[R] Re : CI

[Ethan Johnsons]

Great analogy!
Thanks much for the explanantion.

ej

On 10/19/06, justin bem <justin_bem at yahoo.fr> wrote:
>
>
> Get the the package fortunes in the CRAN and try
> >fortune("Harrell") for experience of and advance R-user. For the second
> question, the binomial distribution converge to normal distribution
> asymptomaticaly. You can not expect get the same result by using the two
> approximations on a finite sample. Watching a football macth in a TV screen
> is not the same thing that see the game in a staduim.
>
>  Justin BEM
> Elève Ingénieur Statisticien Economiste
> BP 294 Yaoundé.
> Tél (00237)9597295.
>
>
>
> ----- Message d'origine ----
> De : Ethan Johnsons <ethan.johnsons at gmail.com>
> À : "Liaw, Andy" <andy_liaw at merck.com>
> Cc : r-help at stat.math.ethz.ch
> Envoyé le : Jeudi, 19 Octobre 2006, 5h43mn 36s
> Objet : Re: [R] CI
>
>
> Thx so much.
>
> I just got into R world for my small research.
> I thought that R is free so doesn't have many features, but it seems I
> was wrong.
>
> Why do these two return different values?
>
> 0.2666456 0.6133544
> 0.2698531 0.6213784
>
> I think the diff is ignorable, but would ask.
>
> ej
>
> On 10/19/06, Liaw, Andy <andy_liaw at merck.com> wrote:
> > You did ask for CI of mean, so that's what you got.  If you want CI for
> > proportion, here are two (non-bootstrap) ways:
> >
> > R> confint(lm(I(x == 1) ~ 1), level=.9)
> >                   5 %      95 %
> > (Intercept) 0.2666456 0.6133544
> > R> binom.test(sum(x == 1), length(x), conf.level=.9)
> >
> >         Exact binomial test
> >
> > data:  sum(x == 1) and length(x)
> > number of successes = 11, number of trials = 25, p-value = 0.69
> > alternative hypothesis: true probability of success is not equal to 0.5
> > 90 percent confidence interval:
> >  0.2698531 0.6213784
> > sample estimates:
> > probability of success
> >                   0.44
> >
> > I hope these are not HW problems?
> >
> > Andy
> >
> > From: Ethan Johnsons
> > >
> > > Thank you so much for the feedback.
> > >
> > > The random numbers are working great.  I have tried
> > > non-random numbers, and the outcome is not correct with confint.
> > >
> > > Is there a way to compute i.e. a 90% confidence interval for
> > > percent of 1?
> > >
> > > i.e. where 1 = apple; 2 = orange
> > >
> > > > x
> > >  [1] 2 2 2 2 2 1 1 2 2 1 2 1 2 2 2 1 1 1 1 1 1 1 2 2 2
> > > > table (x)
> > > x
> > >  1  2
> > > 11 14
> > >
> > > > x =11
> > > > confint(lm(x~1), level=0.90)
> > >             5 % 95 %
> > > (Intercept) NaN  NaN
> > >
> > > ej
> > >
> > > On 10/18/06, Liaw, Andy <andy_liaw at merck.com> wrote:
> > > > Here's one way:
> > > >
> > > > R> x <- c(6,11,5,14,30,11,17,3,9,3,8,8) confint(lm(x~1), level=.9)
> > > >                  5 %    95 %
> > > > (Intercept) 6.546834 14.2865
> > > >
> > > > Andy
> > > >
> > > > From: Ethan Johnsons
> > > > >
> > > > > I have a quick question, please.
> > > > >
> > > > > Does R have function to compute i.e. a 90% confidence
> > > interval for
> > > > > the mean for these numbers?
> > > > >
> > > > > > mean (6,11,5,14,30,11,17,3,9,3,8,8)
> > > > > [1] 6
> > > > >
> > > > > I thought pt or qt would give me the interval, but it seems not.
> > > > >
> > > > > thx much.
> > > > >
> > > > > ej
> > > > >
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