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
reverse
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. S
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, t
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
wrote:
>
>
steve
On Thu, Dec 31, 2015 at 4:16 PM, Rolf Turner
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.
>>
>> Here is an example of left censor
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*
<-matrix(c(1450,1800,1840,1820,1860,1780,1760,1800,1900,1770,1790,1780,1850,1760,1450,1710,1575,1475,1780,1790,1780,1450,1790,1800,
, at 9:00 AM, Steven Stoline 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 to modify yours, but i was not be able to make
> &g
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 tibco.com
>
>
gt; > 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
> wrote:
> > Dear Pe
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 wrote:
> many thanks
>
> steve
>
> On Fri, Nov 27, 2015 at 9:20 AM, peter dalgaard wrote:
>
>> Something like this?
>>
>> f <- function(x) x^3-2*
t - 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 wrote:
>
> > Dear All:
> >
> > I am trying to explain to my students how to c
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]
with
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