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Sagot :
To find the linear regression equation that represents the given set of data, we will follow these steps:
1. Define Variables:
- Let [tex]\( x \)[/tex] represent the homework grade.
- Let [tex]\( y \)[/tex] represent the test grade.
2. Given Data:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 61 & 49 \\ 81 & 67 \\ 86 & 86 \\ 70 & 56 \\ 74 & 66 \\ 74 & 77 \\ 55 & 52 \\ \hline \end{array} \][/tex]
3. Calculate the Means [tex]\( \bar{x} \)[/tex] and [tex]\( \bar{y} \)[/tex]:
[tex]\[ \bar{x} = \frac{61 + 81 + 86 + 70 + 74 + 74 + 55}{7} = \frac{501}{7} \approx 71.6 \][/tex]
[tex]\[ \bar{y} = \frac{49 + 67 + 86 + 56 + 66 + 77 + 52}{7} = \frac{453}{7} \approx 64.7 \][/tex]
4. Calculate the Slope [tex]\( m \)[/tex]:
[tex]\[ m = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sum (x_i - \bar{x})^2} \][/tex]
To find [tex]\( \sum (x_i - \bar{x})(y_i - \bar{y}) \)[/tex] and [tex]\( \sum (x_i - \bar{x})^2 \)[/tex]:
[tex]\[ \sum (x_i - \bar{x})(y_i - \bar{y}) = (61 - 71.6)(49 - 64.7) + (81 - 71.6)(67 - 64.7) + (86 - 71.6)(86 - 64.7) + (70 - 71.6)(56 - 64.7) + (74 - 71.6)(66 - 64.7) + (74 - 71.6)(77 - 64.7) + (55 - 71.6)(52 - 64.7) \][/tex]
[tex]\[ = (-10.6)(-15.7) + 9.4(2.3) + 14.4(21.3) + (-1.6)(-8.7) + 2.4(1.3) + 2.4(12.3) + (-16.6)(-12.7) \][/tex]
[tex]\[ \approx 166.62 + 21.62 + 306.72 + 13.92 + 3.12 + 29.52 + 210.82 = 752.34 \][/tex]
[tex]\[ \sum (x_i - \bar{x})^2 = (61 - 71.6)^2 + (81 - 71.6)^2 + (86 - 71.6)^2 + (70 - 71.6)^2 + (74 - 71.6)^2 + (74 - 71.6)^2 + (55 - 71.6)^2 \][/tex]
[tex]\[ = (-10.6)^2 + 9.4^2 + 14.4^2 + (-1.6)^2 + 2.4^2 + 2.4^2 + (-16.6)^2 \][/tex]
[tex]\[ = 112.36 + 88.36 + 207.36 + 2.56 + 5.76 + 5.76 + 275.56 = 697.68 \][/tex]
[tex]\[ m = \frac{752.34}{697.68} \approx 1.08 \approx 1.1 \, (\text{Rounded to nearest tenth}) \][/tex]
5. Calculate the Intercept [tex]\( b \)[/tex]:
[tex]\[ b = \bar{y} - m \bar{x} \][/tex]
[tex]\[ b = 64.7 - 1.1 \times 71.6 = 64.7 - 78.76 = -14.06 \approx -14.1 \, (\text{Rounded to nearest tenth}) \][/tex]
6. Linear Regression Equation:
[tex]\[ y = 1.1x - 14.1 \][/tex]
7. Estimate Homework Grade for Test Grade of 42:
We need to find [tex]\( x \)[/tex] such that [tex]\( y = 42 \)[/tex]:
[tex]\[ 42 = 1.1x - 14.1 \][/tex]
[tex]\[ 1.1x = 42 + 14.1 \][/tex]
[tex]\[ 1.1x = 56.1 \][/tex]
[tex]\[ x = \frac{56.1}{1.1} \approx 51 \][/tex]
So, the linear regression equation is [tex]\( y = 1.1x - 14.1 \)[/tex] and the estimated homework grade for a test grade of 42 is approximately [tex]\( 51 \)[/tex].
1. Define Variables:
- Let [tex]\( x \)[/tex] represent the homework grade.
- Let [tex]\( y \)[/tex] represent the test grade.
2. Given Data:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 61 & 49 \\ 81 & 67 \\ 86 & 86 \\ 70 & 56 \\ 74 & 66 \\ 74 & 77 \\ 55 & 52 \\ \hline \end{array} \][/tex]
3. Calculate the Means [tex]\( \bar{x} \)[/tex] and [tex]\( \bar{y} \)[/tex]:
[tex]\[ \bar{x} = \frac{61 + 81 + 86 + 70 + 74 + 74 + 55}{7} = \frac{501}{7} \approx 71.6 \][/tex]
[tex]\[ \bar{y} = \frac{49 + 67 + 86 + 56 + 66 + 77 + 52}{7} = \frac{453}{7} \approx 64.7 \][/tex]
4. Calculate the Slope [tex]\( m \)[/tex]:
[tex]\[ m = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sum (x_i - \bar{x})^2} \][/tex]
To find [tex]\( \sum (x_i - \bar{x})(y_i - \bar{y}) \)[/tex] and [tex]\( \sum (x_i - \bar{x})^2 \)[/tex]:
[tex]\[ \sum (x_i - \bar{x})(y_i - \bar{y}) = (61 - 71.6)(49 - 64.7) + (81 - 71.6)(67 - 64.7) + (86 - 71.6)(86 - 64.7) + (70 - 71.6)(56 - 64.7) + (74 - 71.6)(66 - 64.7) + (74 - 71.6)(77 - 64.7) + (55 - 71.6)(52 - 64.7) \][/tex]
[tex]\[ = (-10.6)(-15.7) + 9.4(2.3) + 14.4(21.3) + (-1.6)(-8.7) + 2.4(1.3) + 2.4(12.3) + (-16.6)(-12.7) \][/tex]
[tex]\[ \approx 166.62 + 21.62 + 306.72 + 13.92 + 3.12 + 29.52 + 210.82 = 752.34 \][/tex]
[tex]\[ \sum (x_i - \bar{x})^2 = (61 - 71.6)^2 + (81 - 71.6)^2 + (86 - 71.6)^2 + (70 - 71.6)^2 + (74 - 71.6)^2 + (74 - 71.6)^2 + (55 - 71.6)^2 \][/tex]
[tex]\[ = (-10.6)^2 + 9.4^2 + 14.4^2 + (-1.6)^2 + 2.4^2 + 2.4^2 + (-16.6)^2 \][/tex]
[tex]\[ = 112.36 + 88.36 + 207.36 + 2.56 + 5.76 + 5.76 + 275.56 = 697.68 \][/tex]
[tex]\[ m = \frac{752.34}{697.68} \approx 1.08 \approx 1.1 \, (\text{Rounded to nearest tenth}) \][/tex]
5. Calculate the Intercept [tex]\( b \)[/tex]:
[tex]\[ b = \bar{y} - m \bar{x} \][/tex]
[tex]\[ b = 64.7 - 1.1 \times 71.6 = 64.7 - 78.76 = -14.06 \approx -14.1 \, (\text{Rounded to nearest tenth}) \][/tex]
6. Linear Regression Equation:
[tex]\[ y = 1.1x - 14.1 \][/tex]
7. Estimate Homework Grade for Test Grade of 42:
We need to find [tex]\( x \)[/tex] such that [tex]\( y = 42 \)[/tex]:
[tex]\[ 42 = 1.1x - 14.1 \][/tex]
[tex]\[ 1.1x = 42 + 14.1 \][/tex]
[tex]\[ 1.1x = 56.1 \][/tex]
[tex]\[ x = \frac{56.1}{1.1} \approx 51 \][/tex]
So, the linear regression equation is [tex]\( y = 1.1x - 14.1 \)[/tex] and the estimated homework grade for a test grade of 42 is approximately [tex]\( 51 \)[/tex].
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