Question 6 of 15, Step 1 of 1

The following table gives the data for the grades on the midterm exam and the grades on the final exam. Determine the equation of the regression line, $\hat{y}=b_{0}+b_{1} x$. Round the slope and $y$-intercept to the nearest thousandth.
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|}
\hline \multicolumn{10}{|c|}{ Grades on Midterm and Final Exams } \\
\hline Grades on Midterm & 74 & 66 & 84 & 93 & 89 & 87 & 70 & 93 & 74 & 77 \\
\hline Grades on Final & 78 & 78 & 74 & 94 & 72 & 90 & 62 & 94 & 82 & 74 \\
\hline
\end{tabular}
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\[
\hat{y}=
\]

Step 1 ：Define the data with midterm grades as [74, 66, 84, 93, 89, 87, 70, 93, 74, 77] and final grades as [78, 78, 74, 94, 72, 90, 62, 94, 82, 74].

Step 2 ：Calculate the means of midterm grades and final grades, which are 80.7 and 79.8 respectively.

Step 3 ：Calculate the terms needed for the numerator and denominator of \(b_{1}\). The numerator terms are [12.06, 26.46, -19.14, 174.66, -64.74, 64.26, 190.46, 174.66, -14.74, 21.46] and the denominator terms are [44.89, 216.09, 10.89, 151.29, 68.89, 39.69, 114.49, 151.29, 44.89, 13.69].

Step 4 ：Calculate \(b_{1}\) and \(b_{0}\) using the formula \(b_{1} = \frac{\sum \text{{numerator terms}}}{\sum \text{{denominator terms}}}\) and \(b_{0} = \text{{mean of final grades}} - b_{1} \times \text{{mean of midterm grades}}\). The calculated \(b_{1}\) is 0.660 and \(b_{0}\) is 26.503.

Step 5 ：\(\boxed{\text{{The equation of the regression line is }} \hat{y}=26.503+0.660x}\)