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除了樱花 · 2022年03月21日

考试范围

NO.PZ2020010801000025

问题如下：

A model was estimated using daily data from the S&P 500 from 1977 until 2017 which included five day-of-the-week dummies (n = 10,087). The $R^2$ from this regression was 0.000599. Is there evidence that the mean varies with the day of the week?

选项：

解释：

The model estimated is

$Y_i = \beta_1D_1 + \beta_2D_2 + \beta_3D_3 + \beta_4D_4 + \beta_5D_5 + \epsilon_i$ ,

where Di is a dummy that takes the value 1 if the index of the weekday is i (e.g., Monday = 1, Tuesday = 2, c). The restriction is that

$H_0:\beta_1 = \beta_2 = \beta_3 = \beta_4 = \beta_5$

so there this is no day-of-the-week effect. This model can be equivalently written as

$Y_i = \mu + \delta_2D_2 + \delta_3D_3 + \delta_4D_4 + \delta_5D_5 + \epsilon_i$ ,

therefore, here the null is

$H_0:\delta_2 = \delta_3 = \delta_4 = \delta_5$ .

In the two models, $\mu = \beta_1$ , and $\mu + \delta_i = \beta_i$ . The second form of the model is a more standard null for an F-stat.

The F-stat of the regression is

$\frac{(R^2-0)/4}{(1-R^2)/(n-5)}=\frac{0.000599/4}{(1-0.000599)/(10087-5)}=1.51$

The distribution is an $F_{4,10082}$ and the critical value using a 5% size is 2.37. The test statistic is less than the critical value, therefore, the null that all effects are 0 is not rejected.

这种难度和类型的题考试会考吗

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品职答疑小助手雍 · 2022年03月21日

同学你好，这题比较难，难点是5个哑变量运用哑变量的性质把它变换成了4个，这个内容是计量里比较细的知识点，不在我们考纲范围内了。这题弄清检验几个变量应该用F检验，并记住F检验的公式就够了。

考试时间其实挺紧张的，所以出题的复杂程度不会特别复杂。

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NO.PZ2020010801000025 问题如下 A mol westimateusing ily ta from the S P 500 from 1977 until 2017 whiinclufive y-of-the-week mmies (n = 10,087). The R2R^2R2 from this regression w0.000599. Is there evinththe mevaries with the y of the week? The mol estimateis Yi=β1+β2+β3+β4+β5+ϵiY_i = \beta_11 + \beta_22 + \beta_33 + \beta_44 + \beta_55 + \epsilon_iYi=β1+β2+β3+β4+β5+ϵi, where is a mmy thtakes the value 1 if the inx of the weeky is i (e.g., Mony = 1, Tuesy = 2, c). The restriction is thH0:β1=β2=β3=β4=β5H_0:\beta_1 = \beta_2 = \beta_3 = \beta_4 = \beta_5H0:β1=β2=β3=β4=β5 so there this is no y-of-the-week effect. This mol cequivalently written Yi=μ+δ2+δ3+δ4+δ5+ϵiY_i = \mu + \lta_22 + \lta_33 + \lta_44 + \lta_55 + \epsilon_iYi=μ+δ2+δ3+δ4+δ5+ϵi, therefore, here the null is H0:δ2=δ3=δ4=δ5H_0:\lta_2 = \lta_3 = \lta_4 = \lta_5H0:δ2=δ3=δ4=δ5. In the two mols, μ=β1\mu = \beta_1μ=β1, anμ+δi=βi\mu + \lta_i = \beta_iμ+δi=βi. The seconform of the mol is a more stanrnull for F-stat. The F-stof the regression is(R2−0)/4(1−R2)/(n−5)=0.000599/4(1−0.000599)/(10087−5)=1.51\frac{(R^2-0)/4}{(1-R^2)/(n-5)}=\frac{0.000599/4}{(1-0.000599)/(10087-5)}=1.51(1−R2)/(n−5)(R2−0)/4=(1−0.000599)/(10087−5)0.000599/4=1.51The stribution is F4,10082F_{4,10082}F4,10082 anthe criticvalue using a 5% size is 2.37. The test statistic is less ththe criticvalue, therefore, the null thall effects are 0 is not rejecte 请问一下F检验分子不是应该等于(UnrestricteR 2 - RestricteR2)/q 吗。题目说了unrestricteR2但并没有说restricteR2是多少。为什么公式里面直接就把restricteR2 忽略掉了呢

2024-05-24 11:45 1 · 回答

NO.PZ2020010801000025 问题如下 A mol westimateusing ily ta from the S P 500 from 1977 until 2017 whiinclufive y-of-the-week mmies (n = 10,087). The R2R^2R2 from this regression w0.000599. Is there evinththe mevaries with the y of the week? The mol estimateis Yi=β1+β2+β3+β4+β5+ϵiY_i = \beta_11 + \beta_22 + \beta_33 + \beta_44 + \beta_55 + \epsilon_iYi=β1+β2+β3+β4+β5+ϵi, where is a mmy thtakes the value 1 if the inx of the weeky is i (e.g., Mony = 1, Tuesy = 2, c). The restriction is thH0:β1=β2=β3=β4=β5H_0:\beta_1 = \beta_2 = \beta_3 = \beta_4 = \beta_5H0:β1=β2=β3=β4=β5 so there this is no y-of-the-week effect. This mol cequivalently written Yi=μ+δ2+δ3+δ4+δ5+ϵiY_i = \mu + \lta_22 + \lta_33 + \lta_44 + \lta_55 + \epsilon_iYi=μ+δ2+δ3+δ4+δ5+ϵi, therefore, here the null is H0:δ2=δ3=δ4=δ5H_0:\lta_2 = \lta_3 = \lta_4 = \lta_5H0:δ2=δ3=δ4=δ5. In the two mols, μ=β1\mu = \beta_1μ=β1, anμ+δi=βi\mu + \lta_i = \beta_iμ+δi=βi. The seconform of the mol is a more stanrnull for F-stat. The F-stof the regression is(R2−0)/4(1−R2)/(n−5)=0.000599/4(1−0.000599)/(10087−5)=1.51\frac{(R^2-0)/4}{(1-R^2)/(n-5)}=\frac{0.000599/4}{(1-0.000599)/(10087-5)}=1.51(1−R2)/(n−5)(R2−0)/4=(1−0.000599)/(10087−5)0.000599/4=1.51The stribution is F4,10082F_{4,10082}F4,10082 anthe criticvalue using a 5% size is 2.37. The test statistic is less ththe criticvalue, therefore, the null thall effects are 0 is not rejecte 请问第二个公式是怎么推导出来的？为什么少了一个regressor

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2023-02-05 02:23 1 · 回答