find the value of the correlation coefficient for the data shown in the table. round to the nearest…

Answer:

D. $r = - 0.85$

Explanation:

Step1: Identificar fórmula

La fórmula para el coeficiente de correlación $r$ es $r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^{2}-(\sum x)^{2}][n\sum y^{2}-(\sum y)^{2}]}}$.

Step2: Calcular sumatorias

Sean $x = {4,5,6}$, $y={6000,4500,3500}$. $\sum x=4 + 5+6=15$, $\sum y=6000 + 4500+3500 = 14000$, $\sum xy=4\times6000+5\times4500+6\times3500=24000 + 22500+21000 = 67500$, $\sum x^{2}=4^{2}+5^{2}+6^{2}=16 + 25+36 = 77$, $\sum y^{2}=6000^{2}+4500^{2}+3500^{2}=36000000+20250000+12250000 = 68500000$, $n = 3$.

Step3: Sustituir valores

$r=\frac{3\times67500-15\times14000}{\sqrt{[3\times77 - 15^{2}][3\times68500000-14000^{2}]}}$. $3\times67500=202500$, $15\times14000 = 210000$, $3\times77=231$, $15^{2}=225$, $3\times68500000 = 205500000$, $14000^{2}=196000000$. $r=\frac{202500 - 210000}{\sqrt{(231 - 225)(205500000-196000000)}}=\frac{- 7500}{\sqrt{6\times9500000}}=\frac{-7500}{\sqrt{57000000}}\approx - 0.85$.