Calculated spread and mean using mean results only

Question

Background: We have a large database of measurement data. The individual data points are actually the mean results of a particular sampling. No further information is available against each data point, only the sample mean is available. We can assume that the sample sizes are the same for each sampling. Commissioning a new study to gather new data with the appropriate details is not feasible from a resource, time, and access perspective, therefore we can only perform the analysis to the best of our ability using this information on hand.

Objective: To determine the "spread of the mean" and the "mean of the mean", using which we hope to determine the probability of failure given a target "pass" value. To clarify, the requestor of this analysis does not care whether an individual measurement fails the target value, but the mean of each sampling must exceed the target value (the requirement coming from an old industry standard).

My original concern was that the data do not provide information on sample standard deviation and therefore we can make no further comments on the population distribution given incomplete information. However, because we are not interested in the behaviour of individual measurements, and instead we wish to analyse the distribution of the mean, it has been argued that we should dispense of the notion that each data point is a mean, and just simply treat the analysis as we would if each data point was a single measurement value.

Question to the community: Is this a sound argument? If not, what statistical principle have we fallen afoul of, and what is our alternative?

Answer 1

Sounds fine to me. Here's an illustration which should be close enough to your scenario.

Say we have a set of $H$ datapoints. Each of these is a mean taken from $n$ individual observations, which are identical and independently distributed with some mean $\mu$ and variance $\sigma^2$.

So we have $\bar X_1, \cdots, \bar X_H$ where $E[\bar X_h]=\mu$ and $V(\bar X_h)=\sigma^2/n$. Implicitly, we assume $\bar X_h = \frac{1}{n} \sum_{i=1}^n X_{hi}$, except we don't have access to $X_{hi}$, only $\bar X_h$.

If we want the best guess of the mean overall, we can take a mean of means: $\bar X = \frac{1}{H}\sum_{h=1}^H \bar X_h$. Then, the expectation of this can be shown to be $\mu$ (good, it's unbiased). The variance of this estimator can be shown to be $\frac{\sigma^2}{nH}$. And taking the square root of the variance of the estimator, will give a standard error on the mean.

Technically: under this framework, the mean doesn't have a distribution. It's just a value. But the uncertainty on the mean is what you care about, I think.

Answer 2

"However, because we are not interested in the behaviour of individual measurements, and instead we wish to analyse the distribution of the mean, it has been argued that we should dispense of the notion that each data point is a mean, and just simply treat the analysis as we would if each data point was a single measurement value."

I see no problem with that approach; it is sound.

The one major caveat is that you need to be able to treat all you data points as identically distributed; i.e. no change in the process occured while collecting the data. In particular, $n$ remained constant, the sampling process remain the same (how/when, etc. the samples were collected, the mean computed, etc..). If this holds, no need to worry about not knowing the original data. Basically, you have a sampling distribution of the mean, with a given sample size. The mean of the sampling distribution is the mean of the population, and the variance of your sampling distribution is the variance of the population (unknown, but you can estimate it) divided by $\sqrt{n}$. So you can get estimates of both the population mean and variance...