Question

what is the value of the correlation coefficient (r) for the following data?

(1 point)

-0.97

0.94

-0.94

-0.58

Explanation:

Step1: Calculate the means of \(x\) and \(y\)

\(\bar{x}=\frac{10 + 10+50+60+90+90}{6}=\frac{310}{6}\approx51.67\)
\(\bar{y}=\frac{170 + 200+120+60+20+40}{6}=\frac{610}{6}\approx101.67\)

Step2: Calculate the numerator and denominator of the correlation coefficient formula

The formula for \(r\) is \(r=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})(y_{i}-\bar{y})}{\sqrt{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}\sum_{i = 1}^{n}(y_{i}-\bar{y})^{2}}}\)

For \(i = 1\): \((x_{1}-\bar{x})(y_{1}-\bar{y})=(10 - 51.67)(170-101.67)=(- 41.67)\times68.33=-2848.11\)
\((x_{1}-\bar{x})^{2}=(10 - 51.67)^{2}=(-41.67)^{2}=1736.39\)
\((y_{1}-\bar{y})^{2}=(170 - 101.67)^{2}=68.33^{2}=4668.99\)

For \(i = 2\): \((x_{2}-\bar{x})(y_{2}-\bar{y})=(10 - 51.67)(200 - 101.67)=(-41.67)\times98.33=-4098.11\)
\((x_{2}-\bar{x})^{2}=(10 - 51.67)^{2}=1736.39\)
\((y_{2}-\bar{y})^{2}=(200 - 101.67)^{2}=98.33^{2}=9668.79\)

For \(i = 3\): \((x_{3}-\bar{x})(y_{3}-\bar{y})=(50 - 51.67)(120 - 101.67)=(-1.67)\times18.33=-30.61\)
\((x_{3}-\bar{x})^{2}=(50 - 51.67)^{2}=(-1.67)^{2}=2.79\)
\((y_{3}-\bar{y})^{2}=(120 - 101.67)^{2}=18.33^{2}=336.09\)

For \(i = 4\): \((x_{4}-\bar{x})(y_{4}-\bar{y})=(60 - 51.67)(60 - 101.67)=8.33\times(-41.67)=-347.11\)
\((x_{4}-\bar{x})^{2}=(60 - 51.67)^{2}=8.33^{2}=69.44\)
\((y_{4}-\bar{y})^{2}=(60 - 101.67)^{2}=(-41.67)^{2}=1736.39\)

For \(i = 5\): \((x_{5}-\bar{x})(y_{5}-\bar{y})=(90 - 51.67)(20 - 101.67)=38.33\times(-81.67)=-3130.39\)
\((x_{5}-\bar{x})^{2}=(90 - 51.67)^{2}=38.33^{2}=1469.19\)
\((y_{5}-\bar{y})^{2}=(20 - 101.67)^{2}=(-81.67)^{2}=6669.99\)

For \(i = 6\): \((x_{6}-\bar{x})(y_{6}-\bar{y})=(90 - 51.67)(40 - 101.67)=38.33\times(-61.67)=-2363.71\)
\((x_{6}-\bar{x})^{2}=(90 - 51.67)^{2}=1469.19\)
\((y_{6}-\bar{y})^{2}=(40 - 101.67)^{2}=(-61.67)^{2}=3803.19\)

\(\sum_{i = 1}^{6}(x_{i}-\bar{x})(y_{i}-\bar{y})=-2848.11-4098.11-30.61-347.11-3130.39-2363.71=-12818.04\)

\(\sum_{i = 1}^{6}(x_{i}-\bar{x})^{2}=1736.39+1736.39 + 2.79+69.44+1469.19+1469.19=6483.39\)

\(\sum_{i = 1}^{6}(y_{i}-\bar{y})^{2}=4668.99+9668.79+336.09+1736.39+6669.99+3803.19=26883.44\)

\(r=\frac{-12818.04}{\sqrt{6483.39\times26883.44}}\approx\frac{-12818.04}{\sqrt{174473777.7}}\approx\frac{-12818.04}{13210.37}\approx - 0.97\)

Answer:

-0.97