Question

question #13
determine the r - value for the data set below.

0.871
0.813
0.747
0.930

Explanation:

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

\(\bar{x}=\frac{4 + 6+6+7+9}{5}=\frac{32}{5} = 6.4\)
\(\bar{y}=\frac{10 + 13+11+13+19}{5}=\frac{66}{5}=13.2\)

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 the numerator:
\((4 - 6.4)(10 - 13.2)=(- 2.4)\times(-3.2)=7.68\)
\((6 - 6.4)(13 - 13.2)=(-0.4)\times(-0.2) = 0.08\)
\((6 - 6.4)(11 - 13.2)=(-0.4)\times(-2.2)=0.88\)
\((7 - 6.4)(13 - 13.2)=(0.6)\times(-0.2)=-0.12\)
\((9 - 6.4)(19 - 13.2)=(2.6)\times(5.8)=15.08\)
\(\sum_{i = 1}^{5}(x_{i}-\bar{x})(y_{i}-\bar{y})=7.68+0.08 + 0.88-0.12+15.08=23.5\)

For the denominator:
\((4 - 6.4)^{2}=(-2.4)^{2}=5.76\)
\((6 - 6.4)^{2}=(-0.4)^{2}=0.16\)
\((6 - 6.4)^{2}=(-0.4)^{2}=0.16\)
\((7 - 6.4)^{2}=(0.6)^{2}=0.36\)
\((9 - 6.4)^{2}=(2.6)^{2}=6.76\)
\(\sum_{i = 1}^{5}(x_{i}-\bar{x})^{2}=5.76+0.16+0.16+0.36+6.76 = 13.2\)

\((10 - 13.2)^{2}=(-3.2)^{2}=10.24\)
\((13 - 13.2)^{2}=(-0.2)^{2}=0.04\)
\((11 - 13.2)^{2}=(-2.2)^{2}=4.84\)
\((13 - 13.2)^{2}=(-0.2)^{2}=0.04\)
\((19 - 13.2)^{2}=(5.8)^{2}=33.64\)
\(\sum_{i = 1}^{5}(y_{i}-\bar{y})^{2}=10.24+0.04+4.84+0.04+33.64 = 48.8\)

\(\sqrt{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}\sum_{i = 1}^{n}(y_{i}-\bar{y})^{2}}=\sqrt{13.2\times48.8}=\sqrt{644.16}\approx25.38\)

Step3: Calculate \(r\)

\(r=\frac{23.5}{25.38}\approx0.93\)

Answer:

\(0.930\)