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Regarding the violation of multicollinearity, which of the following description is wrong?

MA_775_DIABLO [31]

Complete question is;

Regarding the violation of multicollinearity, which of the following description is wrong?

a. It changes the intercept of the regression line.

b. It changes the sign of the slope.

c. It changes the slope of the regression line.

d. It changes the value of F-tests.

e. It changes the value of T-tests

Answer:

a. It changes the intercept of the regression line

Step-by-step explanation:

Multicollinearity is a term used in multiple regression analysis to show a high correlation between independent variables of a study.

Since it deals with independent variables correlation, it means it must be found before getting the regression equation.

Now, looking at the options, the one that doesn't relate with multicollinearity is option A because the intercept of the regression line is the value of y that is predicted when x is 0. Meanwhile, multicollinearity from definition above does in no way change the intercept of the regression line because it doesn't predict the y-value when x is zero.

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2 years ago

A bin is constructed from sheet metal with a square base and 4 equal rectangular sides. if the bin is constructed from 48 square

kondaur [170]

This is a problem of maxima and minima using derivative.

In the figure shown below we have the representation of this problem, so we know that the base of this bin is square. We also know that there are four square rectangles sides. This bin is a cube, therefore the volume is:

V = length x width x height

That is:

$V = xxy = x^{2}y$

We also know that the <span>bin is constructed from 48 square feet of sheet metal, s</span>o:

Surface area of the square base = $x^{2}$

Surface area of the rectangular sides = $4xy$

Therefore, the total area of the cube is:

$A = 48 ft^{2} = x^{2} + 4xy$

Isolating the variable y in terms of x:

$y = \frac{48- x^{2} }{4x}$

Substituting this value in V:

$V = x^{2}( \frac{48- x^{2} }{x}) = 48x- x^{3}$

Getting the derivative and finding the maxima. This happens when the derivative is equal to zero:

$\frac{dv}{dx} = 48-3x^{2} =0$

Solving for x:

$x = \sqrt{\frac{48}{3}} = \sqrt{16} = 4$

Solving for y:

$y = \frac{48- 4^{2} }{(4)(4)} = 2$

Then, <span>the dimensions of the largest volume of such a bin is:
</span>
Length = 4 ft
Width = 4 ft
Height = 2 ft

And its volume is:

$V = (4^{2} )(2) = 32 ft^{3}$