To determine the regression equation that models the given data, we can follow these steps:
1. Calculate the means of [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
[tex]\[
\bar{x} = \frac{\sum x}{n} \quad \text{and} \quad \bar{y} = \frac{\sum y}{n}
\][/tex]
Given the values [tex]\(\sum x = 632\)[/tex], [tex]\(\sum y = 404\)[/tex], and [tex]\( n = 5 \)[/tex]:
[tex]\[
\bar{x} = \frac{632}{5} = 126.4 \quad \text{and} \quad \bar{y} = \frac{404}{5} = 80.8
\][/tex]
2. Calculate the slope ( [tex]\( b \)[/tex] ) of the regression line:
[tex]\[
b = \frac{\sum (xy) - n \cdot \bar{x} \cdot \bar{y}}{\sum (x^2) - n \cdot \bar{x}^2}
\][/tex]
Using the given data [tex]\(\sum xy = 51448\)[/tex] and [tex]\(\sum x^2 = 80142\)[/tex]:
[tex]\[
b = \frac{51448 - 5 \cdot 126.4 \cdot 80.8}{80142 - 5 \cdot (126.4)^2}
\][/tex]
Calculating the numerator and denominator separately:
[tex]\[
\text{Numerator} = 51448 - 5 \cdot 126.4 \cdot 80.8 \approx 51448 - 51065.6 = 382.4
\][/tex]
[tex]\[
\text{Denominator} = 80142 - 5 \cdot (126.4)^2 \approx 80142 - 80084.8 = 257.2
\][/tex]
So, the slope [tex]\( b \)[/tex] is:
[tex]\[
b = \frac{382.4}{257.2} \approx 1.48678
\][/tex]
3. Calculate the intercept ( [tex]\( a \)[/tex] ):
[tex]\[
a = \bar{y} - b \cdot \bar{x}
\][/tex]
[tex]\[
a = 80.8 - 1.48678 \cdot 126.4 \approx 80.8 - 187.92908 = -107.12908
\][/tex]
4. Form the regression equation:
The regression equation is of the form [tex]\( y = a + bx \)[/tex].
Given the values calculated:
[tex]\[
y = -107.12908 + 1.48678x
\][/tex]
So, the regression equation that correctly models the data is:
[tex]\[
\boxed{y = -107.12908 + 1.48678x}
\][/tex]
