## Two sample t test for means with unknown and unequal variances

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

## Two sample t test for means with unknown and unequal 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 and unequal.

## 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 and unequal.

## Step by Step Procedure

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

The standard error of difference between means is
```
$$
\begin{eqnarray*}
SE(\overline{x}_1 - \overline{x}_2) &=& \sqrt{ \frac{s_1^2}{n_1}+ \frac{s_2^2}{n_2}}.
\end{eqnarray*}
$$
```

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
```
$$
\begin{eqnarray*}
t & =& \frac{(\overline{x}_1-\overline{x}_2)-(\mu_1-\mu_2)}{SE(\overline{x}_1-\overline{x}_2)}\\\\
& =& \frac{(\overline{x}_1-\overline{x}_2)-(\mu_1-\mu_2)}{\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}}
\end{eqnarray*}
$$
```

The test statistic $t$ follows Students’ $t$ distribution with $\nu$ degrees of freedom, where
```
$$
\begin{eqnarray*}
\nu & =& \frac{\bigg(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\bigg)^2}{\frac{s_1^4}{n_1^2(n_1-1)}+\frac{s_2^4}{n_2^2(n_2-1)}}
\end{eqnarray*}
$$
```

The value of $\nu$ can be rounded to the nearest integer.

### Step 3 Specify the level of significance $\alpha$

### 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
```
$$
\begin{eqnarray*}
t_{obs} &= & \frac{(\overline{x}_1-\overline{x}_2)-0}{\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}}
\end{eqnarray*}
$$
```

### 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.