Question

find the equation for the least squares regression line of the data described below. eduardo is indignant that his favorite author takes too long to write new books. to determine how atypical his favorite authors speed is, eduardo looked at how writing speed varies between different active authors. for his favorite novelists, he recorded the number of years they had been actively writing, x, and how many novels they had written, y.
years novels
9 13
20 13
22 15
22 13
34 18
round your answers to the nearest thousandth.
y = x +

Explanation:

Explicación:

Paso 1: Calcular sumas necesarias

Sean $n = 5$, $x_i$ los valores de años y $y_i$ los valores de novelas.
Calculamos $\sum_{i = 1}^{n}x_i=9 + 20+22 + 22+34=107$, $\sum_{i = 1}^{n}y_i=13 + 13+15+13+18=72$, $\sum_{i = 1}^{n}x_i^2=9^2 + 20^2+22^2 + 22^2+34^2=81+400+484+484+1156 = 2605$, $\sum_{i = 1}^{n}x_iy_i=9\times13 + 20\times13+22\times15+22\times13+34\times18=117+260+330+286+612 = 1605$.

Paso 2: Calcular la pendiente $m$

La fórmula para la pendiente $m$ de la línea de regresión 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: $m=\frac{5\times1605 - 107\times72}{5\times2605-(107)^2}=\frac{8025 - 7704}{13025 - 11449}=\frac{321}{1576}\approx0.204$.

Paso 3: Calcular la intersección $b$

La media de $x$ es $\bar{x}=\frac{\sum_{i = 1}^{n}x_i}{n}=\frac{107}{5} = 21.4$, y la media de $y$ es $\bar{y}=\frac{\sum_{i = 1}^{n}y_i}{n}=\frac{72}{5}=14.4$.
Usando la fórmula $b=\bar{y}-m\bar{x}$, tenemos $b = 14.4-0.204\times21.4=14.4 - 4.366 = 10.034$.

Respuesta:

$y = 0.204x+10.034$

Answer:

Explicación:

Paso 1: Calcular sumas necesarias

Sean $n = 5$, $x_i$ los valores de años y $y_i$ los valores de novelas.
Calculamos $\sum_{i = 1}^{n}x_i=9 + 20+22 + 22+34=107$, $\sum_{i = 1}^{n}y_i=13 + 13+15+13+18=72$, $\sum_{i = 1}^{n}x_i^2=9^2 + 20^2+22^2 + 22^2+34^2=81+400+484+484+1156 = 2605$, $\sum_{i = 1}^{n}x_iy_i=9\times13 + 20\times13+22\times15+22\times13+34\times18=117+260+330+286+612 = 1605$.

Paso 2: Calcular la pendiente $m$

La fórmula para la pendiente $m$ de la línea de regresión 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: $m=\frac{5\times1605 - 107\times72}{5\times2605-(107)^2}=\frac{8025 - 7704}{13025 - 11449}=\frac{321}{1576}\approx0.204$.

Paso 3: Calcular la intersección $b$

La media de $x$ es $\bar{x}=\frac{\sum_{i = 1}^{n}x_i}{n}=\frac{107}{5} = 21.4$, y la media de $y$ es $\bar{y}=\frac{\sum_{i = 1}^{n}y_i}{n}=\frac{72}{5}=14.4$.
Usando la fórmula $b=\bar{y}-m\bar{x}$, tenemos $b = 14.4-0.204\times21.4=14.4 - 4.366 = 10.034$.

Respuesta:

$y = 0.204x+10.034$