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WINWIN8 · 2022年07月10日

题目和经典题有出入

  1. CFA II
  2. Quantitative

* 问题详情，请 查看题干

NO.PZ201709270100000506

问题如下：

6. Based on the regression output in Exhibit 3 and sales data in Exhibit 4, the forecasted value of quarterly sales for March 2016 for PoweredUP is closest to:

选项：

A.

$4.193 billion.

B.

$4.205 billion.

C.

$4.231 billion.

解释：

C is correct. The quarterly sales for March 2016 is calculated as follows:

beginarraylln⁡ Salest-ln Salest-1=b0+b1(ln Salest-1-ln Salest-2)+b2(ln Salest−4-ln Salest-5)ln⁡ Salest−ln⁡3.868=0.0092−0.1279(ln⁡3.868−ln⁡3.780)+0.7239(ln⁡3.836−ln⁡3.418)ln⁡ Salest=1.44251Salest=e1.44251=4.231begin{array}{l}\ln{\text{ Sales}}_\text{t}{\text{-ln Sales}}_\text{t-1}{\text{=b}}_\text{0}{\text{+b}}_\text{1}{\text{(ln Sales}}_\text{t-1}{\text{-ln Sales}}_\text{t-2}{\text{)+b}}_\text{2}\text{(ln Sales}_{t-4}^{}{\text{-ln Sales}}_\text{t-5}\text{)}\\\ln{\text{ Sales}}_\text{t}-\ln3.868=0.0092-0.1279(\ln3.868-\ln3.780)+0.7239(\ln3.836-\ln3.418)\\\ln{\text{ Sales}}_\text{t}=1.44251\\{\text{Sales}}_\text{t}=e^{1.44251}=4.231

在经典题36页明确标明了 用这个model : “ ln Salest​-ln Salest-1​=b0​+b1​(ln Salest-1​-ln Salest-2​)+b2​(ln Salest−4​-ln Salest-5​)”

但是课后题这边是：“ ln Salest​-ln Salest-1​=b0​+b1​(ln Salest-1​-ln Salest-2​)” ，所以我的问题是，哪一边的出错了？ 考试时，会按照哪种来？

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1 个答案

星星_品职助教 · 2022年07月11日

同学你好，

本题用的方程也是 ln Salest​-ln Salest-1​=b0​+b1​(ln Salest-1​-ln Salest-2​)+b2​(ln Salest−4​-ln Salest-5​)，和经典题是一样的。

题目中的“Based on the regression output in Exhibit 3”对应的即是这个方程。


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NO.PZ201709270100000506 问题如下 6. Baseon the regression output in Exhibit 3 ansales ta in Exhibit 4, the forecastevalue of quarterly sales for Mar2016 for PowereP is closest to: A.$4.193 billion. B.$4.205 billion. C.$4.231 billion. C is correct. The quarterly sales for Mar2016 is calculatefollows:beginarraylln⁡ Salest-ln Salest-1=b0+b1(ln Salest-1-ln Salest-2)+b2(ln Salest−4-ln Salest-5)ln⁡ Salest−ln⁡3.868=0.0092−0.1279(ln⁡3.868−ln⁡3.780)+0.7239(ln⁡3.836−ln⁡3.418)ln⁡ Salest=1.44251Salest=e1.44251=4.231begin{array}{l}\ln{\text{ Sales}}_\text{t}{\text{-ln Sales}}_\text{t-1}{\text{=b}}_\text{0}{\text{+b}}_\text{1}{\text{(ln Sales}}_\text{t-1}{\text{-ln Sales}}_\text{t-2}{\text{)+b}}_\text{2}\text{(ln Sales}_{t-4}^{}{\text{-ln Sales}}_\text{t-5}\text{)}\\\ln{\text{ Sales}}_\text{t}-\ln3.868=0.0092-0.1279(\ln3.868-\ln3.780)+0.7239(\ln3.836-\ln3.418)\\\ln{\text{ Sales}}_\text{t}=1.44251\\{\text{Sales}}_\text{t}=e^{1.44251}=4.231beginarraylln Salest​-ln Salest-1​=b0​+b1​(ln Salest-1​-ln Salest-2​)+b2​(ln Salest−4​-ln Salest-5​)ln Salest​−ln3.868=0.0092−0.1279(ln3.868−ln3.780)+0.7239(ln3.836−ln3.418)ln Salest​=1.44251Salest​=e1.44251=4.231 0.0092−0.1279(ln3.868−ln3.780)+0.7239(ln3.836−ln3.418)得出来不是0.089776吗 然后怎么算ln（Sales t）呢？

2024-07-13 18:34 1 · 回答

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2023-03-18 14:02 1 · 回答

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2023-01-28 23:09 2 · 回答

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