Consider the following estimation from Microfit:
Ordinary Least Squares Estimation
- Dependent variable is LC
- (49 observations) used for estimation from (1950-1998):
- LC: Log of real customers expenditure in 1995 prices.
- LY: Log of real personal disposable income in 1995 prices.
- LP: Log of the consumer price index.
a) Briefly explain what the results tell us about the determination of consumption.
From the observations and date, this data is time series. Equation before estimation: LC = a + B1 LC (t-1) + B2 LC (t-2) + B3 LYt + B4 LY (t-1) + B5 LY (t-2) + B6 LP t + B7 LP (t-1) + B8 LP (t-2) + error term t. The equation has (t-2) as the two years before this years consumption can affect the result.
There is a relationship as R squared is close to one. In statistics, the coefficient of determination, denoted R² or r² and pronounced “R squared”, is the proportion of the variance in the dependent variable that is predictable from the independent variable.
Steps:
- Validity Check
| Sig. | |||
| 1% | ? | < | 0.01 |
| 5% | 0.01 | < | 0.05 |
| 10% | 0.05 | < | 0.10 |
First, lets take a 5% significance level, therefore we can see that all variables except LP (-1) and LP (-2) are insignificant.
- Interpret
-Sign: Second, the signs are all positive except for LC (-2), LY (-1), LP, and LP (-2).
-Size: If disposable income grows by one, consumer expenditure grows by .53920 or 53%, variable significantly significant. The graph would have consumer expenditure on the Y axis and Disposable income on the X axis.
The F-test represents the overall validity of the model:
H0 = B1 LC (t-1) = B2 LC (t-2) = B3 LYt = B4 LY (t-1) = B5 LY (t-2) = B6 LP t = B7 LP (t-1) = B8 LP (t-2) = 0
H1: At least one coefficient is unequal to zero.
From the diagnostic tests table (good or bad table), we can see that the [Prob] of A is [.512] so we do not reject the null. As we do not reject the null hypothesis of f-test, none of the variables (coefficients) is statistically significant, so the model has no statistical validity. However, the [Prob] of D is [.020] so there is an error as we have not rejected the null when we should had.