Two Sample t-test (Independent Sample with Unequal Variances)
In this tutorial we will discuss some numerical examples on two sample t test for difference between two population means when the population variances are unknown and unequal.
t-test calculator for two means
The $t$-test calculator for testing two population means makes it easy to calculate the test statistic, $t$ critical value and the $p$-value given the sample information, level of significance and the type of alternative hypothesis (i.e. left-tailed, right-tailed or two-tailed.)
| t test Calculator for two means | ||
|---|---|---|
| Sample 1 | Sample 2 | |
| Mean | ||
| Standard Deviation | ||
| Sample Size | ||
| Variances | Equal | Unequal |
| Level of Significance ($\alpha$) | ||
| Tail | Left tailed Right tailed Two tailed |
|
| Results | ||
| Standard Error of Diff. of Means: | ||
| Test Statistics t: | ||
| Degrees of Freedom: | ||
| t-critical value(s): | ||
| p-value: | ||
How to use $t$-test calculator for testing two means?
Step 1 - Enter the sample mean for first sample $\overline{X}_1$ and second sample $\overline{X}_2$
Step 2 - Enter the sample standard deviations for first sample $s_1$ and second sample $s_2$
Step 3 - Enter the sample size for first sample $n_1$ and second sample $n_2$
Step 4 - Select whether variances are equal or unequal
Step 5 - Enter the level of significance $\alpha$
Step 6 - Select the alternative hypothesis (left-tailed / right-tailed / two-tailed)
Step 7 - Click on "Calculate" button to get the result
Independent sample $t$-test Example 1
A statistician claims that the average score on logical reasoning test taken by students who major in Physics is less than that of students who major in English. The result of the exams, given to 22 Physics students and 33 English students, is shown here. Is there enough evidence to reject the statistician's claim at $\alpha= 0.05$? Assume that the standard deviations for the two populations are not equal.
| . | Physics Major | English Major |
|---|---|---|
| Sample Mean | 85 | 76 |
| sample SD | 19 | 23 |
| Sample size | 22 | 33 |
Solution
Given that the sample size $n_1 = 22$, $n_2 = 33$, sample mean $\overline{x}_1= 85$, $\overline{x}_2= 76$, sample standard deviation $s_1 = 19$ and $s_2 = 23$.
Step 1 State the hypothesis testing problem
The hypothesis testing problem is
$H_0 : \mu_1 = \mu_2$ against $H_1 : \mu_1 < \mu_2$ ($\textit{left-tailed}$)
Step 2 Define test statistic
The test statistic for testing above hypothesis testing problem is
$$ \begin{aligned} t=\frac{(\overline{x}_1 -\overline{x}_1)-(\mu_1 - \mu_2)}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}} \end{aligned} $$
The test statistic $t$ follows Students' $t$ distribution with $\nu$ degrees of freedom, where
$$ \begin{aligned} \nu = \frac{\bigg(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\bigg)^2}{\frac{s_1^4}{n_1^2(n_1-1)}+\frac{s_2^4}{n_2^2(n_2-1)}}=50 \end{aligned} $$
rounded to nearest integer.
Step 3 Level of significance
The significance level is $\alpha = 0.05$.
Step 4 Determine the critical value
As the alternative hypothesis is $\textit{left-tailed}$, the critical value of $t$ using $\alpha = 0.05$ and degrees of freedom $=50$ $\text{is}$ $\text{-1.676}$.
The rejection region (i.e. critical region) is $\text{t < -1.676}$.
Step 5 Computation
The test statistic for testing above hypothesis testing problem under the null hypothesis is
$$ \begin{aligned} t&=\frac{(\overline{x}_1 -\overline{x}_1)-0}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}\\ &= \frac{(85-76)}{\sqrt{\frac{19^2}{22}+\frac{23^2}{33}}}\\ &= 1.5802 \end{aligned} $$
Step 6 Decision (Traditional approach)
The rejection region (i.e. critical region) is $\text{t < -1.676}$. The test statistic is $t =1.5802$ which falls $\text{outside}$ the critical region, we $\textit{fail to reject}$ the null hypothesis.
OR
Step 6 Decision ($p$-value approach)
The test is $\textit{left-tailed}$ test, so p-value is the area to the $\textit{left}$ of the test statistic ($t=1.5802$). That is p-value = $P(t\leq 1.5802 ) = 0.9398$.
The p-value is $0.9398$ which is $\textit{greater than}$ the significance level of $\alpha = 0.05$, we $\textit{fail to reject}$ the null hypothesis.
Independent sample $t$-test Example 2
Suppose we wanted to test the hypothesis that a control group of cancer patients (Group 1) would report higher mean pain ratings than an experimental group receiving special massage treatments (Group 2). Use the following information to test the hypothesis at $\alpha =0.05$:
| . | Control (Group 1) | Treatment (Group 2) |
|---|---|---|
| Sample mean | 78.1 | 75.1 |
| Sample SD | 42.1 | 39.7 |
| Sample size | 25 | 25 |
Solution
Given that the sample size $n_1 = 25$, $n_2 = 25$, sample mean $\overline{x}_1= 78.1$, $\overline{x}_2= 75.1$, sample standard deviation $s_1 = 42.1$ and $s_2 = 39.7$.
Step 1 State the hypothesis testing problem
The hypothesis testing problem is
$H_0 : \mu_1 = \mu_2$ against $H_1 : \mu_1 > \mu_2$ ($\textit{right-tailed}$)
Step 2 Define test statistic
The test statistic for testing above hypothesis testing problem is
$$ \begin{aligned} t=\frac{(\overline{x}_1 -\overline{x}_1)-(\mu_1 - \mu_2)}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}} \end{aligned} $$
The test statistic $t$ follows Students' $t$ distribution with $\nu$ degrees of freedom, where
$$ \begin{aligned} \nu = \frac{\bigg(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\bigg)^2}{\frac{s_1^4}{n_1^2(n_1-1)}+\frac{s_2^4}{n_2^2(n_2-1)}}=48 \end{aligned} $$
rounded to nearest integer.
Step 3 Level of significance
The significance level is $\alpha = 0.05$.
Step 4 Determine the critical value
As the alternative hypothesis is $\textit{right-tailed}$, the critical value of $t$ using $\alpha = 0.05$ and degrees of freedom $=48$ $\text{is}$ $\text{1.677}$.
The rejection region (i.e. critical region) is $\text{t > 1.677}$.
Step 5 Computation
The test statistic for testing above hypothesis testing problem under the null hypothesis is
$$ \begin{aligned} t&=\frac{(\overline{x}_1 -\overline{x}_1)-0}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}\\ &= \frac{(78.1-75.1)}{\sqrt{\frac{42.1^2}{25}+\frac{39.7^2}{25}}}\\ &= 0.2592 \end{aligned} $$
Step 6 Decision (Traditional approach)
The rejection region (i.e. critical region) is $\text{t > 1.677}$. The test statistic is $t =0.2592$ which falls $\text{outside}$ the critical region, we $\textit{fail to reject}$ the null hypothesis.
OR
Step 6 Decision ($p$-value approach)
The test is $\textit{right-tailed}$ test, so p-value is the area to the $\textit{right}$ of the test statistic ($t=0.2592$). That is p-value = $P(t\geq 0.2592 ) = 0.3983$.
The p-value is $0.3983$ which is $\textit{greater than}$ the significance level of $\alpha = 0.05$, we $\textit{fail to reject}$ the null hypothesis.
Independent sample $t$-test Example 3
Two methods were used to measure the brightness of independent clay samples. The following table display the summary statistics for the two independent samples. We are interested in the difference between the two population means for the two methods. Assume that brightness measurements are normally distributed. Also assume that the population variances are unequal.
Method A : $\overline{x}_1= 91.6$, $s_1 = 2.3$ and $n_1 = 12$
Method B : $\overline{x}_2= 92.5$, $s_2 = 1.6$ and $n_2 = 12$
Setup the hypothesis problem and test at $\alpha = 0.01$ level of significance.
Solution
Given that the sample size $n_1 = 12$, $n_2 = 12$, sample mean $\overline{x}_1= 91.6$, $\overline{x}_2= 92.5$, sample standard deviation $s_1 = 2.3$ and $s_2 = 1.6$.
Step 1 State the hypothesis testing problem
The hypothesis testing problem is
$H_0 : \mu_1 = \mu_2$ against $H_1 : \mu_1 \neq \mu_2$ ($\textit{two-tailed}$)
Step 2 Define test statistic
The test statistic for testing above hypothesis testing problem is
$$ \begin{aligned} t=\frac{(\overline{x}_1 -\overline{x}_1)-(\mu_1 - \mu_2)}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}} \end{aligned} $$
The test statistic $t$ follows Students' $t$ distribution with $\nu$ degrees of freedom, where
$$ \begin{aligned} \nu = \frac{\bigg(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\bigg)^2}{\frac{s_1^4}{n_1^2(n_1-1)}+\frac{s_2^4}{n_2^2(n_2-1)}}=20 \end{aligned} $$
rounded to nearest integer.
Step 3 Level of significance
The significance level is $\alpha = 0.01$.
Step 4 Determine the critical value
As the alternative hypothesis is $\textit{two-tailed}$, the critical value of $t$ using $\alpha = 0.01$ and degrees of freedom $=20$ $\text{are}$ $\text{-2.845 and 2.845}$.
The rejection region (i.e. critical region) is $\text{t < -2.845 or t > 2.845}$.
Step 5 Computation
The test statistic for testing above hypothesis testing problem under the null hypothesis is
$$ \begin{aligned} t&=\frac{(\overline{x}_1 -\overline{x}_1)-0}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}\\ &= \frac{(91.6-92.5)}{\sqrt{\frac{2.3^2}{12}+\frac{1.6^2}{12}}}\\ &= -1.1128 \end{aligned} $$
Step 6 Decision (Traditional approach)
The rejection region (i.e. critical region) is $\text{t < -2.845 or t > 2.845}$. The test statistic is $t =-1.1128$ which falls $\text{outside}$ the critical region, we $\textit{fail to reject}$ the null hypothesis.
OR
Step 6 Decision ($p$-value approach)
The test is $\textit{two-tailed}$ test, so p-value is the area to the $\textit{extreme}$ of the test statistic ($t=-1.1128$). That is p-value = $2*P(t\geq 1.1128 ) = 0.279$.
The p-value is $0.279$ which is $\textit{greater than}$ the significance level of $\alpha = 0.01$, we $\textit{fail to reject}$ the null hypothesis.
Endnote
In this tutorial, you learned the about how to solve numerical examples on $t$-test for testing two population means with unknown and unequal variances. You also learned about the step by step procedure to apply $t$-test for testing two population means and how to use $t$-test calculator for testing two population means to get the value of test statistic, p-value, and t-critical value.
To learn more about other hypothesis testing problems, hypothesis testing calculators and step by step procedure, please refer to the following tutorials:
Let me know in the comments if you have any questions on $t$-test calculator for two means (unequal variances) with examples and your thought on this article.
