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Sagot :
To tackle this problem, we need to determine the equation of the regression line for the given data. A regression line is typically expressed in the form:
[tex]\[ \hat{y} = b_0 + b_1 x \][/tex]
where:
- [tex]\( b_0 \)[/tex] is the y-intercept of the line
- [tex]\( b_1 \)[/tex] is the slope of the line
- [tex]\( x \)[/tex] is the independent variable (height in feet)
- [tex]\( \hat{y} \)[/tex] is the dependent variable (number of stories)
From the given data:
[tex]\[ \begin{array}{|c|c|c|c|c|c|c|} \hline \text{Height, } x & 764 & 625 & 520 & 510 & 492 & 484 \\ \hline \text{Stories, } y & 55 & 47 & 44 & 42 & 39 & 37 \\ \hline \end{array} \][/tex]
Step-by-Step Solution:
1. Calculate the means of [tex]\(x\)[/tex] and [tex]\(y\)[/tex]:
[tex]\[ \text{Mean of } x, \overline{x} = \frac{764 + 625 + 520 + 510 + 492 + 484}{6} = \frac{3395}{6} \][/tex]
[tex]\[ \overline{x} = 565.83 \text{ (rounded to 2 decimal places)} \][/tex]
[tex]\[ \text{Mean of } y, \overline{y} = \frac{55 + 47 + 44 + 42 + 39 + 37}{6} = \frac{264}{6} \][/tex]
[tex]\[ \overline{y} = 44 \][/tex]
2. Calculate the slope ([tex]\(b_1\)[/tex]) and intercept ([tex]\(b_0\)[/tex]):
[tex]\[ b_1 = \frac{\sum_{i=1}^{n} (x_i - \overline{x})(y_i - \overline{y})}{\sum_{i=1}^{n} (x_i - \overline{x})^2} \][/tex]
Given the calculations:
[tex]\[ b_1 = 0.057 \text{ (rounded to 3 decimal places)} \][/tex]
[tex]\[ b_0 = \overline{y} - b_1 \cdot \overline{x} = 44 - (0.057 \cdot 565.83) \][/tex]
[tex]\[ b_0 = 11.91 \text{ (rounded to 2 decimal places)} \][/tex]
Thus, the regression equation is:
[tex]\[ \hat{y} = 0.057x + 11.91 \][/tex]
Predictions:
Using the regression equation, we can predict the value of [tex]\(y\)[/tex] for the given [tex]\(x\)[/tex]-values:
(a) For [tex]\( x = 503 \)[/tex] feet:
[tex]\[ \hat{y} = 0.057 \cdot 503 + 11.91 \][/tex]
[tex]\[ \hat{y} \approx 40.44 \][/tex]
(b) For [tex]\( x = 650 \)[/tex] feet:
[tex]\[ \hat{y} = 0.057 \cdot 650 + 11.91 \][/tex]
[tex]\[ \hat{y} \approx 48.77 \][/tex]
(c) For [tex]\( x = 802 \)[/tex] feet:
[tex]\[ \hat{y} = 0.057 \cdot 802 + 11.91 \][/tex]
[tex]\[ \hat{y} \approx 57.40 \][/tex]
(d) For [tex]\( x = 731 \)[/tex] feet:
[tex]\[ \hat{y} = 0.057 \cdot 731 + 11.91 \][/tex]
[tex]\[ \hat{y} \approx 53.37 \][/tex]
So, the regression equation is:
[tex]\[ \hat{y} = 0.057x + 11.91 \][/tex]
and the predicted values of [tex]\(y\)[/tex] for each given [tex]\(x\)[/tex]-value are approximately:
- For [tex]\(x = 503\)[/tex] feet, [tex]\(\hat{y} \approx 40.44\)[/tex]
- For [tex]\(x = 650\)[/tex] feet, [tex]\(\hat{y} \approx 48.77\)[/tex]
- For [tex]\(x = 802\)[/tex] feet, [tex]\(\hat{y} \approx 57.40\)[/tex]
- For [tex]\(x = 731\)[/tex] feet, [tex]\(\hat{y} \approx 53.37\)[/tex]
[tex]\[ \hat{y} = b_0 + b_1 x \][/tex]
where:
- [tex]\( b_0 \)[/tex] is the y-intercept of the line
- [tex]\( b_1 \)[/tex] is the slope of the line
- [tex]\( x \)[/tex] is the independent variable (height in feet)
- [tex]\( \hat{y} \)[/tex] is the dependent variable (number of stories)
From the given data:
[tex]\[ \begin{array}{|c|c|c|c|c|c|c|} \hline \text{Height, } x & 764 & 625 & 520 & 510 & 492 & 484 \\ \hline \text{Stories, } y & 55 & 47 & 44 & 42 & 39 & 37 \\ \hline \end{array} \][/tex]
Step-by-Step Solution:
1. Calculate the means of [tex]\(x\)[/tex] and [tex]\(y\)[/tex]:
[tex]\[ \text{Mean of } x, \overline{x} = \frac{764 + 625 + 520 + 510 + 492 + 484}{6} = \frac{3395}{6} \][/tex]
[tex]\[ \overline{x} = 565.83 \text{ (rounded to 2 decimal places)} \][/tex]
[tex]\[ \text{Mean of } y, \overline{y} = \frac{55 + 47 + 44 + 42 + 39 + 37}{6} = \frac{264}{6} \][/tex]
[tex]\[ \overline{y} = 44 \][/tex]
2. Calculate the slope ([tex]\(b_1\)[/tex]) and intercept ([tex]\(b_0\)[/tex]):
[tex]\[ b_1 = \frac{\sum_{i=1}^{n} (x_i - \overline{x})(y_i - \overline{y})}{\sum_{i=1}^{n} (x_i - \overline{x})^2} \][/tex]
Given the calculations:
[tex]\[ b_1 = 0.057 \text{ (rounded to 3 decimal places)} \][/tex]
[tex]\[ b_0 = \overline{y} - b_1 \cdot \overline{x} = 44 - (0.057 \cdot 565.83) \][/tex]
[tex]\[ b_0 = 11.91 \text{ (rounded to 2 decimal places)} \][/tex]
Thus, the regression equation is:
[tex]\[ \hat{y} = 0.057x + 11.91 \][/tex]
Predictions:
Using the regression equation, we can predict the value of [tex]\(y\)[/tex] for the given [tex]\(x\)[/tex]-values:
(a) For [tex]\( x = 503 \)[/tex] feet:
[tex]\[ \hat{y} = 0.057 \cdot 503 + 11.91 \][/tex]
[tex]\[ \hat{y} \approx 40.44 \][/tex]
(b) For [tex]\( x = 650 \)[/tex] feet:
[tex]\[ \hat{y} = 0.057 \cdot 650 + 11.91 \][/tex]
[tex]\[ \hat{y} \approx 48.77 \][/tex]
(c) For [tex]\( x = 802 \)[/tex] feet:
[tex]\[ \hat{y} = 0.057 \cdot 802 + 11.91 \][/tex]
[tex]\[ \hat{y} \approx 57.40 \][/tex]
(d) For [tex]\( x = 731 \)[/tex] feet:
[tex]\[ \hat{y} = 0.057 \cdot 731 + 11.91 \][/tex]
[tex]\[ \hat{y} \approx 53.37 \][/tex]
So, the regression equation is:
[tex]\[ \hat{y} = 0.057x + 11.91 \][/tex]
and the predicted values of [tex]\(y\)[/tex] for each given [tex]\(x\)[/tex]-value are approximately:
- For [tex]\(x = 503\)[/tex] feet, [tex]\(\hat{y} \approx 40.44\)[/tex]
- For [tex]\(x = 650\)[/tex] feet, [tex]\(\hat{y} \approx 48.77\)[/tex]
- For [tex]\(x = 802\)[/tex] feet, [tex]\(\hat{y} \approx 57.40\)[/tex]
- For [tex]\(x = 731\)[/tex] feet, [tex]\(\hat{y} \approx 53.37\)[/tex]
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