To find decimal part of 350 and 70 we divide one by the other .
The fraction is
We can reduce the fraction to its lowest form.
Dividing by common factor 10.
=
Dividing by common factor 7
0.2 is the decimal part of 350 is 70
Answer to the question is 0.2
Answer:
a) Null hypothesis:
Alternative hypothesis:
b) The scatter plot is on the figure attached.
c)
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:
Let X the Court Income and Y= Justice Salary, for this case we have that:
On this case we got that r =0.874
Part a
Null hypothesis:
Alternative hypothesis:
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:
Where n represent the sample size and the statistic t follows a t distribution with n-2 degrees of freedom:
On this case our value of n = 9 and the statistic is given by:
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:
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.
