Question

In a small town, the relationship between robberies and police officers on duty has been collected. X represents the number of police officers on duty and y represents the number of reported thefts. x 10 15 16 1 4 6 18 12 14 7 y 5 2 1 9 7 8 1 5 3 6. At alpha = 0.01 what is the critical value for r ? (values are +/-)

Answer 1

Answer:

$t=\frac{-0.9690\sqrt{10-2}}{\sqrt{1-(-0.9690)^2}}=-11.107$

Step-by-step explanation:

The correlation coefficient is a "statistical measure that calculates the strength of the relationship between the relative movements of two variables". It's denoted by r and its always between -1 and 1.

And in order to calculate the correlation coefficient we can use this formula:

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

For our case we have this:

n=10 $\sum x = 103, \sum y = 47, \sum xy = 343, \sum x^2 =1347, \sum y^2 =295$

$r=\frac{10(343)-(103)(47)}{\sqrt{[10(1347) -(103)^2][10(295) -(47)^2]}}=-0.9691$

So then the correlation coefficient would be r =-0.9690

In order to test the hypothesis if the correlation coefficient it's significant we have the following hypothesis:

Null hypothesis: $\rho =0$

Alternative hypothesis: $\rho \neq 0$

The statistic to check the hypothesis is given by:

$t=\frac{r \sqrt{n-2}}{\sqrt{1-r^2}}$

And is distributed with n-2 degreed of freedom. df=n-2=10-2=8

In our case the value for the statistic would be:

$t=\frac{-0.9690\sqrt{10-2}}{\sqrt{1-(-0.9690)^2}}=-11.107$

Since $\alpha=0.01$ then $\alpha/2 =0.005$ and using the degrees of freedom we see that the critical values that accumulates 0.005 of the area on each tail are:

$t_{\alpha/2}=-3.36$ and $t_{1-\alpha/2}=3.36$

And in our case since our calculated value it's outside of the interval (-3.36;3.36) we can reject the null hypothesis at 1% the significance.