Background: From a binomial experiment (k successes in n trials), the proportion of successes in the "population" is estimated by p = k/n. This estimate has a standard error of SE = sqrt(p(1-p)/n). Thus, alpha*100% of the intervals p +/- z[0.5*alpha]*SE will include the "population" proportion.
My problem: How is the "population" defined? Consider the following experiment: you have a cell culture plate with tenthousands of cells growing on it, and some of which have a particular feature you are interest in. The frequency of this feauture may be altered by some treatment of the cells. So you may want to know: what frequency or proportion of cells with this feature can one expect (given a treatment)? So you count some n of the cells on the plate and record that k of them do have this feature. Point and interval estimate for the proportion are given as above. But this clearly refers to *this* particular cell culture plate, what is the (statistical) population, but not the population you are actually interested in. Taking another plate, possibly prepared from newly isolated cells, you may get a considerably different estimate, far outside of the confidence interval determined for the first plate. So the estimate from one plate does not really tell you what to expect from another plate.
Now: Using the counts of all used plated together (n1+n2+n3...) and (k1+k2+k3...) *should* give an estimate over the population of cell culture plates, what seem to be the more sensible "population" to look at. However, the confidence interval for estimate will be very tiny, and I wonder if this is correct. I would think that one should better record just the estimates from the plates and use a beta-model to infere the "population of cell plates". If one of these ways is correct, I still would like to understand why. And if the beta-model thing is correct, I would like to know how the information of the quality of the individual, plate-wise estimates can be considered (think of on plate where you counted n=1000 cells and onother where you counted n=10 cells).
I would make one small modification. I think it is the beta-binomial model that would do the trick. I'm guessing that is what Manuel is referring to also. See the link for R function betabin and also some references.
http://cran.r-project.org/web/packages/aod/aod.pdf
Using the binomial, you are assuming that the treatment affect all the cells in the same way (homogenously for each cell and independent of other cells in the same plate). In that case, your population is conformed with all the cells. Looking at your results, probably this assumption does not hold (dependence between cells neighbors or some unknown factors affecting the plates). So, in my view, the beta-model is more correct because it is accounting the variability of preparing the plates. Now, the population is the set of plates. To take into account the quality of the individuals, you should incorporate some weights into your analisis with information about the number of cells in the plate or the standard error of your estimates.
Hope this helps
I would make one small modification. I think it is the beta-binomial model that would do the trick. I'm guessing that is what Manuel is referring to also. See the link for R function betabin and also some references.
http://cran.r-project.org/web/packages/aod/aod.pdf
Nice! Thanks. I was just not aware of the beta-binomial. I think this is the solution.
I found these two interesting sites:
http://www.apsnet.org/EDCENTER/ADVANCED/TOPICS/ECOLOGYANDEPIDEMIOLOGYINR/SPATIALANALYSIS/Pages/AdvancedDiscussionAndIllustration.aspx
http://www.tastingscience.info/publications/BetaBinomial.pdf
Your question is sound and leads us to very primary questions in statistics. I would comment three of them. (a) the estimation of confidence intervals on a proportion based on a binomial sample; (b) the amalgamation of results from different plates; (c) the proposed beta model for proportions.
(a) The use of standard deviation of a binomial random variable to build up a confidence interval is a very dangerous practice if we are far from asymptotic conditions. The problem becomes important because, when estimating small proportions, asymptotic conditions require millions of counts. Exact confidence intervals on proportions from a binomial sample are available: the old, but not well-known, Neyman geometrical method. An (1-a) confidence interval (p_1,p_2), symmetrical in probability, about the probability p, is obtained by solving the equations in p_1, p_2,
F_X( x_o | p_2, n) = a/2 , F_X( x_o | p_1) = 1 - a/2 ,
where X is the n-trial binomial variate; F_X(x|p,n) is the cumulative binomial distribution function for n trials and probability p; x_o is the observed number of counts; and p_1, p_2 are the end-points of the (1-a)-interval you look for.
I would suggest to avoid any confidence interval based on the standard deviation of binomials or normal approximations to the binomial. When the probability to be estimated is small you can get intervals covering nonsensical negative values.
(b) I agree with M. Febrero-Bande that adding positive counts over the total trials you are assuming homogeneity of the whole population, does not matter which is the actual characteristics of the plate. You suggested (k_1 + k_2 + ... ) / (n_1 + n_2 + ...) as an estimator of the proportion. This is the typical scenario of the Simpson paradox. This estimate of the global proportion can be in contradiction with the plate proportions. I agree that this way should be avoided if there is some evidence of non homogeneity and independence of plates (your case).
(c) With respect to the beta-binomial model, I think that there is a more comfortable and flexible model. The beta distribution (the one-dimensional case of the Dirichlet distribution on the simplex) is not natural and it is too rigid for modelling random proportions (see Aitchison 1986 book on compositional data analysis). Alternatively, you can use the logistic-normal or logit-normal model. Let me explain it briefly. Assume that p is a random probability or proportion. If logit(p)=log( p /(1-p) ) is normal(mu, sigma2) distributed then, we say that p has logistic-normal distribution with parameters (mu, sigma2) (see Aitchison and Shen, 1980 (Biometrika) and Aitchison, 1982 (J. Roy. Stat. Soc. series B)). If f(x|mu,sigma2) is the normal density N(mu,sigma2), the logistic-normal density g(y|mu, sigma2) is
g(y|mu,sigma2) = 1/[y(1-y)] f( logit(y) | mu, sigma2) for 0
We call the fraction or percentage of a population which falls into a particular category a population proportion, and denote it by the generic symbolπ .
The sample proportion, p, is an obvious point estimator of the population proportion. In practice, a random sample of size n would be selected, and the number of elements, x, of that sample falling into the category of interest is recorded. x is, of course, a discrete random variable, because its value is the result of counting units (that is, x can have only non-negative whole number values: 0, 1, 2, …, n). In fact, if the sampling process is truly random, x will be a binomial random variable, having the properties
and (LSP-1)
(We will look at the binomial distribution in greater detail later in the course).
For large enough sample sizes, the sample proportion, p = x/n, is approximately normally distributed, and from (LSP-1) we know that
(LSP-2a)
and
(LSP-2b)
This is enough information for us to be able to devise a 100(1-α )% confidence interval formula for estimating the population proportion, π , based on observation of a large enough sample. Recall the general "shape" of such an interval estimate formula when the underlying sampling distribution is symmetric (as are the normal and t-distributions):
population parameter = point estimator ± (probability factor) x (standard error)
This immediately gives us the formula
π = p ± zα /2 σ p @100(1 - α )%
Substituting (LSP-2b) directly into this formula leaves us with the π 's on the right-hand side. To get around this problem, we substitute the value of p as an approximation to π , to get the following working formula:
(LSP-3)
The rule of thumb is that this formula is valid if both np ≥ 5 and n(1 - p) ≥ 5. In practice, this means that you can use formula (LSP-3) to estimate π if you have observed at least five elements of the population in the category of interest, and at least five elements which are not in the category of interest. Such a situation is referred to as the large-sample case when population proportions are being estimated.
If either or both of these conditions are not satisfied (what one might call the "small-sample case"), statistical inferences about π must be based on the binomial distribution directly. Such methods are more complicated, and are used only when there is no alternative. Statistical inferences about population proportions which use small samples tend to give relatively un-useful results in a number of ways. Reasonable precision and reliability in estimating population proportions almost certainly requires the use of large samples.
About Juan Jose Egozcue "logistic-normal or logit-normal model" This distribution is also called Johnsons System Bounded distribution (Johnson, N.L. 1949. Systems of frequency curves generated by methods of translation. Biometrica 36:
149‐176) This is very popular within forestry see (say) Mønness, E. (2011). The power-normal distribution: Application to forest stands. Canadian Journal of Forest Research, 41(4):707–714.
It is very dageros to use normal aproximation for confidence interval construction. Confidence intervals constant width you can find in the article 'Minimax confidence intervals for the binomial parameter'
M.M. Lutsenko, S.G. Maloshevskii
Journal of Statistical Planning and Inference. 113(2003) p.67-77
- Badges
- Science topic
- Similar topics
- Medicine
- Physiology
- Biosignals
- Biomedical Signal Processing
- Statistical Analysis