Question

find the equation for the least squares regression line of the data described below. every summer, the dayton fair has a grass - hopper catching contest. grasshoppers are released into a fenced area, and participants are given five minutes to catch as many as they can. to even the playing field, the organizers developed a handicap system based on the results of this years competition. the system takes into account the different experience levels of the contest participants. to create the system, the organizers recorded the number of times each contestant had previously participated, x, and the number of grasshoppers each contestant had caught this year, y. previous competitions entered grasshoppers caught 0 20 2 16 3 21 3 14 4 25 9 31 round your answers to the nearest thousandth. y = x +

Explanation:

Explicación:

Paso 1: Calcular sumatorias

Sean $x_i$ el número de veces que un participante participó previamente y $y_i$ el número de saltamontes capturados.
Tenemos $n = 6$ datos.
$\sum_{i = 1}^{n}x_i=0 + 2+3 + 3+4 + 9=21$
$\sum_{i = 1}^{n}y_i=20 + 16+21 + 14+25 + 31=127$
$\sum_{i = 1}^{n}x_i^2=0^2+2^2 + 3^2+3^2+4^2 + 9^2=0 + 4+9 + 9+16 + 81=129$
$\sum_{i = 1}^{n}x_iy_i=(0\times20)+(2\times16)+(3\times21)+(3\times14)+(4\times25)+(9\times31)=0 + 32+63+42+100+279=516$

Paso 2: 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:
$m=\frac{6\times516 - 21\times127}{6\times129-21^2}=\frac{3096-2667}{774 - 441}=\frac{429}{333}\approx1.288$

Paso 3: Calcular la intersección $b$

La fórmula para la intersección $b$ es $b=\bar{y}-m\bar{x}$, donde $\bar{x}=\frac{\sum_{i = 1}^{n}x_i}{n}=\frac{21}{6}=3.5$ y $\bar{y}=\frac{\sum_{i = 1}^{n}y_i}{n}=\frac{127}{6}\approx21.167$
$b = 21.167-1.288\times3.5=21.167 - 4.508=16.659$

Respuesta:

$y = 1.288x+16.659$

Answer:

Explicación:

Paso 1: Calcular sumatorias

Sean $x_i$ el número de veces que un participante participó previamente y $y_i$ el número de saltamontes capturados.
Tenemos $n = 6$ datos.
$\sum_{i = 1}^{n}x_i=0 + 2+3 + 3+4 + 9=21$
$\sum_{i = 1}^{n}y_i=20 + 16+21 + 14+25 + 31=127$
$\sum_{i = 1}^{n}x_i^2=0^2+2^2 + 3^2+3^2+4^2 + 9^2=0 + 4+9 + 9+16 + 81=129$
$\sum_{i = 1}^{n}x_iy_i=(0\times20)+(2\times16)+(3\times21)+(3\times14)+(4\times25)+(9\times31)=0 + 32+63+42+100+279=516$

Paso 2: 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:
$m=\frac{6\times516 - 21\times127}{6\times129-21^2}=\frac{3096-2667}{774 - 441}=\frac{429}{333}\approx1.288$

Paso 3: Calcular la intersección $b$

La fórmula para la intersección $b$ es $b=\bar{y}-m\bar{x}$, donde $\bar{x}=\frac{\sum_{i = 1}^{n}x_i}{n}=\frac{21}{6}=3.5$ y $\bar{y}=\frac{\sum_{i = 1}^{n}y_i}{n}=\frac{127}{6}\approx21.167$
$b = 21.167-1.288\times3.5=21.167 - 4.508=16.659$

Respuesta:

$y = 1.288x+16.659$