Question

find the equation for the least squares regression line of the data described below. maura has noticed that her bike ride to work takes longer on some days than others. she is curious to see how the morning temperature is related to the duration of her commute. for the past several mornings, she measured the temperature (in celsius), x, and the time her commute had taken (in minutes), y.
temperature (degrees celsius) | commute time (minutes)
-4.35 | 59.71
1.63 | 51.34
4.08 | 53.40
4.13 | 40.53
4.72 | 31.12
round your answers to the nearest thousandth.
y = x +

Explanation:

Step1: Calcular sumatorias

Sean $n = 5$ los pares de datos.
Calcular $\sum_{i = 1}^{n}x_i=-4.35 + 1.63+4.08 + 4.13+4.72=10.11$
Calcular $\sum_{i = 1}^{n}y_i=59.71+51.34 + 53.40+40.53+31.12=236.1$
Calcular $\sum_{i = 1}^{n}x_i^2=(-4.35)^2+1.63^2 + 4.08^2+4.13^2+4.72^2=19.0225 + 2.6569+16.6464+17.0569+22.2784 = 77.6611$
Calcular $\sum_{i = 1}^{n}x_iy_i=(-4.35)\times59.71+1.63\times51.34 + 4.08\times53.40+4.13\times40.53+4.72\times31.12=-259.7385+83.6842+217.872+167.3889+146.8864 = 356.193$

Step2: Calcular la pendiente $m$

La fórmula para la pendiente $m$ de la recta de regresión lineal es $m=\frac{n\sum_{i = 1}^{n}x_iy_i-\sum_{i = 1}^{n}x_i\sum_{i = 1}^{n}y_i}{n\sum_{i = 1}^{n}x_i^2 - (\sum_{i = 1}^{n}x_i)^2}$
Sustituir valores: $n = 5$, $\sum_{i = 1}^{n}x_i = 10.11$, $\sum_{i = 1}^{n}y_i=236.1$, $\sum_{i = 1}^{n}x_i^2 = 77.6611$ y $\sum_{i = 1}^{n}x_iy_i=356.193$
$m=\frac{5\times356.193-10.11\times236.1}{5\times77.6611-(10.11)^2}$
$=\frac{1780.965 - 2387.971}{388.3055 - 102.2121}=\frac{-607.006}{286.0934}\approx - 2.1217\approx - 2.122$

Step3: Calcular la intersección $b$

La fórmula para la intersección $b$ es $b=\overline{y}-m\overline{x}$, donde $\overline{x}=\frac{\sum_{i = 1}^{n}x_i}{n}=\frac{10.11}{5}=2.022$ y $\overline{y}=\frac{\sum_{i = 1}^{n}y_i}{n}=\frac{236.1}{5}=47.22$
$b = 47.22-(-2.122)\times2.022$
$=47.22 + 4.2907 = 51.5107\approx51.511$

Answer:

$y=-2.122x + 51.511$