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ciaoyy · 2018年12月11日

问一道题：NO.PZ201709270100000506 第6小题

* 问题详情，请查看题干

问题如下图：

选项：

A.

B.

C.

解释：

这道题中时间序列方程的b0，b1的t检验量都不显著，落在接受域里，为什么不是接受原假设，使得b0=0，b1=0？最后原方程只剩下0.7239*（lnSalest-4-lnSalest-5）呢？

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菲菲_品职助教 · 2018年12月17日

其实在二级更多的都是回归分析，我想了一下，之前说的不存在原假设和备择假设其实有点绝对，如果要等同的话，原假设可以是H0：b1=0，这样的形式，如果得出来的结果是拒绝原假设，就相当于是b1≠0，说明这个自变量是可以用来解释因变量的，但是具体的解释力度还是要结合其他数据来看，比如说R²，还有系数本身。

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菲菲_品职助教 · 2018年12月11日

同学你好，这里是回归分析不是假设检验，所以不存在原假设和备择假设，回归分析里面的t统计值只是用来判断这个自变量对于因变量的影响有多大，所以如果题目没有说要剔除某个自变量，就按照原来的模型进行计算。

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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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2022-10-18 22:31 1 · 回答

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2022-07-10 21:50 1 · 回答