Proportionate Reduction In Error Calculator
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PRE, proportional reduction of errorPRE, proportional reduction of error having a mental medical condition financial status relatively bad relatively good
The Proportionate Reduction In Error Is A Measure Of The Quizlet
total yes 390 (97,5 %) 10 (2,5 %) 400 (100 %) proportionate reduction in error symbol no 40 (6,7 %) 560 (93,3 %) 600 (100 %) total 430 (43 %) 570 (57 %) proportional reduction calculator 1000 (100 %) Using one of the illustrations from the previous lecture (where we considered mental health to be the independent variable and financial status to be the
Proportional Reduction In Error Stata
dependent variable), let’s guess the financial status of the individual respondents based on our knowledge of the distribution: 57% have relatively good, 43% have relatively worse financial status.Let’s imagine that the respondents turn up one by one and we have to guess their financial status as accurately as possible. What’s the best way to do that? having
Proportionate Reduction In Error Can Be Symbolized By
a mental medical condition financial status relatively bad relatively good total yes 390 (97,5 %) 10 (2,5 %) 400 (100 %) no 40 (6,7 %) 560 (93,3 %) 600 (100 %) total 430 (43 %) 570 (57 %) 1000 (100 %) Declareing each respondent to have a relatively good financial status is the safest way: thus we are wrong in 430 cases out of 1000.How does the situation change if we already know Table 1 and we can ask each respondent whether or not they have a mental medical condition?In this case we can improve the chances of our guesswork by categorizing everyone with a mental problem as having worse financial status, while those without mental problems as having better financial status. Thus the number of mistakes we make is down to 50.In other words, the guessing error characterizes the relationship of the two variables. Associational indices that work on this principle are called ’proportional reduction of error’ (PRE) indices.Calculating (λ) to get the connection of two n
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Users Badges Unanswered Ask Question _ Cross Validated is a question and answer site for people interested in statistics, machine learning, data proportional error physics analysis, data mining, and data visualization. Join them; it only takes a minute: Sign up Here's how it works: Anybody can ask a question Anybody can answer The best answers are voted up and rise to the http://www.tankonyvtar.hu/en/tartalom/tamop425/0010_2A_21_Nemeth_Renata-Simon_David_Tarsadalomstatisztika_magyar_es_angol_nyelven_eng/ch08s02.html top calculating p-value from PRE (proportional reduction in error) up vote 0 down vote favorite In R, I have created a PRE value to test whether or not modelA is significantly different from modelC. How can I get a significance value for either the PRE distribution or the R^2 distribution (pearson's R)? #modelA<-lm(...) #modelC<-lm(...) PRE=(sum(modelC$residuals^2)-sum(modelA$residuals^2))/sum(modelC$residuals^2) r regression multiple-regression linear-model share|improve this question asked Sep 22 '15 at 17:11 Rilcon42 272111 Are they nested http://stats.stackexchange.com/questions/173699/calculating-p-value-from-pre-proportional-reduction-in-error models? If so a small rescaling of your formula should yield a standard partial F (discussed in a number of questions here). If they're not nested, most people would use some criterion for comparison rather than a test (such as comparing AIC or BIC, for example), but it's possible to use something like Vuong's test. –Glen_b♦ Oct 15 '15 at 2:55 An F test would work, I was hoping someone knew of a way to get the PRE distribution critical values in R though. –Rilcon42 Oct 16 '15 at 3:33 You can back them straight out of the F-test –Glen_b♦ Oct 16 '15 at 6:45 add a comment| 1 Answer 1 active oldest votes up vote -1 down vote There is no PRE distribution. You just have to use your judgement if the reduction in error seems large enough to justify adding complexity to the model. share|improve this answer answered Oct 15 '15 at 1:55 Michael 1 add a comment| Your Answer draft saved draft discarded Sign up or log in Sign up using Google Sign up using Facebook Sign up using Email and Password Post as a guest Name Email Post as a guest Name Email discard By posting your answer, you agree to the privacy policy and terms of service. Not the answer you're looking for?
here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack http://stackoverflow.com/questions/32723604/calculating-significance-of-pre-proportional-reduction-in-error Overflow the company Business Learn more about hiring developers or posting ads with us Stack Overflow Questions Jobs Documentation Tags Users Badges Ask Question x Dismiss Join the Stack Overflow Community Stack Overflow is a http://www.oxfordreference.com/view/10.1093/oi/authority.20110803100349896 community of 6.2 million programmers, just like you, helping each other. Join them; it only takes a minute: Sign up Calculating significance of PRE (proportional reduction in error) up vote 1 down vote favorite I reduction in am trying to calculate the significance of the PRE for two linear models. All of my variables are continuous (so I cant use the DAMisc package). I tried to do it by hand using the following R code, but I keep getting NaN when I test it. What did I do wrong? library(psych) modelC<-lm(reason.4~reason.16,data=iqitems) modelA<-lm(reason.4~reason.16+reason.17,data=iqitems) SSEC<-sum(modelC$residuals^2) #2019 SSEA<-sum(modelA$residuals^2) #1977 PRE=(SSEC-SSEA)/SSEC #0.02 pA<-2 pC<-1 n<-min(summary(modelA)$df)+1 #4 deg_freedom<-min(summary(modelA)$df) #3 Fvalue <- (PRE/(pA-pC))/((1-PRE)*(n-pA)) #0.0105 reduction in error pf(Fvalue,pC-pA,deg_freedom,lower.tail=FALSE) #NaN r statistics share|improve this question asked Sep 22 '15 at 17:52 Rilcon42 1,34831128 Have you checked for missing values in the residuals vectors? –42- Sep 22 '15 at 19:53 add a comment| 1 Answer 1 active oldest votes up vote 0 down vote accepted The second argument of the pf function is the degrees of freedom of the numerator of the f-statistic and the third argument is the degrees of freedom of the denominator of the f-statistic. The degrees of freedom of your second argument is -1 as is right now as pC - pA = 1 - 2 = -1. This is what gives you the error as the degrees of freedom are non-negative.I have no experience in PRE I think you need: #second argument being pA - pC = 1 > pf(0.0105, 1, 3,lower.tail=FALSE) [1] 0.9248493 share|improve this answer edited Sep 22 '15 at 21:11 answered Sep 22 '15 at 20:56 LyzandeR 16.9k112640 add a comment| Your Answer draft saved draft discarded Sign up or log in Sign up using Google Sign up using Facebook Sign up using Email and Password Post as a guest Name Email Post as a guest Name Email discard By posting your answer, you agree to the privacy
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