Answer :
To find the equation of the least-squares regression line and make predictions based on it, we need to follow a series of steps.
### Given Data
We are given the following data points for horsepower ([tex]\(x\)[/tex]) and top speed ([tex]\(y\)[/tex]):
| Horsepower (x) | Top Speed (y) |
|----------------|---------------|
| 60 | 25 |
| 80 | 20 |
| 90 | 40 |
| 115 | 45 |
| 150 | 50 |
| 250 | 55 |
| 300 | 60 |
### Step-by-Step Solution
1. Calculate the mean of [tex]\(x\)[/tex] and [tex]\(y\)[/tex]:
- Mean of [tex]\(x\)[/tex] ([tex]\( \bar{x} \)[/tex]):
[tex]\[ \bar{x} = \frac{60 + 80 + 90 + 115 + 150 + 250 + 300}{7} \approx 149.29 \][/tex]
- Mean of [tex]\(y\)[/tex] ([tex]\( \bar{y} \)[/tex]):
[tex]\[ \bar{y} = \frac{25 + 20 + 40 + 45 + 50 + 55 + 60}{7} \approx 42.14 \][/tex]
2. Calculate the elements for the coefficients [tex]\( \beta \)[/tex] and [tex]\( \alpha \)[/tex]:
- Numerator for [tex]\( \beta \)[/tex]:
[tex]\[ \text{Numerator} = \sum (x_i - \bar{x}) (y_i - \bar{y}) \approx 7085.71 \][/tex]
- Denominator for [tex]\( \beta \)[/tex]:
[tex]\[ \text{Denominator} = \sum (x_i - \bar{x})^2 \approx 50321.43 \][/tex]
3. Calculate the slope ([tex]\( \beta \)[/tex]):
[tex]\[ \beta = \frac{\text{Numerator}}{\text{Denominator}} \approx 0.1408 \][/tex]
4. Calculate the intercept ([tex]\( \alpha \)[/tex]):
[tex]\[ \alpha = \bar{y} - \beta \bar{x} \approx 42.14 - 0.1408 \times 149.29 \approx 21.12 \][/tex]
### Equation of the Least-Squares Regression Line
Now that we have the slope ([tex]\(\beta\)[/tex]) and the intercept ([tex]\(\alpha\)[/tex]), the equation of the least-squares regression line is:
[tex]\[ \hat{y} = \alpha + \beta x \][/tex]
Substituting the values:
[tex]\[ \hat{y} = 21.12 + 0.1408 x \][/tex]
### Predicting Top Speed for 100 Horsepower
To predict the top speed for a boat motor with 100 horsepower ([tex]\(x = 100\)[/tex]):
[tex]\[ \hat{y} = 21.12 + 0.1408 \times 100 \approx 21.12 + 14.08 \approx 35.20 \][/tex]
### Summary
The equation of the least-squares regression line is:
[tex]\[ \hat{y} = 21.12 + 0.1408x \][/tex]
According to the regression equation, a boat motor with 100 horsepower is predicted to have a top speed of approximately:
[tex]\[ 35.20 \text{ miles per hour} \][/tex]
Thus, the filled-in blanks are:
[tex]\[ \boxed{21.12}, \boxed{0.1408}, \boxed{35.20} \][/tex]
### Given Data
We are given the following data points for horsepower ([tex]\(x\)[/tex]) and top speed ([tex]\(y\)[/tex]):
| Horsepower (x) | Top Speed (y) |
|----------------|---------------|
| 60 | 25 |
| 80 | 20 |
| 90 | 40 |
| 115 | 45 |
| 150 | 50 |
| 250 | 55 |
| 300 | 60 |
### Step-by-Step Solution
1. Calculate the mean of [tex]\(x\)[/tex] and [tex]\(y\)[/tex]:
- Mean of [tex]\(x\)[/tex] ([tex]\( \bar{x} \)[/tex]):
[tex]\[ \bar{x} = \frac{60 + 80 + 90 + 115 + 150 + 250 + 300}{7} \approx 149.29 \][/tex]
- Mean of [tex]\(y\)[/tex] ([tex]\( \bar{y} \)[/tex]):
[tex]\[ \bar{y} = \frac{25 + 20 + 40 + 45 + 50 + 55 + 60}{7} \approx 42.14 \][/tex]
2. Calculate the elements for the coefficients [tex]\( \beta \)[/tex] and [tex]\( \alpha \)[/tex]:
- Numerator for [tex]\( \beta \)[/tex]:
[tex]\[ \text{Numerator} = \sum (x_i - \bar{x}) (y_i - \bar{y}) \approx 7085.71 \][/tex]
- Denominator for [tex]\( \beta \)[/tex]:
[tex]\[ \text{Denominator} = \sum (x_i - \bar{x})^2 \approx 50321.43 \][/tex]
3. Calculate the slope ([tex]\( \beta \)[/tex]):
[tex]\[ \beta = \frac{\text{Numerator}}{\text{Denominator}} \approx 0.1408 \][/tex]
4. Calculate the intercept ([tex]\( \alpha \)[/tex]):
[tex]\[ \alpha = \bar{y} - \beta \bar{x} \approx 42.14 - 0.1408 \times 149.29 \approx 21.12 \][/tex]
### Equation of the Least-Squares Regression Line
Now that we have the slope ([tex]\(\beta\)[/tex]) and the intercept ([tex]\(\alpha\)[/tex]), the equation of the least-squares regression line is:
[tex]\[ \hat{y} = \alpha + \beta x \][/tex]
Substituting the values:
[tex]\[ \hat{y} = 21.12 + 0.1408 x \][/tex]
### Predicting Top Speed for 100 Horsepower
To predict the top speed for a boat motor with 100 horsepower ([tex]\(x = 100\)[/tex]):
[tex]\[ \hat{y} = 21.12 + 0.1408 \times 100 \approx 21.12 + 14.08 \approx 35.20 \][/tex]
### Summary
The equation of the least-squares regression line is:
[tex]\[ \hat{y} = 21.12 + 0.1408x \][/tex]
According to the regression equation, a boat motor with 100 horsepower is predicted to have a top speed of approximately:
[tex]\[ 35.20 \text{ miles per hour} \][/tex]
Thus, the filled-in blanks are:
[tex]\[ \boxed{21.12}, \boxed{0.1408}, \boxed{35.20} \][/tex]