Question

Recently, an automobile insurance company performed a study of a random sample of 15 of its customers to determine if there is a positive relationship between the number of miles driven and the age of the driver. The sample correlation coefficient is r = .38. Given this information, and assuming that the test is to be performed at the .05 level of significance, which of the following is the correct test statistic?

Answer 1

`t=(0.38)/(√(1-(0.38)^2)) √(15-2)=1.481`

And the degrees of freedom are given by df=15-2=13

Explanation:

Previous concepts

Pearson correlation coefficient(r), "measures a linear dependence between two variables (x and y). Its a parametric correlation test because it depends to the distribution of the data. And other assumption is that the variables x and y needs to follow a normal distribution".

The t distribution (Student’s t-distribution) is a "probability distribution that is used to estimate population parameters when the sample size is small (n<30) or when the population variance is unknown".

The shape of the t distribution is determined by its degrees of freedom and when the degrees of freedom increase the t distirbution becomes a normal distribution approximately.

Solution to the problem

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]))`

On this case we got that r =0.38

In order to test a hypothesis related to the correlation coeffcient we need to use the following statistic:

`t=(r)/(√(1-r^2)) √(n-2)`

Where n represent the sample size and the statistic t follows a t distribution with n-2 degrees of freedom:

`t \sim t_(n-2)`

On this case our value of n = 15 and the statistic is given by:

`t=(0.38)/(√(1-(0.38)^2)) √(15-2)=1.481`

And the degrees of freedom are given by df=15-2=13