what is the correlation coefficient for the data shown in the table?\n0\n1\n4\n5

Answer

Explanation:

Step1: Recall correlation - coefficient formula

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, find the means of $x$ and $y$ values. For $x$ - values: $x_1 = 0,x_2 = 1,x_3 = 4,x_4 = 5$, $\bar{x}=\frac{0 + 1+4 + 5}{4}=\frac{10}{4}=2.5$. For $y$ - values: $y_1 = 0,y_2 = 1,y_3 = 4,y_4 = 5$, $\bar{y}=\frac{0 + 1+4 + 5}{4}=2.5$.

Step2: Calculate numerator

$(x_1-\bar{x})(y_1 - \bar{y})=(0 - 2.5)(0 - 2.5)=6.25$ $(x_2-\bar{x})(y_2 - \bar{y})=(1 - 2.5)(1 - 2.5)=2.25$ $(x_3-\bar{x})(y_3 - \bar{y})=(4 - 2.5)(4 - 2.5)=2.25$ $(x_4-\bar{x})(y_4 - \bar{y})=(5 - 2.5)(5 - 2.5)=6.25$ $\sum_{i = 1}^{4}(x_i-\bar{x})(y_i - \bar{y})=6.25+2.25+2.25+6.25 = 17$.

Step3: Calculate denominator - part 1

$(x_1-\bar{x})^2=(0 - 2.5)^2 = 6.25$ $(x_2-\bar{x})^2=(1 - 2.5)^2 = 2.25$ $(x_3-\bar{x})^2=(4 - 2.5)^2 = 2.25$ $(x_4-\bar{x})^2=(5 - 2.5)^2 = 6.25$ $\sum_{i = 1}^{4}(x_i-\bar{x})^2=6.25+2.25+2.25+6.25 = 17$.

Step4: Calculate denominator - part 2

$(y_1-\bar{y})^2=(0 - 2.5)^2 = 6.25$ $(y_2-\bar{y})^2=(1 - 2.5)^2 = 2.25$ $(y_3-\bar{y})^2=(4 - 2.5)^2 = 2.25$ $(y_4-\bar{y})^2=(5 - 2.5)^2 = 6.25$ $\sum_{i = 1}^{4}(y_i-\bar{y})^2=6.25+2.25+2.25+6.25 = 17$. $\sqrt{\sum_{i = 1}^{4}(x_i-\bar{x})^2\sum_{i = 1}^{4}(y_i-\bar{y})^2}=\sqrt{17\times17}=17$.

Step5: Calculate correlation - coefficient

$r=\frac{\sum_{i = 1}^{4}(x_i-\bar{x})(y_i - \bar{y})}{\sqrt{\sum_{i = 1}^{4}(x_i-\bar{x})^2\sum_{i = 1}^{4}(y_i-\bar{y})^2}}=\frac{17}{17}=1$.