Perhaps I misunderstand, but ?chisq.test explicitly says:
"If x is a matrix with at least two rows and columns, it is taken as a
two-dimensional contingency table: the entries of x must be
non-negative integers. Otherwise, x and y must be vectors or factors
of the same length; cases with missing v
Hi,
Is there a function where I can specify expected proportions for the
two-way table to
calculate the Chi-square test? chisq.test allows specifying only the
one-way table.
Otherwise, I will have to write the function, but I never trust myself not
to make a mess
programing.
Thanks,
Miloš
Miloš
uot;Sergio Ferreira Cardoso"
> Cc: "R-help list"
> Enviadas: Sábado, 21 De Janeiro de 2017 6:09:22
> Assunto: Re: [R] Chi-square test
> Dear Sergio,
>
> You appear to have asked this question twice on r-help.
>
> Anova() has no specific method for “gls
> On Jan 20, 2017, at 7:36 AM, Sergio Ferreira Cardoso
> wrote:
>
> Dear all,
>
> Anova() for .car package retrieves Chi-square statistics when I'm testing a
> model the significance of a multivariate .gls model
> gls(x~1+2+3+x,corBrownian(phy=tree), ...).
> Is this Chi-square a two-sided
Dear Sergio,
You appear to have asked this question twice on r-help.
Anova() has no specific method for “gls” models (I assume, though you don’t say
so, that the model is fit by gls() in the nlme package), but the default method
works and provides Wald chi-square tests for terms in the model. I
Dear all,
Anova() for .car package retrieves Chi-square statistics when I'm testing a
model the significance of a multivariate .gls model
gls(x~1+2+3+x,corBrownian(phy=tree), ...).
Is this Chi-square a two-sided test?
Thank you.
Best,
Sérgio.
[[alternative HTML version deleted]]
On Feb 20, 2015, at 10:05 AM, pari hesabi wrote:
> Hello,
> If the vector of observed frequencies is:
> f<-c(0,0,0,2,3,6,17,15,21,21,14,10,5,1,5)
> and the vector of probability :p11<-c(7.577864e-06, 1.999541e-04
> ,1.833510e-03, 9.059845e-03, 2.886977e-02, 6.546229e-02 ,1.124083e-01,
> 1.5
Original Message-
From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Berend Hasselman
Sent: Friday, February 20, 2015 12:13 PM
To: pari hesabi
Cc: r-help@r-project.org
Subject: Re: [R] Chi-square test
> On 20-02-2015, at 19:05, pari hesabi wrote:
>
> Hello,
> If the ve
> On 20-02-2015, at 19:05, pari hesabi wrote:
>
> Hello,
> If the vector of observed frequencies is:
> f<-c(0,0,0,2,3,6,17,15,21,21,14,10,5,1,5)
> and the vector of probability :p11<-c(7.577864e-06, 1.999541e-04
> ,1.833510e-03, 9.059845e-03, 2.886977e-02, 6.546229e-02 ,1.124083e-01,
> 1.5
Hello,
If the vector of observed frequencies is:
f<-c(0,0,0,2,3,6,17,15,21,21,14,10,5,1,5)
and the vector of probability :p11<-c(7.577864e-06, 1.999541e-04
,1.833510e-03, 9.059845e-03, 2.886977e-02, 6.546229e-02 ,1.124083e-01,
1.525880e-01, 1.689712e-01, 1.563522e-01, 1.232031e-01, 8.395000
r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Rick Bilonick
Sent: Monday, September 15, 2014 10:18 AM
To: r-help@r-project.org
Subject: Re: [R] chi-square test
On 09/15/2014 10:57 AM, eliza botto wrote:
> Dear useRs of R,
> I have two datasets (TT and SS) and i want
On 09/15/2014 10:57 AM, eliza botto wrote:
Dear useRs of R,
I have two datasets (TT and SS) and i wanted to to see if my data is uniformly
distributed or not?I tested it through chi-square test and results are given at the end
of it.Now apparently P-value has a significant importance but I cant
Dear useRs of R,
I have two datasets (TT and SS) and i wanted to to see if my data is uniformly
distributed or not?I tested it through chi-square test and results are given at
the end of it.Now apparently P-value has a significant importance but I cant
interpret the results and why it says that
On Mon, Jun 17, 2013 at 10:07 AM, peter dalgaard wrote:
>
> On Jun 17, 2013, at 10:36 , R. Michael Weylandt wrote:
>>
>> What do you mean 'results of individual cells'? As documented in
>> ?chisq.test, you might be looking for one or more of
>>
>> data.table$observed
>> data.table$expected
>> data
On Jun 17, 2013, at 10:36 , R. Michael Weylandt wrote:
> On Mon, Jun 17, 2013 at 9:14 AM, Dave Clark wrote:
>> I`m doing the chi square test in R, see below code:
>>
>>> row1 <- c(27,17,13,21,80,24,35,41,18,51) #Category A (1-10) counts
>>> row2 <- c(27,11,26,13,30,28,17,30,10,21) #Category B (
On Mon, Jun 17, 2013 at 9:14 AM, Dave Clark wrote:
> I`m doing the chi square test in R, see below code:
>
>> row1 <- c(27,17,13,21,80,24,35,41,18,51) #Category A (1-10) counts
>> row2 <- c(27,11,26,13,30,28,17,30,10,21) #Category B (1-10) counts
>> data.table <- rbind(row1,row2)
>> data.table
>
>
I`m doing the chi square test in R, see below code:
> row1 <- c(27,17,13,21,80,24,35,41,18,51) #Category A (1-10) counts
> row2 <- c(27,11,26,13,30,28,17,30,10,21) #Category B (1-10) counts
> data.table <- rbind(row1,row2)
> data.table
then:
> chisq.test(data.table)
This gives me the chi
ilto:r-help-bounces@r-
> project.org] On Behalf Of ghe...@mathnmaps.com
> Sent: Tuesday, October 11, 2011 1:32 PM
> To: r-help@r-project.org
> Subject: [R] Chi-Square test and survey results
>
> An organization has asked me to comment on the validity of their
> recent all-employe
George,
Perhaps the site of the RISQ project (Representativity indicators for
Survey Quality) might be of use: http://www.risq-project.eu/ . They
also provide R-code to calculate their indicators.
HTH,
Jan
Quoting ghe...@mathnmaps.com:
An organization has asked me to comment on the val
gheine wrote on 10/11/2011 02:31:46 PM:
>
> An organization has asked me to comment on the validity of their
> recent all-employee survey. Survey responses, by geographic region,
> compared
> with the total number of employees in each region, were as follows:
>
> > ByRegion
>All.Emp
An organization has asked me to comment on the validity of their
recent all-employee survey. Survey responses, by geographic region,
compared
with the total number of employees in each region, were as follows:
ByRegion
All.Employees Survey.Respondents
Region_1735
should not be 0 but we will ignore the
> warning.
> >
> > now it should generate p value for next row taking 35 and 84 (v1 and v2)
>
> > as observed and 22 and 84 (w1 and w2) as expected.so here it will do chi
>
> > square test for all 6 rows and will generate 6 p values.My data frame
>
If your data is d1:
temp <- apply(d1[,1:4], 1, order, decreasing=TRUE)[1:2,]
temp <- rbind(temp, temp+4)
result <- sapply(1:nrow(d1), function(i)
chisq.test(matrix(as.matrix(d1[i,temp[,i]]), ncol=2)))
Uwe Ligges
On 16.08.2011 23:26, Bansal, Vikas wrote:
Dear all,
I have been working on t
rows and will generate 6 p values.My data frame
has
> lot of rows(approx. ).
>
> Can you please help me with this.
>
>
>
> Thanking you,
> Warm Regards
> Vikas Bansal
> Msc Bioinformatics
> Kings College London
> ___________
7:11 PM
To: Bansal, Vikas
Cc: r-help@r-project.org
Subject: Re: [R] Chi square test on data frame
I think everything below is right, but it's all a little helter-skelter so take
it with a grain of salt:
First things first, make your data with dput() for the list.
Y = structure(c(0, 35, 0,
I think everything below is right, but it's all a little helter-skelter so
take it with a grain of salt:
First things first, make your data with dput() for the list.
Y = structure(c(0, 35, 0, 0, 0, 0, 84, 84, 0, 48, 84, 0, 22, 0, 0,
0, 0, 0, 10, 0, 48, 0, 0, 48, 0, 22, 0, 0, 0, 0, 84, 84, 0, 48,
Is there anyone who can help me with chi square test on data frame.I am
struggling from last 2 days.I will be very thankful to you.
Dear all,
I have been working on this problem from so many hours but did not find any
solution.
I have a data frame with 8 columns-
V1 V2 V3
Dear all,
I have been working on this problem from so many hours but did not find any
solution.
I have a data frame with 8 columns-
V1 V2 V3 V4 W1 W2W3 W4
1 084 22 10 0 84 0 0
235840
Hello!
Very sorry for a probably very simple question - I looked but did not
find an answer in the archives.
I have a table "counts" (below) that shows counts by Option within
each of my 2 groups. However, my groups have different sizes (N1=255
and N2=68). Table "prop" shows the resulting proporti
> Next time the launch of an incoming nuclear strike is detected,
> set them to work as follows (following Karl Pearson's historical
> precedent):
>
> "Anti-aircraft guns all day long": Computing for the
> Ministry of Munitions
> JUNE BARROW GREEN (Open University)
>From January 1
On 26-Nov-08 17:57:52, Andrew Choens wrote:
> [...]
> And, since I do work for government, if I ask for a roomful of
> calculators, I might just get them. And really, what am I going
> to do with a roomful of calculators?
>
> --andy
> Insert something humorous here. :-)
Next time the launch of a
On Thu, 2008-11-27 at 00:46 +0800, Berwin A Turlach wrote:
> Chuck explained already the reason for this small difference. I just
> take issue about it being an important difference. In my opinion,
> this difference is not important at all. It would only be important
> to people who are still st
G'day Andy,
On Wed, 26 Nov 2008 14:51:50 +
Andrew Choens <[EMAIL PROTECTED]> wrote:
> I was asked by my boss to do an analysis on a large data set, and I am
> trying to convince him to let me use R rather than SPSS.
Very laudable of you. :)
> This is the output from R:
> > chisq.test(test2
On 11/26/2008 9:51 AM, Andrew Choens wrote:
> I was asked by my boss to do an analysis on a large data set, and I am
> trying to convince him to let me use R rather than SPSS. I think Sweave
> could make my life much much easier. To get me a little closer to this
> goal, I ran my analysis through R
I was asked by my boss to do an analysis on a large data set, and I am
trying to convince him to let me use R rather than SPSS. I think Sweave
could make my life much much easier. To get me a little closer to this
goal, I ran my analysis through R and SPSS and compared the resulting
values. In all
On Tuesday 08 April 2008 17:04:16 Roslina Zakaria wrote:
> Hi R-users,
> I would like to find the goodness of fit using Chi-suare test for my data
> below: xobs=observed data, xtwe=predicted data using tweedie,
> xgam=predicted data using gamma
>
> > xobs <- c(223,46,12,5,7,17)
> > xtwe <- c(217.33
Hi R-users,
I would like to find the goodness of fit using Chi-suare test for my data below:
xobs=observed data, xtwe=predicted data using tweedie, xgam=predicted data
using gamma
> xobs <- c(223,46,12,5,7,17)
> xtwe <- c(217.33,39,14,18.33,6.67,14.67)
> xgam <- c(224.67,37.33,12.33,15.33,5.33,15)
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