Question

what is the correlation coefficient for the data shown in the table? 0 1 4 5

Explanation:

Step1: Recall the formula for correlation coefficient

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

First, calculate \(\bar{x}\) and \(\bar{y}\).
\(\bar{x}=\frac{0 + 1+4+5}{4}=\frac{10}{4} = 2.5\)
\(\bar{y}=\frac{0 + 1+4+5}{4}=2.5\)

Step2: Calculate \((x_{i}-\bar{x})(y_{i}-\bar{y})\), \((x_{i}-\bar{x})^{2}\) and \((y_{i}-\bar{y})^{2}\)

For \(i = 1\): \(x_{1}=0,y_{1}=0\)
\((x_{1}-\bar{x})(y_{1}-\bar{y})=(0 - 2.5)(0 - 2.5)=6.25\)
\((x_{1}-\bar{x})^{2}=(0 - 2.5)^{2}=6.25\)
\((y_{1}-\bar{y})^{2}=(0 - 2.5)^{2}=6.25\)

For \(i = 2\): \(x_{2}=1,y_{2}=1\)
\((x_{2}-\bar{x})(y_{2}-\bar{y})=(1 - 2.5)(1 - 2.5)=2.25\)
\((x_{2}-\bar{x})^{2}=(1 - 2.5)^{2}=2.25\)
\((y_{2}-\bar{y})^{2}=(1 - 2.5)^{2}=2.25\)

For \(i = 3\): \(x_{3}=4,y_{3}=4\)
\((x_{3}-\bar{x})(y_{3}-\bar{y})=(4 - 2.5)(4 - 2.5)=2.25\)
\((x_{3}-\bar{x})^{2}=(4 - 2.5)^{2}=2.25\)
\((y_{3}-\bar{y})^{2}=(4 - 2.5)^{2}=2.25\)

For \(i = 4\): \(x_{4}=5,y_{4}=5\)
\((x_{4}-\bar{x})(y_{4}-\bar{y})=(5 - 2.5)(5 - 2.5)=6.25\)
\((x_{4}-\bar{x})^{2}=(5 - 2.5)^{2}=6.25\)
\((y_{4}-\bar{y})^{2}=(5 - 2.5)^{2}=6.25\)

Step3: Calculate \(\sum_{i = 1}^{n}(x_{i}-\bar{x})(y_{i}-\bar{y})\), \(\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}\) and \(\sum_{i = 1}^{n}(y_{i}-\bar{y})^{2}\)

\(\sum_{i = 1}^{4}(x_{i}-\bar{x})(y_{i}-\bar{y})=6.25+2.25+2.25+6.25 = 17.5\)
\(\sum_{i = 1}^{4}(x_{i}-\bar{x})^{2}=6.25+2.25+2.25+6.25 = 17.5\)
\(\sum_{i = 1}^{4}(y_{i}-\bar{y})^{2}=6.25+2.25+2.25+6.25 = 17.5\)

Step4: Calculate \(r\)

\(r=\frac{17.5}{\sqrt{17.5\times17.5}}=\frac{17.5}{17.5}=1\)

Another way:
We can also observe that \(y=x\) for all data points \((x,y)\) in the table.
When \(y = ax + b\) and \(a>0\), if all data points lie on the line \(y=x\) (a special case of \(y = ax + b\) with \(a = 1\) and \(b=0\)), the correlation coefficient \(r = 1\)

Answer:

1