## Two sample t test with unknown and equal variances

In this tutorial we will discuss two sample t-test for testing difference between two population means when the population variances are unknown but equal.

## Two sample t test with unknown and equal variances

Let $\overline{x}_1$ be the sample mean and $s_1$ be the sample standard deviation of a random sample of size $n_1$ from a population with mean $\mu_1$ and variance $\sigma^2_1$.

Let $\overline{x}_2$ be the sample mean and $s_2$ be the sample standard deviation of a random sample of size $n_2$ from a population with mean $\mu_2$ and variance $\sigma^2_2$.

Suppose the variances $\sigma^2_1$ and $\sigma^2_2$ are unknown but equal.

## Assumptions

Assumptions for two sample $t$-test are as follows:

a. The population from which, the two samples drawn are Normal distributions.

b. The two population variances are unknown but equal.

## Procedure

We wish to test the hypothesis $H_0 : \mu_1 = \mu_2$.

The standard error of difference between means is

$$SE(\overline{x}_1-\overline{x}_2) = s_p\sqrt{\frac{1}{n_1}+ \frac{1}{n_2}}$$

where $s_p$ is the pooled standard deviation and is given by $$s_p =\sqrt{\frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1+n_2-2}}.$$

## Step by Step Procedure

The step by step hypothesis testing procedure is as follows:

### Step 1 State the hypothesis testing problem

The hypothesis testing problem can be structured in any one of the three situations as follows:

Situation Hypothesis Testing Problem
Situation A : $H_0: \mu_1=\mu_2$ against $H_a : \mu_1 < \mu_2$ (Left-tailed)
Situation B : $H_0: \mu_1=\mu_2$ against $H_a : \mu_1 > \mu_2$ (Right-tailed)
Situation C : $H_0: \mu_1=\mu_2$ against $H_a : \mu_1 \neq \mu_2$ (Two-tailed)

### Step 2 Define the test statistic

The test statistic for testing above hypothesis is $$\begin{eqnarray*} t & =& \frac{(\overline{x}_1-\overline{x}_2)-(\mu_1-\mu_2)}{SE(\overline{x}_1-\overline{x}_2)}\\\\ & =& \frac{(\overline{x}_1-\overline{x}_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}} \end{eqnarray*}$$ The test statistic $t$ follows Students’ $t$ distribution with $n_1+n_2-2$ degrees of freedom.

### Step 4 Determine the critical values

For the specified value of $\alpha$ determine the critical region depending upon the alternative hypothesis.

• For left-tailed alternative hypothesis: Find the $t$-critical value using \begin{aligned} P(t<-t_\alpha) = \alpha. \end{aligned}
• For right-tailed alternative hypothesis: $t_\alpha$. \begin{aligned} P(t>t_\alpha) = \alpha. \end{aligned}
• For two-tailed alternative hypothesis: $t_{\alpha/2}$.

$$P(t<-t_{\alpha/2} \text{ or } t> t_{\alpha/2}) = \alpha.$$

### Step 5 Computation

Compute the test statistic under the null hypothesis $H_0$ using equation \begin{aligned} t_{obs} &= \frac{(\overline{x}_1-\overline{x}_2)-0}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}} \end{aligned}

### Step 6 Decision (Traditional Approach)

It is based on the critical values.

• For left-tailed alternative hypothesis: Reject $H_0$ if $t_{obs}\leq -t_\alpha$.
• For right-tailed alternative hypothesis: Reject $H_0$ if $t_{obs}\geq t_\alpha$.
• For two-tailed alternative hypothesis: Reject $H_0$ if $|t_{obs}|\geq t_{\alpha/2}$.

OR

### Step 6 Decision ($p$-value Approach)

It is based on the $p$-value.

Alternative Hypothesis Type of Hypothesis $p$-value
$H_a: \mu_1<\mu_2$ Left-tailed $p$-value $= P(t\leq t_{obs})$
$H_a: \mu_1>\mu_2$ Right-tailed $p$-value $= P(t\geq t_{obs})$
$H_a: \mu_1\neq \mu_2$ Two-tailed $p$-value $= 2P(t\geq abs(t_{obs}))$

If $p$-value is less than $\alpha$, then reject the null hypothesis $H_0$ at $\alpha$ level of significance, otherwise fail to reject $H_0$ at $\alpha$ level of significance.