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3 years ago

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Step 1 of 3: Each year a nationally recognized publication conducts its Survey of America’s Best Graduate and Professional Schoo

ls. An academic advisor wants to predict the typical starting salary of a graduate at a top business school using GMAT score of the school as a predictor variable. A least-squares regression line of SALARY versus GMAT is y = -92040 + 228x, and the correlation, r, is 0.81. If the sample size used to create the regression line is n=20, is r statistically significant at a 0.05 level?

1 answer:

lukranit [14]3 years ago

4 0

Answer:

Step-by-step explanation:

Hello!

Given the variables

Y: Starting salary of a graduate at a top business school

X: GMAT score of the business school

The sample used had a size n=20

The estimated regression equation was ^Y= -92040 + 228Xi

The correlation coefficients were r= 0.81

You need to know if r is statistically significant at a 0.05 level:

Assuming the hypotheses are:

H₀: There is a linear correlation between the variables of interest

H₁: There is no linear correlation between the variables of interest

α: 0.05

This test is two-tailed.

The p-value for a r=0.81 and n=20 is 0.000015

Using the p-value approach, the decision rule is:

If p-value ≤ α, reject the null hypothesis

If p-value > α, do not reject the null hypothesis

In this case, the p-value is less than α, the decision is to reject the null hypothesis.

At a 5% level, you can conclude that the coefficient of correlation is statistically significant.

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Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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3 years ago

(1) (10 points) Find the characteristic polynomial of A (2) (5 points) Find all eigenvalues of A. You are allowed to use your ca

Yuri [45]

Answer:

Step-by-step explanation:

Since this question is lacking the matrix A, we will solve the question with the matrix

$\left[\begin{matrix}4 & -2 \\ 1 & 1 \end{matrix}\right]$

so we can illustrate how to solve the problem step by step.

a) The characteristic polynomial is defined by the equation $det(A-\lambdaI)=0$ where I is the identity matrix of appropiate size and lambda is a variable to be solved. In our case,

$\left|\left[\begin{matrix}4-\lamda & -2 \\ 1 & 1-\lambda \end{matrix}\right]\right|= 0 = (4-\lambda)(1-\lambda)+2 = \lambda^2-5\lambda+4+2 = \lambda^2-5\lambda+6$

So the characteristic polynomial is $\lambda^2-5\lambda+6=0$ .

b) The eigenvalues of the matrix are the roots of the characteristic polynomial. Note that

$\lambda^2-5\lambda+6=(\lambda-3)(\lambda-2) =0$

So $\lambda=3, \lambda=2$

c) To find the bases of each eigenspace, we replace the value of lambda and solve the homogeneus system(equalized to zero) of the resultant matrix. We will illustrate the process with one eigen value and the other one is left as an exercise.

If $\lambda=3$ we get the following matrix

$\left[\begin{matrix}1 & -2 \\ 1 & -2 \end{matrix}\right]$ .

Since both rows are equal, we have the equation

$x-2y=0$ . Thus x=2y. In this case, we get to choose y freely, so let's take y=1. Then x=2. So, the eigenvector that is a base for the eigenspace associated to the eigenvalue 3 is the vector (2,1)

For the case $\lambda=2$ , using the same process, we get the vector (1,1).

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$P=\left[\begin{matrix}2&1 \\ 1 & 1 \end{matrix}\right], D=\left[\begin{matrix}3&0 \\ 0 & 2 \end{matrix}\right]$

This matrices are not unique, since they depend on the order in which we arrange the eigenvalues in the matrix D. Another pair or matrices that diagonalize A is

$P=\left[\begin{matrix}1&2 \\ 1 & 1 \end{matrix}\right], D=\left[\begin{matrix}2&0 \\ 0 & 3 \end{matrix}\right]$

which is obtained by interchanging the eigenvalues on the diagonal and their respective eigenvectors

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