Examlex

Solved

You Have a Limited Dependent Variable (Y)and a Single Explanatory

question 41

Essay

You have a limited dependent variable (Y)and a single explanatory variable (X).You estimate the relationship using the linear probability model,a probit regression,and a logit regression.The results are as follows: You have a limited dependent variable (Y)and a single explanatory variable (X).You estimate the relationship using the linear probability model,a probit regression,and a logit regression.The results are as follows:   = 2.858 - 0.037 × X (0.007) Pr(Y = 1   X)= F (15.297 - 0.236 × X) Pr(Y = 1   X)= Φ (8.900 - 0.137 × X) (0.058) (a)Although you cannot compare the coefficients directly,you are told that  it can be shown  that certain relationships between the coefficients of these models hold approximately.These are for the slope:   ≈ 0.625 ×   ,   ≈ 0.25 ×   .Take the logit result above as a base and calculate the slope coefficients for the linear probability model and the probit regression.Are these values close? (b)For the intercept,the same conversion holds for the logit-to-probit transformation.However,for the linear probability model,there is a different conversion:   ≈ 0.25 ×   + 0.5 Using the logit regression as the base,calculate a few changes in X (temperature in degrees of Fahrenheit)to see how good the approximations are. = 2.858 - 0.037 × X
(0.007)
Pr(Y = 1 You have a limited dependent variable (Y)and a single explanatory variable (X).You estimate the relationship using the linear probability model,a probit regression,and a logit regression.The results are as follows:   = 2.858 - 0.037 × X (0.007) Pr(Y = 1   X)= F (15.297 - 0.236 × X) Pr(Y = 1   X)= Φ (8.900 - 0.137 × X) (0.058) (a)Although you cannot compare the coefficients directly,you are told that  it can be shown  that certain relationships between the coefficients of these models hold approximately.These are for the slope:   ≈ 0.625 ×   ,   ≈ 0.25 ×   .Take the logit result above as a base and calculate the slope coefficients for the linear probability model and the probit regression.Are these values close? (b)For the intercept,the same conversion holds for the logit-to-probit transformation.However,for the linear probability model,there is a different conversion:   ≈ 0.25 ×   + 0.5 Using the logit regression as the base,calculate a few changes in X (temperature in degrees of Fahrenheit)to see how good the approximations are. X)= F (15.297 - 0.236 × X)
Pr(Y = 1 You have a limited dependent variable (Y)and a single explanatory variable (X).You estimate the relationship using the linear probability model,a probit regression,and a logit regression.The results are as follows:   = 2.858 - 0.037 × X (0.007) Pr(Y = 1   X)= F (15.297 - 0.236 × X) Pr(Y = 1   X)= Φ (8.900 - 0.137 × X) (0.058) (a)Although you cannot compare the coefficients directly,you are told that  it can be shown  that certain relationships between the coefficients of these models hold approximately.These are for the slope:   ≈ 0.625 ×   ,   ≈ 0.25 ×   .Take the logit result above as a base and calculate the slope coefficients for the linear probability model and the probit regression.Are these values close? (b)For the intercept,the same conversion holds for the logit-to-probit transformation.However,for the linear probability model,there is a different conversion:   ≈ 0.25 ×   + 0.5 Using the logit regression as the base,calculate a few changes in X (temperature in degrees of Fahrenheit)to see how good the approximations are. X)= Φ (8.900 - 0.137 × X)
(0.058)
(a)Although you cannot compare the coefficients directly,you are told that "it can be shown" that certain relationships between the coefficients of these models hold approximately.These are for the slope: You have a limited dependent variable (Y)and a single explanatory variable (X).You estimate the relationship using the linear probability model,a probit regression,and a logit regression.The results are as follows:   = 2.858 - 0.037 × X (0.007) Pr(Y = 1   X)= F (15.297 - 0.236 × X) Pr(Y = 1   X)= Φ (8.900 - 0.137 × X) (0.058) (a)Although you cannot compare the coefficients directly,you are told that  it can be shown  that certain relationships between the coefficients of these models hold approximately.These are for the slope:   ≈ 0.625 ×   ,   ≈ 0.25 ×   .Take the logit result above as a base and calculate the slope coefficients for the linear probability model and the probit regression.Are these values close? (b)For the intercept,the same conversion holds for the logit-to-probit transformation.However,for the linear probability model,there is a different conversion:   ≈ 0.25 ×   + 0.5 Using the logit regression as the base,calculate a few changes in X (temperature in degrees of Fahrenheit)to see how good the approximations are. ≈ 0.625 × You have a limited dependent variable (Y)and a single explanatory variable (X).You estimate the relationship using the linear probability model,a probit regression,and a logit regression.The results are as follows:   = 2.858 - 0.037 × X (0.007) Pr(Y = 1   X)= F (15.297 - 0.236 × X) Pr(Y = 1   X)= Φ (8.900 - 0.137 × X) (0.058) (a)Although you cannot compare the coefficients directly,you are told that  it can be shown  that certain relationships between the coefficients of these models hold approximately.These are for the slope:   ≈ 0.625 ×   ,   ≈ 0.25 ×   .Take the logit result above as a base and calculate the slope coefficients for the linear probability model and the probit regression.Are these values close? (b)For the intercept,the same conversion holds for the logit-to-probit transformation.However,for the linear probability model,there is a different conversion:   ≈ 0.25 ×   + 0.5 Using the logit regression as the base,calculate a few changes in X (temperature in degrees of Fahrenheit)to see how good the approximations are. , You have a limited dependent variable (Y)and a single explanatory variable (X).You estimate the relationship using the linear probability model,a probit regression,and a logit regression.The results are as follows:   = 2.858 - 0.037 × X (0.007) Pr(Y = 1   X)= F (15.297 - 0.236 × X) Pr(Y = 1   X)= Φ (8.900 - 0.137 × X) (0.058) (a)Although you cannot compare the coefficients directly,you are told that  it can be shown  that certain relationships between the coefficients of these models hold approximately.These are for the slope:   ≈ 0.625 ×   ,   ≈ 0.25 ×   .Take the logit result above as a base and calculate the slope coefficients for the linear probability model and the probit regression.Are these values close? (b)For the intercept,the same conversion holds for the logit-to-probit transformation.However,for the linear probability model,there is a different conversion:   ≈ 0.25 ×   + 0.5 Using the logit regression as the base,calculate a few changes in X (temperature in degrees of Fahrenheit)to see how good the approximations are. ≈ 0.25 × You have a limited dependent variable (Y)and a single explanatory variable (X).You estimate the relationship using the linear probability model,a probit regression,and a logit regression.The results are as follows:   = 2.858 - 0.037 × X (0.007) Pr(Y = 1   X)= F (15.297 - 0.236 × X) Pr(Y = 1   X)= Φ (8.900 - 0.137 × X) (0.058) (a)Although you cannot compare the coefficients directly,you are told that  it can be shown  that certain relationships between the coefficients of these models hold approximately.These are for the slope:   ≈ 0.625 ×   ,   ≈ 0.25 ×   .Take the logit result above as a base and calculate the slope coefficients for the linear probability model and the probit regression.Are these values close? (b)For the intercept,the same conversion holds for the logit-to-probit transformation.However,for the linear probability model,there is a different conversion:   ≈ 0.25 ×   + 0.5 Using the logit regression as the base,calculate a few changes in X (temperature in degrees of Fahrenheit)to see how good the approximations are. .Take the logit result above as a base and calculate the slope coefficients for the linear probability model and the probit regression.Are these values close?
(b)For the intercept,the same conversion holds for the logit-to-probit transformation.However,for the linear probability model,there is a different conversion: You have a limited dependent variable (Y)and a single explanatory variable (X).You estimate the relationship using the linear probability model,a probit regression,and a logit regression.The results are as follows:   = 2.858 - 0.037 × X (0.007) Pr(Y = 1   X)= F (15.297 - 0.236 × X) Pr(Y = 1   X)= Φ (8.900 - 0.137 × X) (0.058) (a)Although you cannot compare the coefficients directly,you are told that  it can be shown  that certain relationships between the coefficients of these models hold approximately.These are for the slope:   ≈ 0.625 ×   ,   ≈ 0.25 ×   .Take the logit result above as a base and calculate the slope coefficients for the linear probability model and the probit regression.Are these values close? (b)For the intercept,the same conversion holds for the logit-to-probit transformation.However,for the linear probability model,there is a different conversion:   ≈ 0.25 ×   + 0.5 Using the logit regression as the base,calculate a few changes in X (temperature in degrees of Fahrenheit)to see how good the approximations are. ≈ 0.25 × You have a limited dependent variable (Y)and a single explanatory variable (X).You estimate the relationship using the linear probability model,a probit regression,and a logit regression.The results are as follows:   = 2.858 - 0.037 × X (0.007) Pr(Y = 1   X)= F (15.297 - 0.236 × X) Pr(Y = 1   X)= Φ (8.900 - 0.137 × X) (0.058) (a)Although you cannot compare the coefficients directly,you are told that  it can be shown  that certain relationships between the coefficients of these models hold approximately.These are for the slope:   ≈ 0.625 ×   ,   ≈ 0.25 ×   .Take the logit result above as a base and calculate the slope coefficients for the linear probability model and the probit regression.Are these values close? (b)For the intercept,the same conversion holds for the logit-to-probit transformation.However,for the linear probability model,there is a different conversion:   ≈ 0.25 ×   + 0.5 Using the logit regression as the base,calculate a few changes in X (temperature in degrees of Fahrenheit)to see how good the approximations are. + 0.5
Using the logit regression as the base,calculate a few changes in X (temperature in degrees of Fahrenheit)to see how good the approximations are.


Definitions:

Durbin-Watson Statistic

A measure used in statistics to detect the presence of autocorrelation in the residuals from a linear regression analysis.

Hypotheses

Proposed explanations for a phenomenon, set forth as a basis for empirical testing.

First-Order Autocorrelation

A statistical measure of the relationship between a variable's current value and its immediate past value.

Durbin-Watson Statistic

A statistical method utilized to identify autocorrelation within the residuals of a regression analysis.

Related Questions