Answer :
To determine the [tex]\( r^2 \)[/tex] value for the given data, we need to follow the steps for performing linear regression. Below, I'll provide a detailed step-by-step solution to find the [tex]\( r^2 \)[/tex] value.
Given data:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & 2 \\ \hline 2 & 10 \\ \hline 5 & 5 \\ \hline 7 & 18 \\ \hline 10 & 25 \\ \hline \end{array} \][/tex]
First, we'll calculate the necessary sums:
- [tex]\( \sum x \)[/tex]
- [tex]\( \sum y \)[/tex]
- [tex]\( \sum xy \)[/tex]
- [tex]\( \sum x^2 \)[/tex]
Using these sums, we can then apply the formulas for the slope [tex]\( m \)[/tex] and the intercept [tex]\( b \)[/tex] of the line [tex]\( y = mx + b \)[/tex]:
### Step 1: Calculate sums
[tex]\[ \sum x = 1 + 2 + 5 + 7 + 10 = 25 \][/tex]
[tex]\[ \sum y = 2 + 10 + 5 + 18 + 25 = 60 \][/tex]
[tex]\[ \sum xy = (1 \cdot 2) + (2 \cdot 10) + (5 \cdot 5) + (7 \cdot 18) + (10 \cdot 25) = 2 + 20 + 25 + 126 + 250 = 423 \][/tex]
[tex]\[ \sum x^2 = 1^2 + 2^2 + 5^2 + 7^2 + 10^2 = 1 + 4 + 25 + 49 + 100 = 179 \][/tex]
### Step 2: Calculate slope [tex]\( m \)[/tex] and intercept [tex]\( b \)[/tex]
Using the formulas for a line [tex]\( y = mx + b \)[/tex]:
[tex]\[ m = \frac{n(\sum xy) - (\sum x)(\sum y)}{n(\sum x^2) - (\sum x)^2} \][/tex]
where [tex]\( n = 5 \)[/tex] (number of data points).
[tex]\[ m = \frac{5(423) - (25)(60)}{5(179) - (25)^2} = \frac{2115 - 1500}{895 - 625} = \frac{615}{270} = \frac{205}{90} \approx 2.278 \][/tex]
[tex]\[ b = \frac{\sum y - m(\sum x)}{n} \][/tex]
[tex]\[ b = \frac{60 - 2.278(25)}{5} = \frac{60 - 56.95}{5} = \frac{3.05}{5} \approx 0.61 \][/tex]
The equation of the regression line is approximately:
[tex]\[ y = 2.278x + 0.61 \][/tex]
### Step 3: Calculate predicted [tex]\( \hat{y} \)[/tex] values
[tex]\[ \hat{y}_1 = 2.278 \cdot 1 + 0.61 \approx 2.888 \][/tex]
[tex]\[ \hat{y}_2 = 2.278 \cdot 2 + 0.61 \approx 5.166 \][/tex]
[tex]\[ \hat{y}_3 = 2.278 \cdot 5 + 0.61 \approx 11.99 \][/tex]
[tex]\[ \hat{y}_4 = 2.278 \cdot 7 + 0.61 \approx 16.546 \][/tex]
[tex]\[ \hat{y}_5 = 2.278 \cdot 10 + 0.61 \approx 23.39 \][/tex]
### Step 4: Calculate [tex]\( r^2 \)[/tex]
The formula to compute [tex]\( r^2 \)[/tex] is:
[tex]\[ r^2 = 1 - \frac{\sum (y_i - \hat{y}_i)^2}{\sum (y_i - \bar{y})^2} \][/tex]
where [tex]\( \bar{y} \)[/tex] is the mean of [tex]\( y \)[/tex]:
[tex]\[ \bar{y} = \frac{\sum y}{n} = \frac{60}{5} = 12 \][/tex]
Calculate the sum of squares:
[tex]\[ SS_{\text{tot}} = \sum (y_i - \bar{y})^2 = (2-12)^2 + (10-12)^2 + (5-12)^2 + (18-12)^2 + (25-12)^2 = 100 + 4 + 49 + 36 + 169 = 358 \][/tex]
[tex]\[ SS_{\text{res}} = \sum (y_i - \hat{y}_i)^2 = (2-2.888)^2 + (10-5.166)^2 + (5-11.99)^2 + (18-16.546)^2 + (25-23.39)^2 = 0.788^2 + 4.834^2 + 6.99^2 + 1.454^2 + 1.61^2 = 0.620 + 23.37 + 48.86 + 2.114 + 2.592 = 77.556 \][/tex]
Finally, compute [tex]\( r^2 \)[/tex]:
[tex]\[ r^2 = 1 - \frac{77.556}{358} \approx 1 - 0.217 = 0.783 \][/tex]
The value of [tex]\( r^2 \)[/tex] to three decimal places is:
\boxed{0.783}
Thus, the correct answer is D. 0.783.
Given data:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & 2 \\ \hline 2 & 10 \\ \hline 5 & 5 \\ \hline 7 & 18 \\ \hline 10 & 25 \\ \hline \end{array} \][/tex]
First, we'll calculate the necessary sums:
- [tex]\( \sum x \)[/tex]
- [tex]\( \sum y \)[/tex]
- [tex]\( \sum xy \)[/tex]
- [tex]\( \sum x^2 \)[/tex]
Using these sums, we can then apply the formulas for the slope [tex]\( m \)[/tex] and the intercept [tex]\( b \)[/tex] of the line [tex]\( y = mx + b \)[/tex]:
### Step 1: Calculate sums
[tex]\[ \sum x = 1 + 2 + 5 + 7 + 10 = 25 \][/tex]
[tex]\[ \sum y = 2 + 10 + 5 + 18 + 25 = 60 \][/tex]
[tex]\[ \sum xy = (1 \cdot 2) + (2 \cdot 10) + (5 \cdot 5) + (7 \cdot 18) + (10 \cdot 25) = 2 + 20 + 25 + 126 + 250 = 423 \][/tex]
[tex]\[ \sum x^2 = 1^2 + 2^2 + 5^2 + 7^2 + 10^2 = 1 + 4 + 25 + 49 + 100 = 179 \][/tex]
### Step 2: Calculate slope [tex]\( m \)[/tex] and intercept [tex]\( b \)[/tex]
Using the formulas for a line [tex]\( y = mx + b \)[/tex]:
[tex]\[ m = \frac{n(\sum xy) - (\sum x)(\sum y)}{n(\sum x^2) - (\sum x)^2} \][/tex]
where [tex]\( n = 5 \)[/tex] (number of data points).
[tex]\[ m = \frac{5(423) - (25)(60)}{5(179) - (25)^2} = \frac{2115 - 1500}{895 - 625} = \frac{615}{270} = \frac{205}{90} \approx 2.278 \][/tex]
[tex]\[ b = \frac{\sum y - m(\sum x)}{n} \][/tex]
[tex]\[ b = \frac{60 - 2.278(25)}{5} = \frac{60 - 56.95}{5} = \frac{3.05}{5} \approx 0.61 \][/tex]
The equation of the regression line is approximately:
[tex]\[ y = 2.278x + 0.61 \][/tex]
### Step 3: Calculate predicted [tex]\( \hat{y} \)[/tex] values
[tex]\[ \hat{y}_1 = 2.278 \cdot 1 + 0.61 \approx 2.888 \][/tex]
[tex]\[ \hat{y}_2 = 2.278 \cdot 2 + 0.61 \approx 5.166 \][/tex]
[tex]\[ \hat{y}_3 = 2.278 \cdot 5 + 0.61 \approx 11.99 \][/tex]
[tex]\[ \hat{y}_4 = 2.278 \cdot 7 + 0.61 \approx 16.546 \][/tex]
[tex]\[ \hat{y}_5 = 2.278 \cdot 10 + 0.61 \approx 23.39 \][/tex]
### Step 4: Calculate [tex]\( r^2 \)[/tex]
The formula to compute [tex]\( r^2 \)[/tex] is:
[tex]\[ r^2 = 1 - \frac{\sum (y_i - \hat{y}_i)^2}{\sum (y_i - \bar{y})^2} \][/tex]
where [tex]\( \bar{y} \)[/tex] is the mean of [tex]\( y \)[/tex]:
[tex]\[ \bar{y} = \frac{\sum y}{n} = \frac{60}{5} = 12 \][/tex]
Calculate the sum of squares:
[tex]\[ SS_{\text{tot}} = \sum (y_i - \bar{y})^2 = (2-12)^2 + (10-12)^2 + (5-12)^2 + (18-12)^2 + (25-12)^2 = 100 + 4 + 49 + 36 + 169 = 358 \][/tex]
[tex]\[ SS_{\text{res}} = \sum (y_i - \hat{y}_i)^2 = (2-2.888)^2 + (10-5.166)^2 + (5-11.99)^2 + (18-16.546)^2 + (25-23.39)^2 = 0.788^2 + 4.834^2 + 6.99^2 + 1.454^2 + 1.61^2 = 0.620 + 23.37 + 48.86 + 2.114 + 2.592 = 77.556 \][/tex]
Finally, compute [tex]\( r^2 \)[/tex]:
[tex]\[ r^2 = 1 - \frac{77.556}{358} \approx 1 - 0.217 = 0.783 \][/tex]
The value of [tex]\( r^2 \)[/tex] to three decimal places is:
\boxed{0.783}
Thus, the correct answer is D. 0.783.
