Fiona wrote the predicted and residual values for a data set using the line of best fit [tex]y = 3.71x - 8.85[/tex].

\begin{tabular}{|c|c|c|c|}
\hline
[tex]$x$[/tex] & Given & Predicted & Residual \\
\hline
1 & -5.1 & -5.14 & 0.04 \\
\hline
2 & -1.3 & -1.43 & -0.13 \\
\hline
3 & 1.9 & 2.28 & -0.38 \\
\hline
4 & 6.2 & 5.99 & 0.21 \\
\hline
\end{tabular}

Which statements are true about the table? Select three options.

A. The data point for [tex]x=1[/tex] is above the line of best fit.

B. The residual value for [tex]x=3[/tex] should be a positive number because the data point is above the line of best fit.

C. Fiona made a subtraction error when she computed the residual value for [tex]x=4[/tex].

D. The residual value for [tex]x=2[/tex] should be a positive number because the given point is above the line of best fit.

E. The residual value for [tex]x=3[/tex] is negative because the given point is below the line of best fit.



Answer :

Let's analyze each statement in detail based on the given data and the information about the residuals from the table.

1. The data point for [tex]\( x = 1 \)[/tex] is above the line of best fit.
- Given value for [tex]\( x = 1 \)[/tex] is [tex]\(-5.1\)[/tex]
- Predicted value for [tex]\( x = 1 \)[/tex] is [tex]\(-5.14\)[/tex]

Since [tex]\(-5.1\)[/tex] (given value) is greater than [tex]\(-5.14\)[/tex] (predicted value), the data point is indeed above the line of best fit. Thus, this statement is True.

2. The residual value for [tex]\( x = 3 \)[/tex] should be a positive number because the data point is above the line of best fit.
- Given value for [tex]\( x = 3 \)[/tex] is [tex]\(1.9\)[/tex]
- Predicted value for [tex]\( x = 3 \)[/tex] is [tex]\(2.28\)[/tex]
- Residual value for [tex]\( x = 3 \)[/tex] is computed as [tex]\(1.9 - 2.28 = -0.38\)[/tex]

Since [tex]\(1.9\)[/tex] (given value) is less than [tex]\(2.28\)[/tex] (predicted value), the data point is below the line of best fit, and the residual should be negative. Thus, this statement is False.

3. Fiona made a subtraction error when she computed the residual value for [tex]\( x = 4 \)[/tex].
- Given value for [tex]\( x = 4 \)[/tex] is [tex]\(6.2\)[/tex]
- Predicted value for [tex]\( x = 4 \)[/tex] is [tex]\(5.99\)[/tex]
- Residual value for [tex]\( x = 4 \)[/tex] is computed as [tex]\(6.2 - 5.99 = 0.21\)[/tex]

Since Fiona's calculated residual value ([tex]\(0.21\)[/tex]) matches the computed residual value, there is no subtraction error. Thus, this statement is False.

4. The residual value for [tex]\( x = 2 \)[/tex] should be a positive number because the given point is above the line of best fit.
- Given value for [tex]\( x = 2 \)[/tex] is [tex]\(-1.3\)[/tex]
- Predicted value for [tex]\( x = 2 \)[/tex] is [tex]\(-1.43\)[/tex]
- Residual value for [tex]\( x = 2 \)[/tex] is computed as [tex]\(-1.3 - (-1.43) = 0.13\)[/tex]

Since [tex]\(-1.3\)[/tex] (given value) is greater than [tex]\(-1.43\)[/tex] (predicted value), the data point is above the line of best fit, and the residual should be positive. Thus, this statement is True.

5. The residual value for [tex]\( x = 3 \)[/tex] is negative because the given point is below the line of best fit.
- Given value for [tex]\( x = 3 \)[/tex] is [tex]\(1.9\)[/tex]
- Predicted value for [tex]\( x = 3 \)[/tex] is [tex]\(2.28\)[/tex]
- Residual value for [tex]\( x = 3 \)[/tex] is computed as [tex]\(1.9 - 2.28 = -0.38\)[/tex]

As mentioned earlier, since [tex]\(1.9\)[/tex] (given value) is less than [tex]\(2.28\)[/tex] (predicted value), the data point is below the line of best fit, and the residual is negative. Thus, this statement is True.

Therefore, the true statements are:

1. The data point for [tex]\( x = 1 \)[/tex] is above the line of best fit.
4. The residual value for [tex]\( x = 2 \)[/tex] should be a positive number because the given point is above the line of best fit.
5. The residual value for [tex]\( x = 3 \)[/tex] is negative because the given point is below the line of best fit.