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) 1.03 1,073 1.17 1,178 1.56 1,341 1.63 1,218 1.87 1,400 1.97 1,334 round your answer to the nearest thousandth. r =

Explanation:

Step1: Calculate the means

Let $x$ be the percent - carbon values and $y$ be the weight - supported values.
$n = 6$
$\bar{x}=\frac{1.03 + 1.17+1.56+1.63+1.87+1.97}{6}=\frac{9.23}{6}\approx1.5383$
$\bar{y}=\frac{1073 + 1178+1341+1218+1400+1334}{6}=\frac{7544}{6}\approx1257.3333$

Step2: Calculate the numerator and denominator components

Calculate $(x_i-\bar{x})(y_i - \bar{y})$, $(x_i-\bar{x})^2$ and $(y_i - \bar{y})^2$ for each $i$ from $1$ to $n$.

$i$	$x_i$	$y_i$	$x_i-\bar{x}$	$y_i - \bar{y}$	$(x_i-\bar{x})(y_i - \bar{y})$	$(x_i-\bar{x})^2$	$(y_i - \bar{y})^2$
2	1.17	1178	- 0.3683	- 79.3333	29.2177	0.1357	6293.7778
3	1.56	1341	0.0217	83.6667	1.8156	0.0005	7000.6667
4	1.63	1218	0.0917	- 39.3333	- 3.6067	0.0084	1547.0000
5	1.87	1400	0.3317	142.6667	47.3227	0.1100	20354.6667
6	1.97	1334	0.4317	76.6667	33.1127	0.1864	5878.6667

$\sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})=93.6877+29.2177 + 1.8156-3.6067+47.3227+33.1127 = 201.5497$
$\sum_{i = 1}^{n}(x_i-\bar{x})^2=0.2584 + 0.1357+0.0005+0.0084+0.1100+0.1864 = 0.6994$
$\sum_{i = 1}^{n}(y_i - \bar{y})^2=34077.7778+6293.7778+7000.6667+1547.0000+20354.6667+5878.6667 = 75152.5557$

The correlation coefficient formula is $r=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})}{\sqrt{\sum_{i = 1}^{n}(x_i-\bar{x})^2\sum_{i = 1}^{n}(y_i - \bar{y})^2}}$
The denominator is $\sqrt{0.6994\times75152.5557}=\sqrt{52475.7977}\approx229.0759$

Step3: Calculate the correlation coefficient

$r=\frac{201.5497}{229.0759}\approx0.880$

Answer:

$0.880$