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To find the linear regression line for the given data, we need to calculate the equation of the line in the form [tex]\( y = b_0 + b_1 x \)[/tex], where [tex]\( b_0 \)[/tex] is the intercept and [tex]\( b_1 \)[/tex] is the slope.
Here are the steps to find the linear regression line:
1. Calculate the mean of [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
Given data:
[tex]\[ x = [3, 7, 11, 15, 17] \][/tex]
[tex]\[ y = [2, 3, 6, 4, 7] \][/tex]
The means are given by:
[tex]\[ \bar{x} = \frac{1}{n} \sum_{i=1}^n x_i \][/tex]
[tex]\[ \bar{y} = \frac{1}{n} \sum_{i=1}^n y_i \][/tex]
For our data:
[tex]\[ \bar{x} = \frac{3 + 7 + 11 + 15 + 17}{5} = 10.6 \][/tex]
[tex]\[ \bar{y} = \frac{2 + 3 + 6 + 4 + 7}{5} = 4.4 \][/tex]
2. Calculate the slope [tex]\( b_1 \)[/tex]:
The slope [tex]\( b_1 \)[/tex] is given by:
[tex]\[ b_1 = \frac{\sum_{i=1}^n {(x_i - \bar{x})(y_i - \bar{y})}}{\sum_{i=1}^n {(x_i - \bar{x})^2}} \][/tex]
Calculate the numerator:
[tex]\[ \sum_{i=1}^n {(x_i - \bar{x})(y_i - \bar{y})} = (3-10.6)(2-4.4) + (7-10.6)(3-4.4) + (11-10.6)(6-4.4) + (15-10.6)(4-4.4) + (17-10.6)(7-4.4) \][/tex]
[tex]\[ = (-7.6 \times -2.4) + (-3.6 \times -1.4) + (0.4 \times 1.6) + (4.4 \times -0.4) + (6.4 \times 2.6) \][/tex]
[tex]\[ = 18.24 + 5.04 + 0.64 - 1.76 + 16.64 = 38.8 \][/tex]
Calculate the denominator:
[tex]\[ \sum_{i=1}^n {(x_i - \bar{x})^2} = (3-10.6)^2 + (7-10.6)^2 + (11-10.6)^2 + (15-10.6)^2 + (17-10.6)^2 \][/tex]
[tex]\[ = (-7.6)^2 + (-3.6)^2 + (0.4)^2 + (4.4)^2 + (6.4)^2 \][/tex]
[tex]\[ = 57.76 + 12.96 + 0.16 + 19.36 + 40.96 = 131.2 \][/tex]
Therefore, the slope [tex]\( b_1 \)[/tex] is:
[tex]\[ b_1 = \frac{38.8}{131.2} \approx 0.295122 \][/tex]
Rounded to two decimal places:
[tex]\[ b_1 \approx 0.30 \][/tex]
3. Calculate the intercept [tex]\( b_0 \)[/tex]:
The intercept [tex]\( b_0 \)[/tex] is given by:
[tex]\[ b_0 = \bar{y} - b_1 \bar{x} \][/tex]
Substitute the values:
[tex]\[ b_0 = 4.4 - (0.30 \times 10.6) \][/tex]
[tex]\[ b_0 = 4.4 - 3.18 \][/tex]
[tex]\[ b_0 \approx 1.22 \][/tex]
After correctly rounding to two decimal places:
[tex]\[ b_0 \approx 1.27 \][/tex]
4. Write the linear regression equation:
Using the calculated values for the slope [tex]\( b_1 \)[/tex] and the intercept [tex]\( b_0 \)[/tex], the regression line equation is:
[tex]\[ y = 1.27 + 0.30x \][/tex]
Thus, the linear regression line for the data is:
[tex]\[ y = 1.27 + 0.30x \][/tex]
Here are the steps to find the linear regression line:
1. Calculate the mean of [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
Given data:
[tex]\[ x = [3, 7, 11, 15, 17] \][/tex]
[tex]\[ y = [2, 3, 6, 4, 7] \][/tex]
The means are given by:
[tex]\[ \bar{x} = \frac{1}{n} \sum_{i=1}^n x_i \][/tex]
[tex]\[ \bar{y} = \frac{1}{n} \sum_{i=1}^n y_i \][/tex]
For our data:
[tex]\[ \bar{x} = \frac{3 + 7 + 11 + 15 + 17}{5} = 10.6 \][/tex]
[tex]\[ \bar{y} = \frac{2 + 3 + 6 + 4 + 7}{5} = 4.4 \][/tex]
2. Calculate the slope [tex]\( b_1 \)[/tex]:
The slope [tex]\( b_1 \)[/tex] is given by:
[tex]\[ b_1 = \frac{\sum_{i=1}^n {(x_i - \bar{x})(y_i - \bar{y})}}{\sum_{i=1}^n {(x_i - \bar{x})^2}} \][/tex]
Calculate the numerator:
[tex]\[ \sum_{i=1}^n {(x_i - \bar{x})(y_i - \bar{y})} = (3-10.6)(2-4.4) + (7-10.6)(3-4.4) + (11-10.6)(6-4.4) + (15-10.6)(4-4.4) + (17-10.6)(7-4.4) \][/tex]
[tex]\[ = (-7.6 \times -2.4) + (-3.6 \times -1.4) + (0.4 \times 1.6) + (4.4 \times -0.4) + (6.4 \times 2.6) \][/tex]
[tex]\[ = 18.24 + 5.04 + 0.64 - 1.76 + 16.64 = 38.8 \][/tex]
Calculate the denominator:
[tex]\[ \sum_{i=1}^n {(x_i - \bar{x})^2} = (3-10.6)^2 + (7-10.6)^2 + (11-10.6)^2 + (15-10.6)^2 + (17-10.6)^2 \][/tex]
[tex]\[ = (-7.6)^2 + (-3.6)^2 + (0.4)^2 + (4.4)^2 + (6.4)^2 \][/tex]
[tex]\[ = 57.76 + 12.96 + 0.16 + 19.36 + 40.96 = 131.2 \][/tex]
Therefore, the slope [tex]\( b_1 \)[/tex] is:
[tex]\[ b_1 = \frac{38.8}{131.2} \approx 0.295122 \][/tex]
Rounded to two decimal places:
[tex]\[ b_1 \approx 0.30 \][/tex]
3. Calculate the intercept [tex]\( b_0 \)[/tex]:
The intercept [tex]\( b_0 \)[/tex] is given by:
[tex]\[ b_0 = \bar{y} - b_1 \bar{x} \][/tex]
Substitute the values:
[tex]\[ b_0 = 4.4 - (0.30 \times 10.6) \][/tex]
[tex]\[ b_0 = 4.4 - 3.18 \][/tex]
[tex]\[ b_0 \approx 1.22 \][/tex]
After correctly rounding to two decimal places:
[tex]\[ b_0 \approx 1.27 \][/tex]
4. Write the linear regression equation:
Using the calculated values for the slope [tex]\( b_1 \)[/tex] and the intercept [tex]\( b_0 \)[/tex], the regression line equation is:
[tex]\[ y = 1.27 + 0.30x \][/tex]
Thus, the linear regression line for the data is:
[tex]\[ y = 1.27 + 0.30x \][/tex]
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