Question

the following table shows the age and selling price of cars sold recently at a used car lot. calculate the correlation coefficient, r, and determine whether r is statistically significant at the 0.01 level of significance. round your answer to the nearest thousandth. critical values of the pearson correlation coefficient car ages and selling prices car age 2 2 3 6 7 8 10 13 price ($) 31,595 19,969 20,195 15,385 6995 7805 17,295 4925

Explanation:

Step1: Calculate the means

Let $x$ be the car - age and $y$ be the price.
$n = 8$
$\bar{x}=\frac{2 + 2+3+6+7+8+10+13}{8}=\frac{51}{8}=6.375$
$\bar{y}=\frac{31595 + 19969+20195+15385+6995+7805+17295+4925}{8}=\frac{124164}{8}=15520.5$

Step2: Calculate the numerator and denominators of the correlation - coefficient formula

The formula for the Pearson correlation coefficient $r$ is:
$r=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})(y_{i}-\bar{y})}{\sqrt{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}\sum_{i = 1}^{n}(y_{i}-\bar{y})^{2}}}$

Calculate $(x_{i}-\bar{x})(y_{i}-\bar{y})$, $(x_{i}-\bar{x})^{2}$ and $(y_{i}-\bar{y})^{2}$ for each $i$:

$x_{i}$	$y_{i}$	$x_{i}-\bar{x}$	$y_{i}-\bar{y}$	$(x_{i}-\bar{x})(y_{i}-\bar{y})$	$(x_{i}-\bar{x})^{2}$	$(y_{i}-\bar{y})^{2}$
2	19969	$2 - 6.375=-4.375$	$19969 - 15520.5 = 4448.5$	$(-4.375)\times4448.5=-19461.1875$	$19.140625$	$19798452.25$
3	20195	$3 - 6.375=-3.375$	$20195 - 15520.5 = 4674.5$	$(-3.375)\times4674.5=-15776.4375$	$11.390625$	$21840870.25$
6	15385	$6 - 6.375=-0.375$	$15385 - 15520.5=-135.5$	$(-0.375)\times(-135.5)=50.8125$	$0.140625$	$18360.25$
7	6995	$7 - 6.375 = 0.625$	$6995 - 15520.5=-8525.5$	$0.625\times(-8525.5)=-5328.4375$	$0.390625$	$72684750.25$
8	7805	$8 - 6.375 = 1.625$	$7805 - 15520.5=-7715.5$	$1.625\times(-7715.5)=-12537.7875$	$2.640625$	$59538930.25$
10	17295	$10 - 6.375 = 3.625$	$17295 - 15520.5 = 1774.5$	$3.625\times1774.5 = 6432.5625$	$13.140625$	$3148800.25$
13	4925	$13 - 6.375 = 6.625$	$4925 - 15520.5=-10595.5$	$6.625\times(-10595.5)=-70295.1875$	$43.890625$	$112264400.25$

$\sum_{i = 1}^{n}(x_{i}-\bar{x})(y_{i}-\bar{y})=-70325.9375-19461.1875-15776.4375 + 50.8125-5328.4375-12537.7875+6432.5625-70295.1875=-186231.5$

$\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}=19.140625+19.140625+11.390625+0.140625+0.390625+2.640625+13.140625+43.890625=119.875$

$\sum_{i = 1}^{n}(y_{i}-\bar{y})^{2}=258389350.25+19798452.25+21840870.25+18360.25+72684750.25+59538930.25+3148800.25+112264400.25=547684914$

$r=\frac{-186231.5}{\sqrt{119.875\times547684914}}\approx\frac{-186231.5}{\sqrt{65627443.4375}}\approx\frac{-186231.5}{8101.077}\approx - 0.976$

Step3: Determine significance

The degrees of freedom $df=n - 2=8 - 2 = 6$.
For a significance level of $\alpha = 0.01$ and $df = 6$, the critical - value from the Pearson correlation coefficient table is approximately $\pm0.834$.
Since $|r|=|-0.976|>0.834$, $r$ is statistically significant at the 0.01 level of significance.

Answer:

$r\approx - 0.976$ and $r$ is statistically significant at the 0.01 level of significance.