tag:blogger.com,1999:blog-4403702246808450549.post1777108851523170449..comments2020-04-25T06:46:00.285-04:00Comments on Consultant Ninja: Calling amateur mathematiciansUnknownnoreply@blogger.comBlogger8125tag:blogger.com,1999:blog-4403702246808450549.post-54361144769326714842009-04-29T16:25:00.000-04:002009-04-29T16:25:00.000-04:00The data is IRS tax data.
http://www.irs.gov/taxs...The data is IRS tax data.<br /><br />http://www.irs.gov/taxstats/indtaxstats/article/0,,id=134951,00.html<br /><br />X-value: Percentile of the US population below a certain income.<br /><br />Y-value: That income value. <br /><br />Interesting stuff here. I'll have to chew through it. Thanks.Consultant Ninjahttps://www.blogger.com/profile/14546884171826437168noreply@blogger.comtag:blogger.com,1999:blog-4403702246808450549.post-88637890570000570622009-04-29T16:22:00.000-04:002009-04-29T16:22:00.000-04:00TheTheConsultant Ninjahttps://www.blogger.com/profile/14546884171826437168noreply@blogger.comtag:blogger.com,1999:blog-4403702246808450549.post-25148822887010019052009-04-29T16:20:00.000-04:002009-04-29T16:20:00.000-04:00From the looks of the data I can see (assumptions ...From the looks of the data I can see (assumptions may be violated/need to be checked based on the data sources), I have a few suggestions.<br /><br />The most immediate models that come to mind are:<br />ln(y)=B0+B1ln(x1)<br />OR<br />ln(y)=B0+B1x1<br /><br />I would also try a piecewise regression--perhaps try a cubic until a point k, and then try either of the log models for >k.<br /><br />As far as the Lorenz curve...I don't think I can give an intelligent answer without knowing the nature of the data, but I would only try it if the other models fail.<br /><br />Hope that helps! I enjoy reading your blog :)Unknownhttps://www.blogger.com/profile/11651484938158482286noreply@blogger.comtag:blogger.com,1999:blog-4403702246808450549.post-18272740850120215882009-04-29T16:13:00.000-04:002009-04-29T16:13:00.000-04:00This comment has been removed by the author.Unknownhttps://www.blogger.com/profile/11651484938158482286noreply@blogger.comtag:blogger.com,1999:blog-4403702246808450549.post-92131689243731910602009-04-29T12:58:00.000-04:002009-04-29T12:58:00.000-04:00Also, I don't know if you need to get a plug-in fo...Also, I don't know if you need to get a plug-in for Excel to do the regression analysis. See this link as one possible source.<br /><br />http://www.jeremymiles.co.uk/regressionbook/extras/appendix2/excel<br /><br />I am on the new Excel. I can barely figure out how to format cells anymore, the interface is so different.Steve Shuhttp://www.jeremymiles.co.uk/regressionbook/extras/appendix2/excelnoreply@blogger.comtag:blogger.com,1999:blog-4403702246808450549.post-40651400846998735572009-04-29T12:53:00.000-04:002009-04-29T12:53:00.000-04:00akznetsov looks like he is on the right track. #1 ...akznetsov looks like he is on the right track. #1 looks better to me (presuming the coefficients are right - seems so since R squared value was given). #2 seems like it could be an overfit regression. Maybe try the regression with fewer independent variables as opposed to 6.Steve Shuhttp://steveshu.typepad.comnoreply@blogger.comtag:blogger.com,1999:blog-4403702246808450549.post-47022586409011605562009-04-29T03:51:00.000-04:002009-04-29T03:51:00.000-04:00Try these:
1) y = 0,3115exp(0,5227x)
(R2 = 0,959...Try these:<br /><br />1) y = 0,3115exp(0,5227x)<br />(R2 = 0,9595)<br /><br />2) y = 0,0398x6 - 1,6883x5 + 27,6x4 - 217,62x3 + 847,63x2 - 1480,7x + 851,89<br />(R2 = 0,9901)<br /><br />Fits good but for the practical use... it depends<br /><br />RegardsAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-4403702246808450549.post-4080541882043034802009-04-28T23:28:00.000-04:002009-04-28T23:28:00.000-04:00Try y = -7/(x-1)
Seems to fit pretty well. Don't ...Try y = -7/(x-1)<br /><br />Seems to fit pretty well. Don't know helpful it will be though.Consultant Insiderhttps://www.blogger.com/profile/02119744125271437775noreply@blogger.com