[R] Correlations Table of Items when compute Cronbach's Alpha

2017-02-21 Thread Steven Stoline
Dear All: I am using *alpha(data, **check.keys=TRUE) or alpha(data, keys = c(1, 1, 1, -1))* to compute the Cronbach's Alpha. I am using *check.keys=TRUE* or *keys = c(1, 1, 1, -1) *to automatically reverse items. *My question is: *how can I get the correlation tables (matrix) of the

[R] Cronbach's Alpha Values

2017-02-21 Thread Steven Stoline
Dear All: I am using *alpha(data, **check.keys=TRUE) *to compute the Cronbach's Alpha. I am using *check.keys=TRUE* to automatically reverse items. *My question is: *how can I get the correlation tables (matrix) of the reverse items as part of the R output. thank you steve -- Steven M.

[R] “log-rank test” and “Kaplan Meier actuarial plots (time to event curve)”

2016-10-08 Thread Steven Stoline
Dear All: I do need your help with the “log-rank test” and “Kaplan Meier actuarial plots (time to event curve)” with adding the confidence intervals bands for the average score for each time to the curve of each group. *Group variable:* 1 = group 1, 2 = group 2 *Time points:* time1, time2,

Re: [R] [FORGED] Histogram for Left Censored Data

2016-01-03 Thread Steven Stoline
abels=myhist$counts, pos = 3, cex = 0.8) ---rebuild the x-axis labelsx<-c("<0.01", "0.01", "0.02", "0.03", "0.04", "0.05", "0.06", "0.40", "0.50") axis(1, at = xx, labels = labelsx,

Re: [R] [FORGED] Histogram for Left Censored Data

2016-01-03 Thread Steven Stoline
ce=0 ,ylim=c(0,20), col=colors) ---rebuild the x-axis , But not work as it should be axis(1, at=c(myhist$mids[1], myhist$breaks[-(1:2)]), labels=c("<0.01", myhist$breaks[-(1:2)])) with many thanks steve On Sat, Jan 2, 2016 at 11:38 AM, David Winsemius <dwins

Re: [R] [FORGED] Histogram for Left Censored Data

2016-01-01 Thread Steven Stoline
steve On Thu, Dec 31, 2015 at 4:16 PM, Rolf Turner <r.tur...@auckland.ac.nz> wrote: > On 31/12/15 23:20, Steven Stoline wrote: > >> Dear All: >> >> I need helps with creating histograms for data that include left >> censored observations. >> &

[R] Histogram for Left Censored Data

2015-12-31 Thread Steven Stoline
Dear All: I need helps with creating histograms for data that include left censored observations. Here is an example of left censored data *Sulfate.Concentration*

Re: [R] Graphing the Area of Definite Integral

2015-12-16 Thread Steven Stoline
wrote: > > > On Dec 16, 2015, at 9:00 AM, Steven Stoline <sstol...@gmail.com> wrote: > > > > Dear William: *Left and Right Riemann Sums* > > > > > > Is there is a way to modify your function to compute Left Riemann Sum and > > Right Riemann Sum. I tried t

Re: [R] Graphing the Area of Definite Integral

2015-12-16 Thread Steven Stoline
n=16) > [1] 42.5 > > showIntegral(f=function(x)x^2, xmin=-4, xmax=4, n=256) > [1] 42.66602 > > showIntegral(f=function(x)x^2, xmin=-4, xmax=4, n=1024) > [1] 42.3 > > > 2*4^3/3 > [1] 42.7 > > showIntegral > Bill Dunlap > TIBCO Software > wdunlap tib

Re: [R] Graphing the Area of Definite Integral

2015-12-01 Thread Steven Stoline
n=256) > [1] 42.66602 > > showIntegral(f=function(x)x^2, xmin=-4, xmax=4, n=1024) > [1] 42.3 > > > 2*4^3/3 > [1] 42.7 > > showIntegral > Bill Dunlap > TIBCO Software > wdunlap tibco.com > > > On Fri, Nov 27, 2015 at 9:50 PM, Steven Stoline <sstol..

Re: [R] Graphing the Area of Definite Integral

2015-11-28 Thread Steven Stoline
5, pch=19) sum(fm*dx) 1/3 * (64+64) with many thanks steve On Fri, Nov 27, 2015 at 3:36 PM, Steven Stoline <sstol...@gmail.com> wrote: > many thanks > > steve > > On Fri, Nov 27, 2015 at 9:20 AM, peter dalgaard <pda...@gmail.com> wrote: > >> Something like this

[R] Graphing the Area of Definite Integral

2015-11-27 Thread Steven Stoline
Dear All: I am trying to explain to my students how to calculate the definite integral using the Riemann sum. Can someone help me to graph the area under the curve of the function, showing the curve as well as the rectangles between 0 and 4.. *f(x) = x^3 - 2*x * over the interval [0 , 4]

Re: [R] Graphing the Area of Definite Integral

2015-11-27 Thread Steven Stoline
dx > mid <- right - dx/2 > fm <- f(mid) > points(mid, fm) > rect(left,0,right,fm) > > sum(fm*dx) > > 1/4 * 4^4 - 4^2 > > > -pd > > > On 27 Nov 2015, at 13:52 , Steven Stoline <sstol...@gmail.com> wrote: > > > Dear All: > > > &g