Respuesta :
Answer:
[tex]t=\frac{0.38}{\sqrt{1-(0.38)^2}} \sqrt{15-2}=1.481[/tex]
And the degrees of freedom are given by df=15-2=13
Step-by-step 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:
[tex]r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]}}[/tex]
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:
[tex]t=\frac{r}{\sqrt{1-r^2}} \sqrt{n-2}[/tex]
Where n represent the sample size and the statistic t follows a t distribution with n-2 degrees of freedom:
[tex]t \sim t_{n-2}[/tex]
On this case our value of n = 15 and the statistic is given by:
[tex]t=\frac{0.38}{\sqrt{1-(0.38)^2}} \sqrt{15-2}=1.481[/tex]
And the degrees of freedom are given by df=15-2=13
