Econometrics Method — Four Questions will be attached as pdf format. Please complete Exercises: Q5.1, Q5.2, Q6.1 an Q6.2. Essay Example
(i) The t-statistic is obtained as follows;
is obtained as follows;2(ii) Standard error for b
is obtained as follows;3(iii) The estimate for b
is obtained as follows;2(iv) The estimate R
n = sample size = 1519
k = the number of coefficients to be estimated in the regression equation including the y-intercept = 4
(iv) The estimate of σ (SE of regression) is obtained as follows;
is the first regression coefficient which is positive. This Implies that the relationship between a households’ budget spent on alcohol and the total expenditure (tot exp) is positive. This means that as the total expenditure increases, the household budget spent on alcohol increases.2(b) The estimate b
is the second regression coefficient which is negative. This implies a negative linear relationship between the households’ budget spent on alcohol and age. Therefore, as age increases, the budget on alcohol decreases significantly.3The estimate b
is the third regression coefficient which is negative. This implies a negative linear relationship between the households’ budget spent on alcohol and the number of children in the household (nk). This means that as the number of children in a household increases, the household budget on alcohol decreases.4The estimate b
is computed as follows;3(c) A 95% confidence interval for b
n > 30, t = 1.96 (for a two sided test) = -0.001448, since3 b
))b3+ (1.96×SE3 b), b3– (1.96×SE3 bThe 95% CI = (
The 95% CI = (-0.001448-(1.96×0.000208), -0.001448 + (1.96×0.000208))
= (-0.001856, -0.001040)
= 0 at the 0.05 level of significance4: b0(d) Testing for H
The prob = 0.0000, which is less than 0.05. Therefore, the null hypothesis is rejected. This implies that the number of children is highly significant in determining the budget proportion for alcohol.
It is highly likely that the budget allocations are determined by the number of dependents in a household; hence the budget proportion for alcohol decreases as the number of children increases.
iWTRANS = -0.031466 + 0.041383LTOTEXP – 0.000058AGE – 0.012965NK + е
(0.032186) (0.007067) (0.000351) (0.005507)
is negative, thus indicating a negative linear relationship between the proportion of household budget on transportation and the number of children.3, the second regression coefficient, is negative. This implies a negative linear relationship between the proportion of household budget spent on transportation and age. Similarly, b3, the first regression coefficient, is positive, which indicates a positive linear relationship between the proportion of household budget spent on transportation and the log of total expenditure. b2(b) b
From both the economic and logical points of view, these results do not make sense, hence they are not tenable.
(c) The variable that might be excluded from the regression equation is ‘AGE’ and ‘NK’. The relationship between age, number of children and the household budget allocation on transportation might be non-linear.
= 0.024273, hence the proportion of variation explained in the budget proportion allocation to transportation is 2.43%.2(d) R
(e) Using the equation;
WTRANS = -0.031466 + 0.041383LTOTEXP – 0.000058AGE – 0.012965NK, where LTOTEXP = 98.7, AGE = 36, NK = 1 and 2
For NK = 1; WTRANS = -0.031466+0.041383(98.7)-0.000058(36)-0.012965(1) = 4.037983
For NK = 2; WTRANS = -0.031466+0.041383(98.7)-0.000058(36)-0.012965(2) = 4.025018
Residual Sum of Squares + Regression Sum of squares = Total Sum of squares
(b) The F-statistic is given by the formula;
Where; n = sample size, k = number of coefficients including the intercept
Prob (F-statistic) = F0.05 (2, 37) = 3.23
In testing H0: b2 = b3 = 0, the value of the F-Statistic (computed) > F0.05 (DF) (tabulated)
Therefore, we reject the null hypothesis, H0.
Q6.2 Using RESET to test for misspecification
(Mayo and Spanos, 2004)
-Squared is the original coefficient of determination0R
-Squared is the new coefficient of determination1R
n = sample size
k = number of coefficients in new model
= 0 against an alternative NOT2 =b1: b0In testing H
Now, since the computed value (154.889466) is greater than the F-Statistic (79.90759), the null hypothesis is rejected. This implies that there is misspecification in the model at the 0.05 level of significance.
, 71, pp. 1007-1025.Philosophy and ScienceMayo, D. G. and Spanos, A., 2004. Methodology in Practice: Statistical Misspecification Testing.
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