tag:blogger.com,1999:blog-2198942534740642384.post8901700387366466867..comments2019-11-09T08:50:33.495-08:00Comments on Econometrics Beat: Dave Giles' Blog: May I Show You My Collection of p-Values?Dave Gileshttp://www.blogger.com/profile/05389606956062019445noreply@blogger.comBlogger12125tag:blogger.com,1999:blog-2198942534740642384.post-34403439215607631582012-12-30T07:42:16.738-08:002012-12-30T07:42:16.738-08:00Can't we say that #2 is correct under certain ...Can't we say that #2 is correct under certain conditions? Say, holding sample size/test power constant? In clinical trials it's common to have extremely low test size (<= 0.001), which implies that there's some truth to #2. However I'd agree that if we're talking about economic significance, there's little to no difference because other factors would become much more important (sample size, misspecification, use of asymptotic results in finite samples, etc).Isaacnoreply@blogger.comtag:blogger.com,1999:blog-2198942534740642384.post-67528574224427697942012-12-06T10:24:32.436-08:002012-12-06T10:24:32.436-08:00OK - I see the point! Thanks to both of you!OK - I see the point! Thanks to both of you!Dave Gileshttps://www.blogger.com/profile/05389606956062019445noreply@blogger.comtag:blogger.com,1999:blog-2198942534740642384.post-30027413844475299862012-12-06T00:49:46.302-08:002012-12-06T00:49:46.302-08:00I also agree that number 2 is correct. Even if it&...I also agree that number 2 is correct. Even if it's not the same problem and not the same data set. The p-value is basically the definition of the statistically significance metric.<br /><br />It just doesn't work as a comparison of effect sizes or practical significance. But I don't see the problem in calling a very small effect on a large data set more statistically significant than a large effect on a small data set, as long as one upholds the distinctions between effect sizes, statistical significance and pratical significance. <br /><br />All this is of course conditional to all the assumptions which have been needed to calculate the p-value being correct.Eriknoreply@blogger.comtag:blogger.com,1999:blog-2198942534740642384.post-29229483868739574132012-12-05T15:40:16.201-08:002012-12-05T15:40:16.201-08:00The one I like is that the p-value is "the pr...The one I like is that the p-value is "the probability that we would have gotten this result by chance."<br /><br />Since the result was obtained by a chance process, the probability that it was obtained by chance is 1.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-2198942534740642384.post-88718849554714594732012-12-05T14:32:19.738-08:002012-12-05T14:32:19.738-08:00So, is your point about P-values or about an alter...So, is your point about P-values or about an alternative meaning for 'significant'? The question is badly incomplete and thus ambiguous. I assumed that you meant to compare two potential results from a single experiment. What would the point of the comparison be if you expand its scope to include different types of experiments with different sample sizes etc.?<br /><br />P-values are being criticised all over the place, but most of the criticisms are based on misunderstandings.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-2198942534740642384.post-9208688699958858052012-12-05T12:04:18.746-08:002012-12-05T12:04:18.746-08:00OK - but the two numbers being compared need to re...OK - but the two numbers being compared need to relate to the same problem and be based on the same information set. That's my point.Dave Gileshttps://www.blogger.com/profile/05389606956062019445noreply@blogger.comtag:blogger.com,1999:blog-2198942534740642384.post-42493085950510067942012-12-05T11:56:13.983-08:002012-12-05T11:56:13.983-08:00Answer 2 is correct. You have misconstrued the P-v...Answer 2 is correct. You have misconstrued the P-values as being error rates from the Neyman-Pearson hypothesis testing framework. When P-values are more properly used within Fisher's significance testing framework, the size of the P-values is the level of significance.<br />It is common to work within a hybrid of the N-P and F approaches, but the hybrid is not consistent. Either test for a test statistic beyond a pre-defined critical value, or calculate the P-value as an index of evidence. Don't do both.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-2198942534740642384.post-4014926974932283862012-09-25T06:01:25.904-07:002012-09-25T06:01:25.904-07:00I don't understand any of this. Maybe I should...I don't understand any of this. Maybe I should of paid attention in high school math class?Mikehttp://www.rawhoney.canoreply@blogger.comtag:blogger.com,1999:blog-2198942534740642384.post-17104849494631671822012-03-09T17:27:34.735-08:002012-03-09T17:27:34.735-08:00This comment has been removed by a blog administrator.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-2198942534740642384.post-33304868222100820592012-03-04T16:35:57.398-08:002012-03-04T16:35:57.398-08:00This comment has been removed by a blog administrator.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-2198942534740642384.post-18033404892955501382011-04-13T10:25:14.578-07:002011-04-13T10:25:14.578-07:00Thanks Mycroft - a silly typo which I have fixed! ...Thanks Mycroft - a silly typo which I have fixed! Much appreciated.Dave Gileshttps://www.blogger.com/profile/05389606956062019445noreply@blogger.comtag:blogger.com,1999:blog-2198942534740642384.post-18025545906845816952011-04-13T10:18:21.713-07:002011-04-13T10:18:21.713-07:00You write
"You should now be able to see why...You write<br /><br />"You should now be able to see why we can't just report a standard deviation of 0.289 (= 1/12)"<br /><br />But 1/12 is the variance of U[0,1], not the SD. I presume you mean sqrt(1/12). The math works out better this way, too ;-)Anonymousnoreply@blogger.com