The predicted values were computed by using the line of fit: [tex]$y=32-2z$[/tex].

\begin{tabular}{|c|c|c|c|}
\hline
[tex]$z$[/tex] & Given & Predicted & Residual \\
\hline
1 & 6.1 & 30 & [tex]$a$[/tex] \\
\hline
2 & 7.4 & 28 & [tex]$b$[/tex] \\
\hline
3 & 11 & 26 & [tex]$c$[/tex] \\
\hline
4 & 15 & 24 & [tex]$d$[/tex] \\
\hline
\end{tabular}

What are the individual values needed to complete the table?

\begin{array}{l}
a= \\
b= \\
c= \\
d=
\end{array}



Answer :

Let's go through the calculation of each residual step by step for the given values in the table. The residual is computed by subtracting the predicted value from the given value.

1. For [tex]\( z = 1 \)[/tex]:
- Given value: [tex]\( 6.1 \)[/tex]
- Predicted value: [tex]\( 52 \)[/tex]
- Residual [tex]\( a \)[/tex]:
[tex]\[ a = 6.1 - 52 = -45.9 \][/tex]

2. For [tex]\( z = 2 \)[/tex]:
- Given value: [tex]\( 7.4 \)[/tex]
- Predicted value: [tex]\( 8.4 \)[/tex]
- Residual [tex]\( b \)[/tex]:
[tex]\[ b = 7.4 - 8.4 = -1.0 \][/tex]

3. For [tex]\( z = 3 \)[/tex]:
- Given value: [tex]\( 11 \)[/tex]
- Predicted value: [tex]\( 11.6 \)[/tex]
- Residual [tex]\( c \)[/tex]:
[tex]\[ c = 11 - 11.6 = -0.6 \][/tex]

4. For [tex]\( z = 4 \)[/tex]:
- Given value: [tex]\( 15 \)[/tex]
- Predicted value: [tex]\( 14.8 \)[/tex]
- Residual [tex]\( d \)[/tex]:
[tex]\[ d = 15 - 14.8 = 0.2 \][/tex]

So, the individual residual values needed for the table are:
[tex]\[ \begin{array}{l} a = -45.9 \\ b = -1.0 \\ c = -0.6 \\ d = 0.2 \end{array} \][/tex]