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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

Babypopo
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1 Answers1

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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.

WaveX
  • 5,440
  • Thanks for your response! The standard error for the first coefficient is $2.25$. So we have: $t_1=\frac{0,78}{0,17}=4,588$ and $t_1^2=21,05$. – Babypopo Oct 29 '17 at 01:10
  • Make sure to check again, for your scenario I originally wrote $3.85$ which was wrong coefficient. – WaveX Oct 29 '17 at 01:12
  • Thanks for your response! The standard error for $\beta_1$ is $0,17$. So we have: $t_1=\frac{0,78}{0,17}=4,588$ and $t_1^2=21,05=F$. Furthermore it states in the example that we shall assume that for the degrees of freedom for the estimation of the distribution $Pr(F>7)=0,01$. That means that we reject $H_0$? – Babypopo Oct 29 '17 at 01:22