Filters
Question type
Study Flashcards

The linear probability model is


A) the application of the multiple regression model with a continuous left-hand side variable and a binary variable as at least one of the regressors.
B) an example of probit estimation.
C) another word for logit estimation.
D) the application of the linear multiple regression model to a binary dependent variable.

E) A) and D)
F) C) and D)

Correct Answer

verifed

verified

(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.

Correct Answer

verifed

verified

The answer should follow the discussion ...

View Answer

Probit coefficients are typically estimated using


A) the OLS method
B) the method of maximum likelihood
C) non-linear least squares (NLLS)
D) by transforming the estimates from the linear probability model

E) All of the above
F) B) and C)

Correct Answer

verifed

verified

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.

Correct Answer

verifed

verified

(a) blured image ≈ 0.625 × 0.236 = 0.148,which is qu...

View Answer

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?

Correct Answer

verifed

verified

Answers will vary by student.This is an ...

View Answer

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?

Correct Answer

verifed

verified

Pr(Y=1 blured image X=40)= 0.997;Pr(Y=1 blured image X...

View Answer

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


A) the change in odds associated with a unit change in X,holding other regressors constant.
B) not all that meaningful since the dependent variable is either 0 or 1.
C) the change in probability that Y=1 associated with a unit change in X,holding others regressors constant.
D) the response in the dependent variable to a percentage change in the regressor.

E) None of the above
F) C) and D)

Correct Answer

verifed

verified

The major flaw of the linear probability model is that


A) the actuals can only be 0 and 1,but the predicted are almost always different from that.
B) the regression R2 cannot be used as a measure of fit.
C) people do not always make clear-cut decisions.
D) the predicted values can lie above 1 and below 0.

E) B) and C)
F) A) and D)

Correct Answer

verifed

verified

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.

Correct Answer

verifed

verified

a. blured image b.The strength is that the regressio...

View Answer

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


A) minimize the sum of squared prediction errors.
B) maximize the likelihood function.
C) come from a probability distribution and hence have to be positive.
D) are typically larger than those from OLS estimation.

E) All of the above
F) C) and D)

Correct Answer

verifed

verified

Showing 41 - 50 of 50

Related Exams

Exam 1

Economic Questions and Data

17 Questions

Exam 2

Review of Probability

71 Questions

Exam 3

Review of Statistics

63 Questions

Exam 4

Linear Regression With One Regressor

65 Questions

Exam 5

Regression With a Single Regressor: Hypothesis Tests and Confidence Intervals

59 Questions

Exam 6

Linear Regression With Multiple Regressors

65 Questions

Exam 7

Hypothesis Tests and Confidence Intervals in Multiple Regression

65 Questions

Exam 8

Nonlinear Regression Functions

62 Questions

Exam 9

Assessing Studies Based on Multiple Regression

65 Questions

Exam 10

Regression With Panel Data

50 Questions

Exam 12

Instrumental Variables Regression

50 Questions

Exam 13

Experiments and Quasi-Experiments

50 Questions

Exam 14

Introduction to Time Series Regression and Forecasting

50 Questions

Exam 15

Estimation of Dynamic Causal Effects

50 Questions

Exam 16

Additional Topics in Time Series Regression

50 Questions

Exam 17

The Theory of Linear Regression With One Regressor

49 Questions

Exam 18

The Theory of Multiple Regression

50 Questions

Show Answer