Respuesta :
Answer:
a) Null hypothesis: [tex] \rho = 0[/tex]
Alternative hypothesis: [tex] \rho \neq 0[/tex]
b) The scatter plot is on the figure attached.
c) [tex]t=\frac{0.874}{\sqrt{1-(0.874)^2}} \sqrt{9-2}=4.758[/tex]
[tex] p_v = P(t_{7} >4.758) = 0.0010[/tex]
Since the p value is lower than the significance level we have enough evidence to reject the null hypothesis at 5% of significance, and we can conclude that the correlation coefficient is significant.
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]
Let X the Court Income and Y= Justice Salary, for this case we have that:
[tex] n=9, \sum X= 3986, \sumY=398. \sumXY=265160, \sum X^2 =4076636, \sumY^2 =22074[/tex]
On this case we got that r =0.874
Part a
Null hypothesis: [tex] \rho = 0[/tex]
Alternative hypothesis: [tex] \rho \neq 0[/tex]
Part b
The scatter plot is on the figure attached.
Part c
In order to test a hypothesis related to the correlation coefficient 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 = 9 and the statistic is given by:
[tex]t=\frac{0.874}{\sqrt{1-(0.874)^2}} \sqrt{9-2}=4.758[/tex]
And the degrees of freedom are given by df=9-2=7
And the p value for this case since we have a bilateral test is given by:
[tex] p_v = P(t_{7} >4.758) = 0.0010[/tex]
Since the p value is lower than the significance level we have enough evidence to reject the null hypothesis at 5% of significance, and we can conclude that the correlation coefficient is significant.
