Respuesta :
Answer:
a. As college debt increases current investment decreases.
b. Y= 68778.2406 - 1.9112X
Every time the college debt increases one dollar, the estimated mean of the current investments decreases 1.9112 dollars.
c. There is a significant linear relationship between college debt and current investment because the P-value is less than 0.1.
d. Y= $59222.2406
e. R²= 0.9818
Step-by-step explanation:
Hello!
You have the information on a random sample of 20 individuals who graduated from college five years ago. The variables of interest are:
Y: Current investment of an individual that graduated from college 5 years ago.
X: Total debt of an individual when he graduated from college 5 years ago.
a)
To see the relationship between the information about the debt and the investment is it best to make a scatterplot with the sample information.
As you can see in the scatterplot (attachment) there is a negative relationship between the current investment and the debt after college, this means that the greater the debt these individuals had, the less they are currently investing.
The statement that best describes it is: As college debt increases current investment decreases.
b)
The population regression equation is Y= α + βX +Ei
To develope the regression equation you have to estimate alpha and beta:
a= Y[bar] -bX[bar]
a= 44248.55 - (-1.91)*12829.70
a= 68778.2406
b= [tex]\frac{sumXY-\frac{(sumX)(sumY)}{n} }{sumX^2-\frac{(sumX)^2}{n} }[/tex]
b=[tex]\frac{9014653088-\frac{(256594)(884971)}{20} }{4515520748-\frac{(256594)^2}{20} }[/tex]
b= -1.9112
∑X= 256594
∑X²= 4515520748
∑Y= 884971
∑Y²= 43710429303
∑XY= 9014653088
n= 20
Means:
Y[bar]= ∑Y/n= 884971/20= 44248.55
X[bar]= ∑X/n= 256594/20= 12829.70
The estimated regression equation is:
Y= 68778.2406 - 1.9112X
Every time the college debt increases one dollar, the estimated mean of the current investments decreases 1.9112 dollars.
c)
The hypotheses to test if there is a linear regression between the two variables are two tailed:
H₀: β = 0
H₁: β ≠ 0
α: 0.01
To make this test you can use either a Student t or the Snedecor's F (ANOVA)
Using t= b - β = -1.91 - 0 = -31.83
Sb 0.06
The critical region and the p-value for this test are two tailed.
The p-value is: 0.0001
The p-value is less than the level of signification, the decision is to reject the null hypothesis.
Using the
[tex]F= \frac{MSTr}{MSEr}= \frac{4472537017.96}{4400485.72} =1016.37[/tex]
The rejection region using the ANOVA is one-tailed to the right, and so is the p-value.
The p-value is: 0.0001
Using this approach, the decision is also to reject the null hypothesis.
The conclusion is that at a 1% significance level, there is a linear regression between the current investment and the college debt.
The correct statement is:
There is a significant linear relationship between college debt and current investment because the P-value is less than 0.1.
d)
To predict what value will take Y to a given value of X you have to replace it in the estimated regression equation.
Y/X=$5000
Y= 68778.2406 - 1.9112*5000
Y= $59222.2406
The current investment of an individual that had a $5000 college debt is $59222.2406.
e)
To estimate the proportion of variation of the dependent variable that is explained/ given by the independent variable you have to calculate the coefficient of determination R².
[tex]R^2= \frac{b^2[sumX^2-\frac{(sumX)^2}{n} ]}{sumY^2-\frac{(sumY)^2}{n} }[/tex]
[tex]R^2= \frac{-1.9112^2[4515520748-\frac{(256594)^2}{20} ]}{43710429303-\frac{(884971)^2}{20} }[/tex]
R²= 0.9818
This means that 98.18% of the variability of the current investments are explained by the college debt at graduation under the estimated regression model: Y= 68778.2406 - 1.9112X
I hope it helps!
Following are the solution to the given points:
For point a:
This relationship between these two variables is better defined by the fact that as college debt has increased, so does current investment. Please find the attached file.
For point b:
Calculating the analysis of regression:
[tex]r^2 \ \ \ \ \ \ \ \ \ \ \ \ \ 0.9826 \ \ \ \ \ \ \ \ \ \ \ \ n \ \ \ \ \ \ \ \ \ \ \ \ 20\\\\r \ \ \ \ \ \ \ \ \ \ \ \ -0.9913 \ \ \ \ \ \ \ \ \ \ \ \ k \ \ \ \ \ \ \ \ \ \ \ \ 1\\\\\text{Std. Error} \ \ \ \ \ \ \ 2097.733 \ \ \ \ \ \ \ \text{Dep. Var.} \ \ \ \ \ \ \text{Invested}[/tex]
Calculating the analysis of ANOVA table:
[tex]Source \ \ \ \ \ \ \ \ \ \ \ SS \ \ \ \ \ \ \ \ \ \ \ df \ \ \ \ \ \ \ \ \ \ \ MS \ \ \ \ \ \ \ \ \ \ \ F \ \ \ \ \ \ \ \ \ \ \ p-value[/tex]
[tex]Regression \ \ \ \ \ 4,472,537,017.9647\ \ \ \ \ 1 \ \ \ \ 4,472,537,017.9647 \ \ \ \ 1016.37\ \ \ \ 2.74E-17\\\\[/tex]
[tex]Residual\ \ \ \ \ \ \ 79,208,742.9853\ \ \ \ \ \ \ 18 \ \ \ \ \ \ \ 4,400,485.7214\\\\Total\ \ \ \ \ \ \ 4,551,745,760.9500 \ \ \ \ \ \ \ 19[/tex]
Please find the attached file.
[tex]Investment = 68778.2406-1.9119\times debt[/tex]
There is indeed a 1.9119 percent drop in current investment for each and every additional dollar of educational debt.
For point c:
In testing for just a significant linear relation in the regression analysis, there is indeed a significant linear association between school loans and present investment because the P-value is less than 0.1, therefore the correct conclusion at the 0.1 significant level.
For point d:
Calculating the Invest:
[tex]= 68778.2406-1.9119\times 5000 \\\\= 59218.7406[/tex]
So, the predicted value [tex]=\$59218.74[/tex]
( Using excel to calculate the out Predicted value [tex]=\$59218.51[/tex])
For point e:
The proportion of the variation explained[tex]=0.9826[/tex]
Learn more about the data set:
brainly.com/question/24205033
