\begin{question}
Using the data provided in \url{regression.csv} estimate a linear regression of
\texttt{y} on \texttt{x1} and \texttt{x2}. Answer the following questions.
\begin{answerlist}
\item Proportion of variance explained (in percent):
\item F-statistic:
\item Characterize in your own words how the response \texttt{y} depends on the regressors \texttt{x1} and \texttt{x2}.
\item Upload the R script you used to analyze the data.
\end{answerlist}
\end{question}
\begin{solution}
The presented results describe a semi-logarithmic regression.
\begin{Schunk}
\begin{Soutput}
Call:
lm(formula = log(y) ~ x1 + x2, data = d)
Residuals:
Min 1Q Median 3Q Max
-2.68802 -0.67816 -0.01803 0.68866 2.35064
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -0.06802 0.13491 -0.504 0.616
x1 1.37863 0.13351 10.326 9.34e-15 ***
x2 -0.21449 0.13995 -1.533 0.131
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Residual standard error: 1.052 on 58 degrees of freedom
Multiple R-squared: 0.6511, Adjusted R-squared: 0.6391
F-statistic: 54.12 on 2 and 58 DF, p-value: 5.472e-14
\end{Soutput}
\end{Schunk}
The mean of the response \texttt{y} increases with increasing \texttt{x1}.
If \texttt{x1} increases by 1 unit then a change of \texttt{y} by about 296.94 percent can be expected.
Also, the effect of \texttt{x1} is significant at the 5 percent level.
Variable \texttt{x2} has no significant influence on the response at 5 percent level.
The R-squared is 0.6511 and thus 65.11 percent of the
variance of the response is explained by the regression.
The F-statistic is 54.12.
\begin{answerlist}
\item Proportion of variance explained: 65.11 percent.
\item F-statistic: 54.12.
\item Characterization: semi-logarithmic.
\item R code.
\end{answerlist}
\end{solution}
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