<<echo=FALSE, results=hide>>=
## DATA GENERATION
n <- sample(35:65,1)
mx <- runif(1, 40, 60)
my <- runif(1, 200, 280)
sx <- runif(1, 9, 12)
sy <- runif(1, 44, 50)
r <- round(runif(1, 0.5, 0.9), 2)
x <- rnorm(n, mx, sd = sx)
y <- (r * x/sx + rnorm(n, my/sy - r * mx/sx, sqrt(1 - r^2))) * sy

mx <- round(mean(x))
my <- round(mean(y))
r <- round(cor(x, y), digits = 2)
varx <- round(var(x))
vary <- round(var(y))

b <- r * sqrt(vary/varx)
a <- my - b * mx

X <- round(runif(1, -10, 10) + mx)

## QUESTION/ANSWER GENERATION
sol <- round(a + b * X, 3)
@

\begin{question}
For \Sexpr{n} firms the number of employees $X$ and the amount of
expenses for continuing education $Y$ (in EUR) were recorded. The
statistical summary of the data set is given by:

\begin{center}
\begin{tabular}{lcc} \hline
& Variable $X$ & Variable $Y$ \\ \hline
Mean & \Sexpr{mx}  & \Sexpr{my}  \\
Variance & \Sexpr{varx}  & \Sexpr{vary} \\ \hline
\end{tabular}
\end{center}

The correlation between $X$ and $Y$ is equal to \Sexpr{r}.

Estimate the expected amount of money spent for continuing education
by a firm with \Sexpr{X} employees using least squares regression.
\end{question}


%% SOLUTIONS
\begin{solution}
First, the regression line $y_i = \beta_0 + \beta_1 x_i +
\varepsilon_i$ is determined.  The regression coefficients are given by:
\begin{eqnarray*}
&& \hat \beta_1 = r \cdot \frac{s_y}{s_x} = 
\Sexpr{r} \cdot \sqrt{\frac{\Sexpr{vary}}{\Sexpr{varx}}} = \Sexpr{round(b,5)}, \\
&& \hat \beta_0 = \bar y - \hat \beta_1 \cdot \bar x = 
\Sexpr{my} - \Sexpr{round(b,5)} \cdot \Sexpr{mx} = \Sexpr{round(a,5)}.
\end{eqnarray*}

The estimated amount of money spent by a firm with
\Sexpr{X} employees is then given by:
\begin{eqnarray*}
\hat y = \Sexpr{round(a, 5)} + \Sexpr{round(b, 5)} \cdot \Sexpr{X} = \Sexpr{sol}.
\end{eqnarray*}
\end{solution}

%% META-INFORMATION
%% \extype{num}
%% \exsolution{\Sexpr{fmt(sol, 3)}}
%% \exname{Prediction}
%% \extol{0.01}