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

Answer

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)