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$\begin{array}{l}{\text { (i) Using the data in WAGEPRC, estimate the distributed lag model from Problem } 5 \text { in Chapter } 11 .} \\ {\text { Use regression }(12.14) \text { to test for } A R(1) \text { serial correlation. }}\end{array}$$\begin{array}{l}{\text { (ii) Reestimate the model using iterated Cochrane- Orcutt estimation. What is your new estimate of }} \\ {\text { the long-run propensity? }}\end{array}$$\begin{array}{l}{\text { (iii) Using iterated CO, find the standard error for the LRP. (This requires you to estimate a modified }} \\ {\text { equation.) Determine whether the estimated LRP is statistically different from one at the } 5 \% \text { level. }}\end{array}$

(i) $\hat{\rho} \approx .503$ and $\mathrm{t}_{\rho} \approx 9.60$ (ii) LRP is about 1.110 (iii) $\mathrm{H}_{0} : \theta_{0}=1$ has $t-stat \approx .58$ so the LRP is not statistically different from 1 at 5$\%$ significance level.

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Chapter 12

Serial Correlation and Heteroskedasticity in Time Series Regressions

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hard one. After estimating the distributed leg model by all l s, we get the residuals U T hat and run the regression of u T had on its first leg. Using 272 observations, we get Road Nut as 0.53 and the T statistic for this road hat is about 9.6, which is a very strong evidence of positive. A Are one correlation are cute When we estimate the model by iterated c o algorithm, The long run propensity is estimated to be about 1.11 in Part three. We estimate this equation by iterated c o we have the growth of price as a function of the growth of wage in the same time period and other explanatory factors are the differences between all the legs of the growth of wish. With its contemporaneous value. We have 12 legs, so there are 12 additional differences. The error term U T is assumed to follow the er one process. The long run propensity is the estimate of they tell one the coefficient of the growth of wish Contemporary ISS. We get fate a zero hat. Yeah, roughly one point 11 and it's that his arrows, it's standard error is 0.191 We will test. They now have policies. They dare not equals one. The T statistic is they cannot add minus one, divided by the standard error of fate or not at. So we get 1.11 minus one, divided by point one night one. And the T statistic is about 0.58 which is very small, and we are unable to reject the null hypothesis at the 5% level. In other words, the long run propensity. It's not thank you, statistically different from one.

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