Week9 tutorial Solution PDF

Title Week9 tutorial Solution
Author rgrge eugbeu
Course Monash foundation year acc
Institution Monash University
Pages 3
File Size 150.5 KB
File Type PDF
Total Downloads 19
Total Views 142

Summary

seesrserrseersers...


Description

ETF2100/5910 Introductory Econometrics Tutorial exercises for week 9: F-test Consider the wage model: ln (WAGE ) = 1 + 2 EDUC + 3 EDUC2 + 4 EXPER + 5 EXPER2 +6 ( EDUC  EXPER) + 7 HRSWK + e

The data are in the file cps4c_small.wf1 and contains 1000 observations on wage educ exper hrswk married female

earnings per hour years of education post education years experience usual hours worked per week = 1 if married = 1 if female

metro midwest south west black Asian

= 1 if lives in metropolitan area = 1 if lives in midwest = 1 if lives in south = 1 if lives in west = 1 if black = 1 if Asian

Estimation results for this equation and for modified versions of it obtained by dropping some of the variables, are displayed in the Table below. Wage Equation Estimates Variable

Coefficient Estimates and (Standard Errors) Eqn (A)

C

Eqn (B)

Eqn (C)

1.055 (0.266)

1.252 (0.190)

1.573 (0.188)

0.0498 (0.0397)

0.0289 (0.0344)

0.0366 (0.0350)

EDUC^2

0.00319 (0.00169)

0.00352 (0.00366)

0.00293 (0.00170)

EXPER

0.0373 (0.0081)

0.0303 (0.0048)

EDUC

EXPER^2

Eqn (D) 1.917 (0.080)

0.904 (0.096) 0.1006 (0.0063)

0.0279 (0.0054)

−0.000485 −0.000456 (0.000090) (0.000086)

Eqn (E)

0.0295 (0.0048)

−0.000470 −0.000440 (0.000096) (0.000086)

EXPER*EDUC −0.000510 (0.000482) HRSWK

0.01145 (0.00137)

0.01156 (0.00137)

0.01345 (0.00136)

0.01524 (0.00151)

0.01188 (0.00136)

SSE

222.4166

222.6674

233.8317

280.5061

223.6716

(a) What restrictions on the coefficients of Eqn (A) give Eqn (C)? Use an F-test to test these restrictions. What question would you be trying to answer by performing this test? The restrictions are 𝛽4 = 𝛽5 = 𝛽6 = 0 1.

𝐻0 : 𝛽4 = 𝛽5 = 𝛽6 = 0 against 𝐻1 : at least one of the 𝛽’s is not zero

2.

The test statistics is 𝐹 =

3.

1000 − 7 = 993 Using =0.05, 𝐹𝑐 = 𝐹0.95,3,993 = 2.6139 (obtained from EViews)

(𝑆𝑆𝐸𝑅−𝑆𝑆𝐸𝑈)/𝐽 𝑆𝑆𝐸𝑈/(𝑁−𝐾)

~𝐹𝑑𝑓1=𝐽,𝑑𝑓2=(𝑁−𝐾) ,where 𝐽 = 3 and 𝑁 − 𝐾 =

(233.8317−222.4166)/3

= 16.9879

4.

The calculate test statistics is 𝐹𝑐𝑎𝑙𝑐 =

5.

Since 𝐹𝑐𝑎𝑙𝑐 is greater than 𝐹𝑐 , we reject 𝐻0 and conclude that at least one of the 𝛽’s is not zero.

222.4166/(1000−7)

Since the restrictions in the null hypothesis require all parameters relating to the variable experience to be zero, we are performing the test to see if experience has any impact on expected ln(wage). (b) What restrictions on the coefficients of Eqn (B) give Eqn (D)? Use an F-test to test these restrictions. What question would you be trying to answer by performing this test? The restrictions are 𝛽2 = 𝛽3 = 0 1.

𝐻0 : 𝛽2 = 𝛽3 = 0 against 𝐻1 : at least one of the 𝛽’s is not zero

2.

The test statistics is 𝐹 =

3.

1000 − 6 = 994 Using =0.05, 𝐹𝑐 = 𝐹0.95,2,994 = 3.0048 (obtained from EViews)

4.

The calculate test statistics is 𝐹𝑐𝑎𝑙𝑐 =

5.

Since 𝐹𝑐𝑎𝑙𝑐 is greater than 𝐹𝑐 , we reject 𝐻0 and conclude that at least one of the 𝛽’s is not zero.

(𝑆𝑆𝐸𝑅−𝑆𝑆𝐸𝑈)/𝐽 𝑆𝑆𝐸𝑈/(𝑁−𝐾)

~𝐹𝑑𝑓1=𝐽,𝑑𝑓2=(𝑁−𝐾) ,where 𝐽 = 2 and 𝑁 − 𝐾 =

(280.5061−222.6674)/2 222.6674/(1000−6)

= 129.0976

Since the restrictions in the null hypothesis require all parameters relating to the variable education to be zero, we perform the test to see if education has any impact on expected ln(wage).

(c) What restrictions on the coefficients of Eqn (A) give Eqn (E)? Use an F-test to test these restrictions. What question would you be trying to answer by performing this test? The restrictions are 𝛽3 = 𝛽6 = 0 1. 2.

𝐻0 : 𝛽3 = 𝛽6 = 0 against 𝐻1 : at least one of the 𝛽’s is not zero (𝑆𝑆𝐸 −𝑆𝑆𝐸𝑈)/𝐽 ~𝐹𝑑𝑓1=𝐽,𝑑𝑓2=(𝑁−𝐾) ,where 𝐽 = 2 and 𝑁 − 𝐾 = The test statistics is 𝐹 = 𝑆𝑆𝐸𝑅 /(𝑁−𝐾)

3.

1000 − 7 = 993 Using =0.05, 𝐹𝑐 = 𝐹0.95,2,993 = 3.0048 (obtained from EViews)

4.

The calculate test statistics is 𝐹𝑐𝑎𝑙𝑐 =

5.

Since 𝐹𝑐𝑎𝑙𝑐 is not greater than 𝐹𝑐 , we do not reject 𝐻0 and conclude that the data is not incompatible with 𝛽3 = 𝛽6 = 0

𝑈

(223.6716−222.4166)/2 222.4166/(1000−7)

= 2.8015

We perform the test to see if linear form of education is enough to explain expected ln(wage) and at the same time if there is an interaction effect between experience and education.

(d) Based on your answers to parts (a) to (c), which model would you prefer? Why? We start by testing Model (A) against other models

Part (a) compares Model (A) and (C) suggests choosing Model (A) Part (c) compares Model (A) and (E) suggests choosing Model (E) So Model (A) and (C) are inferior to (E). Part (b) compares Model (B) and (D) suggests choosing Model (B) Model (B) and (E) are preferred (but we have not directly compared them). Model (E) is simpler (contains smaller number of regressors) than model (B) and does not include any insignificant variables, whereas (B) includes the square of education which has an insignificant coefficient. Model (E) also has signs that are consistent with our expectations based on economic reasoning. So Model (E) is preferred overall. (e) Starting with model (A), suppose you wish to test hypothesis that a year of education has the same effect on ln (WAGE) as a year of experience. What null and alternative hypotheses would you set up? What is the restricted model, assuming that the null hypothesis is true? Given the SSE R = 276.1225 and N = 1000 , test the hypothesis. The marginal effect of education on expected ln(wage) from model (A) is: 𝛽2 + 2𝛽3 𝐸𝐷𝑈𝐶 + 𝛽6 𝐸𝑋𝑃𝐸𝑅 While the marginal effect of experience on expected ln(wage) is: 𝛽4 + 2𝛽5 𝐸𝑋𝑃𝐸𝑅 + 𝛽6 𝐸𝐷𝑈𝐶 If the two marginal effects are the same then 𝛽2 has to be the same as 𝛽4 , 2𝛽3 has to be the same as 𝛽6 , and 2𝛽5 has to be the same as 𝛽6 . The hypotheses are 𝐻0 : 𝛽2 = 𝛽4 , 𝛽3 = 𝛽5 and 2𝛽3 = 𝛽6 The restricted equation is ln(𝑊𝐴𝐺𝐸) = 𝛽1 + 𝛽2 𝐸𝐷𝑈𝐶 + 𝛽3 𝐸𝐷𝑈𝐶 2 + 𝛽2 𝐸𝑋𝑃𝐸𝑅 + 𝛽3 𝐸𝑋𝑃𝐸𝑅 2 +2𝛽3 (𝐸𝐷𝑈𝐶 × 𝐸𝑋𝑃𝐸𝑅) + 𝛽7 𝐻𝑅𝑆𝑊𝐾 + 𝑒 = 𝛽1 + 𝛽2 (𝐸𝐷𝑈𝐶 + 𝐸𝑋𝑃𝐸𝑅) + 𝛽3 (𝐸𝐷𝑈𝐶 2 + 𝐸𝑋𝑃𝐸𝑅 2 + (2 × 𝐸𝐷𝑈𝐶 × 𝐸𝑋𝑃𝐸𝑅) +𝛽7 𝐻𝑅𝑆𝑊𝐾 + 𝑒 To test these restrictions: 1. 2.

𝐻0 : 𝛽2 = 𝛽4 , 𝛽3 = 𝛽5 and 2𝛽3 = 𝛽6 against 𝐻1 : 𝐻0 is not true (𝑆𝑆𝐸𝑅−𝑆𝑆𝐸𝑈)/𝐽 The test statistics is 𝐹 = ~𝐹𝑑𝑓1=𝐽,𝑑𝑓2=(𝑁−𝐾) ,where 𝐽 = 3 and 𝑁 − 𝐾 = 𝑆𝑆𝐸 /(𝑁−𝐾)

3.

1000 − 7 = 993 Using =0.05, 𝐹𝑐 = 𝐹0.95,3,993 = 2.6139 (obtained from EViews)

𝑈

276.1225−222.4166 3 222.4166 1000−7

4.

The calculate test statistics is 𝐹𝑐𝑎𝑙𝑐 =

5.

Since 𝐹𝑐𝑎𝑙𝑐 is greater than 𝐹𝑐 , we reject 𝐻0 and conclude that a year of education does not have the same impact on expected ln(wage) as a year of experience.

= 79.925...


Similar Free PDFs