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

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

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