The estimated regression equation is yt = 448 + 12t + 18 Qtr1 - 26 Qtr2 + 3 Qtr3. The regression model has three quarterly binaries. The model was fitted to 12 periods of quarterly data starting with the first quarter. Why is there no fourth quarterly binary for Qtr4? a. Because the fourth quarter binary is assumed to be the same as the first quarter. b. Because it is unnecessary (its value is implied by the other three binaries). c. Because there is no seasonality in the fourth quarter in most time series. d. Because the researcher made a mistake (we need binaries for all four quarters).
Added by Bradley A.
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Step 1: The estimated regression equation is yt = 448 + 12t + 18 Qtr1 - 26 Qtr2 + 3 Qtr3, with three quarterly binaries included in the model. Show more…
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A model to forecast quarterly sales (in $100,000s) has been estimated as follows: yt = 0.3t + 2Q1 + 0.8Q2 + 5, where Q1 is the dummy variable for quarter 1, Q2 is the dummy variable for quarter 2, and Q3 is the dummy variable for quarter 3. When all three dummy variables are 0, we have quarter 4. Therefore, quarter 4 is the baseline. (a) Assuming that the time series used starts at the 1st quarter of 2000 (where t=1) and ends at the 4th quarter of 2012, what would the forecast for the first and third quarter of 2015 be? Show computation. (b) What can be inferred from the fact that this forecasting model contains no coefficients for Quarter 3?
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$\begin{array}{l}{\text { (i) Use NYSE to estimate equation }(12.48) . \text { Let } \hat{h}_{t} \text { be the fitted values from this equation (the esti-i- }} \\ {\text { mates of the conditional variance). How many } \hat{h}_{t} \text { are negative? }}\end{array}$ $\quad \text {(ii) Add } \text {return}_{t-1}^{2} \text { to }(12.48) \text { and again compute the fitted values, } \hat{h}_{t} \text { . Are any } \hat{h}_{t}$ $\begin{array}{l}{\text { (iii) Use the } \hat{h}_{t} \text { from part (ii) to estimate }(12.47) \text { by weighted least squares (as in Section } 8-4 ) \text { . }} \\ {\text { Compare your estimate of } \beta_{1} \text { with that in equation }(11.16) . \text { Test } \mathrm{H}_{0} : \beta_{1}=0 \text { and compare the }} \\ {\text { outcome when OLS is used. }}\end{array}$ $\begin{array}{l}{\text { (iv) Now, estimate }(12.47) \text { by WLS, using the estimated ARCH model in }(12.51) \text { to obtain the } \hat{h}_{t}} \\ {\text { Does this change your findings from part (iii)? }}\end{array}$
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