Answer :
To find the residual values and determine the residual plot's appropriateness for the line of best fit, follow the steps below:
### Step-by-Step Solution:
1. Identify the Variables:
- [tex]\( x \)[/tex] values are given as [tex]\( x = [1, 2, 3, 4, 5] \)[/tex].
- Given (observed) values are [tex]\( y_{\text{obs}} = [-3.5, -2.9, -1.1, 2.2, 3.4] \)[/tex].
- Predicted values are [tex]\( y_{\text{pred}} = [-1.1, 2, 5.1, 8.2, 11.3] \)[/tex].
2. Calculate Residuals:
Residual for each data point is given by:
[tex]\[ \text{Residual} = y_{\text{obs}} - y_{\text{pred}} \][/tex]
Perform these calculations for each [tex]\( x \)[/tex]:
- For [tex]\( x = 1 \)[/tex]:
[tex]\[ \text{Residual} = -3.5 - (-1.1) = -2.4 \][/tex]
- For [tex]\( x = 2 \)[/tex]:
[tex]\[ \text{Residual} = -2.9 - 2 = -4.9 \][/tex]
- For [tex]\( x = 3 \)[/tex]:
[tex]\[ \text{Residual} = -1.1 - 5.1 = -6.2 \][/tex]
- For [tex]\( x = 4 \)[/tex]:
[tex]\[ \text{Residual} = 2.2 - 8.2 = -6.0 \][/tex]
- For [tex]\( x = 5 \)[/tex]:
[tex]\[ \text{Residual} = 3.4 - 11.3 = -7.9 \][/tex]
The residual values are summarized in the table:
[tex]\[ \begin{tabular}{|c|c|c|c|} \hline $x$ & \text{Given} & \text{Predicted} & \text{Residual} \\ \hline 1 & -3.5 & -1.1 & -2.4 \\ \hline 2 & -2.9 & 2 & -4.9 \\ \hline 3 & -1.1 & 5.1 & -6.2 \\ \hline 4 & 2.2 & 8.2 & -6.0 \\ \hline 5 & 3.4 & 11.3 & -7.9 \\ \hline \end{tabular} \][/tex]
3. Analyze the Residual Plot:
- Create a graph with [tex]\( x \)[/tex] values on the x-axis and the residuals on the y-axis.
- Plot the points [tex]\( (x, \text{residual}) \)[/tex].
### Interpretation:
Based on the residuals calculated:
[tex]\[ \text{Residuals} = [-2.4, -4.9, -6.2, -6.0, -7.9] \][/tex]
The pattern of residuals shows a consistent trend of negative residuals which suggests that the residual points are distributed around a negative slope but not around the x-axis. Therefore, the residual plot does not show the points having no pattern or being in a curved pattern. Instead, the distribution indicates that the residual points are evenly distributed on the negative side, and this shows that the line of best fit is perhaps not the ideal fit for this data.
To summarize, the correct choice is:
[tex]\[ \text{No, the points are evenly distributed about the } x \text{ axis.} \][/tex]
### Step-by-Step Solution:
1. Identify the Variables:
- [tex]\( x \)[/tex] values are given as [tex]\( x = [1, 2, 3, 4, 5] \)[/tex].
- Given (observed) values are [tex]\( y_{\text{obs}} = [-3.5, -2.9, -1.1, 2.2, 3.4] \)[/tex].
- Predicted values are [tex]\( y_{\text{pred}} = [-1.1, 2, 5.1, 8.2, 11.3] \)[/tex].
2. Calculate Residuals:
Residual for each data point is given by:
[tex]\[ \text{Residual} = y_{\text{obs}} - y_{\text{pred}} \][/tex]
Perform these calculations for each [tex]\( x \)[/tex]:
- For [tex]\( x = 1 \)[/tex]:
[tex]\[ \text{Residual} = -3.5 - (-1.1) = -2.4 \][/tex]
- For [tex]\( x = 2 \)[/tex]:
[tex]\[ \text{Residual} = -2.9 - 2 = -4.9 \][/tex]
- For [tex]\( x = 3 \)[/tex]:
[tex]\[ \text{Residual} = -1.1 - 5.1 = -6.2 \][/tex]
- For [tex]\( x = 4 \)[/tex]:
[tex]\[ \text{Residual} = 2.2 - 8.2 = -6.0 \][/tex]
- For [tex]\( x = 5 \)[/tex]:
[tex]\[ \text{Residual} = 3.4 - 11.3 = -7.9 \][/tex]
The residual values are summarized in the table:
[tex]\[ \begin{tabular}{|c|c|c|c|} \hline $x$ & \text{Given} & \text{Predicted} & \text{Residual} \\ \hline 1 & -3.5 & -1.1 & -2.4 \\ \hline 2 & -2.9 & 2 & -4.9 \\ \hline 3 & -1.1 & 5.1 & -6.2 \\ \hline 4 & 2.2 & 8.2 & -6.0 \\ \hline 5 & 3.4 & 11.3 & -7.9 \\ \hline \end{tabular} \][/tex]
3. Analyze the Residual Plot:
- Create a graph with [tex]\( x \)[/tex] values on the x-axis and the residuals on the y-axis.
- Plot the points [tex]\( (x, \text{residual}) \)[/tex].
### Interpretation:
Based on the residuals calculated:
[tex]\[ \text{Residuals} = [-2.4, -4.9, -6.2, -6.0, -7.9] \][/tex]
The pattern of residuals shows a consistent trend of negative residuals which suggests that the residual points are distributed around a negative slope but not around the x-axis. Therefore, the residual plot does not show the points having no pattern or being in a curved pattern. Instead, the distribution indicates that the residual points are evenly distributed on the negative side, and this shows that the line of best fit is perhaps not the ideal fit for this data.
To summarize, the correct choice is:
[tex]\[ \text{No, the points are evenly distributed about the } x \text{ axis.} \][/tex]
