Confidence intervals / significance for percentage multiplied by a constant First post here; hopefully this is the right forum (and apologies if not).
I'm looking at some third-party data analysis that tries to predict food waste in the UK using the following method:
1. Review waste composition analyses (manual sorting) of waste from households (about 250 per authority) in a selection of UK council areas. This gives an estimated percentage of food by weight in the waste (with an unknown margin of error at the invididual authority level). Authorities are not selected by the report author - it is effectively a convenience sample based on which ones had analysis done and were prepared to share the results.
2. They then calculated a mean % of food in the waste. They also created a confidence interval for the percentage by calculating 1.96 times the SE of the percentage values (pretty sure this is ropey; struggling to put into words why).
3. They have then multiplied by national figures (based on operational data) contained in a national database. This is census type data, but we have no way of knowing the level of measurement error in the data.
The result is a central estimate plus an upper and lower bound. I'm convinced there is a lot wrong in the approach - not a random sample; calculating intervals on a percentage; unknown error in the national tonnnage data that the percentages are being applied to.
I have 2 questions:
1) Am I right that these intervals are very far from being robust (and should probably not be used)?
2) Is there any halfway sensible approach to analysing data like this to determine whether an observed change is likely to be "real" is is just a matter of random noise.
Apologies for long question, but keen to see if anyone can set me right on what can be done with these sort of modelled estimates.
Billy.
More info.
There were 129 authorities in the data set (out of 382 UK authorities that collect waste). Basic descriptive stats for the calculated percentages are: mean 0.304; sd 0.0628; min 0.125; 1q 0.262; median 0.313; 3q 0.340; max 0.473. Shapiro-Wilk returns a p value of 0.7, so the percentages values themselves are approximately normally distributed.
 A: The major issue in this situation ends up being how representative each of the samples is: the 250 households sampled per authority versus all households in the authority, and the 129 authorities in the data set versus the 328 authorities that collect waste (plus any waste of interest generated outside those 328 authorities). To start, however, let's assume that these are adequately representative.
Provided that the samples are adequately representative, your concerns are alleviated by the central limit theorem. Under the central limit theorem, it doesn't matter that the values are represented as percentages, so long that the distribution of those values meet certain conditions. In fact, one of the earliest theorems now considered a specific case of the central limit theorem is the De Moivre-Laplace theorem for fractions of successes among repeated success/failure trials: effectively percentage values. 
Under those conditions (almost certainly holding for these types of data) and with this many cases, the mean values within each authority and the mean values among authorities will be distributed closely to normal even if the individual values are not distributed in that way.
Let's go through those concerns as they came up in the question, first assuming that the samples are representative.
Margin of error at the individual authority level: with 250 households per authority, the standard error in the estimate of the mean will be only about 1/16 ($1 / \sqrt {250}$) of the standard deviation (SD) among households. That's probably pretty small, and in any event is included in the variability that is observed among authorities.
Error in the mean among authorities: the variance of the sum of values among the 129 authorities includes the actual variance among the authorities plus the within-authority variance. The standard error in the estimate of the mean (sum/129) among authorities will be only about 1/11 ($1 / \sqrt {129}$) of the SD among authorities. If that SD is 6.28%,* the standard error of the mean among authorities is only 0.55%.
Confidence interval for the mean among authorities: with an underlying normal distribution of values among authorities, estimates of the mean among authorities (if the experiment were to be repeated multiple times) will also follow a normal distribution. With a standard normal distribution (mean 0, variance 1), 1.96 is the value exceeded only 2.5% of the time. So taking the SD, multiplying it by 1.96, and adding and subtracting that value from the observed mean provides 95% confidence limits about that mean (leaving 2.5% probability beyond each limit). From the values you provide, those limits would be 29.3% to 31.5% of waste as food.
Extrapolation to whole UK: Census-type data are typically very reliable, and are generally the best available in any event. Again assuming that samples are representative of the UK as a whole, multiplying the mean fraction of food in waste (and the associated CI) by the total national waste will provide reasonable estimates of the amount of food waste nationwide. If you have estimates of the likely error in those nationwide census-type data, you could in principle use propagation of uncertainty methods to adjust the confidence limits of the final value.
The assumption of representativeness, however, is crucial. The within-authority representativeness would seem to be difficult to assess. For the among-authority results, the fraction of waste that is food seems likely to depend in potentially systematic ways on many geographic, cultural, demographic, and socio-economic factors. A better approach might then have been to use data on those factors to estimate how the fraction of waste is associated with them, and then take those associations into account to get a weighted estimate of the nationwide total food waste.

*I'm assuming that the values you provided are fractions, and am multiplying them by 100 for percentages. If I'm wrong, just divide all my values by 100.
