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. (1,3),(2,2),(3,10),(4,9),(5,10),(6,10),(7,10) answer r =

Explanation:

Step1: Calculate means

Let \(x_i\) be the first - coordinate and \(y_i\) be the second - coordinate of the points \((x_i,y_i)\).
\(n = 7\)
\(\bar{x}=\frac{1 + 2+3 + 4+5 + 6+7}{7}=\frac{28}{7}=4\)
\(\bar{y}=\frac{3 + 2+10 + 9+10 + 10+10}{7}=\frac{54}{7}\approx7.714286\)

Step2: Calculate numerator and denominator components

Let \(S_{xy}=\sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})\), \(S_{xx}=\sum_{i = 1}^{n}(x_i-\bar{x})^2\), \(S_{yy}=\sum_{i = 1}^{n}(y_i-\bar{y})^2\)

For \(i = 1\): \((x_1-\bar{x})(y_1 - \bar{y})=(1 - 4)(3-\frac{54}{7})=(- 3)\times(3-\frac{54}{7})=(-3)\times\frac{21 - 54}{7}=(-3)\times(-\frac{33}{7})=\frac{99}{7}\)
\((x_1-\bar{x})^2=(1 - 4)^2 = 9\)
\((y_1-\bar{y})^2=(3-\frac{54}{7})^2=(\frac{21 - 54}{7})^2=(-\frac{33}{7})^2=\frac{1089}{49}\)

For \(i = 2\): \((x_2-\bar{x})(y_2 - \bar{y})=(2 - 4)(2-\frac{54}{7})=(-2)\times(2-\frac{54}{7})=(-2)\times\frac{14 - 54}{7}=(-2)\times(-\frac{40}{7})=\frac{80}{7}\)
\((x_2-\bar{x})^2=(2 - 4)^2 = 4\)
\((y_2-\bar{y})^2=(2-\frac{54}{7})^2=(\frac{14 - 54}{7})^2=(-\frac{40}{7})^2=\frac{1600}{49}\)

For \(i = 3\): \((x_3-\bar{x})(y_3 - \bar{y})=(3 - 4)(10-\frac{54}{7})=(-1)\times(10-\frac{54}{7})=(-1)\times\frac{70 - 54}{7}=(-1)\times\frac{16}{7}=-\frac{16}{7}\)
\((x_3-\bar{x})^2=(3 - 4)^2 = 1\)
\((y_3-\bar{y})^2=(10-\frac{54}{7})^2=(\frac{70 - 54}{7})^2=(\frac{16}{7})^2=\frac{256}{49}\)

For \(i = 4\): \((x_4-\bar{x})(y_4 - \bar{y})=(4 - 4)(9-\frac{54}{7})=0\times(9-\frac{54}{7}) = 0\)
\((x_4-\bar{x})^2=(4 - 4)^2 = 0\)
\((y_4-\bar{y})^2=(9-\frac{54}{7})^2=(\frac{63 - 54}{7})^2=(\frac{9}{7})^2=\frac{81}{49}\)

For \(i = 5\): \((x_5-\bar{x})(y_5 - \bar{y})=(5 - 4)(10-\frac{54}{7})=(1)\times(10-\frac{54}{7})=(1)\times\frac{70 - 54}{7}=\frac{16}{7}\)
\((x_5-\bar{x})^2=(5 - 4)^2 = 1\)
\((y_5-\bar{y})^2=(10-\frac{54}{7})^2=(\frac{70 - 54}{7})^2=(\frac{16}{7})^2=\frac{256}{49}\)

For \(i = 6\): \((x_6-\bar{x})(y_6 - \bar{y})=(6 - 4)(10-\frac{54}{7})=(2)\times(10-\frac{54}{7})=(2)\times\frac{70 - 54}{7}=\frac{32}{7}\)
\((x_6-\bar{x})^2=(6 - 4)^2 = 4\)
\((y_6-\bar{y})^2=(10-\frac{54}{7})^2=(\frac{70 - 54}{7})^2=(\frac{16}{7})^2=\frac{256}{49}\)

For \(i = 7\): \((x_7-\bar{x})(y_7 - \bar{y})=(7 - 4)(10-\frac{54}{7})=(3)\times(10-\frac{54}{7})=(3)\times\frac{70 - 54}{7}=\frac{48}{7}\)
\((x_7-\bar{x})^2=(7 - 4)^2 = 9\)
\((y_7-\bar{y})^2=(10-\frac{54}{7})^2=(\frac{70 - 54}{7})^2=(\frac{16}{7})^2=\frac{256}{49}\)

\(S_{xy}=\frac{99}{7}+\frac{80}{7}-\frac{16}{7}+0+\frac{16}{7}+\frac{32}{7}+\frac{48}{7}=\frac{99 + 80-16 + 0+16+32+48}{7}=\frac{259}{7}=37\)
\(S_{xx}=9 + 4+1+0+1+4+9 = 28\)
\(S_{yy}=\frac{1089+1600 + 256+81+256+256+256}{49}=\frac{3794}{49}\approx77.428571\)

Step3: Calculate correlation coefficient

The Pearson correlation coefficient \(r=\frac{S_{xy}}{\sqrt{S_{xx}S_{yy}}}\)
\(r=\frac{37}{\sqrt{28\times\frac{3794}{49}}}=\frac{37}{\sqrt{\frac{28\times3794}{49}}}=\frac{37}{\sqrt{\frac{106232}{49}}}=\frac{37}{\frac{\sqrt{106232}}{7}}\approx\frac{37}{\frac{325.932508}{7}}\approx\frac{37\times7}{325.932508}\approx0.790\)

Answer:

\(r = 0.790\)