Question

find the correlation coefficient, r, of the data described below.
a 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.
for each car frame, the engineers noted the percentage of carbon, x, as well as the weight it could support (in kilograms), y.
percent carbon weight supported (in kilograms)
1.06 1,225
1.13 1,147
1.22 1,153
1.32 1,098
1.33 1,257
1.89 1,396
round your answer to the nearest thousandth.

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:

$0.762$