I have to solve the following problem:
After data cleansing and using ordinary least squares we derive the following formula:
$\hat{Y}_i=3,85+0,78X_i$
where $Y_i$ is the price of a stock of the i-th company (in \$) and $X_i$ the percentage that the i-th company reinvests (in %) for $i=1,2,\dots,56$. The standard errors of the two coefficients $3,85/0,78$ are $2,25$ and $0,17$ respectively. Furthermore we have: $\sum_{i=1}^{56}(x_i-\bar{x_i})^2=1008$.
Determine the F-test of this regression, $R^2$, and the values $p$ of each coefficient.
Furthermore it states in the example that we shall assume that for the degrees of freedom for the estimation of the distribution $Pr(F>7)=0,01$ and for the distribution t_Student $Pr(\tau>1,75)=0,04$ and $Pr(\tau>1,96)=0,01$.
I spent hours of googling the F-test for linear regressions and the setting was always different than in this example.
I am grateful for any advice on how to approach this problem.
Best regards,
Tobias
My suggestion:
Start with a T-test!
Find the t-value, $t_1$, for testing the hypothesis $$H_0: \beta_1 = 0$$ $$H_1: \beta_1 \ne 0$$ You can find this easily by dividing the coefficient ($0. 78$ in this case) by its standard error ($0.17$).
Then take the newly found value and square it. There is a known theorem that, for simple linear regression, the $F$ critical value and $t_1$ critical value are related by: $$F=t_1^2$$
The F-test will have $1$ and $n-2$ degrees of freedom, respectively.
Next, to find $R^2$, I would use the following formula and solve: $$F=\frac{R^2}{(1-R^2)/(n-2)}$$ This formula is merely one of the variants for the F-Test.
The values $p$ for each coefficient can be found going back to the T-test and the T-distribution, either by using a chart or using a calculator or other statistical software.