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安在 · 2022年10月20日

这个是考试原题吗 ?老师课上说开放式题目不考

NO.PZ2021061603000047

问题如下:

An economist collected the monthly returns for KDL's portfolio and a diversified stock index. The data collected are shown in the following table:


The economist calculated the correlation between the two returns and found it to be 0.996. The regression results with the KDL return as the dependent variable and the index return as the independent variable are given as follows:


When reviewing the results, Andrea Fusilier suspected that they were unreliable. She found that the returns for Month 2 should have been 7.21% and 6.49%, instead of the large values shown in the first table. Correcting these values resulted in a revised correlation of 0.824 and the following revised regression results:


Explain how the bad data affected the results.


选项:

解释:

The Month 2 data point is an outlier, lying far away from the other data values.

Because this outlier was caused by a data entry error, correcting the outlier improves the validity and reliability of the regression. In this case, revised R2 is lower (from 0.9921 to 0.6784). The outliers created the illusion of a better fit from the higher R2; the outliers altered the estimate of the slope. The standard error of the estimate is lower when the data error is corrected (from 2.861 to 2.0624), as a result of the lower mean square error. However, at a 0.05 level of significance, both models fit well. The difference in the fit is illustrated in Exhibit 1:


请问下这题的来源

1 个答案

星星_品职助教 · 2022年10月22日

本题为教材课后题,题库里没有考试原题。

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NO.PZ2021061603000047问题如下 economist collectethe monthly returnsfor K's portfolio ana versifiestoinx. The ta collecteare shownin the following table:The economist calculatethe correlationbetween the two returns anfounit to 0.996. The regression results withthe K return the pennt variable anthe inx return theinpennt variable are given follows:When reviewing the results, Anea Fusiliersuspecteththey were unreliable. She founththe returns for Month 2shoulhave been 7.21% an6.49%, insteof the large values shown in thefirst table. Correcting these values resultein a revisecorrelation of 0.824anthe following reviseregression results:Explain how the bta affectetheresults. The Month 2 ta point is outlier, lyingfawfrom the other ta values.Because this outlier wcausea taentry error, correcting the outlier improves the vality anreliability ofthe regression. In this case, reviseR2 is lower (from 0.9921 to 0.6784). Theoutliers createthe illusion of a better fit from the higher R2; the outliersalterethe estimate of the slope. The stanrerror of the estimate is lowerwhen the ta error is correcte(from 2.861 to 2.0624), a result of thelower mesquare error. However, a 0.05 level of significance, both molsfit well. The fferenin the fit is illustratein Exhibit 1: 请问决定系数R的平方下面的stanrerror指的是什么?

2023-07-17 13:34 1 · 回答

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2022-10-08 11:34 2 · 回答

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