Given:
The given bivariate data is(1,6),(3,2), and (2,4).
Calculation:
First determine \(\displaystyle\sum{x}_{{i}}{y}_{{i}},\sum{x}_{{i}}{\quad\text{and}\quad}\sum{y}_{{i}}\).
\(\displaystyle{x}_{{i}}\rightarrow\) first coordinates of the ordered pairs.
\(\displaystyle{y}_{{i}}\rightarrow\) second coordinates of the ordered pairs.
\(\displaystyle\sum{x}_{{i}}{y}_{{i}}={1}\cdot{6}+{3}\cdot{2}+{2}\cdot{4}={20}\)
\(\displaystyle\sum{x}_{{i}}={1}+{3}+{2}={6}\)
\(\displaystyle\sum{{x}_{{i}}^{{2}}}={1}^{{2}}+{3}^{{2}}+{2}^{{2}}={14}\)
\(\displaystyle\sum{y}_{{i}}={6}+{2}+{4}={12}\)
\(\displaystyle\sum{{y}_{{i}}^{{2}}}={6}^{{2}}+{2}^{{2}}+{4}^{{2}}={56}\)
Find the sample variance \(\displaystyle{s}^{{2}}\) using the formula \(\displaystyle{s}^{{2}}=\frac{{\sum{{x}_{{i}}^{{2}}}-\frac{{\sum{{x}_{{i}}^{{2}}}}}{{{n}}}}}{{{n}-{1}}}\)
\(\displaystyle{{s}_{{x}}^{{2}}}=\frac{{{14}-\frac{{{6}^{{2}}}}{{3}}}}{{{3}-{1}}}=\frac{{{14}-{12}}}{{2}}={1}\)
\(\displaystyle{{s}_{{y}}^{{2}}}=\frac{{{56}-\frac{{{12}^{{2}}}}{{3}}}}{{{3}-{1}}}=\frac{{{56}-{48}}}{{2}}={4}\)
Find sample standard deviation.
\(\displaystyle{s}_{{x}}=\sqrt{{1}}={1}\)
\(\displaystyle{s}_{{y}}=\sqrt{{4}}={2}\)
For covariance \(\displaystyle{s}_{{{x}{y}}}{u}{\sin{{g}}}{f}{\quad\text{or}\quad}\mu{l}{a}{s}_{{{x}{y}}}=\frac{{\sum{x}_{{i}}{y}_{{i}}-\frac{{\sum{x}_{{i}}-\sum{y}_{{i}}}}{{{n}}}}}{{{n}-{1}}}\).
Where n = number of ordered pairs.
n=3
\(\displaystyle{s}_{{{x}{y}}}=\frac{{{20}-\frac{{{6}\cdot{12}}}{{3}}}}{{{3}-{1}}}\)
\(\displaystyle=\frac{{{20}-\frac{{{72}}}{{3}}}}{{{3}-{1}}}\)
\(\displaystyle=\frac{{{20}-{24}}}{{2}}\)
\(\displaystyle=-\frac{{4}}{{2}}=-{2}\)
Find the correlation coefficient r using the formula \(\displaystyle{r}=\frac{{{s}_{{{x}{y}}}}}{{{s}_{{x}}{s}_{{y}}}}\).
\(\displaystyle{r}=-\frac{{2}}{{{1}\cdot{2}}}=-{1}\)
Hence the value of correlation coefficient r = -1.
The given bivariate data is(1,6),(3,2), and (2,4).
Calculation:
First determine \(\displaystyle\sum{x}_{{i}}{y}_{{i}},\sum{x}_{{i}}{\quad\text{and}\quad}\sum{y}_{{i}}\).
\(\displaystyle{x}_{{i}}\rightarrow\) first coordinates of the ordered pairs.
\(\displaystyle{y}_{{i}}\rightarrow\) second coordinates of the ordered pairs.
\(\displaystyle\sum{x}_{{i}}{y}_{{i}}={1}\cdot{6}+{3}\cdot{2}+{2}\cdot{4}={20}\)
\(\displaystyle\sum{x}_{{i}}={1}+{3}+{2}={6}\)
\(\displaystyle\sum{{x}_{{i}}^{{2}}}={1}^{{2}}+{3}^{{2}}+{2}^{{2}}={14}\)
\(\displaystyle\sum{y}_{{i}}={6}+{2}+{4}={12}\)
\(\displaystyle\sum{{y}_{{i}}^{{2}}}={6}^{{2}}+{2}^{{2}}+{4}^{{2}}={56}\)
Find the sample variance \(\displaystyle{s}^{{2}}\) using the formula \(\displaystyle{s}^{{2}}=\frac{{\sum{{x}_{{i}}^{{2}}}-\frac{{\sum{{x}_{{i}}^{{2}}}}}{{{n}}}}}{{{n}-{1}}}\)
\(\displaystyle{{s}_{{x}}^{{2}}}=\frac{{{14}-\frac{{{6}^{{2}}}}{{3}}}}{{{3}-{1}}}=\frac{{{14}-{12}}}{{2}}={1}\)
\(\displaystyle{{s}_{{y}}^{{2}}}=\frac{{{56}-\frac{{{12}^{{2}}}}{{3}}}}{{{3}-{1}}}=\frac{{{56}-{48}}}{{2}}={4}\)
Find sample standard deviation.
\(\displaystyle{s}_{{x}}=\sqrt{{1}}={1}\)
\(\displaystyle{s}_{{y}}=\sqrt{{4}}={2}\)
For covariance \(\displaystyle{s}_{{{x}{y}}}{u}{\sin{{g}}}{f}{\quad\text{or}\quad}\mu{l}{a}{s}_{{{x}{y}}}=\frac{{\sum{x}_{{i}}{y}_{{i}}-\frac{{\sum{x}_{{i}}-\sum{y}_{{i}}}}{{{n}}}}}{{{n}-{1}}}\).
Where n = number of ordered pairs.
n=3
\(\displaystyle{s}_{{{x}{y}}}=\frac{{{20}-\frac{{{6}\cdot{12}}}{{3}}}}{{{3}-{1}}}\)
\(\displaystyle=\frac{{{20}-\frac{{{72}}}{{3}}}}{{{3}-{1}}}\)
\(\displaystyle=\frac{{{20}-{24}}}{{2}}\)
\(\displaystyle=-\frac{{4}}{{2}}=-{2}\)
Find the correlation coefficient r using the formula \(\displaystyle{r}=\frac{{{s}_{{{x}{y}}}}}{{{s}_{{x}}{s}_{{y}}}}\).
\(\displaystyle{r}=-\frac{{2}}{{{1}\cdot{2}}}=-{1}\)
Hence the value of correlation coefficient r = -1.
