find the correlation coefficient, r, of the data described below.\na french car company wants to cut costs…

find the correlation coefficient, r, of the data described below.\na french car company wants to cut costs by using cheaper low - carbon steel in its car frames. to determine how varying the carbon content will affect strength, company engineers manufactured several car frames.\nfor each car frame, the engineers noted the percentage of carbon, x, as well as the weight it could support (in kilograms), y.\npercent carbon weight supported (in kilograms)\n1.06 1,225\n1.13 1,147\n1.22 1,153\n1.32 1,098\n1.33 1,257\n1.89 1,396\nround your answer to the nearest thousandth.

find the correlation coefficient, r, of the data described below.\na french car company wants to cut costs by using cheaper low - carbon steel in its car frames. to determine how varying the carbon content will affect strength, company engineers manufactured several car frames.\nfor each car frame, the engineers noted the percentage of carbon, x, as well as the weight it could support (in kilograms), y.\npercent carbon weight supported (in kilograms)\n1.06 1,225\n1.13 1,147\n1.22 1,153\n1.32 1,098\n1.33 1,257\n1.89 1,396\nround your answer to the nearest thousandth.

Answer

Explanation:

Step1: Calculate the means

Let $x_i$ be the percent - carbon values and $y_i$ be the weight - supported values. The number of data points $n = 6$. $\bar{x}=\frac{1.06 + 1.13+1.22 + 1.32+1.33+1.89}{6}=\frac{7.95}{6}=1.325$ $\bar{y}=\frac{1225 + 1147+1153+1098+1257+1396}{6}=\frac{7276}{6}\approx1212.667$

Step2: Calculate the numerator of the correlation coefficient formula

$S_{xy}=\sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})$ $(1.06 - 1.325)(1225-1212.667)+(1.13 - 1.325)(1147 - 1212.667)+(1.22-1.325)(1153 - 1212.667)+(1.32-1.325)(1098 - 1212.667)+(1.33-1.325)(1257 - 1212.667)+(1.89-1.325)(1396 - 1212.667)$ $=(- 0.265)\times12.333+(-0.195)\times(-65.667)+(-0.105)\times(-59.667)+(-0.005)\times(-114.667)+(0.005)\times44.333+(0.565)\times183.333$ $=-3.268 + 12.805+6.265 + 0.573+0.222+103.583$ $=120.18$

Step3: Calculate the denominator of the correlation coefficient formula

$S_{xx}=\sum_{i = 1}^{n}(x_i-\bar{x})^2$ $(1.06 - 1.325)^2+(1.13 - 1.325)^2+(1.22-1.325)^2+(1.32-1.325)^2+(1.33-1.325)^2+(1.89-1.325)^2$ $=(-0.265)^2+(-0.195)^2+(-0.105)^2+(-0.005)^2+(0.005)^2+(0.565)^2$ $=0.070225+0.038025+0.011025+0.000025+0.000025+0.319225$ $=0.43855$

$S_{yy}=\sum_{i = 1}^{n}(y_i-\bar{y})^2$ $(1225 - 1212.667)^2+(1147 - 1212.667)^2+(1153 - 1212.667)^2+(1098 - 1212.667)^2+(1257 - 1212.667)^2+(1396 - 1212.667)^2$ $=12.333^2+(-65.667)^2+(-59.667)^2+(-114.667)^2+(44.333)^2+(183.333)^2$ $=152.17+4312.19+3560.27+13147.39+1965.31+33610.89$ $=56748.22$

$S_{yy}=\sqrt{S_{yy}}=\sqrt{56748.22}\approx238.22$ $S_{xx}=\sqrt{S_{xx}}=\sqrt{0.43855}\approx0.662$ The denominator is $S_{xx}S_{yy}=0.662\times238.22\approx157.70$

Step4: Calculate the correlation coefficient

$r=\frac{S_{xy}}{S_{xx}S_{yy}}=\frac{120.18}{157.70}\approx0.762$

Answer:

$0.762$