When I use a table, from a Schaum book, I see that for the 95
percentile, with v_1 = 1 and v_2 = 1 the value is 161. In the modern era,
looking values up in a table is less than ideal. Therefore, I would
expect R to have a function to do this and based upon my
reading of the documentation, I would expect the following call to get
the value I expect:
pf( .95,1, 1)
However, it produces
0.4918373
Therefore, I conclude that I am using the wrong function. What function
should I use?
Thanks
Bob
Dear Bob, You want...> qf( .95,1, 1)[1] 161.4476 Best, Fraser -----Original Message----- From: Robert Sherry [mailto:rsherry8 at comcast.net] Sent: Monday, December 21, 2015 2:51 PM To: R Project Help Subject: [R] F Distribution When I use a table, from a Schaum book, I see that for the 95 percentile, with v_1 = 1 and v_2 = 1 the value is 161. In the modern era, looking values up in a table is less than ideal. Therefore, I would expect R to have a function to do this and based upon my reading of the documentation, I would expect the following call to get the value I expect: pf( .95,1, 1) However, it produces 0.4918373 Therefore, I conclude that I am using the wrong function. What function should I use? Thanks Bob ______________________________________________ R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see 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.
You want to use qf which gives you the value at a given percentile. pf gives you the p-value for a given value of F (inverse)> qf(0.95, 1, 1)[1] 161.4476> pf(161.4476, 1, 1)[1] 0.95 Peter On Mon, Dec 21, 2015 at 11:51 AM, Robert Sherry <rsherry8 at comcast.net> wrote:> > When I use a table, from a Schaum book, I see that for the 95 percentile, > with v_1 = 1 and v_2 = 1 the value is 161. In the modern era, > looking values up in a table is less than ideal. Therefore, I would expect R > to have a function to do this and based upon my > reading of the documentation, I would expect the following call to get the > value I expect: > pf( .95,1, 1) > However, it produces > 0.4918373 > Therefore, I conclude that I am using the wrong function. What function > should I use? > > Thanks > Bob > > ______________________________________________ > R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see > 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.
qf() (as you might have figured from the help page for pf()....) -pd> On 21 Dec 2015, at 20:51 , Robert Sherry <rsherry8 at comcast.net> wrote: > > > When I use a table, from a Schaum book, I see that for the 95 percentile, with v_1 = 1 and v_2 = 1 the value is 161. In the modern era, > looking values up in a table is less than ideal. Therefore, I would expect R to have a function to do this and based upon my > reading of the documentation, I would expect the following call to get the value I expect: > pf( .95,1, 1) > However, it produces > 0.4918373 > Therefore, I conclude that I am using the wrong function. What function should I use? > > Thanks > Bob > > ______________________________________________ > R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see > 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.-- Peter Dalgaard, Professor, Center for Statistics, Copenhagen Business School Solbjerg Plads 3, 2000 Frederiksberg, Denmark Phone: (+45)38153501 Office: A 4.23 Email: pd.mes at cbs.dk Priv: PDalgd at gmail.com