Answer :
Let's proceed step-by-step to complete the residual plot.
Step One: Find [tex]\(\hat{y}\)[/tex] for the given [tex]\(x\)[/tex]-values in the data set using the regression line equation [tex]\(y = 2.9 + 0.34x\)[/tex].
Given [tex]\(x\)[/tex]-values:
- [tex]\(x = 5\)[/tex]
- [tex]\(x = 7\)[/tex]
- [tex]\(x = 9\)[/tex]
- [tex]\(x = 11\)[/tex]
- [tex]\(x = 13\)[/tex]
- [tex]\(x = 15\)[/tex]
Let's compute [tex]\(\hat{y}\)[/tex] for each [tex]\(x\)[/tex]-value using the regression line equation [tex]\(y = 2.9 + 0.34x\)[/tex]. Additionally, we will round each [tex]\(\hat{y}\)[/tex] to the nearest tenth.
1. For [tex]\(x = 5\)[/tex]:
[tex]\[\hat{y} = 2.9 + 0.34 \times 5 = 2.9 + 1.7 = 4.6\][/tex]
2. For [tex]\(x = 7\)[/tex]:
[tex]\[\hat{y} = 2.9 + 0.34 \times 7 = 2.9 + 2.38 \approx 5.3\][/tex]
3. For [tex]\(x = 9\)[/tex]:
[tex]\[\hat{y} = 2.9 + 0.34 \times 9 = 2.9 + 3.06 \approx 6.0\][/tex]
4. For [tex]\(x = 11\)[/tex]:
[tex]\[\hat{y} = 2.9 + 0.34 \times 11 = 2.9 + 3.74 \approx 6.6\][/tex]
5. For [tex]\(x = 13\)[/tex]:
[tex]\[\hat{y} = 2.9 + 0.34 \times 13 = 2.9 + 4.42 \approx 7.3\][/tex]
6. For [tex]\(x = 15\)[/tex]:
[tex]\[\hat{y} = 2.9 + 0.34 \times 15 = 2.9 + 5.1 = 8.0\][/tex]
Now, let’s fill in the table with these calculated [tex]\(\hat{y}\)[/tex] values:
[tex]\[ \begin{tabular}{|c|c|c|} \hline $x$ & \multicolumn{2}{|c|}{$\hat{y}$} \\ \hline 5 & 4.6 & $\checkmark$ \\ \hline 7 & 5.3 & $\checkmark$ \\ \hline 9 & 6.0 & $\checkmark$ \\ \hline 11 & 6.6 & $\checkmark$ \\ \hline 13 & 7.3 & $\checkmark$ \\ \hline 15 & 8.0 & $\checkmark$ \\ \hline \end{tabular} \][/tex]
These are the [tex]\(\hat{y}\)[/tex] values for the given [tex]\(x\)[/tex]-values, as calculated:
- For [tex]\(x = 5\)[/tex], [tex]\(\hat{y} = 4.6\)[/tex]
- For [tex]\(x = 7\)[/tex], [tex]\(\hat{y} = 5.3\)[/tex]
- For [tex]\(x = 9\)[/tex], [tex]\(\hat{y} = 6.0\)[/tex]
- For [tex]\(x = 11\)[/tex], [tex]\(\hat{y} = 6.6\)[/tex]
- For [tex]\(x = 13\)[/tex], [tex]\(\hat{y} = 7.3\)[/tex]
- For [tex]\(x = 15\)[/tex], [tex]\(\hat{y} = 8.0\)[/tex]
Next step will be to plot the residuals, but this table completes Step One.
Step One: Find [tex]\(\hat{y}\)[/tex] for the given [tex]\(x\)[/tex]-values in the data set using the regression line equation [tex]\(y = 2.9 + 0.34x\)[/tex].
Given [tex]\(x\)[/tex]-values:
- [tex]\(x = 5\)[/tex]
- [tex]\(x = 7\)[/tex]
- [tex]\(x = 9\)[/tex]
- [tex]\(x = 11\)[/tex]
- [tex]\(x = 13\)[/tex]
- [tex]\(x = 15\)[/tex]
Let's compute [tex]\(\hat{y}\)[/tex] for each [tex]\(x\)[/tex]-value using the regression line equation [tex]\(y = 2.9 + 0.34x\)[/tex]. Additionally, we will round each [tex]\(\hat{y}\)[/tex] to the nearest tenth.
1. For [tex]\(x = 5\)[/tex]:
[tex]\[\hat{y} = 2.9 + 0.34 \times 5 = 2.9 + 1.7 = 4.6\][/tex]
2. For [tex]\(x = 7\)[/tex]:
[tex]\[\hat{y} = 2.9 + 0.34 \times 7 = 2.9 + 2.38 \approx 5.3\][/tex]
3. For [tex]\(x = 9\)[/tex]:
[tex]\[\hat{y} = 2.9 + 0.34 \times 9 = 2.9 + 3.06 \approx 6.0\][/tex]
4. For [tex]\(x = 11\)[/tex]:
[tex]\[\hat{y} = 2.9 + 0.34 \times 11 = 2.9 + 3.74 \approx 6.6\][/tex]
5. For [tex]\(x = 13\)[/tex]:
[tex]\[\hat{y} = 2.9 + 0.34 \times 13 = 2.9 + 4.42 \approx 7.3\][/tex]
6. For [tex]\(x = 15\)[/tex]:
[tex]\[\hat{y} = 2.9 + 0.34 \times 15 = 2.9 + 5.1 = 8.0\][/tex]
Now, let’s fill in the table with these calculated [tex]\(\hat{y}\)[/tex] values:
[tex]\[ \begin{tabular}{|c|c|c|} \hline $x$ & \multicolumn{2}{|c|}{$\hat{y}$} \\ \hline 5 & 4.6 & $\checkmark$ \\ \hline 7 & 5.3 & $\checkmark$ \\ \hline 9 & 6.0 & $\checkmark$ \\ \hline 11 & 6.6 & $\checkmark$ \\ \hline 13 & 7.3 & $\checkmark$ \\ \hline 15 & 8.0 & $\checkmark$ \\ \hline \end{tabular} \][/tex]
These are the [tex]\(\hat{y}\)[/tex] values for the given [tex]\(x\)[/tex]-values, as calculated:
- For [tex]\(x = 5\)[/tex], [tex]\(\hat{y} = 4.6\)[/tex]
- For [tex]\(x = 7\)[/tex], [tex]\(\hat{y} = 5.3\)[/tex]
- For [tex]\(x = 9\)[/tex], [tex]\(\hat{y} = 6.0\)[/tex]
- For [tex]\(x = 11\)[/tex], [tex]\(\hat{y} = 6.6\)[/tex]
- For [tex]\(x = 13\)[/tex], [tex]\(\hat{y} = 7.3\)[/tex]
- For [tex]\(x = 15\)[/tex], [tex]\(\hat{y} = 8.0\)[/tex]
Next step will be to plot the residuals, but this table completes Step One.