Answer to Question #263909 in Statistics and Probability for Joana Azaña

Question

1. A used car depot wants to study the relationship between the age of a car and its selling price. Listed below is a random

sample of 9 cars sold at the depot during the last 3 months. 

Car123456789Age (Years)3411076725Selling Price (P *000)350320500120190280190400300

Determine the Coefficient Correlation using Pearson Correlation and test of the significance at 0.05. 

Answer 1

The data provided can be summarized as follows,

$\sum(x)=45,\space \sum(y)=2650,\space \sum(xy)=10670,\space \sum(x^2)=289,\sum(y^2)=889900$ and $n=9$

The Pearson correlation coefficient $r$ can be determined using the formula below,

$r=(n\sum(xy)-\sum(x)\sum(y))/\sqrt{(n\sum(x^2)-(\sum(x))^2)(n\sum(y^2)-(\sum(y))^2)}$

$r=(9(10670)-(45*2650))/\sqrt{(9(289)-45)(9(889900)-2650)}= -0.9740481$

Thus the correlation coefficient $r$ is $-0.97(2dp)$.

The hypothesis to be tested are,

$H_0:\rho=0\space against \space H_1:\rho\lt0$

and apply the student's t distribution to make a decision.

The test statistic is given as,

$t^*=r\sqrt{n-2}/\sqrt{1-r^2}=-0.97\sqrt{7}/\sqrt{1-(-0.97)^2}= -11.38586$

$t^*$ is compared with the table value at $\alpha =0.05$ with $n-2=9-2=7$ degrees of freedom.

Now, $t_{0.05,7}=1.895$ and the null hypothesis is rejected if $t^*\lt -t_{0.05,7}$

Since $t^*=-11.38586\lt-t_{0.05,7}=-1.895$, we reject the null hypothesis and conclude that there is sufficient evidence to show that the correlation coefficient of $r=-0.97$ is significant at 5% level of significance.