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8.2: Hypothesis Testing with t

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• Page ID 7127

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

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

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

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

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

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

\(H_0: μ = 30\)

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

\(H_A: μ > 30\)

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

\[df = N – 1 = 4 – 1 = 3 \nonumber \]

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Independent t-test for two samples

Introduction.

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

Null and alternative hypotheses for the independent t-test

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

H 0 : u 1 = u 2

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

H A : u 1 ≠ u 2

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

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

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

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

Unrelated groups

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

Assumption of normality of the dependent variable

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

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

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

What to do when you violate the normality assumption

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

Assumption of homogeneity of variance

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

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

Overcoming a violation of the assumption of homogeneity of variance

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

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

Reporting the result of an independent t-test

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

Fully reporting your results

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

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

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

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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Choose test type

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

Alternative hypothesis H 1

Test details

Significance level α

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

Degrees of freedom

Calculated as sample size minus one.

Test results

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One Sample T Test – Clearly Explained with Examples | ML+

• October 8, 2020
• Selva Prabhakaran

One sample T-Test tests if the given sample of observations could have been generated from a population with a specified mean.

If it is found from the test that the means are statistically different, we infer that the sample is unlikely to have come from the population.

For example: If you want to test a car manufacturer’s claim that their cars give a highway mileage of 20kmpl on an average. You sample 10 cars from the dealership, measure their mileage and use the T-test to determine if the manufacturer’s claim is true.

By end of this, you will know when and how to do the T-Test, the concept, math, how to set the null and alternate hypothesis, how to use the T-tables, how to understand the one-tailed and two-tailed T-Test and see how to implement in R and Python using a practical example.

Introduction

Purpose of one sample t test, how to set the null and alternate hypothesis, procedure to do one sample t test, one sample t test example, one sample t test implementation, how to decide which t test to perform two tailed, upper tailed or lower tailed.

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The ‘One sample T Test’ is one of the 3 types of T Tests . It is used when you want to test if the mean of the population from which the sample is drawn is of a hypothesized value. You will understand this statement better (and all of about One Sample T test) better by the end of this post.

T Test was first invented by William Sealy Gosset, in 1908. Since he used the pseudo name as ‘Student’ when publishing his method in the paper titled ‘Biometrika’, the test came to be know as Student’s T Test.

Since it assumes that the test statistic, typically the sample mean, follows the sampling distribution, the Student’s T Test is considered as a Parametric test.

The purpose of the One Sample T Test is to determine if a sample observations could have come from a process that follows a specific parameter (like the mean).

It is typically implemented on small samples.

For example, given a sample of 15 items, you want to test if the sample mean is the same as a hypothesized mean (population). That is, essentially you want to know if the sample came from the given population or not.

Let’s suppose, you want to test if the mean weight of a manufactured component (from a sample size 15) is of a particular value (55 grams), with a 99% confidence.

How did we determine One sample T-test is the right test for this?

Because, there is only one sample involved and you want to compare the mean of this sample against a particular (hypothesized) value..

To do this, you need to set up a null hypothesis and an alternate hypothesis .

The null hypothesis usually assumes that there is no difference in the sample means and the hypothesized mean (comparison mean). The purpose of the T Test is to test if the null hypothesis can be rejected or not.

Depending on the how the problem is stated, the alternate hypothesis can be one of the following 3 cases:

• Case 1: H1 : x̅ != µ. Used when the true sample mean is not equal to the comparison mean. Use Two Tailed T Test.
• Case 2: H1 : x̅ > µ. Used when the true sample mean is greater than the comparison mean. Use Upper Tailed T Test.
• Case 3: H1 : x̅ < µ. Used when the true sample mean is lesser than the comparison mean. Use Lower Tailed T Test.

Where x̅ is the sample mean and µ is the population mean for comparison. We will go more into the detail of these three cases after solving some practical examples.

Example 1: A customer service company wants to know if their support agents are performing on par with industry standards.

According to a report the standard mean resolution time is 20 minutes per ticket. The sample group has a mean at 21 minutes per ticket with a standard deviation of 7 minutes.

Can you tell if the company’s support performance is better than the industry standard or not?

Example 2: A farming company wants to know if a new fertilizer has improved crop yield or not.

Historic data shows the average yield of the farm is 20 tonne per acre. They decide to test a new organic fertilizer on a smaller sample of farms and observe the new yield is 20.175 tonne per acre with a standard deviation of 3.02 tonne for 12 different farms.

Did the new fertilizer work?

Step 1: Define the Null Hypothesis (H0) and Alternate Hypothesis (H1)

H0: Sample mean (x̅) = Hypothesized Population mean (µ)

H1: Sample mean (x̅) != Hypothesized Population mean (µ)

The alternate hypothesis can also state that the sample mean is greater than or less than the comparison mean.

Step 2: Compute the test statistic (T)

$$t = \frac{Z}{s} = \frac{\bar{X} – \mu}{\frac{\hat{\sigma}}{\sqrt{n}}}$$

where s is the standard error .

Step 3: Find the T-critical from the T-Table

Use the degree of freedom and the alpha level (0.05) to find the T-critical.

Step 4: Determine if the computed test statistic falls in the rejection region.

Alternately, simply compute the P-value. If it is less than the significance level (0.05 or 0.01), reject the null hypothesis.

Problem Statement:

We have the potato yield from 12 different farms. We know that the standard potato yield for the given variety is µ=20.

x = [21.5, 24.5, 18.5, 17.2, 14.5, 23.2, 22.1, 20.5, 19.4, 18.1, 24.1, 18.5]

Test if the potato yield from these farms is significantly better than the standard yield.

Step 1: Define the Null and Alternate Hypothesis

H0: x̅ = 20

H1: x̅ > 20

n = 12. Since this is one sample T test, the degree of freedom = n-1 = 12-1 = 11.

Let’s set alpha = 0.05, to meet 95% confidence level.

Step 2: Calculate the Test Statistic (T) 1. Calculate sample mean

$$\bar{X} = \frac{x_1 + x_2 + x_3 + . . + x_n}{n}$$

$$\bar{x} = 20.175$$

• Calculate sample standard deviation

$$\bar{\sigma} = \frac{(x_1 – \bar{x})^2 + (x_2 – \bar{x})^2 + (x_3 – \bar{x})^2 + . . + (x_n – \bar{x})^2}{n-1}$$

$$\sigma = 3.0211$$

• Substitute in the T Statistic formula

$$T = \frac{\bar{x} – \mu}{se} = \frac{\bar{x} – \mu}{\frac{\sigma}{\sqrt{n}}}$$

$$T = (20.175 – 20)/(3.0211/\sqrt{12}) = 0.2006$$

Step 3: Find the T-Critical

Confidence level = 0.95, alpha=0.05. For one tailed test, look under 0.05 column. For d.o.f = 12 – 1 = 11, T-Critical = 1.796 .

Now you might wonder why ‘One Tailed test’ was chosen. This is because of the way you define the alternate hypothesis. Had the null hypothesis simply stated that the sample means is not equal to 20, then we would have gone for a two tailed test. More details about this topic in the next section.

Step 4: Does it fall in rejection region?

Since the computed T Statistic is less than the T-critical, it does not fall in the rejection region.

Clearly, the calculated T statistic does not fall in the rejection region. So, we do not reject the null hypothesis.

Since you want to perform a ‘One Tailed Greater than’ test (that is, the sample mean is greater than the comparison mean), you need to specify alternative='greater' in the t.test() function. Because, by default, the t.test() does a two tailed test (which is what you do when your alternate hypothesis simply states sample mean != comparison mean).

The P-value computed here is nothing but p = Pr(T > t) (upper-tailed), where t is the calculated T statistic.

In Python, One sample T Test is implemented in ttest_1samp() function in the scipy package. However, it does a Two tailed test by default , and reports a signed T statistic. That means, the reported P-value will always be computed for a Two-tailed test. To calculate the correct P value, you need to divide the output P-value by 2.

Apply the following logic if you are performing a one tailed test:

For greater than test: Reject H0 if p/2 < alpha (0.05). In this case, t will be greater than 0. For lesser than test: Reject H0 if p/2 < alpha (0.05). In this case, t will be less than 0.

Since it is one tailed test, the real p-value is 0.8446/2 = 0.4223. We do not rejecting the Null Hypothesis anyway.

The decision of whether the computed test statistic falls in the rejection region depends on how the alternate hypothesis is defined.

We know the Null Hypothesis is H0: µD = 0. Where, µD is the difference in the means, that is sample mean minus the comparison mean.

You can also write H0 as: x̅ = µ , where x̅ is sample mean and ‘µ’ is the comparison mean.

Case 1: If H1 : x̅ != µ , then rejection region lies on both tails of the T-Distribution (two-tailed). This means the alternate hypothesis just states the difference in means is not equal. There is no comparison if one of the means is greater or lesser than the other.

In this case, use Two Tailed T Test .

Here, P value = 2 . Pr(T > | t |)

Case 2: If H1: x̅ > µ , then rejection region lies on upper tail of the T-Distribution (upper-tailed). If the mean of the sample of interest is greater than the comparison mean. Example: If Component A has a longer time-to-failure than Component B.

In such case, use Upper Tailed based test.

Here, P-value = Pr(T > t)

Case 3: If H1: x̅ < µ , then rejection region lies on lower tail of the T-Distribution (lower-tailed). If the mean of the sample of interest is lesser than the comparison mean.

In such case, use lower tailed test.

Here, P-value = Pr(T < t)

Hope you are now familiar and clear about with the One Sample T Test. If some thing is still not clear, write in comment. Next, topic is Two sample T test . Stay tuned.

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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 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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In statistics, various tests are used to compare different samples or groups and draw conclusions about populations. These tests, known as statistical tests, focus on analyzing the likelihood or probability of obtaining the observed data under specific assumptions or hypotheses. They provide a framework for assessing evidence in support of or against a particular hypothesis.

A statistical test begins by formulating a null hypothesis (H 0 ) and an alternative hypothesis (H a ). The null hypothesis represents the default assumption, typically stating no effect or no difference, while the alternative hypothesis suggests a specific relationship or effect.

There are different statistical tests like Z-test , T-test, Chi-squared tests , ANOVA , Z-test , and F-test , etc. which are used to compute the p-value. In this article, we will learn about the T-test.

Table of Content

What is T-Test?

Assumptions in t-test, prerequisites for t-test, types of t-tests, one sample t-test, independent sample t-test, paired two-sample t-test, frequently asked questions on t-test.

The t-test is named after William Sealy Gosset’s Student’s t-distribution, created while he was writing under the pen name “Student.”

A t-test is a type of inferential statistic test used to determine if there is a significant difference between the means of two groups. It is often used when data is normally distributed and population variance is unknown.

The t-test is used in hypothesis testing to assess whether the observed difference between the means of the two groups is statistically significant or just due to random variation.

• Independence : The observations within each group must be independent of each other. This means that the value of one observation should not influence the value of another observation. Violations of independence can occur with repeated measures, paired data, or clustered data.
• Normality : The data within each group should be approximately normally distributed i.e the distribution of the data within each group being compared should resemble a normal (bell-shaped) distribution. This assumption is crucial for small sample sizes (n < 30).
• Homogeneity of Variances (for independent samples t-test) : The variances of the two groups being compared should be equal. This assumption ensures that the groups have a similar spread of values. Unequal variances can affect the standard error of the difference between means and, consequently, the t-statistic.
• Absence of Outliers: There should be no extreme outliers in the data as outliers can disproportionately influence the results, especially when sample sizes are small.

Let’s quickly review some related terms before digging deeper into the specifics of the t-test.

A t-test is a statistical method used to compare the means of two groups to determine if there is a significant difference between them. The t-test is a parametric test, meaning it makes certain assumptions about the data. Here are the key prerequisites for conducting a t-test.

Hypothesis Testing :

Hypothesis testing is a statistical method used to make inferences about a population based on a sample of data.

The p-value is the probability of observing a test statistic (or something more extreme) given that the null hypothesis is true.

• A small p-value (typically less than the chosen significance level) suggests that the observed data is unlikely to have occurred by random chance alone, leading to the rejection of the null hypothesis.
• A large p-value suggests that the observed data is likely to have occurred by random chance, and there is not enough evidence to reject the null hypothesis.

Degree of freedom (df):

Significance Level :

The significance level is the predetermined threshold that is used to decide whether to reject the null hypothesis. Commonly used significance levels are 0.05, 0.01, or 0.10. A significance level of 0.05 indicates that the researcher is willing to accept a 5% chance of making a Type I error (incorrectly rejecting a true null hypothesis).

T-statistic :

The t-statistic is a measure of the difference between the means of two groups relative to the variability within each group. It is calculated as the difference between the sample means divided by the standard error of the difference. It is also known as the t-value or t-score.

• If the t-value is large => the two groups belong to different groups. 
• If the t-value is small => the two groups belong to the same group.

T-Distribution

The t-distribution , commonly known as the Student’s t-distribution, is a probability distribution with tails that are thicker than those of the normal distribution.

Statistical Significance

Statistical significance is determined by comparing the p-value to the chosen significance level.

• If the p-value is less than or equal to the significance level, the result is considered statistically significant, and the null hypothesis is rejected.
• If the p-value is greater than the significance level, the result is not statistically significant, and there is insufficient evidence to reject the null hypothesis.

In the context of a t-test, these concepts are applied to compare means between two groups. The t-test assesses whether the means are significantly different from each other, taking into account the variability within the groups. The p-value from the t-test is then compared to the significance level to make a decision about the null hypothesis.

A t-table, or a t-distribution table, is a reference table that provides critical values for the t-test. The table is organized by degrees of freedom and significance levels (usually 0.05 or 0.01). The t-table is used to find the critical t-value corresponding to their specific degrees of freedom and chosen significance level. If the calculated t-value is greater than the critical value from the table, it suggests that the observed difference is statistically significant.

There are three types of t-tests, and they are categorized as dependent and independent t-tests.

• One sample t-test test: The mean of a single group against a known mean.
• Independent samples t-test: compares the means for two groups.
• Paired sample t-test: compares means from the same group at different times (say, one year apart).

One sample t-test is one of the widely used t-tests for comparison of the sample mean of the data to a particularly given value. Used for comparing the sample mean to the true/population mean.

We can use this when the sample size is small. (under 30) data is collected randomly and it is approximately normally distributed. It can be calculated as:

• t = t-value
• x_bar = sample mean
• μ = true/population mean
• σ = standard deviation
• n = sample size

Example Problem

Consider the following example. The weights of 25 obese people were taken before enrolling them into the nutrition camp. The population mean weight is found to be 45 kg before starting the camp. After finishing the camp, for the same 25 people, the sample mean was found to be 75 with a standard deviation of 25. Did the fitness camp work?

One-Sample T-test in Python

The T-value of 6.0 is significantly greater than the critical t-value, leading to rejection of the null hypothesis therefore, we can conclude there is a significant difference in weight before and after the fitness camp. The fitness camp had an effect on the weights of the participants.

The results strongly suggest that the fitness camp was effective in producing a statistically significant change in weight for the participants.

• The T-value and p-value both provide consistent evidence for rejecting the null hypothesis.
• The practical significance should also be considered to understand the real-world impact of this weight change.

An Independent sample t-test, commonly known as an unpaired sample t-test is used to find out if the differences found between two groups is actually significant or just a random occurrence. 

We can use this when:

• the population mean or standard deviation is unknown. (information about the population is unknown)
• the two samples are separate/independent. For eg. boys and girls (the two are independent of each other)

It can be calculated using:

Researchers are investigating whether there is a significant difference in the exam scores of two different teaching methods, A and B. Two independent samples, each representing a different teaching method, have been collected. The objective is to determine if there is enough evidence to suggest that one teaching method leads to higher exam scores compared to the other. Suppose, two independent sample data A and B are given, with the following values. We have to perform the Independent samples t-test for this data.

Two-Sample t-test in Python (Independent)

With T-Value, of 0.989 is less than the critical t-value of 2.1009. Therefore, No significant difference is found between the exam scores of Teaching Method A and Teaching Method B based on the T-value.

With P-Value, of 0.336 is greater than the significance level of 0.05. There is no evidence to reject the null hypothesis, indicating no significant difference between the two teaching methods based on the P-value.

In conclusion, The results suggest that, statistically, there is no significant difference in exam scores between Teaching Method A and Teaching Method B. Therefore, based on this analysis, there is no clear evidence to suggest that one teaching method leads to higher exam scores compared to the other.

Paired sample t-test, commonly known as dependent sample t-test is used to find out if the difference in the mean of two samples is 0. The test is done on dependent samples, usually focusing on a particular group of people or things. In this, each entity is measured twice, resulting in a pair of observations. 

We can use this when :

• Two similar (twin like) samples are given. [Eg, Scores obtained in English and Math (both subjects)]
• The dependent variable (data) is continuous.
• The observations are independent of one another.
• The dependent variable is approximately normally distributed.

It can be calculated using,

• (s_d) is the standard deviation of the differences.
• (n) is the number of paired observations.

Consider the following example. Scores (out of 25) of the subjects Math1 and Math2 are taken for a sample of 10 students. We have to perform the paired sample t-test for this data. 

Paired Two-Sample T-test in Python

The paired sample t-test suggests that there is a statistically significant difference in scores between Math1 and Math2 as T-value of -4.95 is less than the critical t-value of -2.2622 and P-value of 0.00079 is less than the significance level of 0.05. Therefore, based on this analysis, it can be concluded that there is evidence to support the claim that the two sets of scores are different, and the difference is not due to random chance.

The above-discussed types of t-tests are widely used in the fields of research in hospitals by experts to gain important information about the medical data given to them about the effects of various medicines and drugs on the population and help them draw out important inferences regarding the same. However, it is the responsibility of the person to see to it that which t-test would bring out the best results and that all the assumptions of that t-test are adhered to. For any doubt/query, comment below. 

In conclusion, t-test, play a crucial role in hypothesis testing, comparing means, and drawing conclusions about populations. The test can be one-sample, independent two-sample, or paired two-sample, each with specific use cases and assumptions. Interpretation of results involves considering T-values, P-values, and critical values.

These tests aid researchers in making informed decisions based on statistical evidence.

Q. What is the t-test for mean in Python?

The t-test for mean in Python is a statistical method used to determine if there is a significant difference between the means of two groups.

Q. What is the t-test function?

The t-test function is a statistical tool used to compare means and assess the significance of differences between groups, considering factors like sample size and variability.

Q. What is the p-value in t-test Python?

The p-value in a t-test Python indicates the probability of observing the data or more extreme results assuming the null hypothesis is true. A small p-value suggests evidence against the null hypothesis.

Q. Why is it called t-test?

The t-test is named after William Sealy Gosset, who published under the pseudonym “Student.” The name “t” refers to the t-distribution used in the test, particularly applicable for small sample sizes.

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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

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

2 Replies to “Two Sample t-test: Definition, Formula, and Example”

I like the detailed information and simplified in the way I can understand and relate easily. Thank you

It seems a couple of parenthesis is missed at the pooled standard deviation formula. Under square root you have (n1-1)s12 + (n2-1)s22 / (n1+n2-2) but it should be [(n1-1)s12 + (n2-1)s22] / (n1+n2-2) I used square bracket

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