# Transforming models and conducting hypothesis tests

The population relationship between $y$ and $x_1, x_2, x_3$ is $$y_t=\beta_0+\beta_1x_{1t}+\beta_{2}x_{2t}+\beta_3x_{3t}+\epsilon_{t}$$ with $\epsilon_t\sim N(0,1)$.

The following is an estimation by OLS: $$y_t=21.2+14.1x_{1t}-14.9x_{2t}+31.6x_{3t}+\epsilon_{t}$$ $$(0.82) \hspace{5mm} (0.65) \hspace{5mm} (0.15) \hspace{5mm} (0.14),$$ $$\textrm{total sum of squares}=29417, \hspace{5mm} RSS=10296, \hspace{5mm} \textrm{and} \hspace{5mm} n=100.$$

Also other estimates are given as follows: \begin{align} y_t=\gamma_0+\gamma_1(x_{1t}-x_{2t})+\gamma_{3}x_{3t}+\epsilon_{t}\\ y_t=\phi_0+\phi_1(x_{1t}-x_{2t}+2x_{3t})+\epsilon_{t}\\ y_t-x_{1t}=\alpha_0+\alpha_1x_{2t}+\alpha_{3}x_{3t}+\epsilon_{t} \end{align} where the $RRS=10334, 110511$ and $27974$ respectively.

a.) Calculate $R^2$ and $\bar{R}^2$.

To do this I use the formula $R^2=1-\frac{RSS}{TSS}$ to obtain $R^2=1-\frac{10296}{29417}=0.65$.

To obtain $\bar{R}^2$ I use the formula $\bar{R}^2=1-\frac{n-1}{n-k-1}$ which gives $\bar{R}^2=1-\frac{100-1}{100-3-1}=-0.0313$.

However I have also seen a formula which gives $R^2=\frac{n-k-1}{n-1}(1-\bar{R}^2)$ giving $R^2=\frac{100-3-1}{100-1}(1-(-0.0313))=1$. Which value of $R^2$ is correct?

b.) Use $R^2$ to perform an F-test for equation (1).

I am unsure of how to define the null and alternative hypothesis?

Also would we use equation 3 to conduct this test with the $F-statistic=\frac{(R^2_{unrestricted}-R^2_{restricted})/q}{(1-R^2_{unrestricted})/(n-k-1)}$? Which values would we use?

c.) Test hypothesis that $\beta_1=15$.

For this, $H_0: \beta_1=15$ and $H_1: \beta_1\neq15$.

Then the t-statistic is $\frac{\hat{\beta}_1-\beta}{\textrm{S.E}(\hat{\beta}_1}=\frac{14.1-15}{0.65}=-1.3846$. Then at the $5\%$ significance level, since $t\epsilon\lbrace-1.96,1.96\rbrace$, we accept $H_0$ and the test is not statistically significant at the $5\%$ significance level. Therefore, $\beta_1=15$.

d.) Test the hypothesis that $\beta_1=-\beta_2$.

So we define $H_0: \beta_1=-\beta_2$ and $H_1: \beta_1\neq-\beta_2$ To conduct this test as a t-test I would transform the model as follows: Let $\beta_1=\gamma-\beta_2$. Then \begin{align*} y_t=\beta_0+(\gamma-\beta_2)x_{1t}+\beta_{2}x_{2t}+\beta_3x_{3t}+\epsilon_{t}\\ =\beta_0+\gamma x_{1t}-\beta_2x_{1t}+\beta_{2}x_{2t}+\beta_3x_{3t}+\epsilon_{t}\\ =\beta_0+\gamma x_{1t}+\beta_2(x_{2t}-x_{1t})+\beta_3x_{3t}+\epsilon_{t}\\ \end{align*} where $z_i=x_{2t}-x_{1t}$.

Then regress $y$ on $z_i$ and test $H_0:\gamma=0$ and $H_1:\gamma\neq0$.

How would you proceed further from here?

e.) Explain how to test the hypothesis that $H_0: \beta_1=-\beta_2=15$.

Here, I think because I transformed $H_0: \beta_1=-\beta_2$ in part d.) to $H_0: \gamma=0$, would I be correct in saying that the null hypothesis for this question would be $H_0: \gamma=15$ and $H_1: \gamma\neq15$?

Would I also be correct in saying that you would expect this test to not be statistically significant because of the results found in part c.)?

Any help will be great. Thank you!

a.) Assuiming that SSR stands for residual sum of squares your formula for $R^2$ is correct. For $\bar R^2$ there are a couple of different ways to write it, one choice is $$\bar R^2 = 1 - (1 - R^2)\frac{n-1}{n-k-1}$$

d.) You are on the right path here, loosely what we want to do is to force the null hypothesis to be true and see how different the $R^2$ is. In this case we force $\beta_2 = -\beta_1$ and rewrite the equation (dropping $t$ subscripts for simplicity)
This is our restricted equation. From $F-statistic=\frac{(R^2_{unrestricted}-R^2_{restricted})/q}{(1-R^2_{unrestricted})/(n-k-1)}$, where $R^2_{unrestricted}$ is from the original model, $R^2_{restricted}$ is from the restricted model that was just derived , and $q$ is the number of restrictions that are imposed. You should be able to calculated all of this from the given information.