1.bert Morris, CFA, is evaluating the results of an estimation of the number of wireless phone minutes used on a quarterly basis within the territory of Car-tel International, Inc. Some of the information is presented below (in billions of minutes):
Wireless Phone Minutes (WPM)t = bo + b1 WPMt-1 + ε t
ANOVA | Degrees of Freedom | Sum of Squares | Mean Square |
Regression | 1 | 7,212.641 | 7,212.641 |
Error | 26 | 3,102.410 | 119.324 |
Total | 27 | 10,315.051 | |
Coefficients | Coefficient | Standard Error of the Coefficient |
Intercept | -8.0237 | 2.9023 |
WPM -1 | 1.0926 | 0.0673 |
The variance of the residuals from one time period within the time series is not dependent on the variance of the residuals in another.
The value for WPM this period is 544 billion. Using the results of the model, the forecast for three periods in the future is:
A) 586.35.
B) 594.37.
C) 691.30.
D) 683.18.
2. the time series of WPM covariance stationary?
A) Yes, because the computed t-statistic for a slope of 1 is not significant.
B) No, because the computed t-statistic for a slope of 1 is not significant.
C) Yes, because the computed t-statistic for a slope of 1 is significant.
D) No, because the computed t-statistic for a slope of 1 is significant.
3.e above model was specified as a(n):
A) Autoregressive (AR) Model.
B) Moving Average (MA) Model.
C) Autoregressive (AR) Model with a seasonal lag.
D) Moving Average (MA) Model with a seasonal lag.
4.sed upon the information provided, Morris would get more meaningful statistical results by:
A) adding more lags to the model.
B) forcing the regression through the origin, i.e., force bo = 0.
C) first differencing the data.
D) doing nothing. No information provided suggests that any of these will improve the specification.
5.nsider the estimated model xt = -6.0 + 1.1 xt-1 + 0.3 xt-2 + εt that is estimated over 50 periods. The value of the time series for the 49th observation is 20 and the value of the time series for the 50th observation is 22. What is the forecast for the 52nd observation?
A) 24.2.
B) 27.22.
C) 23.
D) 42.
6.nsider the estimated model xt = -6.0 + 1.1 xt-1 + 0.3 xt-2 + εt that is estimated over 50 periods. The value of the time series for the 49th observation is 20 and the value of the time series for the 50th observation is 22. What is the forecast for the 51st observation?
A) 30.2.
B) 24.2.
C) 23.
D) 42.
7.oy Dillard, CFA, has estimated the following equation using semiannual data: xt = 44 + 0.1×xt–1 – 0.25×xt–2 - 0.15×xt–3 + et. Given the data in the table below, what is Dillard’s best forecast of the second half of 2007?
Time | Value |
2003: I | 31 |
2003: II | 31 |
2004: I | 33 |
2004: II | 33 |
2005: I | 36 |
2005: II | 35 |
2006: I | 32 |
2006: II | 33 |
A) 34.05.
B) 60.55.
C) 34.36.
D) 35.00.
8.oy Dillard, CFA, has estimated the following equation using quarterly data: xt = 93 - 0.5×xt–1 + 0.1×xt–4 + et. Given the data in the table below, what is Dillard's best estimate of the first quarter of 2007?
Time | Value |
2005: I | 62 |
2005: II | 62 |
2005: III | 66 |
2005: IV | 66 |
2006: I | 72 |
2006: II | 70 |
2006: III | 64 |
2006: IV | 66 |
A) 66.60.
B) 67.20.
C) 66.40.
D) 64.60.
The variance of the residuals from one time period within the time series is not dependent on the variance of the residuals in another.
The value for WPM this period is 544 billion. Using the results of the model, the forecast for three periods in the future is:
A) 586.35.
B) 594.37.
C) 691.30.
D) 683.18.
The correct answer was D)
The one-period forecast is –8.023 + 1.0926*544 = 586.35.
The two-period forecast is then –8.023 + 1.0926*586.35 = 632.62.
Finally, the three-period forecast is then –8.023 + 1.0926*632.62 = 683.18.
2. the time series of WPM covariance stationary?
A) Yes, because the computed t-statistic for a slope of 1 is not significant.
B) No, because the computed t-statistic for a slope of 1 is not significant.
C) Yes, because the computed t-statistic for a slope of 1 is significant.
D) No, because the computed t-statistic for a slope of 1 is significant.
The correct answer was B)
The t-statistic for the test of the slope equal to 1 is computed by subtracting 1.0 from the coefficient ((1.0926 – 1.0) / 0.0673 =) 1.3759, which is not significant at the 5 percent level. The time series has a unit root and is not covariance stationary.
3.e above model was specified as a(n):
A) Autoregressive (AR) Model.
B) Moving Average (MA) Model.
C) Autoregressive (AR) Model with a seasonal lag.
D) Moving Average (MA) Model with a seasonal lag.
The correct answer was A)
The model is specified as an AR Model, but there is no seasonal lag. No moving averages are employed in the estimation of the model.
4.sed upon the information provided, Morris would get more meaningful statistical results by:
A) adding more lags to the model.
B) forcing the regression through the origin, i.e., force bo = 0.
C) first differencing the data.
D) doing nothing. No information provided suggests that any of these will improve the specification.
The correct answer was C)
Since the coefficient on the slope coefficient is greater than one, the process is not covariance stationary. A common technique to correct for this is to first difference the variable to perform the following regression: Δ(WPM)t = bo + b1 Δ(WPM)t-1 + ε t.
5.nsider the estimated model xt = -6.0 + 1.1 xt-1 + 0.3 xt-2 + εt that is estimated over 50 periods. The value of the time series for the 49th observation is 20 and the value of the time series for the 50th observation is 22. What is the forecast for the 52nd observation?
A) 24.2.
B) 27.22.
C) 23.
D) 42.
The correct answer was B)
Using the chain-rule of forecasting,
Forecasted x51 = -6.0 + 1.1 (22) + 0.3 (20) = 24.2.
Forecasted x52 = -6.0 + 1.1 (24.2) + 0.3 (22) = 27.22.
6.nsider the estimated model xt = -6.0 + 1.1 xt-1 + 0.3 xt-2 + εt that is estimated over 50 periods. The value of the time series for the 49th observation is 20 and the value of the time series for the 50th observation is 22. What is the forecast for the 51st observation?
A) 30.2.
B) 24.2.
C) 23.
D) 42.
The correct answer was B)
Forecasted x51 = -6.0 + 1.1 (22) + 0.3 (20) = 24.2.
7.oy Dillard, CFA, has estimated the following equation using semiannual data: xt = 44 + 0.1×xt–1 – 0.25×xt–2 - 0.15×xt–3 + et. Given the data in the table below, what is Dillard’s best forecast of the second half of 2007?
Time | Value |
2003: I | 31 |
2003: II | 31 |
2004: I | 33 |
2004: II | 33 |
2005: I | 36 |
2005: II | 35 |
2006: I | 32 |
2006: II | 33 |
A) 34.05.
B) 60.55.
C) 34.36.
D) 35.00.
The correct answer was C)
To get the answer, Daily must first make the forecast for 2007:I
E[x2007:I]= 44 + 0.1 × xt–1 - 0.25 × xt–2 - 0.15 × xt–3
E[x2007:I] = 44 + 0.1×33 - 0.25×32 - 0.15×35
E[x2007:I] = 34.05
Then, use this forecast in the equation for the first lag:
E[x2007:II] = 44 + 0.1×34.05 - 0.25×33 - 0.15×32
E[x2007:II] = 34.36
8.oy Dillard, CFA, has estimated the following equation using quarterly data: xt = 93 - 0.5×xt–1 + 0.1×xt–4 + et. Given the data in the table below, what is Dillard's best estimate of the first quarter of 2007?
Time | Value |
2005: I | 62 |
2005: II | 62 |
2005: III | 66 |
2005: IV | 66 |
2006: I | 72 |
2006: II | 70 |
2006: III | 64 |
2006: IV | 66 |
A) 66.60.
B) 67.20.
C) 66.40.
D) 64.60.
The correct answer was B)
To get the answer, Dillard will use the data for 2006: IV and 2006: I, xt–1 = 66 and xt–4 = 72 respectively:
E[x2007:I] = 93– 0.5×xt–2 + 0.1×xt–4
E[x2007:I] = 93– 0.5×66 + 0.1×72
E[x2007:I] = 67.20
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