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eyn · 2024年09月05日

没有明白问题(1)(2),请详细解释下

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NO.PZ202212300100014002

问题如下:

Based on Exhibit 2, Vasileva should reject the null hypothesis that:

选项:

A.the slope is less than or equal to 0.15. B.the intercept is less than or equal to zero. C.crude oil returns do not explain Amtex share returns.

解释:

Crude oil returns explain the Amtex share returns if the slope coefficient is statistically different from zero. The slope coefficient is 0.2354, and the calculated t-statistic is

t=(0.2354-0.0000)/0.0760=3.0974,

which is outside the bounds of the critical values of ±2.728.

Therefore, Vasileva should reject the null hypothesis that crude oil returns do not explain Amtex share returns, because the slope coefficient is statistically different from zero.

A is incorrect because the calculated t-statistic for testing the slope against 0.15 is t=(0.2354-0.1500)/0.0760=1.1237,which is less than the critical value of +2.441.

B is incorrect because the calculated t-statistic is t=(0.0095-0.0000)/0.0078=1.2179, which is less than the critical value of +2.441.

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1 个答案

袁园_品职助教 · 2024年09月06日

嗨,从没放弃的小努力你好:


根据给定的信息,回归方程为:Amtex Share Return = Intercept + Oil return * Crude Oil Return。Intercept 为 0.0095,Oil return 系数为 0.2354。所以回归方程为:Amtex Share Return = 0.0095 + 0.2354 * Oil Return。

表一为 “Selected Data for Crude Oil Returns and Amtex Share Returns”,展示了原油收益(Oil Return)、Amtex 股票收益(Amtex Return)、交叉乘积(Cross-Product)、预测的 Amtex 股票收益(Predicted Amtex Return)、回归残差(Regression Residual)和残差平方(Squared Residual)等数据,共包含 36 个月份的数据以及各数据项的总和与平均值。可以用于计算标准误差等统计量。

(1)基于表1,估计的标准误差是多少?

对于只有一个自变量的线性回归模型,估计的标准误差是均方误差的平方根。均方误差通过将残差平方和除以观测值个数减 2 来计算。

从图表 1 中可知,残差平方和是 0.071475。一共有 36 个观测值。

均方误差 = 0.071475÷(36 - 2)= 0.071475÷34 ≈ 0.0021022。

估计的标准误差 = 0.0021022 的平方根≈0.04585。

答案是 B。

(2)根据图表 2,瓦西列娃应该拒绝以下哪个零假设?

A:斜率小于或等于 0.15。

B:截距小于或等于零。

C:原油收益不能解释 Amtex 股票收益。

如果斜率系数在统计上不同于零,那么原油收益可以解释 Amtex 股票收益。斜率系数是 0.2354,计算出的 t 统计量为:

t = (0.2354 - 0.0000)/0.0760 = 3.0974,这个值超出了临界值 ±2.728 的范围。

因此,瓦西列娃应该拒绝原油收益不能解释 Amtex 股票收益这个零假设,因为斜率系数在统计上不同于零。

A 选项错误,因为针对斜率等于 0.15 进行检验的计算出的 t 统计量为 t = (0.2354 - 0.1500)/0.0760 = 1.1237,这个值小于临界值 + 2.441。所以不能拒绝假设。

B 选项错误,因为针对截距的计算出的 t 统计量为 t = (0.0095 - 0.0000)/0.0078 = 1.2179,这个值小于临界值 + 2.441。也不能拒绝假设。

所以最后答案选C。

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虽然现在很辛苦,但努力过的感觉真的很好,加油!

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NO.PZ202212300100014002 问题如下 Baseon Exhibit 2, Vasileva shoulrejethe nullhypothesis that: A.theslope is less thor equto 0.15. B.theintercept is less thor equto zero. C.cruoil returns not explain Amtex share returns. Cru oil returns explain the Amtex share returns if the slope coefficient is statistically fferent from zero. The slope coefficient is 0.2354, anthe calculatet-statistic ist=(0.2354-0.0000)/0.0760=3.0974,whiis outsi the boun of the criticvalues of ±2.728.Therefore, Vasileva shoulrejethe null hypothesis thcru oil returns not explain Amtex share returns, because the slope coefficient is statistically fferent from zero. A is incorrebecause the calculatet-statistic for testing the slope against 0.15 is t=(0.2354-0.1500)/0.0760=1.1237,whiis less ththe criticvalue of +2.441.B is incorrebecause the calculatet-statistic is t=(0.0095-0.0000)/0.0078=1.2179, whiis less ththe criticvalue of +2.441. 请问这一问的知识点在哪里?用到的是什么公式呢? t = (0.2354 - 0.1500)/0.0760 = 1.1237, t = (0.0095 - 0.0000)/0.0078 = 1.2179为什么分母是去除以stanrerror?

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