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广超 · 2024年08月12日

老师这道题没有理解题干

NO.PZ2018101001000014

问题如下:

Harry is a supermarket’s employer and he is interested in using math model to solve some problems. Recently, he wants to know the relationship among commodity demand volume(V), commodity prices(P) and consumer income level(I). The dependent variable is commodity demand volume and the other two is the independent variables. After doing the multiple regression analysis, he gets the following results:

What is the upper confidence interval of consumer income level’s regression coefficient at 0.1 level?

选项:

A.

3.467

B.

8.919

C.

9.798

解释:

B is correct.

考点: Interpret Regression Coefficient, Assumptions and F-test.

解析: Regression Coefficient Confidence Interval的公式是b1^±tcSb1^,这里的b1^等于6.193,Sb1^ 等于1.652。已知n为150,所以自由度(df)=n-k-1=147。当样本容量足够大(大于120)时,在α =0.1的情况下,tc近似于1.65,所以对于本题来说,置信区间上限等于6.193+1.65*1.652=8.919,选择B选项。

  1. 题干里给了每个x变量的coefficient、standard error和T统计量,分别代表什么啊?为什么是给这三个条件?他是想告诉我们什么?
  2. 题干中的0.1又是什么意思?为什么会对应到1.65?
1 个答案

品职助教_七七 · 2024年08月13日

嗨,爱思考的PZer你好:


1)表格中的“coefficient”分别为方程的截距项和自变量的系数。由此可以得到回归方程的形式为:

Y=4990.519-35.666*P+6.193*I


t-statistic为当原假设为该系数为0时,计算出来的t统计量。可以据此判断该系数是否为0。例如自变量I的t统计量为3.749,显然大于一般的Critical value。就可知该假设检验的结果为I的系数不等于0。

根据t-statistic的公式,可以得到表格中的关系为(coefficient-0)/std.error=t-statistic。


2)0.1为significance level。由此可以推出Critical value。正态分布下,10%的置信区间对应的关键值为1.65。

本题为t分布,但因为n很大,所以关键值和1.65相差无几。在没给出精确t关键值的前提下,可以用1.65来近似计算。(但如果题目给了关于t关键值的信息,就还是要用这个信息来做,不能近似)。

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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.PZ2018101001000014 问题如下 Harry is a supermarket’s employer anhe is interestein using math mol to solve some problems. Recently, he wants to know the relationship among commoty manvolume(V), commoty prices(P) anconsumer income level(I). The pennt variable is commoty manvolume anthe other two is the inpennt variables. After ing the multiple regression analysis, he gets the following results:Whis the upper confinintervof consumer income level’s regression coefficient 0.1 level? A.3.467 B.8.919 C.9.798 B is correct.考点: Interpret Regression Coefficient, Assumptions anF-test.解析: Regression Coefficient ConfinInterval的公式是b1^±tcSb1^,这里的b1^等于6.193,Sb1^ 等于1.652。已知n为150,所以自由度()=n-k-1=147。当样本容量足够大(大于120)时,在α =0.1的情况下,tc近似于1.65,所以对于本题来说,置信区间上限等于6.193+1.65*1.652=8.919,选择 为什么计算置信区间的时候不用题干表格里给的t-statistic而是用1.65呢

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