Let $x$ be a random variable that represents the percentage of successful free throws a professional basketball player makes in a season. Let $y$ be a random variable that represents the percentage of successful field goals a professional basketball player makes in a season. A random sample of $n=6$ professional basketball players gave the following information.
\begin{tabular}{l|llllll}
\hline$x$ & 63 & 79 & 70 & 80 & 84 & 87 \\
\hline$y$ & 46 & 49 & 45 & 55 & 57 & 58 \\
\hline
\end{tabular}

Find $S_{e}$.
1.994
2.131
2.302
2.522
2.819

Step 1 ：Let's denote the percentage of successful free throws a professional basketball player makes in a season as \(x\), and the percentage of successful field goals as \(y\). We have a random sample of \(n=6\) professional basketball players with the following data: \(x = [63, 79, 70, 80, 84, 87]\) and \(y = [46, 49, 45, 55, 57, 58]\).

Step 2 ：We first calculate the sums needed for the formulas: \(sum_x = 463\), \(sum_y = 310\), \(sum_xy = 24153\), and \(sum_x2 = 36135\).

Step 3 ：Next, we calculate the slope (\(b1\)) and y-intercept (\(b0\)) of the regression line using the formulas: \(b1 = \frac{n*sum_xy - sum_x*sum_y}{n*sum_x2 - sum_x^2} = 0.5686194182712003\) and \(b0 = \frac{sum_y - b1*sum_x}{n} = 7.78820155673904\).

Step 4 ：We then calculate the residuals for each data point, which are the differences between the observed and predicted values of \(y\): \(residuals = [ 2.38877509, -3.7091356, -2.59156084, 1.72224498, 1.44776731, 0.74190905]\).

Step 5 ：Finally, we calculate the standard error of the estimate (Se) using the formula: \(Se = \sqrt{\frac{\sum residuals^2}{n-2}}\), which gives us \(Se = 2.8192511375627607\).

Step 6 ：So, the standard error of the estimate (Se) for the given data is \(\boxed{2.819}\).