# Re: [R] proportions confidence intervals

From: Clint Bowman <clint_at_ecy.wa.gov>
Date: Tue 13 Jul 2004 - 04:24:39 EST

It seems to me that a transformation is in order since [0,1] can't possibly contain a normal distribution without cutting off both tails.

On Mon, 12 Jul 2004, Rolf Turner wrote:

>
> Darren Shaw wrote:
>
> > this may be a simple question - but i would appreciate any thoughts
> >
> > does anyone know how you would get one lower and one upper confidence
> > interval for a set of data that consists of proportions. i.e. taking a
> > usual confidence interval for normal data would result in the lower
> > confidence interval being negative - which is not possible given the data
> > (which is constrained between 0 and 1)
> >
> > i can see how you calculate a upper and lower confidence interval for a
> > single proportion, but not for a set of proportions
>
>
> (1) Your question appears to be a bit ``off topic''. I.e. it is
> implement methodology in R.
>
> (2) You need to make the scenario clearer. What do your data
> actually consist of? What are you assuming?
>
> The only reasonable scenario that springs to mind (perhaps this is
> merely indicative of poverty of imagination on my part) is that you
> have a number of ***independent*** samples, each yielding a sample
> proportion, and each coming from the same population (or at least
> from populations having the same population proportion ``p''. I.e.
> you have p.hat_1, ..., p.hat_n and from these you wish to calculate a
> confidence interval for p.
>
> You need to know the sample ***sizes*** for each sample. If you
> don't, you're screwed. Full stop. There is absolutely nothing
> sensible you can do. If you ***do*** know the sample sizes (say k_1,
> ..., k_n) then the problem is trivial.
>
> You have p.hat_j = x_j/k_j for j = 1, ..., n.
>
> Let x = x_1 + ... + x_n and k = k_1 + ... + k_n.
>
> Form p.hat = x/k. (I.e. you ***really*** just have one big
> happy sample.) Then calculate the confidence interval for p
> in the usual way:
>
> p.hat +/- (z-value) * sqrt(p.hat * (1 - p.hat)/k)
>
> If this is not the scenario with which you need to cope, then
> you'll have to explain what that scenario actually is.
>
> cheers,
>
> Rolf Turner
> rolf@math.unb.ca
>
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>

```--
Clint Bowman			INTERNET:	clint@ecy.wa.gov
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