Question

find the correlation coefficient, r, of the data described below. the khan toy company is testing the usability of its new spinning top among different age groups. brad, a designer at the company, gave the tops to a group of children of different ages and had each child spin a top once. brad recorded the age of each child, x, and how long he or she got the top to spin (in seconds), y. age time (in seconds) 5 32 7 56 8 9 9 76 10 75 round your answer to the nearest thousandth. r =

Explanation:

Step1: Calcular sumas básicas

Sean $x_i$ las edades y $y_i$ los tiempos. Tenemos $n = 5$.
$\sum_{i = 1}^{n}x_i=5 + 7+8 + 9+10=39$
$\sum_{i = 1}^{n}y_i=32 + 56+9+76+75=248$
$\sum_{i = 1}^{n}x_i^2=5^2 + 7^2+8^2 + 9^2+10^2=25 + 49+64+81+100=319$
$\sum_{i = 1}^{n}y_i^2=32^2 + 56^2+9^2 + 76^2+75^2=1024+3136 + 81+5776+5625=15642$
$\sum_{i = 1}^{n}x_iy_i=(5\times32)+(7\times56)+(8\times9)+(9\times76)+(10\times75)=160+392+72+684+750=2058$

Step2: Calcular la media de $x$ y $y$

$\bar{x}=\frac{\sum_{i = 1}^{n}x_i}{n}=\frac{39}{5}=7.8$
$\bar{y}=\frac{\sum_{i = 1}^{n}y_i}{n}=\frac{248}{5}=49.6$

Step3: Calcular la covarianza

$S_{xy}=\frac{\sum_{i = 1}^{n}x_iy_i - n\bar{x}\bar{y}}{n - 1}=\frac{2058-5\times7.8\times49.6}{4}=\frac{2058 - 1934.4}{4}=\frac{123.6}{4}=30.9$

Step4: Calcular la desviación estándar de $x$ y $y$

$S_x=\sqrt{\frac{\sum_{i = 1}^{n}x_i^2 - n\bar{x}^2}{n - 1}}=\sqrt{\frac{319-5\times7.8^2}{4}}=\sqrt{\frac{319 - 304.2}{4}}=\sqrt{\frac{14.8}{4}}=\sqrt{3.7}\approx1.924$
$S_y=\sqrt{\frac{\sum_{i = 1}^{n}y_i^2 - n\bar{y}^2}{n - 1}}=\sqrt{\frac{15642-5\times49.6^2}{4}}=\sqrt{\frac{15642-12300.8}{4}}=\sqrt{\frac{3341.2}{4}}=\sqrt{835.3}\approx28.902$

Step5: Calcular el coeficiente de correlación

$r=\frac{S_{xy}}{S_xS_y}=\frac{30.9}{1.924\times28.902}=\frac{30.9}{55.607}\approx0.556$

Answer:

$r\approx0.556$