Question 5 of 15, Step 1 of 1
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The following table gives the data for the hours students spent on homework and their grades on the first test. The equation of the regression line for this data is $\hat{y}=46.027+1.061 x$. This equation is appropriate for making predictions at the 0.01 level of significance. If a student spent 14 hours on their homework, make a prediction for their grade on the first test. Round your prediction to the nearest whole number.
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|}
\hline \multicolumn{10}{|c|}{ Hours Spent on Homework and Test Grades } \\
\hline Hours Spent on Homework & 28 & 11 & 38 & 50 & 15 & 39 & 40 & 35 & 10 & 26 \\
\hline Grade on Test & 75 & 62 & 83 & 93 & 54 & 84 & 92 & 89 & 50 & 88 \\
\hline
\end{tabular}
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So, the predicted grade for a student who spent 14 hours on their homework is \(\boxed{61}\).
Step 1 ：The question is asking to predict the grade of a student who spent 14 hours on their homework using the given regression line equation. The regression line equation is a linear equation in the form of \(y = mx + c\), where \(m\) is the slope of the line, \(x\) is the independent variable (in this case, hours spent on homework), and \(c\) is the y-intercept.
Step 2 ：Given that the slope \(m = 1.061\), the intercept \(c = 46.027\), and the student spent 14 hours on their homework, we substitute these values into the equation.
Step 3 ：The predicted grade is then calculated as \(46.027 + 1.061 \times 14\).
Step 4 ：Rounding this to the nearest whole number, we get the predicted grade as 61.
Step 5 ：So, the predicted grade for a student who spent 14 hours on their homework is \(\boxed{61}\).
