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

`t=(-0.9690√(10-2))/(√(1-(-0.9690)^2))=-11.107`

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=(n(\sum xy)-(\sum x)(\sum y))/(√([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=(10(343)-(103)(47))/(√([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 \\eq 0`

The statistic to check the hypothesis is given by:

`t=(r √(n-2))/(√(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=(-0.9690√(10-2))/(√(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.