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
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.
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,
abels=myhist$counts, pos = 3, cex = 0.8)
---rebuild the x-axis
labelsx<-c("<0.01", "0.01", "0.02", "0.03", "0.04", "0.05", "0.06", "0.40",
"0.50")
axis(1, at = xx, labels = labelsx,
ce=0 ,ylim=c(0,20), col=colors)
---rebuild the x-axis , But not work as it should be
axis(1, at=c(myhist$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
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.
>>
&
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*
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
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
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..
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
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]
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
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