Question

7. Consider the following models: Model A: y/50 = 3.65 + 0.75 x1 R_A^2 = 0.28 Model B: log(y) = 0.34 + 0.25 x1 R_B^2 = 0.35 Model C: log(y) = -2.01 - 0.30 log(x1) + 6.24 log(x2) R_C^2 = 0.22 Model D: y = -3.4 + 6.24 log(x1) - 3.4? R_D^2 = 0.38 According to Model A: If ?x1 = 10 units, then, ceteris paribus, ?y = (a) In Model B, interpret ?1. (b) In Model C, interpret ?1. (c) In Model A, in order for the increase in y to be 3 units we must have ?x1 = (d) In Model D, interpret ?1. (e) In Model A, interpret ?0. (f) In Model A, interpret the coefficient of determination. (g) Can we use R^2 to compare these models? Why or why not?

          7. Consider the following models:

Model A:    y/50 = 3.65 + 0.75 x1    R_A^2 = 0.28
Model B:    log(y) = 0.34 + 0.25 x1    R_B^2 = 0.35
Model C:    log(y) = -2.01 - 0.30 log(x1) + 6.24 log(x2)    R_C^2 = 0.22
Model D:    y = -3.4 + 6.24 log(x1) - 3.4?    R_D^2 = 0.38

According to Model A: If ?x1 = 10 units, then, ceteris paribus, ?y = __________

(a) In Model B, interpret ?1.
(b) In Model C, interpret ?1.
(c) In Model A, in order for the increase in y to be 3 units we must have ?x1 = __________
(d) In Model D, interpret ?1.
(e) In Model A, interpret ?0.
(f) In Model A, interpret the coefficient of determination.
(g) Can we use R^2 to compare these models? Why or why not?

7. Consider the following models:

Model A: y/50 = 3.65 + 0.75 x1 RA^2 = 0.28
Model B: log(y) = 0.34 + 0.25 x1 RB^2 = 0.35
Model C: log(y) = -2.01 - 0.30 log(x1) + 6.24 log(x2) RC^2 = 0.22
Model D: y = -3.4 + 6.24 log(x1) - 3.4? RD^2 = 0.38

According to Model A: If ?x1 = 10 units, then, ceteris paribus, ?y =

(a) In Model B, interpret ?1.
(b) In Model C, interpret ?1.
(c) In Model A, in order for the increase in y to be 3 units we must have ?x1 =
(d) In Model D, interpret ?1.
(e) In Model A, interpret ?0.
(f) In Model A, interpret the coefficient of determination.
(g) Can we use R^2 to compare these models? Why or why not?

Added by Francisco Jose O.

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