Question

find the pearson correlation coefficient r for the given points. round any intermediate calculations to no less than six decimal places, and round your final answer to three decimal places.
(1, 2), (2, 3), (3, 9), (4, 3), (5, 10), (6, 9), (7, 9)

Explanation:

Step1: Calculate the means

Let \(x = [1,2,3,4,5,6,7]\) and \(y=[2,3,9,3,10,9,9]\).
The mean of \(x\), \(\bar{x}=\frac{1 + 2+3+4+5+6+7}{7}=\frac{28}{7}=4\).
The mean of \(y\), \(\bar{y}=\frac{2 + 3+9+3+10+9+9}{7}=\frac{45}{7}\approx6.428571\).

Step2: Calculate the numerator \(\sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})\)

\((1 - 4)(2-\frac{45}{7})+(2 - 4)(3-\frac{45}{7})+(3 - 4)(9-\frac{45}{7})+(4 - 4)(3-\frac{45}{7})+(5 - 4)(10-\frac{45}{7})+(6 - 4)(9-\frac{45}{7})+(7 - 4)(9-\frac{45}{7})\)
\(=(- 3)(2-\frac{45}{7})+(-2)(3-\frac{45}{7})+(-1)(9-\frac{45}{7})+0+(1)(10-\frac{45}{7})+(2)(9-\frac{45}{7})+(3)(9-\frac{45}{7})\)
\(=(-3)\times\frac{14 - 45}{7}+(-2)\times\frac{21 - 45}{7}+(-1)\times\frac{63 - 45}{7}+0+(1)\times\frac{70 - 45}{7}+(2)\times\frac{63 - 45}{7}+(3)\times\frac{63 - 45}{7}\)
\(=(-3)\times(-\frac{31}{7})+(-2)\times(-\frac{24}{7})+(-1)\times\frac{18}{7}+0+\frac{25}{7}+(2)\times\frac{18}{7}+(3)\times\frac{18}{7}\)
\(=\frac{93}{7}+\frac{48}{7}-\frac{18}{7}+0+\frac{25}{7}+\frac{36}{7}+\frac{54}{7}=\frac{93 + 48-18+0+25+36+54}{7}=\frac{238}{7}=34\).

Step3: Calculate the denominator \(\sqrt{\sum_{i = 1}^{n}(x_i-\bar{x})^2\sum_{i = 1}^{n}(y_i - \bar{y})^2}\)

\(\sum_{i = 1}^{n}(x_i-\bar{x})^2=(1 - 4)^2+(2 - 4)^2+(3 - 4)^2+(4 - 4)^2+(5 - 4)^2+(6 - 4)^2+(7 - 4)^2\)
\(=(-3)^2+(-2)^2+(-1)^2+0^2+1^2+2^2+3^2=9 + 4+1+0+1+4+9 = 28\).
\(\sum_{i = 1}^{n}(y_i-\bar{y})^2=(2-\frac{45}{7})^2+(3-\frac{45}{7})^2+(9-\frac{45}{7})^2+(3-\frac{45}{7})^2+(10-\frac{45}{7})^2+(9-\frac{45}{7})^2+(9-\frac{45}{7})^2\)
\(=(\frac{14 - 45}{7})^2+(\frac{21 - 45}{7})^2+(\frac{63 - 45}{7})^2+(\frac{21 - 45}{7})^2+(\frac{70 - 45}{7})^2+(\frac{63 - 45}{7})^2+(\frac{63 - 45}{7})^2\)
\(=(-\frac{31}{7})^2+(-\frac{24}{7})^2+(\frac{18}{7})^2+(-\frac{24}{7})^2+(\frac{25}{7})^2+(\frac{18}{7})^2+(\frac{18}{7})^2\)
\(=\frac{961}{49}+\frac{576}{49}+\frac{324}{49}+\frac{576}{49}+\frac{625}{49}+\frac{324}{49}+\frac{324}{49}=\frac{961+576+324+576+625+324+324}{49}=\frac{3710}{49}\approx75.714286\).
The denominator \(\sqrt{28\times\frac{3710}{49}}=\sqrt{\frac{103880}{49}}\approx\sqrt{2119.999999}\approx46.043454\).

Step4: Calculate the correlation coefficient \(r\)

\(r=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})}{\sqrt{\sum_{i = 1}^{n}(x_i-\bar{x})^2\sum_{i = 1}^{n}(y_i - \bar{y})^2}}=\frac{34}{46.043454}\approx0.738\)

Answer:

\(0.738\)