T-Test to use in various situations
• Example Situations of when to use which test
• Formula of T-Statistic (in various situations)
• T Test Solved Exercise

T-Test was first invented by ‘William Sealy Gosset’ , when he was working at Guinness Brewery. The original goal was to select the best barley grains from small samples, when the population standard deviation was not known.

If Mr. Gossett invented it, why is it called Student’s T-Test instead of ‘Gosett’s T-Test’?

It’s said that Guinness didn’t want him to reveal some sort of industry secret which may be detrimental from a business point of view. However, they allowed him to publish it under a different name.

So he published his work in a paper title ‘Biometrika’ in 1908 under the pseudoname ‘Student’ and the name caught on. As a result this test is often referred to as ‘Student’s T Test’ .

So, effectively this test has been around for more than a century and is in active use today.

This test assumes that the test statistic, which is often the ‘sample mean’, should follow a normal distribution. Let’s look at the types of T-Tests.

What are the three types of T-Tests?

Input: Small numeric array (vector) and hypothesized population mean.

Output: T-Statistic, P value.

When to use (example): Test if the mean weight of adult male mice is 10g (population mean), given a numeric array of mouse weights.

Input: Two numeric arrays which may or may not be of the same size. They belong to two different groups.

When to use (example): Test of the mean weight of mangoes from Farm A equals mean weight of mangoes from Farm B.

Input: Two numeric arrays of same size and observations from the two samples are paired.

Output: T-Statistic, P-Value

When to use (example): Given the numeric ratings of same menu items from two different restaurants (rated by same connoisseur), determine which restaurant’s food tastes better?

How to perform a T Test?

The main part of performing the T-Test is in computing the T Statistic, which is the test statistic in this case.

The way you compute the t-statistic is different depending on the type of t-test. Let’s see how to compute it one by one.

1. One-Sample Student’s t-test:

You perform a one sample T Test when you want to test if the population mean is of a specified value μ given by a sample mean x̄.

Let me put it plainly: you have a sample of observations for which you know the mean (sample mean), and you want to test if the sample came from a population with given (hypothesized) mean.

Step 1: Define the Null and Alternate Hypothesis

H0 (Null Hypothesis): Sample Mean (x̄) = Hypothesized Population Mean(μ)

H1 (Alternate Hypothesis): Sample Mean (x̄) != Hypothesized Population Mean(μ)

Step 2: Compute the T statistic

Use the following formula to compute the t-statistic:

$$t = \frac{x̄ – μ}{s/\sqrt{n}}$$

where, x̄=sample mean, μ=population mean, n=sample size, s=sample standard error.

The only difference with the z-statistic formula is that instead of population standard deviation, it uses the sample’s standard error.

Step 3: Lookup T critical and check if computed T statistic falls in rejection region.

Based on the degrees of freedom (n-1) and the alpha level (typically 0.05), lookup the critical T value from the T-Table . If the absolute value of the computed T-statistic is greater than the critical T-value, that is, it falls in the rejection region, then we reject the null hypothesis that the population mean is of the specified value.

Alternately, you can check the output P-value. P-value is typically an output from the software as a result of the T Test. If the P-value is lesser that the significance level (typically 0.05), then reject the null hypothesis and conclude that the population mean is different from what is stated.

The t-critical score is dependent on:

The shape of the T Distribution depends on the number of degrees of freedom (d.o.f), which is in turn dependent on the number of observations in the sample. It is simply (n-1). T distribution usually has fatter tail, which gets narrower as the degree of freedom increases.

As the number of observations (or d.o.f) increase, the T distribution becomes closer in shape to the standard normal distribution.

2. Independent Two-Sample T-Test:

When trying to compare two samples with a two-sample t-test, you can think of the t-test as a ratio of signal (sample means) to noise (sample variability).

$$t = \frac{signal}{noise} = \frac{difference \; in \; means}{sample \; variability} = \frac{\bar{x_1} – \bar{x_2}}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}$$

where, x̄1, x̄2 = sample means, s_1^1, s_2^2 = sample variances and n_1, n_2 = sample sizes.

Once t-statistic is computed, we then compare it to the critical t-score for a given degrees of freedom df and significance level (α).

How to compute the degrees of freedom?

Since the sample sizes of the two samples can be unequal with different variances, determining the degrees of freedom is not as straight forward as you saw in 1-sample t-test.

The formula for calculating df was given by Welch-Satterthwaite as shown below.

$$df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{1}{n_1-1}\left( \frac{s_1^2}{n1}\right)^2 + {\frac{1}{n_2-1}\left( \frac{s_2^2}{n2}\right)^2}}$$

If you have approximately same variances in each sample, though that may not always be the case, the above equation can be generalized as below:

$$df = n1 + n2 – 2$$

Once degree of freedom is calculated, for the given alpha level (say 0.05) you will lookup the T-table for the critical T value. Then, based on the T-critical, see of the computed T value falls in the rejection region. If it does, we reject the null hypothesis and understand that the means from the two samples are not the same.

Let’s solve an example problem.

Problem Statement

A global fast food chain wants to venture into a new city by setting up a store in a popular mall in the city. It has shortlisted 2 popular locations A and B for the same and wants to choose one of them based on the number of foot falls per week.

The company has obtained the data for the same and wants to know which of the two locations to choose and if there is significant difference between the two.

Here’s how the footfalls per day (in 1000’s) looks like for 10 randomly chosen days:

First, let’s compare the sample means.

$$\bar{x}_{Loc A} = 121.72$$

$$\bar{x}_{Loc B} = 112.84$$

x̄_LocA – x̄_LocB = 8.88

Looks like, Location A gets 8.88k more footfalls on an average over a period of 10 days.

However, the average number of footfalls is in these locations in much more and tend to vary on a day to day basis. So, the question remains: Is an additional avg footfall of 8.88k measured over 10 days, enough to say that Loc A receives different footfalls than Loc B? Is it statistically significant?

How to find this out?

Let’s begin by framing the null and alternate hypothesis.

Step 0: Identify the type of T Test.

Well, here are the facts: There are two samples. Each sample contains the number of footfalls but since both are random samples the observations are not paired. That is, the number of footfalls of 116.2 in location A does not correspond to 110.0 in location B because they may not be captured on the same day. So, the observations are not paired, as a result, the T-Test to perform is the Two Sample Independent T Test .

Step 1: Frame the null and alternate hypothesis Null Hypothesis H0: x̄_LocA = x̄_LocB

Alternate Hypothesis H1: x̄_LocA != x̄_LocB

Step 2: Degrees of freedom Since this is a 2 sample t-Test, the degrees of freedom = 10 + 10 – 2 = 18

Ideally we should use Welch-Satterthwaite’s formula. But for simplicity of manual calculation, I am using this one for now.

Step 3: Compute the t-Statistic

Let’s start by computing the variance Variance of Loc_A = 80.508 Variance of Loc_B = 95.584

$$t = \frac{\bar{x_1} – \bar{x_2}}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} = \frac{8.88}{\sqrt{\frac{80.51}{10} + \frac{95.58}{10}}} = 2.1161$$

Step 4: Lookup the critical value from t-table

If the computed T-statistic falls in the rejection

Since its a two-tailed test with 18 degrees of freedom. And assuming a default 95% confidence, the critical value from the t-table is 2.101

In our case, the t-statistic (2.1161) > t-critical (2.101)

This means, the t-statistic does falls in the rejection zone and so, we reject the null hypothesis and conclude that the means are in fact different.

Below is a reproducible code to implement the t-test for the problem we just discussed.

How to do T Test In R?

How to do T Test In Python?

We have seen a lot of details both T-Test and the different types. Next, we will go through each type of T-Test in more detail with worked out examples. Stay tuned.

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AP®︎/College Statistics

Course: ap®︎/college statistics   >   unit 11.

• Hypotheses for a two-sample t test
• Example of hypotheses for paired and two-sample t tests
• Writing hypotheses to test the difference of means

Two-sample t test for difference of means

• Test statistic in a two-sample t test
• P-value in a two-sample t test
• Conclusion for a two-sample t test using a P-value
• Conclusion for a two-sample t test using a confidence interval
• Making conclusions about the difference of means

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Video transcript

Practice Problems: t-tests

Need a shortcut? Means, standard deviations, and variances can be found at the end of this page.

Means, Standard Deviations, and Variances for the Problems Provided Above

T-test Formula

The t-test formula helps us to compare the average values of two data sets and determine if they belong to the same population or are they different. The t-score is compared with the critical value obtained from the t-table. The large t-score indicates that the groups are different and a small t-score indicates that the groups are similar.

What Is the T-test Formula?

The t-test formula is applied to the sample population. The t-test formula depends on the mean , variance, and  standard deviation of the data being compared. There are 3 types of t-tests that could be performed on the n number of samples collected.

• One-sample test,
• Independent sample t-test and
• Paired samples t-test

The critical value is obtained from the t-table looking for the degree of freedom(df = n-1) and the corresponding α value(usually 0.05 or 0.1). If the t-test obtained statistically > CV then the initial hypothesis is wrong and we conclude that the results are significantly different.

One-Sample T-Test Formula

For comparing the mean of a population \(\overline{x}\) from n samples, with a specified theoretical mean μ, we use a one-sample t-test.

\(t= \dfrac{\overline{x}- μ}{\dfrac{\sigma}{\sqrt{n}}}\)

where σ/√n is the standard error

Independent Sample T-Test

Students t-test is used to compare the mean of two groups of samples. It helps evaluate if the means of the two sets of data are statistically significantly different from each other.

\(t = \dfrac{\overline{x_{1}}-\overline{x_{2}}}{\sqrt{(\dfrac{s_{1}^2}{n_{1}}+\dfrac{s_{2}^2}{{n_{2}}}})}\)

t = Student's t-test

• \(x_{1}\) = mean of first group
• \(x_{2}\)= mean of second group
• \(s_{1}\) = standard deviation of group 1
• \(s_{2}\) = standard deviation of group 1
• \(n_{1}\)= number of observations in group 1
• \(n_{2}\)= number of observations in group 2

Paired Samples T-Test

Whenever two distributions of the variables are highly correlated, they could be pre and post test results from the same people. In such cases, we use the paired samples t-test.

\(t = \dfrac{Σ(x_{1}-x_{2})}{\dfrac{s}{\sqrt{n}}}\)

\(x_{1}-x_{2}\) = Difference mean of the pairs

s= standard deviation

n = sample size

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Examples Using t-test Formula

Example 1: Calculate a t-test for the following data of the number of times people prefer coffee or tea in five time intervals.

Solution: let \(x_{1}\) be the sample of data that prefers coffee and \(x_{2}\) be the sample of data that prefers tea.

let us find the mean, variance and the SD

\(\overline{x_{1}}\) = 31/ 5 = 6.2

\(\overline{x_{2}}\) = 28/5 = 5.6

Σ(x 1 -\(\overline{x_{1}}\)) 2 = 14.8

Σ(x 2 -\(\overline{x_{2}}\)) 2 = 17.2

S 1= 14.8/4 = 3.7

S 2 = 17.2/4 = 4.3

According to the t-test formula,

Applying the known values in the t-test formula, we get

\(t = \dfrac{6.2-5.6}{\sqrt{(\dfrac{3.7}{5}+\dfrac{4.3}{5})}}\)

\(=\dfrac{0.6}{\sqrt{1.6}}\)= 0.6/1.26 = 0.47

Example 2: A company wants to improve its sales. The previous sales data indicated that the average sale of 25 salesmen was $50 per transaction. After training, the recent data showed an average sale of $80 per transaction. If the standard deviation is $15, find the t-score. Has the training provided improved the sales?

\(H_{0}\)accepted hypothesis:the population mean = the claimed value⇒ μ = μ 0

\(H_{0}\)alternate hypothesis: the population mean not equal to the claimed value⇒ μ ≠ μ 0

t - test formula for independent test is \(t= \dfrac{m- μ}{\dfrac{s}{\sqrt{n}}}\)

Mean sale = 80, μ = 50, s= 15 and n= 25

substituting the values, we get t= (80-50)/(15/√25)

t = (30 ×5)/10 = 10

looking at the t-table we find 10 > 1.711 . (I.e. CV for α = 0.05). ∴ the accepted hypothesis is not true. Thus we conclude that the training boosted the sales.

Example 3: A pre-test and post-test conducted during a survey to find the study hours of Patrick on weekends. Calculate the t-score and determine (for α = 0.25) if the pre-test and post-test surveys are significantly different?

According to the t-test formula, we know that \(t = \dfrac{ΣX-Y}{\dfrac{s}{\sqrt{n}}}\)

Σ(X-Y)= -3 = 3

s= Σ(X-Y) 2 /(n-1) = 5 2 /1 = 25

t= 3/(25/2) = 6/25 = 0.24

here degree of freedom is n-1 = 2-1 =1 and the corresponidng critical value in the t-table for α= 0.25, is 1.

Therefore the scores are not significantly different.

FAQs on T-test Formula

How do you calculate the t-test.

The following steps are followed to calculate the t-test.

• Get the data. Find the mean.
• Subtract the mean score from each individual score
• Square the differences.
• Add up all the squared differences.
• Find the variance and standard deviation.
• Key-in the values in the formula: \(t = \dfrac{Σx_{1}- mean}{\dfrac{s}{\sqrt{n}}}\)

What is the Formula for Finding The Independent T-test?

Students t-test is used to compare the mean of two groups of samples.

t = Student's t-test score

\(x_{1}\) = mean of first group and \(x_{2}\)= mean of second group

\(s_{1}\) = standard deviation of group 1 and \(s_{2}\) = standard deviation of group 1

\(n_{1}\)= number of observations in group 1 and \(n_{2}\)= number of observations in group 2

What is a One-Sample t-test?

The one-sample t-test is the statistical test used to determine whether an unknown population mean is different from a specific value. For example, comparing the mean height of the students with respect to the national average height of an adult.

What is a T-test Formula Used For?

We use the T-test Formula to statistically determine if there is a significant difference between the means of two groups that are related in certain aspects. Examples: a gym center tests the weight loss from a few samples, a company hiring candidates is set to determine the skills of 2 candidates from two different universities at the interview, and so on.

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The Two-Sample t -Test

What is the two-sample t -test.

The two-sample t -test (also known as the independent samples t -test) is a method used to test whether the unknown population means of two groups are equal or not.

Is this the same as an A/B test?

Yes, a two-sample t -test is used to analyze the results from A/B tests.

When can I use the test?

You can use the test when your data values are independent, are randomly sampled from two normal populations and the two independent groups have equal variances.

What if I have more than two groups?

Use a multiple comparison method. Analysis of variance (ANOVA) is one such method. Other multiple comparison methods include the Tukey-Kramer test of all pairwise differences, analysis of means (ANOM) to compare group means to the overall mean or Dunnett’s test to compare each group mean to a control mean.

What if the variances for my two groups are not equal?

You can still use the two-sample t- test. You use a different estimate of the standard deviation. 

What if my data isn’t nearly normally distributed?

If your sample sizes are very small, you might not be able to test for normality. You might need to rely on your understanding of the data. When you cannot safely assume normality, you can perform a nonparametric test that doesn’t assume normality.

See how to perform a two-sample t -test using statistical software

• Download JMP to follow along using the sample data included with the software.
• To see more JMP tutorials, visit the JMP Learning Library .

Using the two-sample t -test

The sections below discuss what is needed to perform the test, checking our data, how to perform the test and statistical details.

What do we need?

For the two-sample t -test, we need two variables. One variable defines the two groups. The second variable is the measurement of interest.

We also have an idea, or hypothesis, that the means of the underlying populations for the two groups are different. Here are a couple of examples:

• We have students who speak English as their first language and students who do not. All students take a reading test. Our two groups are the native English speakers and the non-native speakers. Our measurements are the test scores. Our idea is that the mean test scores for the underlying populations of native and non-native English speakers are not the same. We want to know if the mean score for the population of native English speakers is different from the people who learned English as a second language.
• We measure the grams of protein in two different brands of energy bars. Our two groups are the two brands. Our measurement is the grams of protein for each energy bar. Our idea is that the mean grams of protein for the underlying populations for the two brands may be different. We want to know if we have evidence that the mean grams of protein for the two brands of energy bars is different or not.

Two-sample t -test assumptions

To conduct a valid test:

• Data values must be independent. Measurements for one observation do not affect measurements for any other observation.
• Data in each group must be obtained via a random sample from the population.
• Data in each group are normally distributed .
• Data values are continuous.
• The variances for the two independent groups are equal.

For very small groups of data, it can be hard to test these requirements. Below, we'll discuss how to check the requirements using software and what to do when a requirement isn’t met.

Two-sample t -test example

One way to measure a person’s fitness is to measure their body fat percentage. Average body fat percentages vary by age, but according to some guidelines, the normal range for men is 15-20% body fat, and the normal range for women is 20-25% body fat.

Our sample data is from a group of men and women who did workouts at a gym three times a week for a year. Then, their trainer measured the body fat. The table below shows the data.

Table 1: Body fat percentage data grouped by gender

You can clearly see some overlap in the body fat measurements for the men and women in our sample, but also some differences. Just by looking at the data, it's hard to draw any solid conclusions about whether the underlying populations of men and women at the gym have the same mean body fat. That is the value of statistical tests – they provide a common, statistically valid way to make decisions, so that everyone makes the same decision on the same set of data values.

Checking the data

Let’s start by answering: Is the two-sample t -test an appropriate method to evaluate the difference in body fat between men and women?

• The data values are independent. The body fat for any one person does not depend on the body fat for another person.
• We assume the people measured represent a simple random sample from the population of members of the gym.
• We assume the data are normally distributed, and we can check this assumption.
• The data values are body fat measurements. The measurements are continuous.
• We assume the variances for men and women are equal, and we can check this assumption.

Before jumping into analysis, we should always take a quick look at the data. The figure below shows histograms and summary statistics for the men and women.

The two histograms are on the same scale. From a quick look, we can see that there are no very unusual points, or outliers . The data look roughly bell-shaped, so our initial idea of a normal distribution seems reasonable.

Examining the summary statistics, we see that the standard deviations are similar. This supports the idea of equal variances. We can also check this using a test for variances.

Based on these observations, the two-sample t -test appears to be an appropriate method to test for a difference in means.

How to perform the two-sample t -test

For each group, we need the average, standard deviation and sample size. These are shown in the table below.

Table 2: Average, standard deviation and sample size statistics grouped by gender

Without doing any testing, we can see that the averages for men and women in our samples are not the same. But how different are they? Are the averages “close enough” for us to conclude that mean body fat is the same for the larger population of men and women at the gym? Or are the averages too different for us to make this conclusion?

We'll further explain the principles underlying the two sample t -test in the statistical details section below, but let's first proceed through the steps from beginning to end. We start by calculating our test statistic. This calculation begins with finding the difference between the two averages:

$ 22.29 - 14.95 = 7.34 $

This difference in our samples estimates the difference between the population means for the two groups.

Next, we calculate the pooled standard deviation. This builds a combined estimate of the overall standard deviation. The estimate adjusts for different group sizes. First, we calculate the pooled variance:

$ s_p^2 = \frac{((n_1 - 1)s_1^2) + ((n_2 - 1)s_2^2)} {n_1 + n_2 - 2} $

$ s_p^2 = \frac{((10 - 1)5.32^2) + ((13 - 1)6.84^2)}{(10 + 13 - 2)} $

$ = \frac{(9\times28.30) + (12\times46.82)}{21} $

$ = \frac{(254.7 + 561.85)}{21} $

$ =\frac{816.55}{21} = 38.88 $

Next, we take the square root of the pooled variance to get the pooled standard deviation. This is:

$ \sqrt{38.88} = 6.24 $

We now have all the pieces for our test statistic. We have the difference of the averages, the pooled standard deviation and the sample sizes.  We calculate our test statistic as follows:

$ t = \frac{\text{difference of group averages}}{\text{standard error of difference}} = \frac{7.34}{(6.24\times \sqrt{(1/10 + 1/13)})} = \frac{7.34}{2.62} = 2.80 $

To evaluate the difference between the means in order to make a decision about our gym programs, we compare the test statistic to a theoretical value from the t- distribution. This activity involves four steps:

• We decide on the risk we are willing to take for declaring a significant difference. For the body fat data, we decide that we are willing to take a 5% risk of saying that the unknown population means for men and women are not equal when they really are. In statistics-speak, the significance level, denoted by α, is set to 0.05. It is a good practice to make this decision before collecting the data and before calculating test statistics.
• We calculate a test statistic. Our test statistic is 2.80.
• We find the theoretical value from the t- distribution based on our null hypothesis which states that the means for men and women are equal. Most statistics books have look-up tables for the t- distribution. You can also find tables online. The most likely situation is that you will use software and will not use printed tables. To find this value, we need the significance level (α = 0.05) and the degrees of freedom . The degrees of freedom ( df ) are based on the sample sizes of the two groups. For the body fat data, this is: $ df = n_1 + n_2 - 2 = 10 + 13 - 2 = 21 $ The t value with α = 0.05 and 21 degrees of freedom is 2.080.
• We compare the value of our statistic (2.80) to the t value. Since 2.80 > 2.080, we reject the null hypothesis that the mean body fat for men and women are equal, and conclude that we have evidence body fat in the population is different between men and women.

Statistical details

Let’s look at the body fat data and the two-sample t -test using statistical terms.

Our null hypothesis is that the underlying population means are the same. The null hypothesis is written as:

$ H_o:  \mathrm{\mu_1} =\mathrm{\mu_2} $

The alternative hypothesis is that the means are not equal. This is written as:

$ H_o:  \mathrm{\mu_1} \neq \mathrm{\mu_2} $

We calculate the average for each group, and then calculate the difference between the two averages. This is written as:

$\overline{x_1} -  \overline{x_2} $

We calculate the pooled standard deviation. This assumes that the underlying population variances are equal. The pooled variance formula is written as:

The formula shows the sample size for the first group as n 1 and the second group as n 2 . The standard deviations for the two groups are s 1 and s 2 . This estimate allows the two groups to have different numbers of observations. The pooled standard deviation is the square root of the variance and is written as s p .

What if your sample sizes for the two groups are the same? In this situation, the pooled estimate of variance is simply the average of the variances for the two groups:

$ s_p^2 = \frac{(s_1^2 + s_2^2)}{2} $

The test statistic is calculated as:

$ t = \frac{(\overline{x_1} -\overline{x_2})}{s_p\sqrt{1/n_1 + 1/n_2}} $

The numerator of the test statistic is the difference between the two group averages. It estimates the difference between the two unknown population means. The denominator is an estimate of the standard error of the difference between the two unknown population means. 

Technical Detail: For a single mean, the standard error is $ s/\sqrt{n} $  . The formula above extends this idea to two groups that use a pooled estimate for s (standard deviation), and that can have different group sizes.

We then compare the test statistic to a t value with our chosen alpha value and the degrees of freedom for our data. Using the body fat data as an example, we set α = 0.05. The degrees of freedom ( df ) are based on the group sizes and are calculated as:

$ df = n_1 + n_2 - 2 = 10 + 13 - 2 = 21 $

The formula shows the sample size for the first group as n 1 and the second group as n 2 .  Statisticians write the t value with α = 0.05 and 21 degrees of freedom as:

$ t_{0.05,21} $

The t value with α = 0.05 and 21 degrees of freedom is 2.080. There are two possible results from our comparison:

• The test statistic is lower than the t value. You fail to reject the hypothesis of equal means. You conclude that the data support the assumption that the men and women have the same average body fat.
• The test statistic is higher than the t value. You reject the hypothesis of equal means. You do not conclude that men and women have the same average body fat.

t -Test with unequal variances

When the variances for the two groups are not equal, we cannot use the pooled estimate of standard deviation. Instead, we take the standard error for each group separately. The test statistic is:

$ t = \frac{ (\overline{x_1} -  \overline{x_2})}{\sqrt{s_1^2/n_1 + s_2^2/n_2}} $

The numerator of the test statistic is the same. It is the difference between the averages of the two groups. The denominator is an estimate of the overall standard error of the difference between means. It is based on the separate standard error for each group.

The degrees of freedom calculation for the t value is more complex with unequal variances than equal variances and is usually left up to statistical software packages. The key point to remember is that if you cannot use the pooled estimate of standard deviation, then you cannot use the simple formula for the degrees of freedom.

Testing for normality

The normality assumption is more important   when the two groups have small sample sizes than for larger sample sizes.

Normal distributions are symmetric, which means they are “even” on both sides of the center. Normal distributions do not have extreme values, or outliers. You can check these two features of a normal distribution with graphs. Earlier, we decided that the body fat data was “close enough” to normal to go ahead with the assumption of normality. The figure below shows a normal quantile plot for men and women, and supports our decision.

You can also perform a formal test for normality using software. The figure above shows results of testing for normality with JMP software. We test each group separately. Both the test for men and the test for women show that we cannot reject the hypothesis of a normal distribution. We can go ahead with the assumption that the body fat data for men and for women are normally distributed.

Testing for unequal variances

Testing for unequal variances is complex. We won’t show the calculations in detail, but will show the results from JMP software. The figure below shows results of a test for unequal variances for the body fat data.

Without diving into details of the different types of tests for unequal variances, we will use the F test. Before testing, we decide to accept a 10% risk of concluding the variances are equal when they are not. This means we have set α = 0.10.

Like most statistical software, JMP shows the p -value for a test. This is the likelihood of finding a more extreme value for the test statistic than the one observed. It’s difficult to calculate by hand. For the figure above, with the F test statistic of 1.654, the p- value is 0.4561. This is larger than our α value: 0.4561 > 0.10. We fail to reject the hypothesis of equal variances. In practical terms, we can go ahead with the two-sample t -test with the assumption of equal variances for the two groups.

Understanding p-values

Using a visual, you can check to see if your test statistic is a more extreme value in the distribution. The figure below shows a t- distribution with 21 degrees of freedom.

Since our test is two-sided and we have set α = .05, the figure shows that the value of 2.080 “cuts off” 2.5% of the data in each of the two tails. Only 5% of the data overall is further out in the tails than 2.080. Because our test statistic of 2.80 is beyond the cut-off point, we reject the null hypothesis of equal means.

Putting it all together with software

The figure below shows results for the two-sample t -test for the body fat data from JMP software.

The results for the two-sample t -test that assumes equal variances are the same as our calculations earlier. The test statistic is 2.79996. The software shows results for a two-sided test and for one-sided tests. The two-sided test is what we want (Prob > |t|). Our null hypothesis is that the mean body fat for men and women is equal. Our alternative hypothesis is that the mean body fat is not equal. The one-sided tests are for one-sided alternative hypotheses – for example, for a null hypothesis that mean body fat for men is less than that for women.

We can reject the hypothesis of equal mean body fat for the two groups and conclude that we have evidence body fat differs in the population between men and women. The software shows a p -value of 0.0107. We decided on a 5% risk of concluding the mean body fat for men and women are different, when they are not. It is important to make this decision before doing the statistical test.

The figure also shows the results for the t- test that does not assume equal variances. This test does not use the pooled estimate of the standard deviation. As was mentioned above, this test also has a complex formula for degrees of freedom. You can see that the degrees of freedom are 20.9888. The software shows a p- value of 0.0086. Again, with our decision of a 5% risk, we can reject the null hypothesis of equal mean body fat for men and women.

Other topics

If you have more than two independent groups, you cannot use the two-sample t- test. You should use a multiple comparison   method. ANOVA, or analysis of variance, is one such method. Other multiple comparison methods include the Tukey-Kramer test of all pairwise differences, analysis of means (ANOM) to compare group means to the overall mean or Dunnett’s test to compare each group mean to a control mean.

What if my data are not from normal distributions?

If your sample size is very small, it might be hard to test for normality. In this situation, you might need to use your understanding of the measurements. For example, for the body fat data, the trainer knows that the underlying distribution of body fat is normally distributed. Even for a very small sample, the trainer would likely go ahead with the t -test and assume normality.

What if you know the underlying measurements are not normally distributed? Or what if your sample size is large and the test for normality is rejected? In this situation, you can use nonparametric analyses. These types of analyses do not depend on an assumption that the data values are from a specific distribution. For the two-sample t ­-test, the Wilcoxon rank sum test is a nonparametric test that could be used.

Learn Math and Stats with Dr. G

A shortcut is the longest distance between two points.

Independent Samples T-Test By Hand

An independent samples t-test hypothesis test example By Hand

A coffee chain has two locations, one in Queens and one in NYC. The coffee chain owner wants to make sure that all lattes are consistent between the two locations.

A sample of 20 lattes is collected from the Queens store and is determined to have a sample mean of 4.1 oz. of espresso per latte with a sample standard deviation of .12 oz.

A sample of 22 lattes is then taken from the NYC store and is determined to have a sample mean of 4.2 oz. if espresso per latte and a standard deviation of .11 oz.

Use alpha = .05 and run an independent samples t-test to determine if there is a significant difference between the amount of espresso between the two coffee shop lattes.

Step 1: What is Ho and Ha?

Ho: mean espresso in a latte in Queens = mean espresso in a latte in NYC Ho: mean espresso in a latte in Queens ≠ mean espresso in a latte in NYC

Should you use t-test or z-test and why? This is best as a t-test because we are comparing two sample means. Is this a one or two tailed test?

Notice that this is a TWO-TAILED test . We will determine if there is a significant difference.

Our sample size of the Queens sample is n1 = 20 and the std dev s1 is .12

Our sample size of the NYC sample is n2 = 22 and the std dev s2 is .11

Step 2: Calculating the t-test statistic for an independent samples t-test

NOTE: There are three types of t-tests. There is the one sample t-test that compares a single sample to a known population value. There is an independent samples t-test (this example) that compares two samples to each other. There is a paired data (also called correlated data) t-test that compares two samples from data that is related (like pretest score and post test score).

t -test = (sample mean 1 – sample mean 2)/[ sqrt ( s1^2/n1 + s2^2/n2) ]

sample mean1 = 4.1, s1 = .12, n1 = 20

sample mean2 = 4.2 , s2 = .11, n2 = 22

NOTE: s1^2 = (.12)^2 = .12 * .12 = .0144

NOTE: s2^2 = (.11)^2 = .11 * .11 = .0121

t-test = (4.1 – 4.2 ) / [ sqrt (( .12)^2 / 20   + (.11)^2/22) ] =

t-test = (4.1 – 4.2 ) / [ sqrt (.0144 / 20   + .0121/22) ] =

t-test = (4.1 – 4.2 ) / [ sqrt ( .00072    + .00055) ] =

t-test = (4.1 – 4.2 ) / [ sqrt ( .00127 ) ] =

t-test = (4.1 – 4.2 ) / [ .03564 ]

t-test = ( – 0.1 ) / [ .03564 ]

t test = – 2.8058

NOTE: You must use the order of operations when solving any expression.

But – this is not the end of the test!

Step 3: Determine if this value is in a rejection region (reject Ho) or not (do not reject Ho)

Next, using any t-table (these tables are always on the internet) we can get the critical values (tc) for the two tailed test.

Our degrees of freedom for this one sample t-test is :

df = n1 + n2   – 2 = 20 + 22 – 2 = 40

Our alpha value is .05

Our test is two-tailed

Therefore, using any t-table, the two “critical values” that represent the cut-off points for rejection are:

tc = +/- 2.042

This tells is that if our t-test result (which in this case is 13.6) is either bigger than 2.042 or less than -2.042 then we CAN reject the null because we ARE in the rejection region.

-2.8058 <   – 2.042

Therefore, we must Reject Ho

Step 4: Understanding and writing the conclusion – what does this all mean

Recall that Ho says that there is no sig diff between the lattes in Queens and the lattes in NYC.

However, in this case, we REJECT Ho. Our t-test result WAS in the rejection region because the value was smaller than the cut-off of -2.042. In other words, we do not agree with Ho. We do not think that Ho is correct (with a .05 error margin).

Because we reject Ho (do not choose Ho) we then choose Ha.

Ha tells us that there IS A SIG DIFF between the amount of espresso in the Queens lattes and the amount in the NYC lattes.

Statistics Made Easy

Two Sample t-test: Definition, Formula, and Example

A two sample t-test is used to determine whether or not two population means are equal.

This tutorial explains the following:

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

Two Sample t-test: Motivation

Suppose we want to know whether or not the mean weight between two different species of turtles is equal. Since there are thousands of turtles in each population, it would be too time-consuming and costly to go around and weigh each individual turtle.

Instead, we might take a simple random sample of 15 turtles from each population and use the mean weight in each sample to determine if the mean weight is equal between the two populations:

However, it’s virtually guaranteed that the mean weight between the two samples will be at least a little different. The question is whether or not this difference is statistically significant . Fortunately, a two sample t-test allows us to answer this question.

Two Sample t-test: Formula

A two-sample t-test always uses the following null hypothesis:

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

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

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

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

Test statistic:  ( x 1  –  x 2 )  /  s p (√ 1/n 1  + 1/n 2 )

where  x 1  and  x 2 are the sample means, n 1 and n 2  are the sample sizes, and where s p is calculated as:

s p = √  (n 1 -1)s 1 2  +  (n 2 -1)s 2 2  /  (n 1 +n 2 -2)

where s 1 2  and s 2 2  are the sample variances.

If the p-value that corresponds to the test statistic t with (n 1 +n 2 -1) degrees of freedom is less than your chosen significance level (common choices are 0.10, 0.05, and 0.01) then you can reject the null hypothesis.

Two Sample t-test: Assumptions

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

• The observations in one sample should be independent of the observations in the other sample.
• The data should be approximately normally distributed.
• The two samples should have approximately the same variance. If this assumption is not met, you should instead perform Welch’s t-test .
• The data in both samples was obtained using a random sampling method .

Two Sample t-test : Example

Suppose we want to know whether or not the mean weight between two different species of turtles is equal. To test this, will perform a two sample t-test at significance level α = 0.05 using the following steps:

Step 1: Gather the sample data.

Suppose we collect a random sample of turtles from each population with the following information:

• Sample size n 1 = 40
• Sample mean weight  x 1  = 300
• Sample standard deviation s 1 = 18.5
• Sample size n 2 = 38
• Sample mean weight  x 2  = 305
• Sample standard deviation s 2 = 16.7

Step 2: Define the hypotheses.

We will perform the two sample t-test with the following hypotheses:

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

Step 3: Calculate the test statistic  t .

First, we will calculate the pooled standard deviation s p :

s p = √  (n 1 -1)s 1 2  +  (n 2 -1)s 2 2  /  (n 1 +n 2 -2)  = √  (40-1)18.5 2  +  (38-1)16.7 2  /  (40+38-2)  = 17.647

Next, we will calculate the test statistic  t :

t = ( x 1  –  x 2 )  /  s p (√ 1/n 1  + 1/n 2 ) =  (300-305) / 17.647(√ 1/40 + 1/38 ) =  -1.2508

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

According to the T Score to P Value Calculator , the p-value associated with t = -1.2508 and degrees of freedom = n 1 +n 2 -2 = 40+38-2 = 76 is  0.21484 .

Step 5: Draw a conclusion.

Since this p-value is not less than our significance level α = 0.05, we fail to reject the null hypothesis. We do not have sufficient evidence to say that the mean weight of turtles between these two populations is different.

Note:  You can also perform this entire two sample t-test by simply using the Two Sample t-test Calculator .

Additional Resources

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

How to Perform a Two Sample t-test in Excel How to Perform a Two Sample t-test in SPSS How to Perform a Two Sample t-test in Stata How to Perform a Two Sample t-test in R How to Perform a Two Sample t-test in Python How to Perform a Two Sample t-test on a TI-84 Calculator

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

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