Student's t-test

If $C_k\sim {\chi }_k^2$ and $Z\sim N(0,1)$ are independently distributed, then

$$T_k=\frac{Z}{\sqrt{\frac{C_k}{k}}}$$

Is $t_k$ distributed.

We could consider Student’s t-test as an extension of Normal Distribution

Construct the z-score:

$$z=\frac{b-{\beta }_0}{\sqrt{Var(b)}}$$

As a test tool to test the hypothesis that $\beta ={\beta }_0$ (See Hypothesis test). But this is valid only when we know the disturbance Variance, which is ${\sigma }^2$.

Application

The question is we usually don’t know the disturbance variance, so we have to estimate it. An unbiased estimator of the error variance, ${\sigma }^2$, is:

$$\widehat{{\sigma }^2}=s_e^2=\frac{\sum e_i^2}{n-2}=\frac{SSR}{n-2}$$

Using this we could obtain the unbiased estimator of the variance of $b$ as:

$$s_b^2=\frac{s_e^2}{\sum (x_i-\bar{x}{)}^2}$$

We set up a formal test:

• $H_0:\beta ={\beta }_0$: The null hypothesis
• $H_1:\beta \ne {\beta }_0$: The alternative hypothesis.

Recall the the test statistic of Student’s t-test: ^acd1c6

$$z=\frac{b-{\beta }_0}{\sqrt{Var(b)}}$$

Now becomes:

$$t=\frac{b-\beta }{s_b}$$

Recall that the Student’s t-test follow the distribution with $n-2$ degrees of freedom. That is $t\sim t_{n-2}$

Confidence Interval

The $95\%$ Confidence Interval for $\beta$ is given by:

$$b\pm s_b\times c_{.025}$$

Recall that we it is a two-sided test, so we use .025