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一只可爱的猪 · 2021年04月25日

为什么不可以用文中表格给出的standard error

  1. CFA II
  2. Quantitative

* 问题详情，请 查看题干

NO.PZ201512020300000902

问题如下：

Based on Exhibit 1, Olabudo should calculate a prediction interval for the actual US CPI closest to:

选项：

A.

2.7506 to 2.7544

B.

2.7521 to 2.7529

C.

2.7981 to 2.8019.

解释：

A is correct. The prediction interval for inflation is calculated in three steps:

Step 1 – Make the prediction given the US CPI forecast of 2.8:

Y^=b0+b1X=0.0001+(0.9830×2.8)=2.7525\widehat{Y}=b_0+b_1X=0.0001+(0.9830×2.8)=2.7525

Step 2 – Compute the variance of the prediction error:

sf2=s2[1+1n+(X−X‾)2(n−1)×sx2]s^2_f=s^2[1+\frac{1}{n}+\frac{{(X-\overline{X})}^2}{(n-1)\times{s^2_x}}]

sf2=0.00092[1+160+(2.8−1.3350)2(60−1)×0.75392]=0.00000088s^2_f=0.0009^2[1+\frac{1}{60}+\frac{{(2.8-1.3350)}^2}{(60-1)\times{0.7539^2}}]=0.00000088

sf=0.0009s_f=0.0009

Step 3 – Compute the prediction interval:

Y^±tc×sf\widehat{Y}\pm{t_c}\times{s_f}

2.7525±(2.0×0.0009)2.7525\pm{(2.0\times0.0009)}

2.7525–(2.0×0.0009)=2.75062.7525–(2.0\times0.0009)=2.7506; lower bound

2.7525+(2.0×0.0009)=2.75442.7525+(2.0\times0.0009)=2.7544; upper bound

So, given the US CPI forecast of 2.8, the 95% prediction interval is 2.7506 to 2.7544.

为什么不可以用文中表格给出的standard error

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1 个答案

星星_品职助教 · 2021年04月26日

同学你好，

表格中给出的是系数估计量b0和b1的standard error。

而这道题要求的是Y的置信区间，所以要用的是Y的standard error，这个需要代入公式计算的。具体可以参考答案解析里的步骤，表格中notes给的条件2和3都是算这个用的。

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NO.PZ201512020300000902 老师，公式里的S，为什么代入得是0.7539，而不是0.0155？

2021-03-15 23:58 1 · 回答

NO.PZ201512020300000902 2.7521 to 2.7529 2.7981 to 2.8019. A is correct. The prection intervfor inflation is calculatein three steps: Step 1 – Make the prection given the US CPI forecast of 2.8: Y^=b0+b1X=0.0001+(0.9830×2.8)=2.7525\wihat{Y}=b_0+b_1X=0.0001+(0.9830×2.8)=2.7525Y =b0​+b1​X=0.0001+(0.9830×2.8)=2.7525 Step 2 – Compute the varianof the prection error: sf2=s2[1+1n+(X−X‾)2(n−1)×sx2]s^2_f=s^2[1+\frac{1}{n}+\frac{{(X-\overline{X})}^2}{(n-1)\times{s^2_x}}]sf2​=s2[1+n1​+(n−1)×sx2​(X−X)2​] sf2=0.00092[1+160+(2.8−1.3350)2(60−1)×0.75392]=0.00000088s^2_f=0.0009^2[1+\frac{1}{60}+\frac{{(2.8-1.3350)}^2}{(60-1)\times{0.7539^2}}]=0.00000088sf2​=0.00092[1+601​+(60−1)×0.75392(2.8−1.3350)2​]=0.00000088 sf=0.0009s_f=0.0009sf​=0.0009 Step 3 – Compute the prection interval: Y^±tc×sf\wihat{Y}\pm{t_c}\times{s_f}Y ±tc​×sf​ 2.7525±(2.0×0.0009)2.7525\pm{(2.0\times0.0009)}2.7525±(2.0×0.0009) 2.7525–(2.0×0.0009)=2.75062.7525–(2.0\times0.0009)=2.75062.7525–(2.0×0.0009)=2.7506; lower boun2.7525+(2.0×0.0009)=2.75442.7525+(2.0\times0.0009)=2.75442.7525+(2.0×0.0009)=2.7544; upper bounSo, given the US CPI forecast of 2.8, the 95% prection intervis 2.7506 to 2.7544. 这个题目的考点没有明白..而且计算好想很复杂

2021-03-14 13:37 1 · 回答

NO.PZ201512020300000902 求第二步骤的公式也是需要背下来的么？

2021-02-09 21:49 1 · 回答

2.7521 to 2.7529 2.7981 to 2.8019. A is correct. The prection intervfor inflation is calculatein three steps: Step 1 – Make the prection given the US CPI forecast of 2.8: Y^=b0+b1X=0.0001+(0.9830×2.8)=2.7525\wihat{Y}=b_0+b_1X=0.0001+(0.9830×2.8)=2.7525Y =b0​+b1​X=0.0001+(0.9830×2.8)=2.7525 Step 2 – Compute the varianof the prection error: sf2=s2[1+1n+(X−X‾)2(n−1)×sx2]s^2_f=s^2[1+\frac{1}{n}+\frac{{(X-\overline{X})}^2}{(n-1)\times{s^2_x}}]sf2​=s2[1+n1​+(n−1)×sx2​(X−X)2​] sf2=0.00092[1+160+(2.8−1.3350)2(60−1)×0.75392]=0.00000088s^2_f=0.0009^2[1+\frac{1}{60}+\frac{{(2.8-1.3350)}^2}{(60-1)\times{0.7539^2}}]=0.00000088sf2​=0.00092[1+601​+(60−1)×0.75392(2.8−1.3350)2​]=0.00000088 sf=0.0009s_f=0.0009sf​=0.0009 Step 3 – Compute the prection interval: Y^±tc×sf\wihat{Y}\pm{t_c}\times{s_f}Y ±tc​×sf​ 2.7525±(2.0×0.0009)2.7525\pm{(2.0\times0.0009)}2.7525±(2.0×0.0009) 2.7525–(2.0×0.0009)=2.75062.7525–(2.0\times0.0009)=2.75062.7525–(2.0×0.0009)=2.7506; lower boun2.7525+(2.0×0.0009)=2.75442.7525+(2.0\times0.0009)=2.75442.7525+(2.0×0.0009)=2.7544; upper bounSo, given the US CPI forecast of 2.8, the 95% prection intervis 2.7506 to 2.7544. 老师好，请问一下讲义里Sf的公式是直接用SEE的，题目里SEE是0.0009，答案解析中为何还要把0.0009 平方后再去和后面去相乘

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