1. ## Hypothesis Testing Regression

A sample of data from 40 households in a particular area in 2007 yielded the following information, where X is disposeable income per head and Y is disposeable expenditure per head.

$\sum X_i = 200[$

$\sum Y_i= 160$

$\sum X_iY_i=875$

$\sum X^{2}_i =1100$

$\sum(Y_i - \bar{Y})^{2} = 75.25$

I am asked to show that the nul hypothesis $H_0 : \beta = 0.9$ will not be rejected against either a one or two tailed test.

This is what i have so far, but i think i might have got lost a little bit along the way ! Hopefully someone can point me in the right direction, thanks.

$H_0 : \beta = 0.9$

$H_1 : \beta \not = 0.9$

Significence level is 5 %

For my critical value i will use the t statistic with n - 2 degress of freedom, so for the one tailed test, t = 1.684, two-tailed, t = 2.021. (my statistical table dont give the value for 38 so i have used the value for 40.)

Calculate test statistic

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

Where $s_b^{2} = \frac {s_e^{2}}{sum(X-\bar{X})^{2}}$

Where $s_e^{2} = \frac {ESS}{n-2}$

Solving these gives $s_e^2 = \frac{75.25}{38} = 1.98$

$s_b^{2} = \frac {1.98}{1100 -1000} = 0.0198$

$\sqrt {0.0198} = 0.1407$

$t = \frac {0.75 - 0.9}{0.1407} = -1.066$

I reach this point and then i guess i have to compare the t score with some value to see if it is rejected or not ?

2. You reject the null hypothesis if the test statistic is inside either tail.

With a level of significance of 5% a two-tailed test means that the integration of each tail will be equal to 0.025, so the 2.021 is the y such that:

Integral (0 to y) of t-student(n-2)(x) dx = 0.975

Since the t-student is symmetric the other tail goes from -infinity to -2.021. And our value is out of there.

A similar reasoning is applied to a one-tailed test.

3. Thanks for that, however ive just realised that my calculations are horribly wrong.