Question

find the correlation coefficient, r, of the data described below. deions landscape architecture firm won a contract to design a new public playground in seville. to decide how many swings to include, deion collected information about other playgrounds in the city. for each playground, he recorded the area (in square meters), x, and the number of swings, y. area (in square meters) swings 18 3 34 3 36 5 40 12 46 12 51 10 round your answer to the nearest thousandth. r =

Explanation:

Step1: Calcular sumatorias

Sean $x_i$ las áreas y $y_i$ el número de columpios.
Tenemos $n = 6$.
$\sum_{i = 1}^{n}x_i=18 + 34+36 + 40+46+51=225$
$\sum_{i = 1}^{n}y_i=3 + 3+5 + 12+12+10=45$
$\sum_{i = 1}^{n}x_i^2=18^2+34^2+36^2+40^2+46^2+51^2=324 + 1156+1296+1600+2116+2601=9093$
$\sum_{i = 1}^{n}y_i^2=3^2+3^2+5^2+12^2+12^2+10^2=9 + 9+25+144+144+100=431$
$\sum_{i = 1}^{n}x_iy_i=18\times3+34\times3+36\times5+40\times12+46\times12+51\times10=54+102+180+480+552+510=1878$

Step2: Calcular desviaciones estándar y covarianza

La media de $x$, $\bar{x}=\frac{\sum_{i = 1}^{n}x_i}{n}=\frac{225}{6}=37.5$
La media de $y$, $\bar{y}=\frac{\sum_{i = 1}^{n}y_i}{n}=\frac{45}{6}=7.5$
La covarianza $S_{xy}=\frac{\sum_{i = 1}^{n}x_iy_i - n\bar{x}\bar{y}}{n - 1}=\frac{1878-6\times37.5\times7.5}{5}=\frac{1878 - 1687.5}{5}=\frac{190.5}{5}=38.1$
La desviación estándar de $x$, $S_x=\sqrt{\frac{\sum_{i = 1}^{n}x_i^2 - n\bar{x}^2}{n - 1}}=\sqrt{\frac{9093-6\times37.5^2}{5}}=\sqrt{\frac{9093 - 8437.5}{5}}=\sqrt{\frac{655.5}{5}}=\sqrt{131.1}\approx11.45$
La desviación estándar de $y$, $S_y=\sqrt{\frac{\sum_{i = 1}^{n}y_i^2 - n\bar{y}^2}{n - 1}}=\sqrt{\frac{431-6\times7.5^2}{5}}=\sqrt{\frac{431 - 337.5}{5}}=\sqrt{\frac{93.5}{5}}=\sqrt{18.7}\approx4.33$

Step3: Calcular coeficiente de correlación

El coeficiente de correlación $r=\frac{S_{xy}}{S_xS_y}=\frac{38.1}{11.45\times4.33}=\frac{38.1}{49.5785}\approx0.768$

Answer:

$r = 0.768$