similar to: Help on function adf.test

Dear R helpers, I am trying to read a CSV file in R called EUROPE (originally an Excel file which I have saved as a CSV file) using the command EUROPEDATA <- read.csv("EUROPE.csv") EUROPE.csv is basically a matrix of dimension 440*44, and has a line of headers, i.e. each column has a name. Using read.csv I can't load the data into R properly. Although the first 20 columns or

Hello! I do not understand what is meant by: "aic" and "bic" follow a top-down strategy based on the Akaike's and Schwarz's information criteria in the datails to the ADF.test function. What does a "top-down strategy" mean? Probably the respective criterion is minimized and the mode vector contains the lag orders at which the criterion attains it

Hi, I am quite new to R and would appreciate some guidance, if possible. I have imported a csv file: spread <- read.csv("Spread.csv") I get the following error when I try to run adf.test: > adf.test(spread,alternative = c("stationary", "explosive"),0) Error in embed(y, k) : 'x' is not a vector or matrix Why is this? -- View this message in context:

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,

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

this command is used in tseries adf.test(x, alternative = c("stationary", "explosive"), k = trunc((length(x)-1)^(1/3))) this command apply adf test on data given in x .here general equatiuon that is, equation with constant and trend is used.if i did not include constant or trend in the equation and run the command then how i can run this command in tseries.

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) >

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

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

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]

Hi all R-users, I?m trying to assign colors on those p-value in my table output that fall above a certain critical value, let?s say a p-value >0.05. My table looks like this: Assets ADF-Level P-Value ADF-First D P-Value ADF-Second D P-Value [1,] Liabilities -2.3109 0.1988 -3.162 0.025 -6.0281

hi there, i have posted earlier on the list but got no satisfying answer. the problem is not big. I have asterisk server directly connected with internet (79.80.x.x) and clients are behind router. clients/users are registered with asterisk and are using sipura and xlite softphone. Now problem is that when a user calls other by dialing his IP:Port (sip uri), call is connected fine and he can

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

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) >

Hello! I am applying the ADF.test function from package uroot to a time series of data. When I apply the full test, incorporating drift and trend terms, the regressor estimate of the drift term is not significantly different from zero. So I apply the test to a model without drift term, with deterministic trend only. But then I always get the following error:

Dear all: I am having some problems to use the function "sink()". Basically I am doing a loop over two files which contain unit-root variables. Then on a loop, I extract every i element of both files to create an object called z. If z meets some requirements, then I perform a unit root test (ADF test), otherwise not. As this process is repeated several times, for each i I want to get

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

Dear R gurus, As part of finding initial values for a much more complicated fit I want to fit a function of the form y ~ a + bx + cx^d to fairly "noisy" data and have hit some problems. To demonstrate the specific R-related problem, here is an idealised data set, smaller and better fitting than reality: # idealised data set aDF <- data.frame( x= c(1.80, 9.27, 6.48, 2.61, 9.86,

Hi, I am trying to dial a registered user via his IP:Port mechanism, but problem is that the audio data is not reaching to dialed user. here is the scenario. caller and callee both are registered at asterisk server. asterisk server is on public ip so no port forwarding and natting necessary there. however caller and callee both are behind router and there is port forwarding enabled and nat=yes,

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)