Textbook Notes (363,135)
Economics (727)
ECON 3U03 (5)
Chapter 4

# Chapter 4 Notes

4 Pages
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School
McMaster University
Department
Economics
Course
ECON 3U03
Professor
James Bruce
Semester
Winter

Description
4.14 A: Neither WAGE nor ln(WAGE) appear normally distributed. The distribution for WAGE is positively skewed and that for ln(WAGE) is too flat at the top. However, ln(WAGE) more closely resembles a normal distribution. b) Regression results for linear model: WAGE = -4.9122 + 1.1385EDUC R = 0.2024 _____ The estimated return to education at the mean = b /2WAGE * 100 = 11.15% 2 The results for the log-linear model are ln(WAGE) = 0.7884 + 0.1038EDUC R = 0.2146 Estimated return to education = 10.38% c) The Jarque-Bera test results are JB= 3023 (p-value = 0.0000) for the residuals from the linear model and JB= 3.48(p-value = 0.1754) for the residuals from the log-linear model. Both the histograms and the Jarque-Bera test results suggest the residuals from the log-linear model are more compatible with normality. In the log-linear model a null hypothesis of normality is not rejected at a 10% level of significance. In the linear regression model it is rejected at a 1% level of significance. (d) Linear model: R 0.2024 Log-linear model: R = ([corr(y,y)] )/(var(y)var(y)) = (6.87196 )/(38.9815*5.39435) = 0.2246 g 2 2 Since, R >gR We conclude that the log-linear model fits the data better. e) Absolute value of residuals increases in magnitude as EDUC increases. f) Prediction for simple linear model: WAGE = -0.9122 + 1.1385 * 16 = 13.30 2 Prediction for log-linear model: WAGE = exp C0.7884 + 0.1038 * 16 + (0.4902 ) = 13.05 g) log-linear function is preferred because it has higher fit value and its residuals are consistent with normality. The linear model had a smaller prediction error. 5.3: a) (i) t-statistic for1b = 0.476 (ii) Standard error for b 2s se(b )2= 0.00418 (iii) The estimate for β i3 b = 30.0014 2 (iv) To compute R , we need SSE and SST. From the output, SSE = 5.752896. To find SST: 2 2 SST = 1518 * (0.0633) = 6.08246 thus, R = 1 – SSE/SST = 1 – 5.75290/6.08246 = 0.054 (v) The estimated error standard deviation is 0.061622 b) The value b =20.0276 implies that if ln(TOTEXP) increases by 1 unit the alcohol share will increase by 0.0276. The change in the alcohol share from a 1-unit change in total expenditure depends on level of tota
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