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babymother [125]
3 years ago
10

Consider the following regression output where the dependent variable is testscores and the two explanatory variables are the st

udent-teacher ratio and the percent of English learners:
TestScore = 698.9 - 1.10 x STR - 0.650 x PctEL.
You are told that the t-statistic on the student-teacher ratio coefficient is 2.56.
1. The standard error therefore is approximately ____________.
A) 0.43
B) 0.650
C) 0.25
D) 1.96
Mathematics
1 answer:
jeyben [28]3 years ago
4 0

Answer:

The correct option is (A).

Step-by-step explanation:

The multiple linear regression equation is given by,  y=\alpha +\beta _{1} x_{1}+\beta _{2}x_{2}, where <em>α</em> = constant and <em>β</em>_{i} = slope coefficients of regression line.

To test if there is a important relationship amid <em>X</em>_{i} and <em>Y</em>, we use the <em>t</em>-statistic test.

The <em>t</em>-statistic for regression coefficient analysis is given by,

t=\frac{\beta_{i}}{S.E._{\beta_{i}}}

The regression equation for test scores dependent upon the two explanatory variables, the student-teacher ratio and the percent of English learners is:

\text{TestScore} = 698.9 - 1.10 \text{ STR} - 0.650 \text{ PctEL}

A <em>t</em>-test for the significance of the regression coefficient of variable student-teacher ratio (STR) is conducted.

The test statistic is found to be, <em>t</em> = 2.56.

The regression coefficient of variable STR is, 1.10.

Compute the standard error of the regression coefficient as follows:

        t=\frac{\beta_{i}}{S.E._{\beta_{i}}}

S.E._{\beta_{i}}=\frac{\beta_{i}}{t}

          =\frac{1.10}{2.56}\\\\=0.4296875\\\\\approx 0.43

The standard error of the regression coefficient is 0.43.

Thus, the correct option is (A).

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Answer:

See explanation

Step-by-step explanation:

Solution:-

- The demand function of a certain brand is given as price P a function of x quantity of goods ( in hundred ) demanded per month. The relation is:

                           P ( x ) = 50 Ln ( 50 / x ).

- The point price elasticity ( E ) of demand is given by:

                           E = \frac{P}{x}*\frac{dP}{dx}  

- Where, dP / dx : is the rate of change of price ( P ) with each hundred unit of good ( x ) is demanded.

- To determine the " dP / dx " by taking the first derivative of the given relation:

                          P ( x ) = 50 Ln ( 50 / x ).

                          d P ( x ) / dx = [ 50*x / 50 ] * [ -1*50 / x^2 ]

                                              = - 50 / x

- Hence the point price elasticity of demand is given by:

                          E = - ( P / x ) * ( 50 / x )

                          E = -50*P / x^2    

- For an inelastic demand, ! E ! is < 1:

                          ! -50*P / x^2 ! < 1

                          50*P / x^2 < 1

                          P < x^2 / 50

- For an unitary demand, ! E ! is =  1:

                          ! -50*P / x^2 !  = 1

                          50*P / x^2 = 1

                          P = x^2 / 50

- For an inelastic demand, ! E ! is > 1:

                          ! -50*P / x^2 ! > 1

                          50*P / x^2 > 1

                          P > x^2 / 50

2)

If the unit price is increased slightly from $50, will the revenue increase or decrease?

- We see from the calculated demand sensitivity d P / dt:

                          d P ( x ) / dx = - 50 / x

- We see that as P increases the from P = $50, the quantity of goods demanded would be:

                          50 = 50 ln(50/x)

                           1 = Ln ( 50 / x )

                           50/x = e

                           x = 50 / e

Then,

                          d P ( x ) / dx = - 50 / ( 50 / e )

                          d P ( x ) / dx = - e

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Then the product of price and quantity goods also decreases; hence, revenue decreases.

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Question 3
Fiesta28 [93]
Answer: y = -6x + 2

m = y_2 - y_1
/ x_2 - x_1

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y = mx + b
y = -6x +b

Find b using (-1,8)
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Final answer: y = -6x + 2

To check using one of the points: (-1,8) or (2,-10)
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