The concrete problem is that I am refereeing
a paper where a confidence interval is presented for the risk ratio and I do not find it credible. I show below my attempts to do this in R. The example is slightly changed from the authors'. I can obtain a confidence interval for the odds ratio from fisher.test of course === fisher.test example === > outcome <- matrix(c(500, 0, 500, 8), ncol = 2, byrow = TRUE) > fisher.test(outcome) Fisher's Exact Test for Count Data data: outcome p-value = 0.00761 alternative hypothesis: true odds ratio is not equal to 1 95 percent confidence interval: 1.694792 Inf sample estimates: odds ratio Inf === end example === but in epidemiology authors often prefer to present risk ratios. Using the facility on CRAN to search the site I find packages epitools and Epi which both offer confidence intervals for the risk ratio === Epi example === > library(Epi) > twoby2(outcome[c(2,1),c(2,1)]) 2 by 2 table analysis: ------------------------------------------------------ Outcome : Col 1 Comparing : Row 1 vs. Row 2 Col 1 Col 2 P(Col 1) 95% conf. interval Row 1 8 500 0.0157 0.0079 0.0312 Row 2 0 500 0.0000 0.0000 NaN 95% conf. interval Relative Risk: Inf NaN Inf Sample Odds Ratio: Inf NaN Inf Conditional MLE Odds Ratio: Inf 1.6948 Inf Probability difference: 0.0157 0.0027 0.0337 Exact P-value: 0.0076 Asymptotic P-value: NaN ------------------------------------------------------ === end example === So Epi gives me a lower limit of NaN but the same confidence interval and p-value as fisher.test === epitools example === > library(epitools) > riskratio(outcome) $data Outcome Predictor Disease1 Disease2 Total Exposed1 500 0 500 Exposed2 500 8 508 Total 1000 8 1008 $measure risk ratio with 95% C.I. Predictor estimate lower upper Exposed1 1 NA NA Exposed2 Inf NaN Inf $p.value two-sided Predictor midp.exact fisher.exact chi.square Exposed1 NA NA NA Exposed2 0.00404821 0.007610478 0.004843385 $correction [1] FALSE attr(,"method") [1] "Unconditional MLE & normal approximation (Wald) CI" Warning message: Chi-squared approximation may be incorrect in: chisq.test(xx, correct = correction) === end example === And epitools also gives a lower limit of NaN. === end all examples === I would prefer not to have to tell the authors of the paper I am refereeing that I think they are wrong unless I can help them with what they should have done. Is there another package I should have tried? Is there some other way of doing this? Am I doing something fundamentally wrong-headed? Michael Dewey http://www.aghmed.fsnet.co.uk ______________________________________________ [hidden email] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. |
Hello,
A common way to calculate the CI for a relative risk is the following. Given a 2x2 table of the form: a b c d the log of the risk ratio is given by: lrr = log[ a/(a+b) / ( c/(c+d) ) ] which is asymptotically normal with variance: vlrr = 1/a - 1/(a+b) + 1/c - 1/(c+d). So an approximate 95% CI for the risk ratio is given by: exp[ lrr - 1.96*sqrt(vlrr) ], exp[ lrr + 1.96*sqrt(vlrr) ]. A common convention is to add 1/2 to each cell when there are zeros. So, for the table: Col 1 Col 2 Row 1 8 500 Row 2 0 500 lrr = log[ 8.5/509 / ( 0.5/501 ) ] = 2.817 vllr = 1/8.5 - 1/509 + 1/0.5 - 1/501 = 2.1137 exp[ 2.817-1.96*sqrt(2.1137) ] = .97 exp[ 2.817+1.96*sqrt(2.1137) ] = 289.04 Maybe that is what the authors did. Best, -- Wolfgang Viechtbauer Department of Methodology and Statistics University of Maastricht, The Netherlands http://www.wvbauer.com/ > -----Original Message----- > From: [hidden email] [mailto:r-help- > [hidden email]] On Behalf Of Michael Dewey > Sent: Friday, November 10, 2006 15:43 > To: [hidden email] > Subject: [R] Confidence interval for relative risk > > The concrete problem is that I am refereeing > a paper where a confidence interval is > presented for the risk ratio and I do not find > it credible. I show below my attempts to > do this in R. The example is slightly changed > from the authors'. > > I can obtain a confidence interval for > the odds ratio from fisher.test of > course > > === fisher.test example === > > > outcome <- matrix(c(500, 0, 500, 8), ncol = 2, byrow = TRUE) > > fisher.test(outcome) > > Fisher's Exact Test for Count Data > > data: outcome > p-value = 0.00761 > alternative hypothesis: true odds ratio is not equal to 1 > 95 percent confidence interval: > 1.694792 Inf > sample estimates: > odds ratio > Inf > > === end example === > > but in epidemiology authors often > prefer to present risk ratios. > > Using the facility on CRAN to search > the site I find packages epitools and Epi > which both offer confidence intervals > for the risk ratio > > === Epi example === > > > library(Epi) > > twoby2(outcome[c(2,1),c(2,1)]) > 2 by 2 table analysis: > ------------------------------------------------------ > Outcome : Col 1 > Comparing : Row 1 vs. Row 2 > > Col 1 Col 2 P(Col 1) 95% conf. interval > Row 1 8 500 0.0157 0.0079 0.0312 > Row 2 0 500 0.0000 0.0000 NaN > > 95% conf. interval > Relative Risk: Inf NaN Inf > Sample Odds Ratio: Inf NaN Inf > Conditional MLE Odds Ratio: Inf 1.6948 Inf > Probability difference: 0.0157 0.0027 0.0337 > > Exact P-value: 0.0076 > Asymptotic P-value: NaN > ------------------------------------------------------ > > === end example === > > So Epi gives me a lower limit of NaN but the same confidence > interval and p-value as fisher.test > > === epitools example === > > > library(epitools) > > riskratio(outcome) > $data > Outcome > Predictor Disease1 Disease2 Total > Exposed1 500 0 500 > Exposed2 500 8 508 > Total 1000 8 1008 > > $measure > risk ratio with 95% C.I. > Predictor estimate lower upper > Exposed1 1 NA NA > Exposed2 Inf NaN Inf > > $p.value > two-sided > Predictor midp.exact fisher.exact chi.square > Exposed1 NA NA NA > Exposed2 0.00404821 0.007610478 0.004843385 > > $correction > [1] FALSE > > attr(,"method") > [1] "Unconditional MLE & normal approximation (Wald) CI" > Warning message: > Chi-squared approximation may be incorrect in: chisq.test(xx, correct = > correction) > > === end example === > > And epitools also gives a lower limit > of NaN. > > === end all examples === > > I would prefer not to have to tell the authors of the > paper I am refereeing that > I think they are wrong unless I can help them with what they > should have done. > > Is there another package I should have tried? > > Is there some other way of doing this? > > Am I doing something fundamentally wrong-headed? > > > > Michael Dewey > http://www.aghmed.fsnet.co.uk ______________________________________________ [hidden email] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. |
At 15:54 10/11/2006, Viechtbauer Wolfgang (STAT) wrote:
Thanks for the suggestion Wolfgang, but whatever the original authors did that is not it. >Hello, > >A common way to calculate the CI for a relative risk is the >following. Given a 2x2 table of the form: > >a b >c d > >the log of the risk ratio is given by: > >lrr = log[ a/(a+b) / ( c/(c+d) ) ] > >which is asymptotically normal with variance: > >vlrr = 1/a - 1/(a+b) + 1/c - 1/(c+d). > >So an approximate 95% CI for the risk ratio is given by: > >exp[ lrr - 1.96*sqrt(vlrr) ], exp[ lrr + 1.96*sqrt(vlrr) ]. > >A common convention is to add 1/2 to each cell when there are zeros. > >So, for the table: > > Col 1 Col 2 >Row 1 8 500 >Row 2 0 500 > >lrr = log[ 8.5/509 / ( 0.5/501 ) ] = 2.817 >vllr = 1/8.5 - 1/509 + 1/0.5 - 1/501 = 2.1137 > >exp[ 2.817-1.96*sqrt(2.1137) ] = .97 >exp[ 2.817+1.96*sqrt(2.1137) ] = 289.04 > >Maybe that is what the authors did. > >Best, > >-- >Wolfgang Viechtbauer > Department of Methodology and Statistics > University of Maastricht, The Netherlands > http://www.wvbauer.com/ > > > > -----Original Message----- > > From: [hidden email] [mailto:r-help- > > [hidden email]] On Behalf Of Michael Dewey > > Sent: Friday, November 10, 2006 15:43 > > To: [hidden email] > > Subject: [R] Confidence interval for relative risk > > > > The concrete problem is that I am refereeing > > a paper where a confidence interval is > > presented for the risk ratio and I do not find > > it credible. I show below my attempts to > > do this in R. The example is slightly changed > > from the authors'. > > > > I can obtain a confidence interval for > > the odds ratio from fisher.test of > > course > > > > === fisher.test example === > > > > > outcome <- matrix(c(500, 0, 500, 8), ncol = 2, byrow = TRUE) > > > fisher.test(outcome) > > > > Fisher's Exact Test for Count Data > > > > data: outcome > > p-value = 0.00761 > > alternative hypothesis: true odds ratio is not equal to 1 > > 95 percent confidence interval: > > 1.694792 Inf > > sample estimates: > > odds ratio > > Inf > > > > === end example === > > > > but in epidemiology authors often > > prefer to present risk ratios. > > > > Using the facility on CRAN to search > > the site I find packages epitools and Epi > > which both offer confidence intervals > > for the risk ratio > > > > === Epi example === > > > > > library(Epi) > > > twoby2(outcome[c(2,1),c(2,1)]) > > 2 by 2 table analysis: > > ------------------------------------------------------ > > Outcome : Col 1 > > Comparing : Row 1 vs. Row 2 > > > > Col 1 Col 2 P(Col 1) 95% conf. interval > > Row 1 8 500 0.0157 0.0079 0.0312 > > Row 2 0 500 0.0000 0.0000 NaN > > > > 95% conf. interval > > Relative Risk: Inf NaN Inf > > Sample Odds Ratio: Inf NaN Inf > > Conditional MLE Odds Ratio: Inf 1.6948 Inf > > Probability difference: 0.0157 0.0027 0.0337 > > > > Exact P-value: 0.0076 > > Asymptotic P-value: NaN > > ------------------------------------------------------ > > > > === end example === > > > > So Epi gives me a lower limit of NaN but the same confidence > > interval and p-value as fisher.test > > > > === epitools example === > > > > > library(epitools) > > > riskratio(outcome) > > $data > > Outcome > > Predictor Disease1 Disease2 Total > > Exposed1 500 0 500 > > Exposed2 500 8 508 > > Total 1000 8 1008 > > > > $measure > > risk ratio with 95% C.I. > > Predictor estimate lower upper > > Exposed1 1 NA NA > > Exposed2 Inf NaN Inf > > > > $p.value > > two-sided > > Predictor midp.exact fisher.exact chi.square > > Exposed1 NA NA NA > > Exposed2 0.00404821 0.007610478 0.004843385 > > > > $correction > > [1] FALSE > > > > attr(,"method") > > [1] "Unconditional MLE & normal approximation (Wald) CI" > > Warning message: > > Chi-squared approximation may be incorrect in: chisq.test(xx, correct = > > correction) > > > > === end example === > > > > And epitools also gives a lower limit > > of NaN. > > > > === end all examples === > > > > I would prefer not to have to tell the authors of the > > paper I am refereeing that > > I think they are wrong unless I can help them with what they > > should have done. > > > > Is there another package I should have tried? > > > > Is there some other way of doing this? > > > > Am I doing something fundamentally wrong-headed? > > > > > > > > Michael Dewey > > http://www.aghmed.fsnet.co.uk Michael Dewey http://www.aghmed.fsnet.co.uk ______________________________________________ [hidden email] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. |
Michael Dewey <[hidden email]> writes:
> At 15:54 10/11/2006, Viechtbauer Wolfgang (STAT) wrote: > > Thanks for the suggestion Wolfgang, but whatever the original authors > did that is not it. Did you ever say what result they got? -p > >Hello, > > > >A common way to calculate the CI for a relative risk is the > >following. Given a 2x2 table of the form: > > > >a b > >c d > > > >the log of the risk ratio is given by: > > > >lrr = log[ a/(a+b) / ( c/(c+d) ) ] > > > >which is asymptotically normal with variance: > > > >vlrr = 1/a - 1/(a+b) + 1/c - 1/(c+d). > > > >So an approximate 95% CI for the risk ratio is given by: > > > >exp[ lrr - 1.96*sqrt(vlrr) ], exp[ lrr + 1.96*sqrt(vlrr) ]. > > > >A common convention is to add 1/2 to each cell when there are zeros. > > > >So, for the table: > > > > Col 1 Col 2 > >Row 1 8 500 > >Row 2 0 500 > > > >lrr = log[ 8.5/509 / ( 0.5/501 ) ] = 2.817 > >vllr = 1/8.5 - 1/509 + 1/0.5 - 1/501 = 2.1137 > > > >exp[ 2.817-1.96*sqrt(2.1137) ] = .97 > >exp[ 2.817+1.96*sqrt(2.1137) ] = 289.04 > > > >Maybe that is what the authors did. > > > >Best, > > > >-- > >Wolfgang Viechtbauer > > Department of Methodology and Statistics > > University of Maastricht, The Netherlands > > http://www.wvbauer.com/ > > > > > > > -----Original Message----- > > > From: [hidden email] [mailto:r-help- > > > [hidden email]] On Behalf Of Michael Dewey > > > Sent: Friday, November 10, 2006 15:43 > > > To: [hidden email] > > > Subject: [R] Confidence interval for relative risk > > > > > > The concrete problem is that I am refereeing > > > a paper where a confidence interval is > > > presented for the risk ratio and I do not find > > > it credible. I show below my attempts to > > > do this in R. The example is slightly changed > > > from the authors'. > > > > > > I can obtain a confidence interval for > > > the odds ratio from fisher.test of > > > course > > > > > > === fisher.test example === > > > > > > > outcome <- matrix(c(500, 0, 500, 8), ncol = 2, byrow = TRUE) > > > > fisher.test(outcome) > > > > > > Fisher's Exact Test for Count Data > > > > > > data: outcome > > > p-value = 0.00761 > > > alternative hypothesis: true odds ratio is not equal to 1 > > > 95 percent confidence interval: > > > 1.694792 Inf > > > sample estimates: > > > odds ratio > > > Inf > > > > > > === end example === > > > > > > but in epidemiology authors often > > > prefer to present risk ratios. > > > > > > Using the facility on CRAN to search > > > the site I find packages epitools and Epi > > > which both offer confidence intervals > > > for the risk ratio > > > > > > === Epi example === > > > > > > > library(Epi) > > > > twoby2(outcome[c(2,1),c(2,1)]) > > > 2 by 2 table analysis: > > > ------------------------------------------------------ > > > Outcome : Col 1 > > > Comparing : Row 1 vs. Row 2 > > > > > > Col 1 Col 2 P(Col 1) 95% conf. interval > > > Row 1 8 500 0.0157 0.0079 0.0312 > > > Row 2 0 500 0.0000 0.0000 NaN > > > > > > 95% conf. interval > > > Relative Risk: Inf NaN Inf > > > Sample Odds Ratio: Inf NaN Inf > > > Conditional MLE Odds Ratio: Inf 1.6948 Inf > > > Probability difference: 0.0157 0.0027 0.0337 > > > > > > Exact P-value: 0.0076 > > > Asymptotic P-value: NaN > > > ------------------------------------------------------ > > > > > > === end example === > > > > > > So Epi gives me a lower limit of NaN but the same confidence > > > interval and p-value as fisher.test > > > > > > === epitools example === > > > > > > > library(epitools) > > > > riskratio(outcome) > > > $data > > > Outcome > > > Predictor Disease1 Disease2 Total > > > Exposed1 500 0 500 > > > Exposed2 500 8 508 > > > Total 1000 8 1008 > > > > > > $measure > > > risk ratio with 95% C.I. > > > Predictor estimate lower upper > > > Exposed1 1 NA NA > > > Exposed2 Inf NaN Inf > > > > > > $p.value > > > two-sided > > > Predictor midp.exact fisher.exact chi.square > > > Exposed1 NA NA NA > > > Exposed2 0.00404821 0.007610478 0.004843385 > > > > > > $correction > > > [1] FALSE > > > > > > attr(,"method") > > > [1] "Unconditional MLE & normal approximation (Wald) CI" > > > Warning message: > > > Chi-squared approximation may be incorrect in: chisq.test(xx, correct = > > > correction) > > > > > > === end example === > > > > > > And epitools also gives a lower limit > > > of NaN. > > > > > > === end all examples === > > > > > > I would prefer not to have to tell the authors of the > > > paper I am refereeing that > > > I think they are wrong unless I can help them with what they > > > should have done. > > > > > > Is there another package I should have tried? > > > > > > Is there some other way of doing this? > > > > > > Am I doing something fundamentally wrong-headed? > > > > > > > > > > > > Michael Dewey > > > http://www.aghmed.fsnet.co.uk > > Michael Dewey > http://www.aghmed.fsnet.co.uk > > ______________________________________________ > [hidden email] mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > -- O__ ---- Peter Dalgaard Øster Farimagsgade 5, Entr.B c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~~~~~~~~~ - ([hidden email]) FAX: (+45) 35327907 ______________________________________________ [hidden email] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. |
At 12:35 12/11/2006, Peter Dalgaard wrote:
>Michael Dewey <[hidden email]> writes: > > > At 15:54 10/11/2006, Viechtbauer Wolfgang (STAT) wrote: > > > > Thanks for the suggestion Wolfgang, but whatever the original authors > > did that is not it. > >Did you ever say what result they got? > > -p > No, because I did not want to use the original numbers in the request. So as the snippet below indicates I changed the numbers. If I apply Wolfgang's suggestion (which I had already thought of but discarded) I get about 13 for the real example where the authors quote about 5. My question still remains though as to how I can get a confidence interval for the risk ratio without adding a constant to each cell. > > > > it credible. I show below my attempts to > > > > do this in R. The example is slightly changed > > > > from the authors'. Michael Dewey http://www.aghmed.fsnet.co.uk ______________________________________________ [hidden email] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. |
When I have refereed manuscripts for publication and have been
unable to get their answers, I have told the authors they need more explanation of their methodology -- while summarizing what I tried in a few lines. I've even told some that it would make it vastly easier for the referees and increase the potential readership for their article if they make R code available -- at least downloadable somewhere and preferably in a contributed R package, where it could attract interest in the article from audiences who would not likely find it any other way. I've done this for articles that were NOT written to specifically describe statistical software. I have not followed all of this thread. However, one suggestion I saw looked like it used the "delta method", if I understood correctly from skimming without studying the details carefully. Have you also considered 2*log(likelihood ratio) being approximately chi-square? Just my 2e-9 Euros (or 2e-7 Yen or Yuan). Spencer Graves Michael Dewey wrote: > At 12:35 12/11/2006, Peter Dalgaard wrote: > >> Michael Dewey <[hidden email]> writes: >> >> >>> At 15:54 10/11/2006, Viechtbauer Wolfgang (STAT) wrote: >>> >>> Thanks for the suggestion Wolfgang, but whatever the original authors >>> did that is not it. >>> >> Did you ever say what result they got? >> >> -p >> >> > > No, because I did not want to use the original numbers in the > request. So as the snippet below indicates I changed the numbers. If > I apply Wolfgang's suggestion (which I had already thought of but > discarded) I get about 13 for the real example where the authors quote about 5. > > My question still remains though as to how I can get a confidence > interval for the risk ratio without adding a constant to each cell. > > >>>>> it credible. I show below my attempts to >>>>> do this in R. The example is slightly changed >>>>> from the authors'. >>>>> > > Michael Dewey > http://www.aghmed.fsnet.co.uk > > ______________________________________________ > [hidden email] mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > ______________________________________________ [hidden email] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. |
In reply to this post by Michael Dewey
At 14:43 10/11/2006, Michael Dewey wrote:
After considerable help from list members and some digging of my own I have prepared a summary of the findings which I have posted (see link below). Broadly there were four suggestions 1 - Wald-type intervals, 2 - transforming the odds ratio confidence interval, 3 - method based on score test, 4 - method based on likelihood. Method 3 was communicated to me off-list =========================== I haven't followed all of the thread either but has someone already pointed out to you confidence intervals that use the score method? For example Agresti (Categorical Data Analysis 2nd edition, p. 77-78) note that 'although computationally more complex, these methods often perform better'. However, they also note that 'currently they are not available in standard software'. But then again, R is not standard software: the code (riskscoreci) can be found from here: http://www.stat.ufl.edu/~aa/cda/R/two_sample/R2/index.html best regards, Jukka Jokinen ================================================ and so I reproduce it here. Almost a candidate for the fortunes package there. The other three can be found from the archive under the same subject although not all in the same thread. Methods 3 and 4 seem to have more going for them as far as I can judge. Thanks to David Duffy, Spencer Graves, Jukka Jokinen, Terry Therneau, and Wolfgan Viechtbauer. The views and calculations presented here and in the summary are my own and any errors are my responsibility not theirs. The summary document is available from here http://www.zen103156.zen.co.uk/rr.pdf Original post follows. ==================================== >The concrete problem is that I am refereeing >a paper where a confidence interval is >presented for the risk ratio and I do not find >it credible. I show below my attempts to >do this in R. The example is slightly changed >from the authors'. > >I can obtain a confidence interval for >the odds ratio from fisher.test of >course > >=== fisher.test example === > > > outcome <- matrix(c(500, 0, 500, 8), ncol = 2, byrow = TRUE) > > fisher.test(outcome) > > Fisher's Exact Test for Count Data > >data: outcome >p-value = 0.00761 >alternative hypothesis: true odds ratio is not equal to 1 >95 percent confidence interval: > 1.694792 Inf >sample estimates: >odds ratio > Inf > >=== end example === > >but in epidemiology authors often >prefer to present risk ratios. > >Using the facility on CRAN to search >the site I find packages epitools and Epi >which both offer confidence intervals >for the risk ratio > >=== Epi example === > > > library(Epi) > > twoby2(outcome[c(2,1),c(2,1)]) >2 by 2 table analysis: >------------------------------------------------------ >Outcome : Col 1 >Comparing : Row 1 vs. Row 2 > > Col 1 Col 2 P(Col 1) 95% conf. interval >Row 1 8 500 0.0157 0.0079 0.0312 >Row 2 0 500 0.0000 0.0000 NaN > > 95% conf. interval > Relative Risk: Inf NaN Inf > Sample Odds Ratio: Inf NaN Inf >Conditional MLE Odds Ratio: Inf 1.6948 Inf > Probability difference: 0.0157 0.0027 0.0337 > > Exact P-value: 0.0076 > Asymptotic P-value: NaN >------------------------------------------------------ > >=== end example === > >So Epi gives me a lower limit of NaN but the same confidence >interval and p-value as fisher.test > >=== epitools example === > > > library(epitools) > > riskratio(outcome) >$data > Outcome >Predictor Disease1 Disease2 Total > Exposed1 500 0 500 > Exposed2 500 8 508 > Total 1000 8 1008 > >$measure > risk ratio with 95% C.I. >Predictor estimate lower upper > Exposed1 1 NA NA > Exposed2 Inf NaN Inf > >$p.value > two-sided >Predictor midp.exact fisher.exact chi.square > Exposed1 NA NA NA > Exposed2 0.00404821 0.007610478 0.004843385 > >$correction >[1] FALSE > >attr(,"method") >[1] "Unconditional MLE & normal approximation (Wald) CI" >Warning message: >Chi-squared approximation may be incorrect in: chisq.test(xx, correct = >correction) > >=== end example === > >And epitools also gives a lower limit >of NaN. > >=== end all examples === > >I would prefer not to have to tell the authors of the >paper I am refereeing that >I think they are wrong unless I can help them with what they >should have done. > >Is there another package I should have tried? > >Is there some other way of doing this? > >Am I doing something fundamentally wrong-headed? > > > Michael Dewey http://www.aghmed.fsnet.co.uk ______________________________________________ [hidden email] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. |
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