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 & 71 & 62 & 78 & 94 & 83 & 81 & 80 & 94 & 85 & 62 \\ \hline Grades on Final & 88 & 79 & 88 & 91 & 80 & 70 & 71 & 93 & 65 & 77 \\ \hline \end{tabular} Copy Data Answer Keypad Keyboard Shortcuts \[ \hat{y}=\square \] Submit Answer 02023 Hawkes Learning
Solution
Step 1 ：Given data for grades on midterm and final exams as follows: Midterm grades: [71, 62, 78, 94, 83, 81, 80, 94, 85, 62] and Final grades: [88, 79, 88, 91, 80, 70, 71, 93, 65, 77].
Step 2 ：Calculate the number of data points, which is 10.
Step 3 ：Calculate the sums of midterm grades, final grades, product of midterm and final grades, and square of midterm grades. The sums are: sum of midterm grades = 790, sum of final grades = 802, sum of product of midterm and final grades = 63595, sum of square of midterm grades = 63560.
Step 4 ：Calculate the slope (b1) and y-intercept (b0) of the regression line using the formulas: \(b1 = \frac{n*sum\_xy - sum\_x*sum\_y}{n*sum\_x2 - sum\_x^2}\) and \(b0 = \frac{sum\_y - b1*sum\_x}{n}\). The calculated slope and y-intercept are approximately 0.206 and 63.919, respectively.
Step 5 ：Round the slope and y-intercept to the nearest thousandth to get 0.206 and 63.919, respectively.
Step 6 ：Substitute the slope and y-intercept into the equation of the regression line, \(\hat{y}=b_{0}+b_{1} x\), to get the final equation: \(\hat{y}=63.919+0.206x\).
Step 7 ：Final Answer: \(\boxed{\hat{y}=63.919+0.206x}\)
