In: Economics
You have an Excel file attached to this link. You are to first transpose the data so that, instead of being in the horizontal format, it will be converted to the vertical form. Then to run a Multiple Regression and see which of the independent variables are significant and whether the overall model is significant. You also should be able to comment on the goodness of the fit. All of these require that you have thoroughly watched the video lectures on Multiple Regression first.
y | 1786 | 2146 | 2358 | 2218 | 2038 | 2456 | 1624 | 2386 | 1966 | 2562 | 2312 | 2064 | 1712 | 1956 | 2034 | 2182 | 2096 |
x1 | 27 | 32.8 | 36.6 | 34.8 | 31.4 | 37.6 | 26.1 | 36.2 | 41 | 37 | 36.2 | 32.3 | 44.1 | 28.4 | 28.8 | 40.6 | 40.3 |
x2 | 7 | 7.7 | 7.4 | 13.2 | 16 | 10 | 4.5 | 11.6 | 7.1 | 9.7 | 12.7 | 14.8 | 6.9 | 11.4 | 6.4 | 7.4 | 6.4 |
x3 | 284 | 509.04 | 692.96 | 602.24 | 448.56 | 746.16 | 255.11 | 672.24 | 942 | 714 | 672.24 | 486.99 | 1140.71 | 332.16 | 346.64 | 917.76 | 899.79 |
Consider the given problem here “Y” be the dependent variable and “X1, X2 and X3” are the independent variable. Consider the following table shows the regression result.
So, here the regression equation is given below.
=> Y = (-7876.42) + 457.33*X1 – 16.67*X2 – 9.18*X3.
Here the “p value” of X1 is close to zero implied X1 is a significant variable in explaining Y. Similarly, the “p value” of X2 is “0.1640 = 16%” which is more than 0.05, implied X2 is a not significant variable in explaining Y at 5 level of significance.
Finally, the “p value” of X3 is close to zero implied X3 is a significant variable in explaining Y. Now, the value of coefficient of determination is “0.84”, => the regression model can explain 84% variation in Y and 16% variation remain unexplained. So, here the 0.84 is quite higher, => the regression model is really a very good fitted model.