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Sagot :
Let's solve this problem by performing linear regression step-by-step.
First, let's make a table of the given data points for clarity:
| Height (in.) [tex]\(x\)[/tex] | Points per game [tex]\(y\)[/tex] |
|--------------------|-----------------------|
| 64 | 10 |
| 66 | 12 |
| 68 | 14 |
| 70 | 26.219 |
| 72 | 17 |
Step 1: Compute the Mean of [tex]\(x\)[/tex] (heights) and [tex]\(y\)[/tex] (points per game):
[tex]\[ \bar{x} = \frac{64 + 66 + 68 + 70 + 72}{5} = \frac{340}{5} = 68 \][/tex]
[tex]\[ \bar{y} = \frac{10 + 12 + 14 + 26.219 + 17}{5} = \frac{79.219}{5} \approx 15.8438 \][/tex]
Step 2: Calculate the Sum of Squares:
[tex]\( SS_{xy} = \sum (x_i - \bar{x}) (y_i - \bar{y}) \)[/tex]
[tex]\( SS_{xx} = \sum (x_i - \bar{x})^2 \)[/tex]
| [tex]\(x_i\)[/tex] | [tex]\(y_i\)[/tex] | [tex]\(x_i - \bar{x}\)[/tex] | [tex]\(y_i - \bar{y}\)[/tex] | [tex]\((x_i - \bar{x})(y_i - \bar{y})\)[/tex] | [tex]\((x_i - \bar{x})^2\)[/tex] |
|---------|---------|---------------------|-------------------|--------------------------------|------------------------|
| 64 | 10 | -4 | -5.8438 | 23.3752 | 16 |
| 66 | 12 | -2 | -3.8438 | 7.6876 | 4 |
| 68 | 14 | 0 | -1.8438 | 0 | 0 |
| 70 | 26.219 | 2 | 10.3752 | 20.7504 | 4 |
| 72 | 17 | 4 | 1.1562 | 4.6248 | 16 |
[tex]\[ SS_{xy} = 23.3752 + 7.6876 + 0 + 20.7504 + 4.6248 = 56.438 \][/tex]
[tex]\[ SS_{xx} = 16 + 4 + 0 + 4 + 16 = 40 \][/tex]
Step 3: Compute the slope [tex]\(b_1\)[/tex] and intercept [tex]\(b_0\)[/tex]:
[tex]\[ b_1 = \frac{SS_{xy}}{SS_{xx}} = \frac{56.438}{40} \approx 1.41095 \][/tex]
[tex]\[ b_0 = \bar{y} - b_1 \cdot \bar{x} = 15.8438 - 1.41095 \cdot 68 = 15.8438 - 95.9457 \approx -80.1019 \][/tex]
Thus, the linear regression equation is:
[tex]\[ y = 1.41095x - 80.1019 \][/tex]
Step 4: Predict the average points per game for a player who is 70 inches tall:
[tex]\[ \hat{y} = 1.41095 \times 70 - 80.1019 \][/tex]
[tex]\[ \hat{y} = 98.7665 - 80.1019 \approx 18.6646 \][/tex]
Thus, based on our calculations, the predicted average points per game for a basketball player who is 70 inches tall is approximately:
[tex]\[ \boxed{18.6646} \][/tex]
First, let's make a table of the given data points for clarity:
| Height (in.) [tex]\(x\)[/tex] | Points per game [tex]\(y\)[/tex] |
|--------------------|-----------------------|
| 64 | 10 |
| 66 | 12 |
| 68 | 14 |
| 70 | 26.219 |
| 72 | 17 |
Step 1: Compute the Mean of [tex]\(x\)[/tex] (heights) and [tex]\(y\)[/tex] (points per game):
[tex]\[ \bar{x} = \frac{64 + 66 + 68 + 70 + 72}{5} = \frac{340}{5} = 68 \][/tex]
[tex]\[ \bar{y} = \frac{10 + 12 + 14 + 26.219 + 17}{5} = \frac{79.219}{5} \approx 15.8438 \][/tex]
Step 2: Calculate the Sum of Squares:
[tex]\( SS_{xy} = \sum (x_i - \bar{x}) (y_i - \bar{y}) \)[/tex]
[tex]\( SS_{xx} = \sum (x_i - \bar{x})^2 \)[/tex]
| [tex]\(x_i\)[/tex] | [tex]\(y_i\)[/tex] | [tex]\(x_i - \bar{x}\)[/tex] | [tex]\(y_i - \bar{y}\)[/tex] | [tex]\((x_i - \bar{x})(y_i - \bar{y})\)[/tex] | [tex]\((x_i - \bar{x})^2\)[/tex] |
|---------|---------|---------------------|-------------------|--------------------------------|------------------------|
| 64 | 10 | -4 | -5.8438 | 23.3752 | 16 |
| 66 | 12 | -2 | -3.8438 | 7.6876 | 4 |
| 68 | 14 | 0 | -1.8438 | 0 | 0 |
| 70 | 26.219 | 2 | 10.3752 | 20.7504 | 4 |
| 72 | 17 | 4 | 1.1562 | 4.6248 | 16 |
[tex]\[ SS_{xy} = 23.3752 + 7.6876 + 0 + 20.7504 + 4.6248 = 56.438 \][/tex]
[tex]\[ SS_{xx} = 16 + 4 + 0 + 4 + 16 = 40 \][/tex]
Step 3: Compute the slope [tex]\(b_1\)[/tex] and intercept [tex]\(b_0\)[/tex]:
[tex]\[ b_1 = \frac{SS_{xy}}{SS_{xx}} = \frac{56.438}{40} \approx 1.41095 \][/tex]
[tex]\[ b_0 = \bar{y} - b_1 \cdot \bar{x} = 15.8438 - 1.41095 \cdot 68 = 15.8438 - 95.9457 \approx -80.1019 \][/tex]
Thus, the linear regression equation is:
[tex]\[ y = 1.41095x - 80.1019 \][/tex]
Step 4: Predict the average points per game for a player who is 70 inches tall:
[tex]\[ \hat{y} = 1.41095 \times 70 - 80.1019 \][/tex]
[tex]\[ \hat{y} = 98.7665 - 80.1019 \approx 18.6646 \][/tex]
Thus, based on our calculations, the predicted average points per game for a basketball player who is 70 inches tall is approximately:
[tex]\[ \boxed{18.6646} \][/tex]
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