Eviews exercise - Assignment Example

The p-value for the F-statistic is 0.000 (a value less than 5% significance level), we therefore reject the null hypothesis and conclude that the model is fit at 5% significance level. The value of R-squared is 0.5738; this implies that 57.38% of variation in the dependent variable (children) is explained by the six explanatory variables in the model. It is important also to note that the coefficients for the six explanatory variables are all significant at 5% significance level. The coefficient for electric () is -0.

2401; this implies that for any unit increase in electric we expect the dependent variable (children) to decrease by 0.2401. Similarly, the coefficient for TV () is -0.1837; this implies that for any unit increase in TV values we expect the dependent variable (children) to decrease by 0.1837. The p-values for the coefficients () are 0.0019 and 0.045 respectively. This shows that the two variables are significant at 5% significance level. The mean for the number of children among the urban residents is 1.

988 while that of non-urban residents is 2.567. To check for significant difference we look at the t-test; the p-value is 0.000 (a value less than 5% significance level) we therefore reject the null hypothesis and conclude that there is significant difference in the fertility among the urban and ono-urban residents. Fertility rate is high among the non-urban residents as compared to the urban residents. The table below represents the regression results dropping the variable, urban. By dropping the urban, we observe some significant changes in the coefficients.

However, all the coefficients remain to be significant at 5% significance