find the pearson correlation coefficient r for the given points. round any intermediate calculations to no…

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)