Results and interpretation

Figure 8
Dependent variable is MOR5	OLS	Fixed Effect	Random Effect
Independent Variables
UNMPL	-.6475793 (-1.52)	2.113166 (8.11)	2.585086 (9.72)
LNPERCAPGDP	13.35622 (2.79)	-28.44485 (-4.45)	-19.60362 (-3.12)
lgPUD	-133.6972 (-14.76)	-27.18544 (-1.73)	-78.65961 (-5.61)
lgURPOP	57.21655 (5.90)	-132.1327 (-5.19)	-53.04207 (-2.34)
Constant	325.7052 (12.85)	868.808 (21.41)	718.1875 (19.82)
Number of observations	216	216	216
R-square Adj R square Within R square Between R square Overall R square	0.5651 0.5569 - - -	- - 0.8507 0.0761 0.1276	- - 0.8398 0.1765 0.2558
t statistics in parentheses
* p<0.05, ** p<0.01, *** p<0.001

Figure 9. source: Research findings
In figure 9, the results of Hausman Test are given. Since chi2 is higher than probability we have reject null hypothesis and accept alternative hypothesis which states fixed effect best describes model. Now we choose fixed effect model out of three models and we have to check other problems such as serial correlation, heteroscedasticity and so on. In order to serial correlation problem Wooldridge Test for autocorrelation was used. The result of this test is shown in figure 10.
Figure 10. Source: research findings
. Hypothesis of this test is following:
H_0: There is no serial correlation in chosen FE model
H_A: There is serial correlation in chosen FE model
Calculated F is equal to 2554.042 while probability is equal to 0.000 Since F is higher than probability value we have to reject null hypothesis and accept alternative hypothesis. Therefore, in fixed effect model there is serial correlation problem. Furthermore, we have to check heteroskedasticity problem and modified Wald test can be used to check whether residuals are homoscedastic or heteroscedastic. Hypotheses of this test are following:
H_0: Residuals are homoscedastic
H_A: Residuals are heteroscedastic

Figure 11. Source: research findings
Based on results we found that there is heteroscedasticity problem, because chi2 is 5441.13 and probability is 0.00. Therefore, we have to reject null hypothesis and accept alternative hypothesis which states residuals are heteroscedastic. We can relax this problem by using robust.

Figure 12. Source: research findings
In this test, we have R squared is equal to 05661 and F(4, 211) is 103.42.
However, there is another problem, namely, serial correlation problem. We solve this problem and we have to use another model in order to lose these two problems. Now we will do GLS (generally least squares) model in order to retain our model far from potential problems such as heteroscedasticity and serial correlation.