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# Quantitative Analysis【Reading 13】Sample

David Wellington, CFA, has estimated the following log-linear trend model: LN(xt) = b0 + b1t + εt. Using six years of quarterly observations, 2001:I to 2006:IV, Wellington gets the following estimated equation: LN(xt) = 1.4 + 0.02t. The first out-of-sample forecast of xt for 2007:I is closest to:
 A) 1.88.
 B) 4.14.
 C) 6.69.

Wellington’s out-of-sample forecast of LN(xt) is 1.9 = 1.4 + 0.02 × 25, and e1.9 = 6.69.

Modeling the trend in a time series of a variable that grows at a constant rate with continuous compounding is best done with:
 A) a log-linear transformation of the time series.
 B) a moving average model.
 C) simple linear regression.

The log-linear transformation of a series that grows at a constant rate with continuous compounding (exponential growth) will cause the transformed series to be linear.
In the time series model: yt=b0 + b1 t + εt, t=1,2,…,T, the:
 A) change in the dependent variable per time period is b1.
 B) disturbance terms are autocorrelated.
 C) disturbance term is mean-reverting.

The slope is the change in the dependent variable per unit of time. The intercept is the estimate of the value of the dependent variable before the time series begins. The disturbance term should be independent and identically distributed. There is no reason to expect the disturbance term to be mean-reverting, and if the residuals are autocorrelated, the research should correct for that problem.
Yolanda Seerveld is an analyst studying the growth of sales of a new restaurant chain called Very Vegan. The increase in the public’s awareness of healthful eating habits has had a very positive effect on Very Vegan’s business. Seerveld has gathered quarterly data for the restaurant’s sales for the past three years. Over the twelve periods, sales grew from \$17.2 million in the first quarter to \$106.3 million in the last quarter. Because Very Vegan has experienced growth of more than 500% over the three years, the Seerveld suspects an exponential growth model may be more appropriate than a simple linear trend model. However, she begins by estimating the simple linear trend model:
(sales)t = α + β × (Trend)t + εt
Where the Trend is 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12.
 Regression Statistics Multiple R 0.952640 R2 0.907523 Adjusted R2 0.898275 Standard Error 8.135514 Observations 12 1st order autocorrelation coefficient of the residuals: −0.075

 ANOVA df SS Regression 1 6495.203 Residual 10 661.8659 Total 11 7157.069

 Coefficients Standard Error Intercept 10.0015 5.0071 Trend 6.7400 0.6803

The analyst then estimates the following model:
(natural logarithm of sales)t = α + β × (Trend)t + εt
 Regression Statistics Multiple R 0.952028 R2 0.906357 Adjusted R2 0.896992 Standard Error 0.166686 Observations 12 1st order autocorrelation coefficient of the residuals: −0.348

 ANOVA df SS Regression 1 2.6892 Residual 10 0.2778 Total 11 2.9670

 Coefficients Standard Error Intercept 2.9803 0.1026 Trend 0.1371 0.0140

Seerveld compares the results based upon the output statistics and conducts two-tailed tests at a 5% level of significance. One concern is the possible problem of autocorrelation, and Seerveld makes an assessment based upon the first-order autocorrelation coefficient of the residuals that is listed in each set of output. Another concern is the stationarity of the data. Finally, the analyst composes a forecast based on each equation for the quarter following the end of the sample. Are either of the slope coefficients statistically significant?
 A) Yes, both are significant.
 B) The simple trend regression is, but not the log-linear trend regression.
 C) The simple trend regression is not, but the log-linear trend regression is.

The respective t-statistics are 6.7400 / 0.6803 = 9.9074 and 0.1371 / 0.0140 = 9.7929. For 10 degrees of freedom, the critical t-value for a two-tailed test at a 5% level of significance is 2.228, so both slope coefficients are statistically significant. (Study Session 3, LOS 13.a)

Based upon the output, which equation explains the cause for variation of Very Vegan’s sales over the sample period?
 A) Both the simple linear trend and the log-linear trend have equal explanatory power.
 B) The simple linear trend.
 C) The cause cannot be determined using the given information.

To actually determine the explanatory power for sales itself, fitted values for the log-linear trend would have to be determined and then compared to the original data. The given information does not allow for such a comparison. (Study Session 3, LOS 13.b)

With respect to the possible problems of autocorrelation and nonstationarity, using the log-linear transformation appears to have:
 A) improved the results for autocorrelation but not nonstationarity.
 B) improved the results for nonstationarity but not autocorrelation.
 C) not improved the results for either possible problems.

The fact that there is a significant trend for both equations indicates that the data is not stationary in either case. As for autocorrelation, the analyst really cannot test it using the Durbin-Watson test because there are fewer than 15 observations, which is the lower limit of the DW table. Looking at the first-order autocorrelation coefficient, however, we see that it increased (in absolute value terms) for the log-linear equation. If anything, therefore, the problem became more severe. (Study Session 3, LOS 13.b)

The primary limitation of both models is that:
 A) each uses only one explanatory variable.
 B) the results are difficult to interpret.
 C) regression is not appropriate for estimating the relationship.

The main problem with a trend model is that it uses only one variable so the underlying dynamics are really not adequately addressed. A strength of the models is that the results are easy to interpret. The levels of many economic variables such as the sales of a firm, prices, and gross domestic product (GDP) have a significant time trend, and a regression is an appropriate tool for measuring that trend. (Study Session 3, LOS 13.b)

Using the simple linear trend model, the forecast of sales for Very Vegan for the first out-of-sample period is:
 A) \$97.6 million.
 B) \$113.0 million.
 C) \$123.0 million.

The forecast is 10.0015 + (13 × 6.7400) = 97.62. (Study Session 3, LOS 13.a)

Using the log-linear trend model, the forecast of sales for Very Vegan for the first out-of-sample period is:
 A) \$117.0 million.
 B) \$121.2 million.
 C) \$109.4 million.

The forecast is e2.9803 + (13 × 0.1371) = 117.01. (Study Session 3, LOS 13.a)
Clara Holmes, CFA, is attempting to model the importation of an herbal tea into the United States which last year was \$ 54 million. She gathers 24 years of annual data, which is in millions of inflation-adjusted dollars.
She computes the following equation:

(Tea Imports)t = 3.8836 + 0.9288 × (Tea Imports)t − 1 + et
 t-statistics (0.9328) (9.0025)

R2 = 0.7942
Adj. R2 = 0.7844
SE = 3.0892
N = 23

Holmes and her colleague, John Briars, CFA, discuss the implication of the model and how they might improve it. Holmes is fairly satisfied with the results because, as she says “the model explains 78.44 percent of the variation in the dependent variable.” Briars says the model actually explains more than that.
Briars asks about the Durbin-Watson statistic. Holmes said that she did not compute it, so Briars reruns the model and computes its value to be 2.1073. Briars says “now we know serial correlation is not a problem.” Holmes counters by saying “rerunning the model and computing the Durbin-Watson statistic was unnecessary because serial correlation is never a problem in this type of time-series model.”
Briars and Holmes decide to ask their company’s statistician about the consequences of serial correlation. Based on what Briars and Holmes tell the statistician, the statistician informs them that serial correlation will only affect the standard errors and the coefficients are still unbiased. The statistician suggests that they employ the Hansen method, which corrects the standard errors for both serial correlation and heteroskedasticity.
Given the information from the statistician, Briars and Holmes decide to use the estimated coefficients to make some inferences. Holmes says the results do not look good for the future of tea imports because the coefficient on (Tea Import)t − 1 is less than one. This means the process is mean reverting. Using the coefficients in the output, says Holmes, “we know that whenever tea imports are higher than 41.810, the next year they will tend to fall. Whenever the tea imports are less than 41.810, then they will tend to rise in the following year.” Briars agrees with the general assertion that the results suggest that imports will not grow in the long run and tend to revert to a long-run mean, but he says the actual long-run mean is 54.545. Briars then computes the forecast of imports three years into the future.With respect to the statements made by Holmes and Briars concerning serial correlation and the importance of the Durbin-Watson statistic:
 A) they were both incorrect.
 B) Holmes was correct and Briars was incorrect.
 C) Briars was correct and Holmes was incorrect.

Briars was incorrect because the DW statistic is not appropriate for testing serial correlation in an autoregressive model of this sort. Holmes was incorrect because serial correlation can certainly be a problem in such a model. They need to analyze the residuals and compute autocorrelation coefficients of the residuals to better determine if serial correlation is a problem. (Study Session 3, LOS 12.i)

With respect to the statement that the company’s statistician made concerning the consequences of serial correlation, assuming the company’s statistician is competent, we would most likely deduce that Holmes and Briars did not tell the statistician:
 A) the sample size.
 B) the value of the Durbin-Watson statistic.
 C) the model’s specification.

Serial correlation will bias the standard errors. It can also bias the coefficient estimates in an autoregressive model of this type. Thus, Briars and Holmes probably did not tell the statistician the model is an AR(1) specification. (Study Session 3, LOS 12.k)

The statistician’s statement concerning the benefits of the Hansen method is:
 A) not correct, because the Hansen method only adjusts for problems associated with serial correlation but not heteroskedasticity.
 B) not correct, because the Hansen method only adjusts for problems associated with heteroskedasticity but not serial correlation.
 C) correct, because the Hansen method adjusts for problems associated with both serial correlation and heteroskedasticity.

The statistician is correct because the Hansen method adjusts for problems associated with both serial correlation and heteroskedasticity. (Study Session 3, LOS 12.i)

Using the model’s results, Briar’s forecast for three years into the future is:
 A) \$54.543 million.
 B) \$47.151 million.
 C) \$54.108 million.

Briars’ forecasts for the next three years would be:
year one: 3.8836 + 0.9288 × 54 = 54.0388
year two: 3.8836 + 0.9288 × (54.0388) = 54.0748
year three: 3.8836 + 0.9288 × (54.0748) = 54.1083
(Study Session 3, LOS 13.a)

With respect to the comments of Holmes and Briars concerning the mean reversion of the import data, the long-run mean value that:
 A) Briars computes is correct.
 B) Briars computes is not correct, and his conclusion is probably not accurate.
 C) Briars computes is not correct, but his conclusion is probably accurate.

Briars has computed a value that would be correct if the results of the model were reliable. The long-run mean would be 3.8836 / (1 − 0.9288)= 54.5450. (Study Session 3, LOS 13.a)

Given the nature of their analysis, the most likely potential problem that Briars and Holmes need to investigate is:
 A) autocorrelation.
 B) unit root.
 C) multicollinearity.

Multicollinearity cannot be a problem because there is only one independent variable. For a time series AR model, autocorrelation is a bigger worry. The model may have been misspecified leading to statistically significant autocorrelations. Unit root does not seem to be a problem given the value of b1<1. (Study Session 3, LOS 13.e)
Dianne Hart, CFA, is considering the purchase of an equity position in Book World, Inc, a leading seller of books in the United States. Hart has obtained monthly sales data for the past seven years, and has plotted the data points on a graph. Which of the following statements regarding Hart’s analysis of the data time series of Book World’s sales is most accurate? Hart should utilize a:
 A) log-linear model to analyze the data because it is likely to exhibit a compound growth trend.
 B) mean-reverting model to analyze the data because the time series pattern is covariance stationary.
 C) linear model to analyze the data because the mean appears to be constant.

A log-linear model is more appropriate when analyzing data that is growing at a compound rate. Sales are a classic example of a type of data series that normally exhibits compound growth.
 Trend models can be useful tools in the evaluation of a time series of data. However, there are limitations to their usage. Trend models are not appropriate when which of the following violations of the linear regression assumptions is present? A) Heteroskedasticity. B) Model misspecification. C) Serial correlation. -------------------------------------------------------------------------------- One of the primary assumptions of linear regression is that the residual terms are not correlated with each other. If serial correlation, also called autocorrelation, is present, then trend models are not an appropriate analysis tool.
Rhonda Wilson, CFA, is analyzing sales data for the TUV Corp, a current equity holding in her portfolio. She observes that sales for TUV Corp. have grown at a steadily increasing rate over the past ten years due to the successful introduction of some new products. Wilson anticipates that TUV will continue this pattern of success. Which of the following models is most appropriate in her analysis of sales for TUV Corp.?
 A) A linear tend model, because the data series is equally distributed above and below the line and the mean is constant.
 B) A log-linear trend model, because the data series can be graphed using a straight, upward-sloping line.
 C) A log-linear trend model, because the data series exhibits a predictable, exponential growth trend.

The log-linear trend model is the preferred method for a data series that exhibits a trend or for which the residuals are predictable. In this example, sales grew at an exponential, or increasing rate, rather than a steady rate.
To qualify as a covariance stationary process, which of the following does not have to be true?
 A) Covariance(xt, xt-1) = Covariance(xt, xt-2).
 B) E[xt] = E[xt+1].
 C) Covariance(xt, xt-2) = Covariance(xt, xt+2).

If a series is covariance stationary then the unconditional mean is constant across periods. The unconditional mean or expected value is the same from period to period: E[xt] = E[xt+1]. The covariance between any two observations equal distance apart will be equal, e.g., the t and t-2 observations with the t and t+2 observations. The one relationship that does not have to be true is the covariance between the t and t-1 observations equaling that of the t and t-2 observations
To qualify as a covariance stationary process, which of the following does not have to be true?
 A) Covariance(xt, xt-1) = Covariance(xt, xt-2).
 B) E[xt] = E[xt+1].
 C) Covariance(xt, xt-2) = Covariance(xt, xt+2).

If a series is covariance stationary then the unconditional mean is constant across periods. The unconditional mean or expected value is the same from period to period: E[xt] = E[xt+1]. The covariance between any two observations equal distance apart will be equal, e.g., the t and t-2 observations with the t and t+2 observations. The one relationship that does not have to be true is the covariance between the t and t-1 observations equaling that of the t and t-2 observations
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