 # Find the correlatin coefficient r using bivariate data. sodni3 2021-01-07 Answered
Find the correlatin coefficient r using bivariate data.
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Given:
The given bivariate data is(1,6),(3,2), and (2,4).
Calculation:
First determine $\sum {x}_{i}{y}_{i},\sum {x}_{i}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\sum {y}_{i}$.
${x}_{i}\to$ first coordinates of the ordered pairs.
${y}_{i}\to$ second coordinates of the ordered pairs.
$\sum {x}_{i}{y}_{i}=1\cdot 6+3\cdot 2+2\cdot 4=20$
$\sum {x}_{i}=1+3+2=6$
$\sum {x}_{i}^{2}={1}^{2}+{3}^{2}+{2}^{2}=14$
$\sum {y}_{i}=6+2+4=12$
$\sum {y}_{i}^{2}={6}^{2}+{2}^{2}+{4}^{2}=56$
Find the sample variance ${s}^{2}$ using the formula ${s}^{2}=\frac{\sum {x}_{i}^{2}-\frac{\sum {x}_{i}^{2}}{n}}{n-1}$
${s}_{x}^{2}=\frac{14-\frac{{6}^{2}}{3}}{3-1}=\frac{14-12}{2}=1$
${s}_{y}^{2}=\frac{56-\frac{{12}^{2}}{3}}{3-1}=\frac{56-48}{2}=4$
Find sample standard deviation.
${s}_{x}=\sqrt{1}=1$
${s}_{y}=\sqrt{4}=2$
For covariance .
Where n = number of ordered pairs.
n=3
${s}_{xy}=\frac{20-\frac{6\cdot 12}{3}}{3-1}$
$=\frac{20-\frac{72}{3}}{3-1}$
$=\frac{20-24}{2}$
$=-\frac{4}{2}=-2$
Find the correlation coefficient r using the formula $r=\frac{{s}_{xy}}{{s}_{x}{s}_{y}}$.
$r=-\frac{2}{1\cdot 2}=-1$
Hence the value of correlation coefficient r = -1.