Model: OLS, using observations 1953-2004 (T = 52)
Dependent variable: PerCap
Coefficient |
Std. Error |
t-ratio |
p-value |
||
const |
1181.58 |
313.899 |
3.7642 |
0.0005 |
*** |
Inc |
0.218651 |
0.0358441 |
6.1001 |
<0.0001 |
*** |
Gp |
−7.87342 |
1.57167 |
−5.0096 |
<0.0001 |
*** |
Pnc |
1.26823 |
2.36958 |
0.5352 |
0.5951 |
|
Ps |
−18.0738 |
1.58039 |
−11.4363 |
<0.0001 |
*** |
T |
75.298 |
12.1329 |
6.2061 |
<0.0001 |
*** |
Mean dependent var |
4935.619 |
S.D. dependent var |
1059.105 |
||
Sum squared resid |
569458.3 |
S.E. of regression |
111.2633 |
||
R-squared |
0.990046 |
Adjusted R-squared |
0.988964 |
||
F(5, 46) |
915.0172 |
P-value(F) |
7.97e-45 |
||
Log-likelihood |
−315.6159 |
Akaike criterion |
643.2319 |
||
Schwarz criterion |
654.9393 |
Hannan-Quinn |
647.7202 |
||
rho |
0.692419 |
Durbin-Watson |
0.615673 |
Therefore, PerCapt = 1181.58 + 0.218651Inct – 7.87342Gpt + 1.26823Pnct – 18.0738Ps + 75.298T +?t
Per household gas consumption is positively related to per capital disposable income, inversely related to the price index for gasoline, positively related to the price index of new cars, negatively associated to the aggregate price index of the consumer services, and positively related to time trend. When the price index for gasoline increases, per household gas consumption would decrease and vice versa.
In the same way, when price for new cars rises, the household consumption will also increase, and vice versa. The rise in the aggregate price for consumer services would also decrease household gas consumption, and vice versa. The estimated signs of these coefficients do agree with my expectations. For instance, since cars are a luxury commodity, the price index for new cars would be expected to be inversely related to household gas consumption.
From the scatter graph above, it is clear that as per capita disposable income (Inc) increases, per household gas consumption among consumer increases, because now consumers have huge disposable incomes to spend on gas. However, any decrease in the per capital disposal increase would decrease household gas consumption.
Using the critical value method to test individual significance of the coefficients of Inc and Gp at 5% significance level
The coefficient Inc is 0.218651
Therefore,
lb = Coefficient (Inc) – Critical value of t = 0.218651 – 2.0129 = -1.79425
ub = Coefficient (Inc) + Critical value of t = 0.218651 + 2.0129 = 2.23155
Thus,
-1.79425≤ t ≤2.23155
The t-test for the regression coefficient is statistically significant, since it lies within the range.
Per Capital Disposable Income (Inc) Test
The coefficient Inc is −7.87342
Therefore,
lb = Coefficient (Inc) – Critical value of t = −7.87342 – 2.0129 = -9.88632
ub = Coefficient (Inc) + Critical value of t = −7.87342 + 2.0129 = -5.86052
-9.88632≤ t ≤-5.86052
The t-test for the regression coefficient is statistically significant, since it lies within the range.
Gas price elasticity of per household gas consumption (PerCap) and income elasticity of per household gas consumption (PerCap) at the sample mean values
Price elasticity of per capita gas consumption
Mean Median Minimum Maximum
Inc 16805. 16692. 8685.0 27113.
Gp 51.343 47.502 16.668 123.90
Pnc 87.567 78.800 44.800 141.70
Ps 89.777 64.150 19.400 222.80
T 26.500 26.500 1.0000 52.000
The price elasticity is computed using the following formula
PE = (ΔQuantity/ΔPprice) * (Price/Qquantity)
The (ΔQuantity/ΔPrice) is presented in the above equation by the coefficient – 7.87342.
To determine (Price/Quantity), the mean of price (51.343) is used as well as the mean of Ps which is 89.777.
Therefore we have Price elasticity = – 7.87342 * 51.343/89.777= -4.50277
This means that an increase in the price by 1 unit will decrease quantity demanded by 4.50277 units.
Income elasticity of per capita gas consumption
Income elasticity = (ΔQuantity/ΔPprice) * (Price/Qquantity)
The (ΔQuantity/ΔPrice) is presented in the above equation by the coefficient – 7.87342.
To determine (Price/Quantity), the mean of price (51.343) is used as well as the mean of Ps which is 87.567.
Therefore we have PE = – 7.87342 * 51.343/87.567= -4.616
This means that a decrease in the income by 1 unit will decrease quantity demanded by 4.616 units.
The null hypothesis here is represented as β4 – 0 against its alternative, which is positive; thus meaning that β4 >0. Therefore, the test statistic is given as follows:
t= (b4−0)/se (b4) ∼ t46
Provided that β4 – 0 (and that the null hypothesis is true), we use the confidence alpha= 0.05; thus, making the critical value for the one sided alternative (β4>0) equal to 1.686.
From here, the decision rule is not accepting H0 in favor of the H1 if the computed value of t-statistic is within the rejection region of the test; especially, if it is greater than 1.686.
Model 4: OLS, using observations 1953-2004 (T = 52)
Dependent variable: PerCap
Coefficient |
Std. Error |
t-ratio |
p-value |
||
const |
1181.58 |
313.899 |
3.7642 |
0.0005 |
*** |
Inc |
0.218651 |
0.0358441 |
6.1001 |
<0.0001 |
*** |
Gp |
−7.87342 |
1.57167 |
−5.0096 |
<0.0001 |
*** |
Pnc |
1.26823 |
2.36958 |
0.5352 |
0.5951 |
|
Ps |
−18.0738 |
1.58039 |
−11.4363 |
<0.0001 |
*** |
T |
75.298 |
12.1329 |
6.2061 |
<0.0001 |
*** |
Mean dependent var |
4935.619 |
S.D. dependent var |
1059.105 |
||
Sum squared resid |
569458.3 |
S.E. of regression |
111.2633 |
||
R-squared |
0.990046 |
Adjusted R-squared |
0.988964 |
||
F(5, 46) |
915.0172 |
P-value(F) |
7.97e-45 |
||
Log-likelihood |
−315.6159 |
Akaike criterion |
643.2319 |
||
Schwarz criterion |
654.9393 |
Hannan-Quinn |
647.7202 |
||
rho |
0.692419 |
Durbin-Watson |
0.615673 |
The calculation is given as:
t- (b4−0)/se (b4) – (-18.0738−0)/1.58039 – (-11.4363)
The critical value of t is 2.0129, substituting
= 2.0129 – (-18.0738−0)/ 1.58039- (-18.0738−0)/1.58039 – (-11.4363)
= 2.0129 -6.637509 /13.01669
= 2.0129 -0.509923 = 1.502977
Therefore, given that this value is within the rejection area, it then means that there is sufficient evidence at the 5 percent level of significance to induce that the null hypothesis is not correct; therefore, the null hypothesis is not accepted at the 5% level of significance.
(Log-Log Model)
Ln (PerCapt) = α1 + α2 ln (Inct) + α3 ln (Gpt) + α4 Pnct + α5 Pst + ϒT + wt
Model 5: OLS, using observations 1953-2004 (T = 52)
Dependent variable: l_PerCap
Coefficient |
Std. Error |
t-ratio |
p-value |
||
const |
3.1243 |
0.984099 |
3.1748 |
0.0027 |
*** |
l_Inc |
0.563582 |
0.109962 |
5.1252 |
<0.0001 |
*** |
l_Gp |
−0.0851467 |
0.0167173 |
−5.0933 |
<0.0001 |
*** |
Pnc |
−0.000164518 |
0.000482611 |
−0.3409 |
0.7347 |
|
Ps |
−0.00364812 |
0.000326725 |
−11.1657 |
<0.0001 |
*** |
T |
0.0212205 |
0.00334601 |
6.3420 |
<0.0001 |
*** |
Mean dependent var |
8.478230 |
S.D. dependent var |
0.238811 |
||
Sum squared resid |
0.022347 |
S.E. of regression |
0.022041 |
||
R-squared |
0.992317 |
Adjusted R-squared |
0.991482 |
||
F(5, 46) |
1188.246 |
P-value(F) |
2.07e-47 |
||
Log-likelihood |
127.7756 |
Akaike criterion |
−243.5511 |
||
Schwarz criterion |
−231.8437 |
Hannan-Quinn |
−239.0628 |
||
rho |
0.745304 |
Durbin-Watson |
0.492542 |
Ln (PerCapt) = 3.1243 + 0.563582 ln (Inct) -0.0851467 ln (Gpt) – 0.000164518Pnct – 0.00364812 Pst + 0.0212205T + wt
α1 = A value of 3.1243 represents the unconditional expected mean of log PerCap.
α2 = A 100 percent change in Inc leads to a 100*0.563582 percentage point change in the probability that PerCab occurs.
α3 = A 100 percent change in Inc leads to a 100*0.0851467 percentage point change in the probability that PerCab occurs
α4 = the movement of 0.000164518 from 0 to 1 produces a 100*0.000164518 percent point change in the probability that PerCab occurs
α5 = the movement of 0.00364812 from 0 to 1 produces a 100*0.00364812 percent point change in the probability that PerCab occurs
To calculate the 99% confidence interval for ln (Gpt), I will use the following parameters;
Number of predictors = 5
Regression coefficient of ln (Gpt) = -0.0851
Sample size = 52
Standard error of ln (Gpt) = 0.01671
Therefore,
99% confidence interval for ln (Gpt): -0.13000 ≤ α ≤ -0.04020
This part concerned about testing the overall significance of α3. As always performed, I will go ahead and test to establish if the slope of the parameter is different from zero. If it is not, then the variable related to it could not belong in the model.
Thus, the null hypothesis is H0: α3 – 0 and the alternative hypothesis is H1: α3≠0. Then the statistic is given as t = (α3−0)/se (α3) ∼t46. Moreover, if the null hypothesis is indeed true, and this is computed as t= (α3−0)/se (α3) = (−0.0851467−0)/ 0.0167173= -5.093. Furthermore, the two-sided, alpha= 0.05 critical value is 2.013 and therefore, -5.093 is within the 5% rejection region of this test.
Given the following values:
Then,
PerCapt = 3.1243 + 0.563582 ($28000) -0.0851467 (135) – 0.000164518 (130) – 0.00364812 (230) + 0.0212205 (53)
PerCapt = $15772.19
The critical value at 95% confidence interval is 0.0630498
Critical evaluation and comparison the two models to establish which model is best and why.
First model:
PerCapt = 1181.58 + 0.218651Inct – 7.87342Gpt + 1.26823Pnct – 18.0738Ps + 75.298T +?t
Second model:
Ln (PerCapt) = 3.1243 + 0.563582 ln (Inct) -0.0851467 ln (Gpt) – 0.000164518Pnct – 0.00364812 Pst + 0.0212205T + wt
A linear equation such as the first model, which specifies a linear relationship among the its variables provides a good description of the economic behaviors compared to a nonlinear model. An alternative approach is considering a linear relationship among log-transformed variables, which is referred to as log-log model – In this model, as seen from above, the dependent variable and some of the explanatory variables are transformed to logarithms.
Sometimes a variable can be logged because it has a long tail (or it is skewed to the right) whereas its logged value may look like it is normally distributed.
Logged variables are not a good move, since there is no econometric reason to make a variable appear normal.
In essence, the only requirement for OLS regression is that the residuals always appear to be normal. In class, we examined plots of residuals as well as log residuals to try to figure which ones looked more normal. And the OLS regression model had looked more normal compared to the log model. The same case is for the above models.
Besides, taking logs also affect the economic interpretations of the variables. Therefore, liner regression is the best, since it does not alter the economic interpretations of the economic variables as compared to log-log model.
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