what is the correlation coefficient, r, for this data?\n|x|y|\n|4.5|1.3|\n|4.6|1.4|\n|4.6|1.7|\n|5.1|1.6|\n|5…

Explanation:

Step1: Calcular sumas

Sean $n = 5$. Calculemos $\sum x$, $\sum y$, $\sum xy$, $\sum x^{2}$ y $\sum y^{2}$. $\sum x=4.5 + 4.6+4.6+5.1+5.2 = 24$ $\sum y=1.3 + 1.4+1.7+1.6+1.8 = 7.8$ $\sum xy=(4.5\times1.3)+(4.6\times1.4)+(4.6\times1.7)+(5.1\times1.6)+(5.2\times1.8)=4.5\times1.3 + 6.44+7.82+8.16+9.36 = 36.33$ $\sum x^{2}=4.5^{2}+4.6^{2}+4.6^{2}+5.1^{2}+5.2^{2}=20.25+21.16+21.16+26.01+27.04 = 115.62$ $\sum y^{2}=1.3^{2}+1.4^{2}+1.7^{2}+1.6^{2}+1.8^{2}=1.69+1.96+2.89+2.56+3.24 = 12.34$

Step2: Aplicar fórmula de coeficiente de correlación

La fórmula para el coeficiente de correlación $r$ es: [r=\frac{n\sum xy-\sum x\sum y}{\sqrt{n\sum x^{2}-(\sum x)^{2}}\sqrt{n\sum y^{2}-(\sum y)^{2}}}] Sustituimos los valores: [n\sum xy = 5\times36.33=181.65] [\sum x\sum y=24\times7.8 = 187.2] [n\sum x^{2}=5\times115.62 = 578.1] [(\sum x)^{2}=24^{2}=576] [n\sum y^{2}=5\times12.34 = 61.7] [(\sum y)^{2}=7.8^{2}=60.84]

[r=\frac{181.65 - 187.2}{\sqrt{578.1-576}\sqrt{61.7 - 60.84}}] [r=\frac{- 5.55}{\sqrt{2.1}\sqrt{0.86}}] [r=\frac{-5.55}{\sqrt{2.1\times0.86}}=\frac{-5.55}{\sqrt{1.806}}] [r=\frac{-5.55}{1.344}\approx - 0.8]

Answer:

$-0.8$