Exam 11: Regression With a Binary Dependent Variable

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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.

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(Requires Appendix material)Briefly describe the difference between the following models: censored and truncated regression model,count data,ordered responses,and discrete choice data.Try to be specific in terms of describing the data involved.

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The major flaw of the linear probability model is that

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Equation (11.3)in your textbook presents the regression results for the linear probability model. a.Using a spreadsheet program such as Excel,plot the fitted values for whites and blacks in the same graph,for P/I ratios ranging from 0 to 1 (use 0.05 increments). b.Explain some of the strengths and shortcomings of the linear probability model using this graph.

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In the linear probability model,the interpretation of the slope coefficient is

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The linear probability model is

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Probit coefficients are typically estimated using

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Your task is to model students' choice for taking an additional economics course after the first principles course.Describe how to formulate a model based on data for a large sample of students.Outline several estimation methods and their relative advantage over other methods in tackling this problem.How would you go about interpreting the resulting output? What summary statistics should be included?

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Consider the following logit regression: Pr(Y = 1 Consider the following logit regression: Pr(Y = 1   X)= F (15.3 - 0.24 × X) Calculate the change in probability for X increasing by 10 for X = 40 and X = 60.Why is there such a large difference in the change in probabilities? X)= F (15.3 - 0.24 × X) Calculate the change in probability for X increasing by 10 for X = 40 and X = 60.Why is there such a large difference in the change in probabilities?

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(Requires Advanced material)Maximum likelihood estimation yields the values of the coefficients that

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