question 5 1 pts rounded to three decimal places, what is the value of r for this data set? variable x 8 15…

Explanation:

Step1: Recall correlation - coefficient formula

The formula for the correlation coefficient $r$ is $r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^{2}-(\sum x)^{2}][n\sum y^{2}-(\sum y)^{2}]}}$

First, calculate the necessary sums: Let $n = 7$ $\sum x=8 + 15+3+7+2+14+20=69$ $\sum y=50 + 75+23+31+18+68+96=361$ $\sum xy=(8\times50)+(15\times75)+(3\times23)+(7\times31)+(2\times18)+(14\times68)+(20\times96)$ $=400+1125 + 69+217+36+952+1920=4719$ $\sum x^{2}=8^{2}+15^{2}+3^{2}+7^{2}+2^{2}+14^{2}+20^{2}$ $=64 + 225+9+49+4+196+400=947$ $\sum y^{2}=50^{2}+75^{2}+23^{2}+31^{2}+18^{2}+68^{2}+96^{2}$ $=2500+5625+529+961+324+4624+9216=23789$

Step2: Substitute into the formula

$n(\sum xy)-(\sum x)(\sum y)=7\times4719-69\times361$ $=33033-24909 = 8124$

$n\sum x^{2}-(\sum x)^{2}=7\times947-69^{2}$ $=6629 - 4761=1868$

$n\sum y^{2}-(\sum y)^{2}=7\times23789-361^{2}$ $=166523-130321 = 36202$

$\sqrt{[n\sum x^{2}-(\sum x)^{2}][n\sum y^{2}-(\sum y)^{2}]}=\sqrt{1868\times36202}$ $=\sqrt{67625336}\approx8223.46$

$r=\frac{8124}{8223.46}\approx0.988$

Answer:

B. 0.989