## What is the two-sample t-test?

The two-sample t-test (additionally known as the independent samples t-test) is a technique offered to test whether the unknown populace suggests of two teams are equal or not.

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## 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 have the right to I use the test?

You have the right to usage the test once your data worths are independent, are randomly sampled from 2 normal populations and also the 2 independent teams have equal variances.

## What if I have more than 2 groups?

Use a multiple compariboy approach. Analysis of variance (ANOVA) is one such strategy. Other multiple compariboy approaches include the Tukey-Kramer test of all pairwise differences, evaluation of suggests (ANOM) to compare team suggests to the overall expect or Dunnett’s test to compare each team expect to a regulate mean.

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

You have the right to still use the two-sample t-test. You use a various estimate of the typical deviation.

## What if my data isn’t almost typically distributed?

If your sample sizes are exceptionally little, you could not have the ability to test for normality. You might should depend on your expertise of the data. When you cannot safely assume normality, you deserve to perdevelop a nonparametric test that doesn’t assume normality.

### What execute we need?

For the two-sample t-test, we require two variables. One variable specifies the two groups. The second variable is the measurement of interemainder.

We also have an concept, or hypothesis, that the means of the underlying populaces for the 2 teams are different. Here are a couple of examples:

We have actually students who sheight English as their initially language and students who do not. All students take a analysis test. Our two teams are the native English speakers and the non-aboriginal speakers. Our measurements are the test scores. Our principle is that the mean test scores for the underlying populations of native and non-indigenous English speakers are not the very same. We desire to recognize if the intend score for the populace of aboriginal English speakers is various from the human being who learned English as a second language.We meacertain the grams of protein in 2 various brands of power bars. Our 2 teams are the two brands. Our measurement is the grams of protein for each energy bar. Our concept is that the mean grams of protein for the underlying populations for the 2 brands may be different. We want to recognize if we have actually proof that the suppose grams of protein for the 2 brands of energy bars is various or not.Two-sample t-test assumptions

To conduct a valid test:

Data worths have to be independent. Measurements for one monitoring do not influence measurements for any type of other monitoring.Data in each team have to be acquired via a random sample from the populace.File in each team are typically distributed.Data worths are consistent.The variances for the two independent teams are equal.

For extremely tiny groups of information, it have the right to be difficult to test these demands. Below, we"ll talk about just how to check the requirements using software program and also what to execute once a requirement isn’t met.

### Checking the data

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

The information worths are independent. The body fat for any kind of one perboy does not depfinish on the body fat for an additional person.We assume the civilization measured represent a straightforward random sample from the population of members of the gym.We assume the information are generally dispersed, and also we can examine this assumption.The data values are body fat measurements. The measurements are consistent.We assume the variances for guys and women are equal, and also we deserve to check this assumption. The 2 histograms are on the same range. From a quick look, we have the right to see that there are no extremely inexplicable points, or outliers. The information look around bell-shaped, so our initial concept of a normal distribution appears reasonable.

Evaluating the summary statistics, we watch that the typical deviations are similar. This supports the concept of equal variances. We have the right to likewise examine this utilizing a test for variances.

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

Without doing any kind of trial and error, we have the right to check out that the averperiods for men and also womales in our samples are not the exact same. But how various are they? Are the avereras “close enough” for us to conclude that intend body fat is the same for the larger populace of males and women at the gym? Or are the averages as well various for us to make this conclusion?

We"ll even more explain the values underlying the two sample t-test in the statistical details section below, but let"s first continue with the actions from start to end. We start by calculating our test statistic. This calculation starts with finding the distinction between the 2 averages:

\$ 22.29 - 14.95 = 7.34 \$

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

Next off, we calculate the pooled typical deviation. This builds an unified estimate of the in its entirety typical 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 imes28.30) + (12 imes46.82)21 \$

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

\$ =frac816.5521 = 38.88 \$

Next off, we take the square root of the pooled variance to gain the pooled conventional deviation. This is:

\$ sqrt38.88 = 6.24 \$

We currently have actually all the pieces for our test statistic. We have the distinction of the averages, the pooled conventional deviation and also the sample sizes. We calculate our test statistic as follows:

\$ t = frac extdistinction of team averages exttypical error of difference = frac7.34(6.24 imes sqrt(1/10 + 1/13)) = frac7.342.62 = 2.80 \$

To evaluate the difference between the indicates in order to make a decision around 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 hazard we are willing to take for proclaiming a far-reaching distinction. For the body fat data, we decide that we are willing to take a 5% risk of saying that the unknown population implies for males and also women are not equal when they really are. In statistics-sheight, the meaning level, deprovided by α, is collection to 0.05. It is an excellent practice to make this decision before collecting the data and also prior to calculating test statistics.We calculate a test statistic. Our test statistic is 2.80.We uncover the theoretical value from the t-distribution based on our null hypothesis which states that the indicates for guys and also womales are equal. Many statistics books have actually look-up tables for the t-distribution. You deserve to also discover tables digital. The a lot of most likely instance is that you will certainly usage software and also will certainly not use printed tables.To find this worth, we require the significance level (α = 0.05) and the levels of freedom. The levels of freedom (df) are based upon the sample sizes of the 2 teams. For the body fat information, this is:\$ df = n_1 + n_2 - 2 = 10 + 13 - 2 = 21 \$The t value via α = 0.05 and also 21 degrees of freedom is 2.080.We compare the worth of our statistic (2.80) to the t value. Due to the fact that 2.80 > 2.080, we disapprove the null hypothesis that the intend body fat for men and also woguys are equal, and also conclude that we have proof body fat in the population is various in between men and womales.

### Statistical details

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

Our null hypothesis is that the underlying population suggests are the exact same. The null hypothesis is composed as:

\$ H_o: mathrmmu_1 =mathrmmu_2 \$

The different hypothesis is that the indicates are not equal. This is written as:

\$ H_o: mathrmmu_1 eq mathrmmu_2 \$

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

\$overlinex_1 - overlinex_2 \$

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

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

The formula mirrors the sample dimension for the initially group as n1 and also the second team as n2. The conventional deviations for the 2 groups are s1 and s2. This estimate enables the two teams to have different numbers of observations. The pooled standard deviation is the square root of the variance and is composed as sp.

What if your sample sizes for the two groups are the same? In this situation, the pooled estimate of variance is sindicate the average of the variances for the 2 groups:

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

The test statistic is calculated as:

\$ t = frac(overlinex_1 -overlinex_2)s_psqrt1/n_1 + 1/n_2 \$

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

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

The formula mirrors the sample dimension for the initially group as n1 and also the second team as n2. Statisticians create the t value via α = 0.05 and also 21 levels of freedom as:

\$ t_0.05,21 \$

The t worth with α = 0.05 and 21 levels of freedom is 2.080. Tbelow are two possible outcomes from our comparison:

The test statistic is reduced than the t value. You fail to reject the hypothesis of equal means. You conclude that the data assistance the assumption that the men and also women have the exact same average body fat.The test statistic is higher than the t worth. You reject the hypothesis of equal means. You do not conclude that men and also woguys have the exact same average body fat.

### t-Test through unequal variances

When the variances for the 2 teams are not equal, we cannot usage the pooled estimate of standard deviation. Instead, we take the traditional error for each team separately. The test statistic is:

\$ t = frac (overlinex_1 - overlinex_2)sqrts_1^2/n_1 + s_2^2/n_2 \$

The numerator of the test statistic is the exact same. It is the difference between the averperiods of the two groups. The denominator is an estimate of the as a whole typical error of the difference in between implies. It is based on the sepaprice traditional error for each team.

The degrees of liberty calculation for the t worth is more facility via unequal variances than equal variances and also is typically left up to statistical software packperiods. The essential allude to remember is that if you cannot use the pooled estimate of standard deviation, then you cannot use the easy formula for the levels of flexibility.

### Testing for normality

The normality presumption is more important once the two groups have small sample sizes than for bigger sample sizes.

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Common distributions are symmetric, which means they are “even” on both sides of the center. Typical distributions execute not have excessive worths, or outliers. You can inspect these two attributes of a normal circulation via graphs. Earlier, we chose that the body fat data was “close enough” to normal to go ahead with the presumption of normality. The number listed below reflects a normal quantile plot for males and also woguys, and supports our decision.