Answer :
To determine which point is farthest from the line of best fit, we need to look at the residuals. The residual for a data point is the difference between the observed value and the value predicted by the line of best fit.
We'll follow these steps:
1. Residuals from the table:
[tex]\[ \begin{array}{|c|c|c|} \hline x & y & \text{Residual} \\ \hline 1 & 3.3 & 0.68 \\ \hline 2 & 5 & 0.04 \\ \hline 3 & 6.2 & -1.1 \\ \hline 4 & 9 & -0.64 \\ \hline 5 & 13 & 1.02 \\ \hline \end{array} \][/tex]
2. Absolute residuals:
To find the point farthest from the line, we consider the magnitude (absolute value) of the residuals:
- For [tex]\(x = 1\)[/tex], the residual is [tex]\( \left|0.68\right| = 0.68 \)[/tex]
- For [tex]\(x = 2\)[/tex], the residual is [tex]\( \left|0.04\right| = 0.04 \)[/tex]
- For [tex]\(x = 3\)[/tex], the residual is [tex]\( \left|1.1\right| = 1.1 \)[/tex]
- For [tex]\(x = 4\)[/tex], the residual is [tex]\( \left|0.64\right| = 0.64 \)[/tex]
- For [tex]\(x = 5\)[/tex], the residual is [tex]\( \left|1.02\right| = 1.02 \)[/tex]
3. Compare the absolute residuals:
Now, we compare the absolute residuals to find the maximum value:
- [tex]\( 0.68 \)[/tex]
- [tex]\( 0.04 \)[/tex]
- [tex]\( 1.1 \)[/tex]
- [tex]\( 0.64 \)[/tex]
- [tex]\( 1.02 \)[/tex]
The maximum absolute residual is [tex]\( 1.1 \)[/tex].
4. Identify the corresponding point:
The point with the residual of [tex]\( -1.1 \)[/tex] corresponds to [tex]\( x = 3 \)[/tex] and [tex]\( y = 6.2 \)[/tex].
Thus, the point [tex]\((3, 6.2)\)[/tex] is farthest from the line of best fit with a residual value of [tex]\(-1.1\)[/tex].
We'll follow these steps:
1. Residuals from the table:
[tex]\[ \begin{array}{|c|c|c|} \hline x & y & \text{Residual} \\ \hline 1 & 3.3 & 0.68 \\ \hline 2 & 5 & 0.04 \\ \hline 3 & 6.2 & -1.1 \\ \hline 4 & 9 & -0.64 \\ \hline 5 & 13 & 1.02 \\ \hline \end{array} \][/tex]
2. Absolute residuals:
To find the point farthest from the line, we consider the magnitude (absolute value) of the residuals:
- For [tex]\(x = 1\)[/tex], the residual is [tex]\( \left|0.68\right| = 0.68 \)[/tex]
- For [tex]\(x = 2\)[/tex], the residual is [tex]\( \left|0.04\right| = 0.04 \)[/tex]
- For [tex]\(x = 3\)[/tex], the residual is [tex]\( \left|1.1\right| = 1.1 \)[/tex]
- For [tex]\(x = 4\)[/tex], the residual is [tex]\( \left|0.64\right| = 0.64 \)[/tex]
- For [tex]\(x = 5\)[/tex], the residual is [tex]\( \left|1.02\right| = 1.02 \)[/tex]
3. Compare the absolute residuals:
Now, we compare the absolute residuals to find the maximum value:
- [tex]\( 0.68 \)[/tex]
- [tex]\( 0.04 \)[/tex]
- [tex]\( 1.1 \)[/tex]
- [tex]\( 0.64 \)[/tex]
- [tex]\( 1.02 \)[/tex]
The maximum absolute residual is [tex]\( 1.1 \)[/tex].
4. Identify the corresponding point:
The point with the residual of [tex]\( -1.1 \)[/tex] corresponds to [tex]\( x = 3 \)[/tex] and [tex]\( y = 6.2 \)[/tex].
Thus, the point [tex]\((3, 6.2)\)[/tex] is farthest from the line of best fit with a residual value of [tex]\(-1.1\)[/tex].