x=c(3.05176E-05,0.000457764,0.003204346,0.0138855,0.04165649,0.09164429,0.1527405,0.1963806,0.1963806,0.1527405,0.09164429,0.04165649,0.0138855,0.003204346,0.000457764,3.05176E-05)
// R 2.3.1
Can someone please explain why this error returns?
y=numeric(100)
x=matrix(runif(16),4,4)
for(i in 2:100)
+{
+ y[i]=which(rmultinom(1, size = 1, prob = x[y[i-1], ])==1)
+}
Error in rmultinom(n, size, prob) : too few positive probabilities
thx much
ej
An apparatus exists whereby a collection of balls is displaced to the
top of a stack by suction. A top level (Level 1) each ball is shifted
1 unit to the left or 1 unit to the right at random with equal
probability. The ball then drops down to level 2. At Level 2, each
ball is again shifted 1 unit
-square distribution (pchisq).
I am a newbie in R. Your help will be greatly appreciated.
Thx
ej
On 12/5/06, Don McKenzie [EMAIL PROTECTED] wrote:
Ethan Johnsons wrote:
If we use this data as an example, does ks.test still valid?
E.Coli GroupObservedExpected
A5777.9
B
On 12/4/06, Bernardo Rangel tura [EMAIL PROTECTED] wrote:
At 03:55 AM 12/3/2006, Ethan Johnsons wrote:
I have the clinical study data.
Year 0 Year 3
Retinol (nmol/L)N Mean +-sd Mean +-sd
Vitamin A group 73
Do you know/have a function that takes a vector x and provides a
returned p-value that uses the Chi-Square Goodness-of-Fit test to test
the goodness of fit of a standard normal distribution.
Awaiting your positive reply.
Thx
ej
__
want to do this test.
HTH,
Simon.
Ethan Johnsons wrote:
Do you know/have a function that takes a vector x and provides a
returned p-value that uses the Chi-Square Goodness-of-Fit test to test
the goodness of fit of a standard normal distribution.
Awaiting your positive reply.
Thx
ej
I don't find any other test avail for this?
Am I missing something?
thx
ej
On 12/3/06, Michael Dewey [EMAIL PROTECTED] wrote:
At 05:55 03/12/2006, Ethan Johnsons wrote:
I have the clinical study data.
Year 0 Year 3
Retinol (nmol/L
I am getting Chi-squared approximation may be incorrect in:
chisq.test(x) with the data bleow.
Frequency distribution of number of male offspring in families of size 5.
Number of Male OffspringN
0 518
12245
2
I have the clinical study data.
Year 0 Year 3
Retinol (nmol/L)N Mean +-sd Mean +-sd
Vitamin A group 73 1.89+-0.36 2.06+-0.53
Trace group57 1.83+-0.31 1.78+-0.30
where N is the number
Does anyone have/know how to write the power function for z tests
something like below?
function(a,m0,m1,n,s){
t1 = -qnorm(1-a)
num = abs(m0-m1) * sqrt(n)
t2 = num/s
pow = pnorm(t1 + t2)
}
thx much
ej
__
R-help@stat.math.ethz.ch mailing list
What would be the R formulae for a two-sided test?
I have a formula for a one-sided test:
powertest - function(a,m0,m1,n,s){
t1 = -qnorm(1-a)
num = abs(m0-m1) * sqrt(n)
t2 = num/s
pow = pnorm(t1 + t2)
}
Would you pls let me know if you know of?
Thank you,
ej
)
Thank you,
ej
On 27 Oct 2006 16:37:08 +0200, Peter Dalgaard [EMAIL PROTECTED] wrote:
Ethan Johnsons [EMAIL PROTECTED] writes:
What would be the R formulae for a two-sided test?
I have a formula for a one-sided test:
powertest - function(a,m0,m1,n,s){
t1 = -qnorm(1-a)
num = abs(m0-m1
considered what they do not say. ~William W. Watt
A statistical analysis, properly conducted, is a delicate dissection of
uncertainties, a surgery of suppositions. ~M.J.Moroney
-Oorspronkelijk bericht-
Van: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Namens Ethan Johnsons
Verzonden
PROTECTED] [mailto:[EMAIL PROTECTED] Namens Ethan Johnsons
Verzonden: vrijdag 27 oktober 2006 16:59
Aan: Peter Dalgaard
CC: r-help@stat.math.ethz.ch
Onderwerp: Re: [R] Power of test
Thank you so mcuh for the explanation, Chuck Peter.
Two quick questions,please.
It states that delta
Thank you so much.
I see your point.
ej
On 10/20/06, Petr Pikal [EMAIL PROTECTED] wrote:
Hi
On 19 Oct 2006 at 13:19, Ethan Johnsons wrote:
Date sent: Thu, 19 Oct 2006 13:19:40 -0400
From: Ethan Johnsons [EMAIL PROTECTED]
To: Chuck Cleland
A quick question, please.
46 e coli lab samples are tested, 6 of them returned positive.
So, the best point estimate for p is 6/46 = 0.1304348.
For a 95% CI for p, I thought binom.test would give me the correct
result, but it seems it is not the right function to use. What is
the R
From: Ethan Johnsons [EMAIL PROTECTED]
To: r-help@stat.math.ethz.ch
Subject: [R] binom.test
Date: Fri, 20 Oct 2006 17:18:02 -0400
A quick question, please.
46 e coli lab samples are tested, 6 of them returned positive.
So, the best point estimate for p is 6/46 = 0.1304348
.
- Message d'origine
De : Ethan Johnsons [EMAIL PROTECTED]
À : Liaw, Andy [EMAIL PROTECTED]
Cc : r-help@stat.math.ethz.ch
Envoyé le : Jeudi, 19 Octobre 2006, 5h43mn 36s
Objet : Re: [R] CI
Thx so much.
I just got into R world for my small research.
I thought that R is free so
R-experts:
A quick question, please.
From a lab exp, I got 12 positives out of 50.
To get 90% CI for this , I think binom.test might be the one to be used.
Is there a better way or function to calculate this?
binom.test(x=12, n=50, p=12/50, conf.level = 0.90)
Exact binomial test
Please let me ask you another quick question.
I have results for e coli, and am trying to get 95% CI with the sd (1.783956).
I got the result from another tool as (1.21, 3.42).
But, I like to verify it with R. What function do you use for this?
Thx much for the feedback.
It is a big help.
ej
On 10/19/06, Ted Harding [EMAIL PROTECTED] wrote:
On 19-Oct-06 Ethan Johnsons wrote:
R-experts:
A quick question, please.
From a lab exp, I got 12 positives out of 50.
To get 90% CI for this , I think binom.test might be
the one
Thx so much, Chuck.
R is a reallt sweet tool to use..
I can try many things with i.e. intervals (gls(e.coli ~ 1), level=0.90).
It is a great help.
ej
On 10/19/06, Chuck Cleland [EMAIL PROTECTED] wrote:
Ethan Johnsons wrote:
Please let me ask you another quick question.
I have results
On 10/19/06, Chuck Cleland [EMAIL PROTECTED] wrote:
Ethan Johnsons wrote:
Please let me ask you another quick question.
I have results for e coli, and am trying to get 95% CI with the sd
(1.783956).
I got the result from another tool as (1.21, 3.42).
But, I like to verify it with R. What
I have a quick question, please.
Does R have function to compute i.e. a 90% confidence interval for the
mean for these numbers?
mean (6,11,5,14,30,11,17,3,9,3,8,8)
[1] 6
I thought pt or qt would give me the interval, but it seems not.
thx much.
ej
of success
0.44
I hope these are not HW problems?
Andy
From: Ethan Johnsons
Thank you so much for the feedback.
The random numbers are working great. I have tried
non-random numbers, and the outcome is not correct with confint.
Is there a way to compute i.e
What is the way to calculate the probability that between 1 and 5?
i.e.
We can do:
pbinom(q=1,size=1000,prob=0.0005,lower.tail=FALSE)
[1] 0.09016608
pbinom(q=5,size=1000,prob=0.0005,lower.tail=FALSE)
[1] 1.398798e-05
pbinom(q=1:5,size=1000,prob=0.0005,lower.tail=FALSE)
[1] 9.016608e-02
33 died overa 10-year period from COPD, whereas only 24 such deaths
could be expected based onstatewide mortality rates.
I came up with ppois(q = 33, lambda = 24, lower.tail = FALSE),
but it seems it is ppois(q = 32, lambda = 24, lower.tail = FALSE).
I am confused with the q value. Why is it 32
thx so much..Maciej .
ej
On 9/23/06, Maciej Bliziński [EMAIL PROTECTED] wrote:
On Sat, 2006-09-23 at 01:37 -0400, Ethan Johnsons wrote:
How do you change the Y-axis from 0 ~ 0.6?
plot(..., ylim = c(0, 0.6))
--
Maciej Bliziński
http://automatthias.wordpress.com
Is there a way to plug in the random samples, rnorm(100), in this plot?
I am using 100, but want it to be randome samples.
plot(seq(from=0,to=100,by=1),1-pbinom
(seq(from=0,to=100,by=1),size=100,prob=0.05),pch=15)
thx much
__
R-help@stat.math.ethz.ch
With this:
plot(seq(from=0,to=10,by=1),1-pbinom
(seq(from=0,to=10,by=1),size=6,prob=0.50),pch=15)
How do you change the Y-axis from 0 ~ 0.6?
I only get 0.0 ~1.0.
thx much
__
R-help@stat.math.ethz.ch mailing list
I have this data and calculated binom, but am very confused with the value
(at least 5, q =4, q=5, q =6??) and parameters? Can someone explain how to
plug in the va;ues and parameters?
Hospital # Tested # Positive # Positive (per 1000) A 3741 30 8 C 5006
11 2.2
If 500 newborns are
The expected number of bladder cancer over next 20 years a tire
industry is 1.8. Poission distribution is assumed to hold and 6
reported deaths are caused by bladder cancer among the employees.
Trying to find how unusual this event is.
ppois(q=6, lambda=1.8, lower.tail = TRUE, log.p = FALSE)
A quick question!
The number of episodes per year of otitis media follows a Possion
distribution with lambda = 1.6 episodes per year. Wouldn't the
probability of getting 3 or more episodes of otitis media in the first
2 years of life be:
ppois(q=3, lambda=1.6*2, lower.tail = TRUE, log.p =
A quick question, please.
x = c(0.0001, 0.0059, 0.0855, 0.9082)
y = c(0.54, 0.813, 0.379, 0.35)
where A = 1st set, B = 2nd set, C = 3rd set, D = 4th set respectivley.
How do you make hist plot side by side for x y?
i.e. 0.0001 and then to the right 0.54, 0.0059 and then to the right 0.813,
thx so much, Marc.
ej
On 9/14/06, Marc Schwartz [EMAIL PROTECTED] wrote:
On Thu, 2006-09-14 at 19:37 -0400, Ethan Johnsons wrote:
A quick question, please.
x = c(0.0001, 0.0059, 0.0855, 0.9082)
y = c(0.54, 0.813, 0.379, 0.35)
where A = 1st set, B = 2nd set, C = 3rd set, D = 4th set
1 1 4
2 2 5
3 3 6
For a data frame, 'names(x) - c(Apple, Orange)' also works, because
a dataframe is stored internally as a list of columns.
-- Tony Plate
Ethan Johnsons wrote:
A quick question please!
How do you rename column names? i.e. V1 -- Apple
A quick question, please!
How do you extract a certain value of vector?
i.e. x = c(2,5,3,6,21,3,6,24, )
How do you get the 1st one (which is 2); the 5th one (which is 21); etc?
thx much,
ej
[[alternative HTML version deleted]]
__
A quick question please!
How do you rename column names? i.e. V1 -- Apple; V2 -- Orange, etc.
thx much
ej
[[alternative HTML version deleted]]
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do
# R 2.3.1
I read the data set into R, but the data set have two rows per record.
i.e.
.
105 1 1 11031 5 4 8 5 22
105 2 2 2 2 1 2 17 16 26
106 1 2 606 1 5
I have used the following successfully:
xls = odbcConnectExcel(fname)
Rawdata.temp = sqlQuery( chan, select * from [sheet1$], max=2800 )
close(xls)
Presuming your data is in the Excel tab sheet1.
Of course, this assumes that column headers are in the first row.
Greg Johnson
-Original
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