Question

based on the data shown below, calculate the correlation coefficient (to three decimal places)

x	y
4	47.6
5	51
6	51.8
7	55.6
8	51.6
9	52
10	55.8
11	55.8
12	58.2

r =

Explanation:

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

Let \(n = 9\).
\(\bar{x}=\frac{4 + 5+6+7+8+9+10+11+12}{9}=\frac{72}{9}=8\)
\(\bar{y}=\frac{47.6+51+51.8+55.6+51.6+52+55.8+55.8+58.2}{9}=\frac{489.4}{9}\approx54.378\)

Step2: Calculate the numerator of the correlation - coefficient formula

\(\sum_{i = 1}^{n}(x_{i}-\bar{x})(y_{i}-\bar{y})=(4 - 8)(47.6-54.378)+(5 - 8)(51 - 54.378)+(6 - 8)(51.8-54.378)+(7 - 8)(55.6-54.378)+(8 - 8)(51.6-54.378)+(9 - 8)(52 - 54.378)+(10 - 8)(55.8-54.378)+(11 - 8)(55.8-54.378)+(12 - 8)(58.2-54.378)\)
\(=(-4)(-6.778)+(-3)(-3.378)+(-2)(-2.578)+(-1)(1.222)+0\times(-2.778)+1\times(-2.378)+2\times1.422+3\times1.422+4\times3.822\)
\(=27.112 + 10.134+5.156-1.222+0 - 2.378+2.844+4.272+15.288\)
\(=61.216\)

Step3: Calculate the denominator of the correlation - coefficient formula

\(\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}=(4 - 8)^{2}+(5 - 8)^{2}+(6 - 8)^{2}+(7 - 8)^{2}+(8 - 8)^{2}+(9 - 8)^{2}+(10 - 8)^{2}+(11 - 8)^{2}+(12 - 8)^{2}\)
\(=(-4)^{2}+(-3)^{2}+(-2)^{2}+(-1)^{2}+0^{2}+1^{2}+2^{2}+3^{2}+4^{2}\)
\(=16 + 9+4 + 1+0+1+4+9+16=60\)
\(\sum_{i = 1}^{n}(y_{i}-\bar{y})^{2}=(47.6-54.378)^{2}+(51 - 54.378)^{2}+(51.8-54.378)^{2}+(55.6-54.378)^{2}+(51.6-54.378)^{2}+(52 - 54.378)^{2}+(55.8-54.378)^{2}+(55.8-54.378)^{2}+(58.2-54.378)^{2}\)
\(\approx(-6.778)^{2}+(-3.378)^{2}+(-2.578)^{2}+(1.222)^{2}+(-2.778)^{2}+(-2.378)^{2}+(1.422)^{2}+(1.422)^{2}+(3.822)^{2}\)
\(\approx45.94 + 11.41+6.65+1.49+7.72+5.66+2.02+2.02+14.61\)
\(\approx97.52\)
The denominator is \(\sqrt{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}\sum_{i = 1}^{n}(y_{i}-\bar{y})^{2}}=\sqrt{60\times97.52}\approx\sqrt{5851.2}\approx76.48\)

Step4: Calculate the correlation coefficient \(r\)

\(r=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})(y_{i}-\bar{y})}{\sqrt{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}\sum_{i = 1}^{n}(y_{i}-\bar{y})^{2}}}=\frac{61.216}{76.48}\approx0.800\)

Answer:

\(0.800\)