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To determine the sum of the residuals and understand its implications for the line of best fit, we start by considering the given linear equation for the line of best fit:
[tex]\[ y = 1.46x - 0.76 \][/tex]
We will calculate the residuals for each data point. The residual for a data point [tex]\((x, y)\)[/tex] is defined as:
[tex]\[ \text{residual} = y - y_{\text{pred}} \][/tex]
where [tex]\( y_{\text{pred}} \)[/tex] is the predicted value from the line of best fit.
For each given data point [tex]\((x, y)\)[/tex], we predict the [tex]\( y \)[/tex]-value ([tex]\( y_{\text{pred}} \)[/tex]) using the line of best fit and then calculate the residual:
1. For [tex]\((4, 6)\)[/tex]:
[tex]\[ y_{\text{pred}} = 1.46 \cdot 4 - 0.76 = 5.84 \][/tex]
[tex]\[ \text{residual} = 6 - 5.84 = 0.16 \][/tex]
2. For [tex]\((6, 7)\)[/tex]:
[tex]\[ y_{\text{pred}} = 1.46 \cdot 6 - 0.76 = 7.00 \][/tex]
[tex]\[ \text{residual} = 7 - 7.00 = 0.00 \][/tex]
3. For [tex]\((7, 8)\)[/tex]:
[tex]\[ y_{\text{pred}} = 1.46 \cdot 7 - 0.76 = 9.46 \][/tex]
[tex]\[ \text{residual} = 8 - 9.46 = -1.46 \][/tex]
4. For [tex]\((9, 13)\)[/tex]:
[tex]\[ y_{\text{pred}} = 1.46 \cdot 9 - 0.76 = 12.50 \][/tex]
[tex]\[ \text{residual} = 13 - 12.50 = 0.50 \][/tex]
5. For [tex]\((10, 14)\)[/tex]:
[tex]\[ y_{\text{pred}} = 1.46 \cdot 10 - 0.76 = 13.84 \][/tex]
[tex]\[ \text{residual} = 14 - 13.84 = 0.16 \][/tex]
6. For [tex]\((11, 15)\)[/tex]:
[tex]\[ y_{\text{pred}} = 1.46 \cdot 11 - 0.76 = 15.30 \][/tex]
[tex]\[ \text{residual} = 15 - 15.30 = -0.30 \][/tex]
Now, sum up all these residuals:
[tex]\[ \text{total residual sum} = 0.16 + 0 + (-1.46) + 0.50 + 0.16 + (-0.30) \approx -1.06 \][/tex]
The sum of the residuals is approximately [tex]\(-1.06\)[/tex].
Given that the sum of residuals is [tex]\(-1.06\)[/tex], which is very close to zero but not exactly zero, we can make a couple of observations:
- A sum of residuals equal to zero indicates a perfect fit.
- Since the sum here is very close to zero, it suggests that the line of best fit is quite accurate.
- Therefore, the option that describes this situation is:
[tex]\[ \boxed{-1.06 ; \text{This indicates that the line of best fit is accurate and a good model for prediction.}} \][/tex]
[tex]\[ y = 1.46x - 0.76 \][/tex]
We will calculate the residuals for each data point. The residual for a data point [tex]\((x, y)\)[/tex] is defined as:
[tex]\[ \text{residual} = y - y_{\text{pred}} \][/tex]
where [tex]\( y_{\text{pred}} \)[/tex] is the predicted value from the line of best fit.
For each given data point [tex]\((x, y)\)[/tex], we predict the [tex]\( y \)[/tex]-value ([tex]\( y_{\text{pred}} \)[/tex]) using the line of best fit and then calculate the residual:
1. For [tex]\((4, 6)\)[/tex]:
[tex]\[ y_{\text{pred}} = 1.46 \cdot 4 - 0.76 = 5.84 \][/tex]
[tex]\[ \text{residual} = 6 - 5.84 = 0.16 \][/tex]
2. For [tex]\((6, 7)\)[/tex]:
[tex]\[ y_{\text{pred}} = 1.46 \cdot 6 - 0.76 = 7.00 \][/tex]
[tex]\[ \text{residual} = 7 - 7.00 = 0.00 \][/tex]
3. For [tex]\((7, 8)\)[/tex]:
[tex]\[ y_{\text{pred}} = 1.46 \cdot 7 - 0.76 = 9.46 \][/tex]
[tex]\[ \text{residual} = 8 - 9.46 = -1.46 \][/tex]
4. For [tex]\((9, 13)\)[/tex]:
[tex]\[ y_{\text{pred}} = 1.46 \cdot 9 - 0.76 = 12.50 \][/tex]
[tex]\[ \text{residual} = 13 - 12.50 = 0.50 \][/tex]
5. For [tex]\((10, 14)\)[/tex]:
[tex]\[ y_{\text{pred}} = 1.46 \cdot 10 - 0.76 = 13.84 \][/tex]
[tex]\[ \text{residual} = 14 - 13.84 = 0.16 \][/tex]
6. For [tex]\((11, 15)\)[/tex]:
[tex]\[ y_{\text{pred}} = 1.46 \cdot 11 - 0.76 = 15.30 \][/tex]
[tex]\[ \text{residual} = 15 - 15.30 = -0.30 \][/tex]
Now, sum up all these residuals:
[tex]\[ \text{total residual sum} = 0.16 + 0 + (-1.46) + 0.50 + 0.16 + (-0.30) \approx -1.06 \][/tex]
The sum of the residuals is approximately [tex]\(-1.06\)[/tex].
Given that the sum of residuals is [tex]\(-1.06\)[/tex], which is very close to zero but not exactly zero, we can make a couple of observations:
- A sum of residuals equal to zero indicates a perfect fit.
- Since the sum here is very close to zero, it suggests that the line of best fit is quite accurate.
- Therefore, the option that describes this situation is:
[tex]\[ \boxed{-1.06 ; \text{This indicates that the line of best fit is accurate and a good model for prediction.}} \][/tex]
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