Question

find the equation for the least squares regression line of the data described below. todd is an editor reviewing short story submissions for a literary magazine. he collected data showing that hes an efficient reviewer and wants to leverage this to get a raise. for a week, todd recorded the number of words in each short story submission, x, and how long it took him to review that story (in minutes), y. words minutes 1,998 31.04 3,538 40.38 3,553 41.37 4,669 31.98 5,077 35.15 6,782 59.09 round your answers to the nearest thousandth. y = x +

Explanation:

Explicación:

Paso 1: Calcular sumatorias

Sean $x_i$ el número de palabras y $y_i$ el tiempo en minutos.
Sea $n = 6$.
Calculamos $\sum_{i = 1}^{n}x_i=1998 + 3538+3553 + 4669+5077+6782=25617$.
$\sum_{i = 1}^{n}y_i=31.04 + 40.38+41.37+31.98+35.15+59.09 = 239.01$.
$\sum_{i = 1}^{n}x_i^2=1998^2+3538^2+3553^2+4669^2+5077^2+6782^2$
$=3992004+12417444+12623809+21799561+25775929+46005524 = 122614271$.
$\sum_{i = 1}^{n}x_iy_i=1998\times31.04+3538\times40.38+3553\times41.37+4669\times31.98+5077\times35.15+6782\times59.09$
$=61917.92+142864.44+146987.61+148314.62+178456.55+400838.38 = 1079379.52$.

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:
$m=\frac{6\times1079379.52 - 25617\times239.01}{6\times122614271-(25617)^2}$
$=\frac{6476277.12-6123769.17}{735685626 - 656230689}$
$=\frac{352507.95}{79454937}\approx0.004$.

Paso 3: Calcular la intersección $b$

La media de $x$ es $\bar{x}=\frac{\sum_{i = 1}^{n}x_i}{n}=\frac{25617}{6}=4269.5$.
La media de $y$ es $\bar{y}=\frac{\sum_{i = 1}^{n}y_i}{n}=\frac{239.01}{6}=39.835$.
La fórmula para la intersección $b$ es $b=\bar{y}-m\bar{x}$.
$b = 39.835-0.004\times4269.5$
$=39.835 - 17.078=22.757$.

Respuesta:

$y = 0.004x+22.757$

Answer:

Explicación:

Paso 1: Calcular sumatorias

Sean $x_i$ el número de palabras y $y_i$ el tiempo en minutos.
Sea $n = 6$.
Calculamos $\sum_{i = 1}^{n}x_i=1998 + 3538+3553 + 4669+5077+6782=25617$.
$\sum_{i = 1}^{n}y_i=31.04 + 40.38+41.37+31.98+35.15+59.09 = 239.01$.
$\sum_{i = 1}^{n}x_i^2=1998^2+3538^2+3553^2+4669^2+5077^2+6782^2$
$=3992004+12417444+12623809+21799561+25775929+46005524 = 122614271$.
$\sum_{i = 1}^{n}x_iy_i=1998\times31.04+3538\times40.38+3553\times41.37+4669\times31.98+5077\times35.15+6782\times59.09$
$=61917.92+142864.44+146987.61+148314.62+178456.55+400838.38 = 1079379.52$.

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:
$m=\frac{6\times1079379.52 - 25617\times239.01}{6\times122614271-(25617)^2}$
$=\frac{6476277.12-6123769.17}{735685626 - 656230689}$
$=\frac{352507.95}{79454937}\approx0.004$.

Paso 3: Calcular la intersección $b$

La media de $x$ es $\bar{x}=\frac{\sum_{i = 1}^{n}x_i}{n}=\frac{25617}{6}=4269.5$.
La media de $y$ es $\bar{y}=\frac{\sum_{i = 1}^{n}y_i}{n}=\frac{239.01}{6}=39.835$.
La fórmula para la intersección $b$ es $b=\bar{y}-m\bar{x}$.
$b = 39.835-0.004\times4269.5$
$=39.835 - 17.078=22.757$.

Respuesta:

$y = 0.004x+22.757$