上一主题:Reading 12: Multiple Regression and Issues in Regression Analy
下一主题:Quantitative Analysis 【Reading 12】Sample
返回列表 发帖
Which of the following statements regarding seasonality is least accurate?
A)
Not correcting for seasonality when, in fact, seasonality exists in the time series results in a violation of an assumption of linear regression.
B)
The presence of seasonality makes it impossible to forecast using a time-series model.
C)
A time series that is first differenced can be adjusted for seasonality by incorporating the first-differenced value for the previous year's corresponding period.



Forecasting is no different in the case of seasonal component in the time-series model than any other forecasting.

TOP

Which of the following is a seasonally adjusted model?
A)
Salest = b0 + b1 Sales t-1 + b2 Sales t-2 + εt.
B)
Salest = b1 Sales t-1+ εt.
C)
(Salest - Sales t-1)= b0 + b1 (Sales t-1 - Sales t-2) + b2 (Sales t-4 - Sales t-5) + εt.



This model is a seasonal AR with first differencing.

TOP

The table below shows the autocorrelations of the lagged residuals for the first differences of the natural logarithm of quarterly motorcycle sales that were fit to the AR(1) model: (ln salest − ln salest − 1) = b0 + b1(ln salest − 1 − ln salest − 2) + εt. The critical t-statistic at 5% significance is 2.0, which means that there is significant autocorrelation for the lag-4 residual, indicating the presence of seasonality. Assuming the time series is covariance stationary, which of the following models is most likely to CORRECT for this apparent seasonality?

Lagged Autocorrelations of First Differences in the Log of Motorcycle Sales

Lag

Autocorrelation

Standard Error

t-Statistic

1


−0.0738

0.1667

−0.44271

2


−0.1047

0.1667

−0.62807

3


−0.0252

0.1667

−0.15117

4


0.5528

0.1667

3.31614
A)
ln salest = b0 + b1(ln salest − 1) − b2(ln salest − 4) + εt.
B)
(ln salest − ln salest − 4) = b0 + b1(ln salest − 1 − ln salest − 2) + εt.
C)
(ln salest − ln salest − 1) = b0 + b1(ln salest − 1 − ln salest − 2) + b2(ln salest − 4 − ln salest − 5) + εt.



Seasonality is taken into account in an autoregressive model by adding a seasonal lag variable that corresponds to the seasonality. In the case of a first-differenced quarterly time series, the seasonal lag variable is the first difference for the fourth time period. Recognizing that the model is fit to the first differences of the natural logarithm of the time series, the seasonal adjustment variable is (ln salest − 4 − ln salest − 5).

TOP

Which of the following is least likely a consequence of a model containing ARCH(1) errors? The:
A)
model's specification can be corrected by adding an additional lag variable.
B)
variance of the errors can be predicted.
C)
regression parameters will be incorrect.



The presence of autoregressive conditional heteroskedasticity (ARCH) indicates that the variance of the error terms is not constant. This is a violation of the regression assumptions upon which time series models are based. The addition of another lag variable to a model is not a means for correcting for ARCH (1) errors.

TOP

Suppose you estimate the following model of residuals from an autoregressive model:
εt2 = 0.25 + 0.6ε2t-1 + µt, where ε = ε^

If the residual at time t is 0.9, the forecasted variance for time t+1 is:
A)
0.790.
B)
0.736.
C)
0.850.



The variance at t = t + 1 is 0.25 + [0.60 (0.9)2] = 0.25 + 0.486 = 0.736. See also, ARCH models.

TOP

The data below yields the following AR(1) specification: xt = 0.9 – 0.55xt-1 + Et , and the indicated fitted values and residuals.
Timextfitted valuesresiduals
11 --
2-10.35-1.35
321.450.55
4-1-0.2-0.8
501.45-1.45
620.91.1
70-0.20.2
810.90.1
920.351.65

The following sets of data are ordered from earliest to latest. To test for ARCH, the researcher should regress:
A)
(1.8225, 0.3025, 0.64, 2.1025, 1.21, 0.04, 0.01) on (0.3025, 0.64, 2.1025, 1.21, 0.04, 0.01, 2.7225).
B)
(-1.35, 0.55, -0.8, -1.45, 1.1, 0.2, 0.1, 1.65) on (0.35, 1.45, -0.2, 1.45, 0.9, -0.2, 0.9, 0.35)
C)
(1, 4, 1, 0, 4, 0, 1, 4) on (1, 1, 4, 1, 0, 4, 0, 1)



The test for ARCH is based on a regression of the squared residuals on their lagged values. The squared residuals are (1.8225, 0.3025, 0.64, 2.1025, 1.21, 0.04, 0.01, 2.7225). So, (1.8225, 0.3025, 0.64, 2.1025, 1.21, 0.04, 0.01) is regressed on (0.3025, 0.64, 2.1025, 1.21, 0.04, 0.01, 2.7225). If coefficient a1 in: is statistically different from zero, the time series exhibits ARCH(1).

TOP

One choice a researcher can use to test for nonstationarity is to use a:
A)
Breusch-Pagan test, which uses a modified t-statistic.
B)
Dickey-Fuller test, which uses a modified χ2 statistic.
C)
Dickey-Fuller test, which uses a modified t-statistic.



The Dickey-Fuller test estimates the equation (xt – xt-1) = b0 + (b1 - 1) * xt-1 + et and tests if H0: (b1 – 1) = 0. Using a modified t-test, if it is found that (b1–1) is not significantly different from zero, then it is concluded that b1 must be equal to 1.0 and the series has a unit root.

TOP

Consider the following estimated model:
(Salest - Sales t-1) = 30 + 1.25 (Sales t-1 - Sales t-2) + 1.1 (Sales t-4 - Sales t-5) t=1,2,.. T

and Sales for the periods 1999.1 through 2000.2:
tPeriodSales
T2000.2$2,000
T-12000.1$1,800
T-21999.4$1,500
T-31999.3$1,400
T-41999.2$1,900
T-51999.1$1,700

The forecasted Sales amount for 2000.3 is closest to:
A)
$2,625.
B)
$2,270.
C)
$1,730.



Note that since we are forecasting 2000.3, the numbering of the "t" column has changed.
Change in sales = $30 + 1.25 ($2,000-1,800) + 1.1 ($1,400-1,900)
Change in sales = $30 + 250 - 550 = -$270
Sales = $2,000 – 270 = $1,730

TOP

Consider the following estimated model:
(Salest - Sales t-1)= 100 - 1.5 (Sales t-1 - Sales t-2) + 1.2 (Sales t-4 - Sales t-5) t=1,2,.. T
and Sales for the periods 1999.1 through 2000.2:

tPeriodSales
T2000.2$1,000
T-12000.1$900
T-21999.4$1,200
T-31999.3$1,400
T-41999.2$1,000
T-51999.1$800

The forecasted Sales amount for 2000.3 is closest to:
A)
$1,730.
B)
$1,430.
C)
$730.



Change in sales = $100 - 1.5 ($1,000-900) + 1.2 ($1,400-1,000)
Change in sales = $100 - 150 + 480 =$430
Sales = $1,000 + 430 = $1,430

TOP

Bill Johnson, CFA, has prepared data concerning revenues from sales of winter clothing made by Polar Corporation for presentation in the following table (in $ millions):



Change In Sales

Lagged Change
In Sales

Seasonal Lagged
Change In Sales

Quarter

Sales

Y

Y + (−1)

Y + (−4)

2006.1

182




2006.2

74

−108



2006.3

78

4

−108


2006.4

242

164

4


2007.1

194

−48

164


2007.2

79

−115

−48

−108

2007.3

90

11

−115

4

2007.4

260

170

11

w


The preceding table will be used by Johnson to forecast values using:
A)
a log-linear trend model with a seasonal lag.
B)
a serially correlated model with a seasonal lag.
C)
an autoregressive model with a seasonal lag.



Johnson will use the table to forecast values using an autoregressive model for periods in succession since each following forecast relies on the forecast for the preceding period. The seasonal lag is introduced to account for seasonal variations in the observed data.

The value that Johnson should enter in the table in place of "w" is:
A)
−48.
B)
164.
C)
−115.



The seasonal lagged change in sales shows the change in sales from the period 4 quarters before the current period. Sales in the year 2006 quarter 4 increased $164 million over the prior period.

Assume that Johnson prepares a change in sales regression analysis model with seasonality which includes the following:

Coefficients

Intercept

−6.032

Lag 1

0.017

Lag 4

0.983

Based on the model, expected sales in the first quarter of 2008 will be closest to:

A)
155.
B)
190.
C)
210.



Substituting the 1-period lagged data from 2007.4 and the 4-period lagged data from 2007.1 into the model formula, change in sales is predicted to be −6.032 + (0.017 × 170) + (0.983 × −48) = −50.326. Expected sales are 260 + (−50.326) = 209.674.

TOP

返回列表
上一主题:Reading 12: Multiple Regression and Issues in Regression Analy
下一主题:Quantitative Analysis 【Reading 12】Sample