Time Series Regression

Question 1

A regression model is obtained as follows: Ŷ’_t = -2.31 + 2.81X’_t with a Durbin-Watson statistic = 1.74. The model consists of 20 time series’ being evaluated.

If the generalized differences are denoted by Y’_t and X’_t , what is the new statistical conclusion based on this newly obtained DW statistic at α = 0.05?

Answer 1

Using the Durbin-Watson statistical table, we can discern that, using the following inputs:

**k = 1
n = 20
α = 0.05**

The DW statistical value we’re interested in is d_U = 1.411.

Because our DW test statistic is = 1.74, we do not reject the null hypothesis of:
H₀: ρ = 0.

1.74 > d_U (1.411).

The autocorrelation in this model, is not strong enough to have an effect on the least squares estimate of the slope coefficient.

Question 2

You are provided an estimated value of: p̂ = 0.585 for the first time lag autocorrelation from an ACF plot.

You decide to run a new regression based on the generalized difference,

What equations should you use for calculating the generalized difference for Y_t and X_t?

Answer 2

The generalized differences for Y_t and X_t are as follows:

Y’_t = Y_t - 0.585Y_t-1

X’_{t =} X_t - 0.585X_t-1

Question 3

Use the statistical output below.

If there were 20 time series’ being evaluated (n = 20), based on the Durbin-Watson statistic and a significance level of α = 0.05, what conclusion can you make in terms of the serial correlation of residuals?

Model Coefficient Results

Predictor   Coef   SE Coef   T    P
Constant   -6.4011   0.8435   -7.59 0.000
Industry   2.83585 0.02284   124.14 0.000

S = 0.319059   R² = 99.9%   R² (adj) = 99.9%

Durbin-Watson statistic = 0.8237

Answer 3

The following values are needed to apply to the Durbin-Watson statistical tables:

**k = 1
n = 20
α = 0.05**

The Durbin-Watson statistic with these inputs, results in d_L = 1.201.

Since the DW test statistic in the coefficient results is 0.8237, we have evidence to reject the null hypothesis of: H₀ : ρ = 0 and support the alternative hypothesis of: H₁ : ρ > 0.

0.8237 < d_L (1.201).

Question 4

Using the statistical output below, what is the linear regression model?

What is the significance of the predictor at a 5% significance level?

Model Coefficient Results

Predictor   Coef   SE Coef   T    P
Constant   -6.4011   0.8435   -7.59 0.000
Industry   2.83585 0.02284   124.14 0.000

S = 0.319059   R² = 99.9%   R² (adj) = 99.9%

Durbin-Watson statistic = 0.8237

Answer 4

The linear regression model is:

E(Y<sub>t</sub>) = Constant Coef + Industry Coef x X<sub>t</sub>
E(Y<sub>t</sub>) = -6.4011 + 2.83585*X*<sub><em>t</em></sub>

Since the p-value for the predictor (Industry) in the output = 0, at a 5% significance level, it is very significant.

Answer 5

When the Durbin-Watson test statistic lies between the lower and upper bounds of of the critical values, the test results are deemed to be inconclusive.