Displaying 20 results from an estimated 40000 matches similar to: "F test"
2009 Mar 24
0
Unit root
I am confused by obtaining different results when testing for unit root when using different packages. I have 2625 price entries for which I want to determine whether they exhibit unit root. First I test using adf.test from tseries package by running:
> adf.test(P, k=30)
Augmented Dickey-Fuller Test
data: P
Dickey-Fuller = -4.685, Lag order = 30, p-value = 0.01
alternative hypothesis:
2010 Oct 29
3
Dickey Fuller Test
Dear Users, please help with the following DF test:
=====
library(tseries)
library(timeSeries)
Y=c(3519,3803,4332,4251,4661,4811,4448,4451,4343,4067,4001,3934,3652,3768
,4082,4101,4628,4898,4476,4728,4458,4004,4095,4056,3641,3966,4417,4367
,4821,5190,4638,4904,4528,4383,4339,4327,3856,4072,4563,4561,4984,5316
,4843,5383,4889,4681,4466,4463,4217,4322,4779,4988,5383,5591,5322,5404
2003 Jun 03
1
tseries "adf.test"
I have a question regarding the adf.test command in the tseries library.
I have a vector of time series observations (2265 daily log prices for the
OEX to be exact). I also have this same data in first-differenced form. I
want to test both vectors individually for staionarity with an Augmented
Dickey-Fuller test. I noticed when I use the adf.test command from the
tseries library, the general
2013 Jun 23
1
Scaling Statistical
Short question: Is it possible to use statistical tests, like the Augmented
Dickey-Fuller test, in functions with for-loops? If not, are there any
alternative ways to scale measures?
Detailed explanation: I am working with time-series, and I want to flag
curves that are not stationary and which display pulses, trends, or level
shifts.
>df
DATE ID VALUE2012-03-06 1
2013 Apr 30
1
ADF test --time series
Hi all,
I was running the adf test in R.
CODE 1:
adf.test(data$LOSS)
Augmented Dickey-Fuller Test
data: data$LOSS
Dickey-Fuller = -1.9864, Lag order = 2, p-value = 0.5775
alternative hypothesis: stationary
CODE 2:
adf.test(diff(diff(data$LOSS)))
Augmented Dickey-Fuller Test
data: diff(diff(data$LOSS))
Dickey-Fuller = -6.9287, Lag order = 2, p-value = 0.01
alternative
2004 Jan 13
3
How can I test if a not independently and not identically distributed time series residuals' are uncorrelated ?
I'm analizing the Argentina stock market (merv)
I download the data from yahoo
library(tseries)
Argentina <- get.hist.quote(instrument="^MERV","1996-10-08","2003-11-03", quote="Close")
merv <- na.remove(log(Argentina))
I made the Augmented Dickey-Fuller test to analyse
if merv have unit root:
adf.test(merv,k=13)
Dickey-Fuller = -1.4645,
2009 Jun 05
1
ADF test
Hi,
While doing the ADF test in R using the following command I am getting the
error and the result..">
x.ct=ur.df(rev$REVENUE,start=1,end=length(rev$REVENUE),frequency=1)
Error in ur.df(rev$REVENUE, start = 1, end = length(rev$REVENUE), frequency
= 1) :
unused argument(s) (start = 1, end = 4, frequency = 1)
>
2013 Mar 22
4
error while extracting the p-value from adf.test
Hello all,
I tried to extract the p-value from adf.test in tseries; however, I got the error message such as
> ht=adf.test(list.var$aa)
> ht$p-value
Error in ht$p - value : non-numeric argument to binary operator
> ht
Augmented Dickey-Fuller Test
data: list.var$aa
Dickey-Fuller = -2.3147, Lag order = 4, p-value = 0.4461
alternative hypothesis: stationary
> ht$data
[1]
2004 Jan 14
0
How can I test if a not independently and not identicallydistributed time series residuals' are uncorrelated ?
I'm analizing the Argentina stock market (merv)
I download the data from yahoo
library(tseries)
Argentina <- get.hist.quote(instrument="^MERV","1996-10-08","2003-11-03", quote="Close")
merv <- na.remove(log(Argentina))
I made the Augmented Dickey-Fuller test to analyse
if merv have unit root:
adf.test(merv,k=13)
Dickey-Fuller = -1.4645,
2007 Dec 08
2
time series tests
Hi all,
Can anyone clear my doubts about what conclusions to take with the following what puts of some time series tests:
> adf.test(melbmax)
Augmented Dickey-Fuller Test
data: melbmax
Dickey-Fuller = -5.4075, Lag order = 15, p-value = 0.01
alternative hypothesis: stationary
Warning message:
p-value smaller than printed p-value in: adf.test(melbmax)
2011 Jun 06
2
Problem in R documentation
I am not able to run Dickey-Fuller test.
adf.test() function is not working. It is showing 'Error: could not find
function "adf.test"
Can any tell how to call "time series" library?
--
Siddharth Arun,
4th Year Undergraduate student
Industrial Engineering and Management,
IIT Kharagpur
[[alternative HTML version deleted]]
2007 Oct 24
0
Different results in the unit root test. Why?
Situation:
I had tired a 1000-data generated by random error(i.i.d.), then I sub it
into different unit root tests. I got different results among the tests. The
following are the test statistics I got:
adf.test @ tseries ~ -10.2214 (lag = 9)
ur.df @ urca ~ -21.8978
ur.sp @ urca ~ -27.68
pp.test @ tseries ~ -972.3343 (truncation lag =7)
ur.pp @ urca ~ -973.2409
ur.kpss @ urca ~ 0.1867
kpss.test @
2008 Dec 08
1
About adf.test
Dear sir,
I am a new user of R statistical package. I want to perform
adf.test(augmented dickey fuller test), which packages I need to install in
order to perform it. I am getting following message on my monitor.
*x<-rnorm(1000)
> adf.test(x)
Error: could not find function "adf.test"
*I am waiting for your response.
Kamlesh Kumar.
--
Kamlesh Kumar
Appt. No. - QQ420,
2010 Feb 17
0
adf.test help
Hi,
I am trying to test whether a series is return series stationary, but
before proceeding I wanted to make sure I understand correctly how to
use the adf.test function and interpret its output... Could you please
let me know whether I am correct in my interpretations?
ex: I take x such as I know it doesn't have a unit root, and is
therefore stationary
1/
> x <- rnorm(1000)
>
1999 Sep 20
0
Updated tseries package
Fritz just put the updated tseries package to CRAN. I mainly removed
(and corrected) code such that tseries fits together with package ts.
New code is White's and Teraesvirta's tests for neglected non-linearity
(also for the regression case). From the INDEX file:
NelPlo Nelson-Plosser Macroeconomic Time Series
adf.test Augmented Dickey-Fuller Test
amif
1999 Sep 20
0
Updated tseries package
Fritz just put the updated tseries package to CRAN. I mainly removed
(and corrected) code such that tseries fits together with package ts.
New code is White's and Teraesvirta's tests for neglected non-linearity
(also for the regression case). From the INDEX file:
NelPlo Nelson-Plosser Macroeconomic Time Series
adf.test Augmented Dickey-Fuller Test
amif
2009 Mar 20
1
Dickey Fuller test of a time series (problem)
Hi all,
I tried to do a Dickey Fuller test with R using adf.test with a time series of german stock prices. I have 10 stocks from 1985 to 2009 with monthly stock prices. So if you do the math I have 289 values for each stock.
I tried to do the test for each stock alone and had the 289 values of my first stock listed in R.
When I tried to do the test with command adf.test(x, k=1) the following
2008 Feb 26
2
Obtaining values from adfstat objects
Hi,
I'm using the ADF.test function in the uroot package to obtain an
adfstat-class object. I'm wondering how I can extract the values (test
statistic, p value, etc.) from this class, since it doesn't seem to have
usual values. I get the following summary, but I'm not sure how to do
anything with these values -- how can I put the number into another
variable?
--------- ------ -
2007 Aug 16
2
ADF test
Hi all,
Hope you people do not feel irritated for repeatedly sending mail on Time series.
Here I got another problem on the same, and hope I would get some answer from you.
I have following dataset:
data[,1]
[1] 4.96 4.95 4.96 4.96 4.97 4.97 4.97 4.97 4.97 4.98 4.98 4.98 4.98 4.98 4.99 4.99 5.00 5.01
[19] 5.01 5.00 5.01 5.01 5.01 5.01 5.02 5.01 5.02 5.02 5.03 5.03 5.03
2009 May 15
1
Dickey-Fuller Tests with no constant and no trend
R has a Dickey-Fuller Test implementation (adf.test) that tests for unit
roots in an autoregressive process with a constant and linear trend. Is
there a DF implementation that doesn't use the constant or trend?
Thanks,
Jake.
--
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