Question

find the equation for the least - squares regression line of the data described below. isabella wants to ensure that the new board game shes designing is accessible to a wide age range. to test this, she had several children of various ages play the board game against each other. each child played the game the same number of times. for each child, isabella recorded his or her age, x, and the number of games that child had won, y. age games won 7 0 7 0 10 2 12 10 12 9 round your answers to the nearest thousandth. y = x +

Explanation:

Step1: Calcular sumatorias básicas

Sean $x_i$ las edades y $y_i$ el número de partidas ganadas. Tenemos $n = 5$ datos.
$\sum_{i = 1}^{n}x_i=7 + 7+10 + 12+12=48$
$\sum_{i = 1}^{n}y_i=0 + 0+2 + 10+9=21$
$\sum_{i = 1}^{n}x_i^2=7^2 + 7^2+10^2 + 12^2+12^2=49+49 + 100+144+144 = 486$
$\sum_{i = 1}^{n}x_iy_i=7\times0+7\times0 + 10\times2+12\times10+12\times9=0+0 + 20+120+108 = 248$

Step2: Calcular la pendiente $m$

La fórmula para la pendiente $m$ de la recta de regresión lineal es $m=\frac{n\sum_{i = 1}^{n}x_iy_i-\sum_{i = 1}^{n}x_i\sum_{i = 1}^{n}y_i}{n\sum_{i = 1}^{n}x_i^2-(\sum_{i = 1}^{n}x_i)^2}$
Sustituyendo los valores:
\[

$$\begin{align*} m&=\frac{5\times248-48\times21}{5\times486 - 48^2}\\ &=\frac{1240-1008}{2430 - 2304}\\ &=\frac{232}{126}\\ &\approx1.841 \end{align*}$$

\]

Step3: Calcular la intersección $b$

La fórmula para la intersección $b$ es $b=\overline{y}-m\overline{x}$, donde $\overline{x}=\frac{\sum_{i = 1}^{n}x_i}{n}=\frac{48}{5} = 9.6$ y $\overline{y}=\frac{\sum_{i = 1}^{n}y_i}{n}=\frac{21}{5}=4.2$
\[

$$\begin{align*} b&=4.2-1.841\times9.6\\ &=4.2 - 17.674\\ &\approx - 13.474 \end{align*}$$

\]

Answer:

$y = 1.841x-13.474$