[R] chi-Squared distribution
Dear Rs: outer(1:3, 1:3, function(df1, df2) qf(0.95, df1, df2)) I compare this F distribution results with the table, the answers were perfect. But I need to see for chi-sqaured distribution. When I employed the similar formula outer(1:3, 1:3, function(df1, df2) qchisq(0.95, df1, df2)) , I am getting unexpected results. I need to see the following values: p=0.750 . 1 1.323 2 2.773 3 4.108 Thanking you Prasanna Prasanna Balaprakash, Université Libre de Bruxelles, 50, Av. F. Roosevelt, CP 194/6, B-1050 Brussels, Belgium. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] R: chi-Squared distribution
Hi, Attention chi-squared distribution, unlike F distribution, has only df1 as parameter, not df1 and df2. So correct into: outer(1:3, 1:3, function(df1, df2) qchisq(0.95, df1, df2)) outer(1:3, 1:3, function(df1, df2) qchisq(0.95, df1)) Regards, Vito you wrote: Dear Rs: outer(1:3, 1:3, function(df1, df2) qf(0.95, df1, df2)) I compare this F distribution results with the table, the answers were perfect. But I need to see for chi-sqaured distribution. When I employed the similar formula outer(1:3, 1:3, function(df1, df2) qchisq(0.95, df1, df2)) , I am getting unexpected results. I need to see the following values: p=0.750 . 1 1.323 2 2.773 3 4.108 Thanking you Prasanna = Diventare costruttori di soluzioni Became solutions' constructors The business of the statistician is to catalyze the scientific learning process. George E. P. Box Top 10 reasons to become a Statistician 1. Deviation is considered normal 2. We feel complete and sufficient 3. We are 'mean' lovers 4. Statisticians do it discretely and continuously 5. We are right 95% of the time 6. We can legally comment on someone's posterior distribution 7. We may not be normal, but we are transformable 8. We never have to say we are certain 9. We are honestly significantly different 10. No one wants our jobs Visitate il portale http://www.modugno.it/ e in particolare la sezione su Palese http://www.modugno.it/archivio/palese/ __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] chi-Squared distribution
if you check ?qchisq, you'll see that the second argument of the function denotes the non-centrality parameter! Try qchisq(0.75, 1:3) to get your answer. Best, Dimitris Dimitris Rizopoulos Ph.D. Student Biostatistical Centre School of Public Health Catholic University of Leuven Address: Kapucijnenvoer 35, Leuven, Belgium Tel: +32/16/336899 Fax: +32/16/337015 Web: http://www.med.kuleuven.ac.be/biostat http://www.student.kuleuven.ac.be/~m0390867/dimitris.htm - Original Message - From: Prasanna Balaprakash [EMAIL PROTECTED] To: r-help@stat.math.ethz.ch Sent: Friday, January 21, 2005 12:27 PM Subject: [R] chi-Squared distribution Dear Rs: outer(1:3, 1:3, function(df1, df2) qf(0.95, df1, df2)) I compare this F distribution results with the table, the answers were perfect. But I need to see for chi-sqaured distribution. When I employed the similar formula outer(1:3, 1:3, function(df1, df2) qchisq(0.95, df1, df2)) , I am getting unexpected results. I need to see the following values: p=0.750 . 1 1.323 2 2.773 3 4.108 Thanking you Prasanna Prasanna Balaprakash, Université Libre de Bruxelles, 50, Av. F. Roosevelt, CP 194/6, B-1050 Brussels, Belgium. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] chi-Squared distribution
- Original Message - From: Dimitris Rizopoulos [EMAIL PROTECTED] To: Prasanna Balaprakash [EMAIL PROTECTED] Cc: r-help@stat.math.ethz.ch Sent: Friday, January 21, 2005 1:05 PM Subject: Re: [R] chi-Squared distribution if you check ?qchisq, you'll see that the second My mistake, the third argument argument of the function denotes the non-centrality parameter! Try qchisq(0.75, 1:3) to get your answer. Best, Dimitris Dimitris Rizopoulos Ph.D. Student Biostatistical Centre School of Public Health Catholic University of Leuven Address: Kapucijnenvoer 35, Leuven, Belgium Tel: +32/16/336899 Fax: +32/16/337015 Web: http://www.med.kuleuven.ac.be/biostat http://www.student.kuleuven.ac.be/~m0390867/dimitris.htm - Original Message - From: Prasanna Balaprakash [EMAIL PROTECTED] To: r-help@stat.math.ethz.ch Sent: Friday, January 21, 2005 12:27 PM Subject: [R] chi-Squared distribution Dear Rs: outer(1:3, 1:3, function(df1, df2) qf(0.95, df1, df2)) I compare this F distribution results with the table, the answers were perfect. But I need to see for chi-sqaured distribution. When I employed the similar formula outer(1:3, 1:3, function(df1, df2) qchisq(0.95, df1, df2)) , I am getting unexpected results. I need to see the following values: p=0.750 . 1 1.323 2 2.773 3 4.108 Thanking you Prasanna Prasanna Balaprakash, Université Libre de Bruxelles, 50, Av. F. Roosevelt, CP 194/6, B-1050 Brussels, Belgium. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] chi-Squared distribution in Friedman test
Dear R helpers: Thanks for the previous reply. I am using Friedman racing test. According the the book Pratical Nonprametric Statistic by WJ Conover, after computing the statistics, he suggested to use chi-squared or F distribution to accept or reject null hypothesis. After looking into the source code, I found that R uses chi-sqaured distribution as below: PVAL - pchisq(STATISTIC, PARAMETER, lower = FALSE) but still I cant figure out why they are using this pschisq insted of dchisq. Sorry I am wrong!! Thanking you truly Prasanna - Original Message - From: Dimitris Rizopoulos [EMAIL PROTECTED] To: Prasanna Balaprakash [EMAIL PROTECTED] Cc: r-help@stat.math.ethz.ch Sent: Friday, January 21, 2005 1:05 PM Subject: Re: [R] chi-Squared distribution if you check ?qchisq, you'll see that the second My mistake, the third argument argument of the function denotes the non-centrality parameter! Try qchisq(0.75, 1:3) to get your answer. Best, Dimitris Dimitris Rizopoulos Ph.D. Student Biostatistical Centre School of Public Health Catholic University of Leuven Address: Kapucijnenvoer 35, Leuven, Belgium Tel: +32/16/336899 Fax: +32/16/337015 Web: http://www.med.kuleuven.ac.be/biostat http://www.student.kuleuven.ac.be/~m0390867/dimitris.htm - Original Message - From: Prasanna Balaprakash [EMAIL PROTECTED] To: r-help@stat.math.ethz.ch Sent: Friday, January 21, 2005 12:27 PM Subject: [R] chi-Squared distribution Dear Rs: outer(1:3, 1:3, function(df1, df2) qf(0.95, df1, df2)) I compare this F distribution results with the table, the answers were perfect. But I need to see for chi-sqaured distribution. When I employed the similar formula outer(1:3, 1:3, function(df1, df2) qchisq(0.95, df1, df2)) , I am getting unexpected results. I need to see the following values: p=0.750 . 1 1.323 2 2.773 3 4.108 Thanking you Prasanna Prasanna Balaprakash, Université Libre de Bruxelles, 50, Av. F. Roosevelt, CP 194/6, B-1050 Brussels, Belgium. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] R: chi-Squared distribution in Friedman test
Hi, pchisq - distribution function dchisq - density function pval is the area under the curve, to calculte it you use distribution function which is the integral of density function. See: http://www.itl.nist.gov/div898/handbook/eda/section3/eda362.htm http://mathworld.wolfram.com/DistributionFunction.html f(x) density function F(x) distribution function =Pr(Xx)= integral(f(x)) Hoping I helped you! Regards Vito you wrote: Dear R helpers: Thanks for the previous reply. I am using Friedman racing test. According the the book Pratical Nonprametric Statistic by WJ Conover, after computing the statistics, he suggested to use chi-squared or F distribution to accept or reject null hypothesis. After looking into the source code, I found that R uses chi-sqaured distribution as below: PVAL - pchisq(STATISTIC, PARAMETER, lower = FALSE) but still I cant figure out why they are using this pschisq insted of dchisq. Sorry I am wrong!! Thanking you truly Prasanna = Diventare costruttori di soluzioni Became solutions' constructors The business of the statistician is to catalyze the scientific learning process. George E. P. Box Top 10 reasons to become a Statistician 1. Deviation is considered normal 2. We feel complete and sufficient 3. We are 'mean' lovers 4. Statisticians do it discretely and continuously 5. We are right 95% of the time 6. We can legally comment on someone's posterior distribution 7. We may not be normal, but we are transformable 8. We never have to say we are certain 9. We are honestly significantly different 10. No one wants our jobs Visitate il portale http://www.modugno.it/ e in particolare la sezione su Palese http://www.modugno.it/archivio/palese/ __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] chi-Squared distribution in Friedman test
Prasanna Balaprakash wrote: Dear R helpers: Thanks for the previous reply. I am using Friedman racing test. According the the book Pratical Nonprametric Statistic by WJ Conover, after computing the statistics, he suggested to use chi-squared or F distribution to accept or reject null hypothesis. After looking into the source code, I found that R uses chi-sqaured distribution as below: PVAL - pchisq(STATISTIC, PARAMETER, lower = FALSE) but still I cant figure out why they are using this pschisq insted of dchisq. Sorry I am wrong!! You want to get the distribution function rather than the density. Uwe Ligges Thanking you truly Prasanna - Original Message - From: Dimitris Rizopoulos [EMAIL PROTECTED] To: Prasanna Balaprakash [EMAIL PROTECTED] Cc: r-help@stat.math.ethz.ch Sent: Friday, January 21, 2005 1:05 PM Subject: Re: [R] chi-Squared distribution if you check ?qchisq, you'll see that the second My mistake, the third argument argument of the function denotes the non-centrality parameter! Try qchisq(0.75, 1:3) to get your answer. Best, Dimitris Dimitris Rizopoulos Ph.D. Student Biostatistical Centre School of Public Health Catholic University of Leuven Address: Kapucijnenvoer 35, Leuven, Belgium Tel: +32/16/336899 Fax: +32/16/337015 Web: http://www.med.kuleuven.ac.be/biostat http://www.student.kuleuven.ac.be/~m0390867/dimitris.htm - Original Message - From: Prasanna Balaprakash [EMAIL PROTECTED] To: r-help@stat.math.ethz.ch Sent: Friday, January 21, 2005 12:27 PM Subject: [R] chi-Squared distribution Dear Rs: outer(1:3, 1:3, function(df1, df2) qf(0.95, df1, df2)) I compare this F distribution results with the table, the answers were perfect. But I need to see for chi-sqaured distribution. When I employed the similar formula outer(1:3, 1:3, function(df1, df2) qchisq(0.95, df1, df2)) , I am getting unexpected results. I need to see the following values: p=0.750 . 1 1.323 2 2.773 3 4.108 Thanking you Prasanna Prasanna Balaprakash, Université Libre de Bruxelles, 50, Av. F. Roosevelt, CP 194/6, B-1050 Brussels, Belgium. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] chi-Squared distribution in Friedman test
Easy: First letter d = density, p = probability = cumulative distribution function (cdf), q = quantile = inverse cdf, r = pseudo-random number generation. This is in, for example, An Introduction to R. Are you familiar with help.start()? From an R command prompt (at least with recent versions under Windows), this opens a web browser with a page offering manuals, reference, and Miscellaneous Material. An Intro to R is in the upper left. Probability Distributions is section 8. hope this helps. spencer graves Prasanna Balaprakash wrote: Dear R helpers: Thanks for the previous reply. I am using Friedman racing test. According the the book Pratical Nonprametric Statistic by WJ Conover, after computing the statistics, he suggested to use chi-squared or F distribution to accept or reject null hypothesis. After looking into the source code, I found that R uses chi-sqaured distribution as below: PVAL - pchisq(STATISTIC, PARAMETER, lower = FALSE) but still I cant figure out why they are using this pschisq insted of dchisq. Sorry I am wrong!! Thanking you truly Prasanna - Original Message - From: Dimitris Rizopoulos [EMAIL PROTECTED] To: Prasanna Balaprakash [EMAIL PROTECTED] Cc: r-help@stat.math.ethz.ch Sent: Friday, January 21, 2005 1:05 PM Subject: Re: [R] chi-Squared distribution if you check ?qchisq, you'll see that the second My mistake, the third argument argument of the function denotes the non-centrality parameter! Try qchisq(0.75, 1:3) to get your answer. Best, Dimitris Dimitris Rizopoulos Ph.D. Student Biostatistical Centre School of Public Health Catholic University of Leuven Address: Kapucijnenvoer 35, Leuven, Belgium Tel: +32/16/336899 Fax: +32/16/337015 Web: http://www.med.kuleuven.ac.be/biostat http://www.student.kuleuven.ac.be/~m0390867/dimitris.htm - Original Message - From: Prasanna Balaprakash [EMAIL PROTECTED] To: r-help@stat.math.ethz.ch Sent: Friday, January 21, 2005 12:27 PM Subject: [R] chi-Squared distribution Dear Rs: outer(1:3, 1:3, function(df1, df2) qf(0.95, df1, df2)) I compare this F distribution results with the table, the answers were perfect. But I need to see for chi-sqaured distribution. When I employed the similar formula outer(1:3, 1:3, function(df1, df2) qchisq(0.95, df1, df2)) , I am getting unexpected results. I need to see the following values: p=0.750 . 1 1.323 2 2.773 3 4.108 Thanking you Prasanna Prasanna Balaprakash, Université Libre de Bruxelles, 50, Av. F. Roosevelt, CP 194/6, B-1050 Brussels, Belgium. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
