Question

the tables contain two sets of data. compute the correlation for both sets of data. give your answers to three decimal places. for data set a, r = 0.664 for data set b, r =

Explanation:

Step1: Recall correlation formula

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

Step2: Calculate sums for data - set A

For data - set A:
Let $n = 4$.
$\sum x=1 + 2+3 + 4=10$, $\sum y=1 + 1.5+0.5 + 4=7$, $\sum xy=1\times1+2\times1.5 + 3\times0.5+4\times4=1 + 3+1.5 + 16=21.5$, $\sum x^{2}=1^{2}+2^{2}+3^{2}+4^{2}=1 + 4+9 + 16=30$, $\sum y^{2}=1^{2}+1.5^{2}+0.5^{2}+4^{2}=1+2.25 + 0.25+16=19.5$.
$n(\sum xy)=4\times21.5 = 86$, $(\sum x)(\sum y)=10\times7 = 70$, $n\sum x^{2}=4\times30 = 120$, $(\sum x)^{2}=10^{2}=100$, $n\sum y^{2}=4\times19.5 = 78$, $(\sum y)^{2}=7^{2}=49$.
$r_A=\frac{4\times21.5-10\times7}{\sqrt{(4\times30 - 10^{2})(4\times19.5 - 7^{2})}}=\frac{86 - 70}{\sqrt{(120 - 100)(78 - 49)}}=\frac{16}{\sqrt{20\times29}}=\frac{16}{\sqrt{580}}\approx0.664$.

Step3: Calculate sums for data - set B

For data - set B:
Let $n = 7$.
$\sum x=1\times3+2 + 3+4\times3=3 + 2+3 + 12=20$, $\sum y=1\times3+1.5 + 0.5+4\times3=3 + 1.5+0.5 + 12=17$, $\sum xy=1\times1\times3+2\times1.5+3\times0.5+4\times4\times3=3+3 + 1.5+48=55.5$, $\sum x^{2}=1^{2}\times3+2^{2}+3^{2}+4^{2}\times3=3 + 4+9 + 48=64$, $\sum y^{2}=1^{2}\times3+1.5^{2}+0.5^{2}+4^{2}\times3=3+2.25 + 0.25+48=53.5$.
$n(\sum xy)=7\times55.5 = 388.5$, $(\sum x)(\sum y)=20\times17 = 340$, $n\sum x^{2}=7\times64 = 448$, $(\sum x)^{2}=20^{2}=400$, $n\sum y^{2}=7\times53.5 = 374.5$, $(\sum y)^{2}=17^{2}=289$.
$r_B=\frac{7\times55.5-20\times17}{\sqrt{(7\times64 - 20^{2})(7\times53.5 - 17^{2})}}=\frac{388.5 - 340}{\sqrt{(448 - 400)(374.5 - 289)}}=\frac{48.5}{\sqrt{48\times85.5}}=\frac{48.5}{\sqrt{4104}}\approx0.757$.

Answer:

For data - set A, $r = 0.664$
For data - set B, $r = 0.757$