# Calculated spread and mean using mean results only

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?

• Will the future means have the same sample size as the past data? Alternatively are the two same sizes known? Commented Feb 28 at 3:28
• Yes, we are expecting any future values to retain the same sample size. Commented Feb 28 at 4:06

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.

• Thanks for the comments, I do believe you have the correct understanding of the problem we're trying to solve. I follow your explanation for most of the post, however I'm unsure on the calculation of the variance of the estimator that you've given. Sigma-squared is the variance of the underlying population, which is a piece of information that I do not have. In this particular case, I do not even have the sample standard deviations to use as an estimate of the population standard deviation. How do I get to the variance of the estimator, and therefore the SE of mean? Commented Feb 28 at 18:47
• Ah. I was just trying to illustrate that your concern "My original concern was that the data do not provide information on sample standard deviation" was unfounded - the means themselves do provide information on the standard deviation. Just taking the standard error on the mean of means (i.e. treating the means themselves as individual data points) is OK, and gives the uncertainty on the mean of the original ($X_{hi}$) variables. Commented Feb 28 at 22:44
• @AlexJ , Why is there $H$ in the denominator of the formula for the standard error estimate? I think it should be $SE(\bar{X_h}) = \frac{\hat{\sigma}_h}{\sqrt{n}}$. SEM (standard error of the mean doesn't depend on $H$ which is the number of samples) Commented Feb 29 at 11:17
• @HappyCretine I don't understand the edits you made, can you explain your intentions? If the distribution are iid, then $\mu_i = \mu$ and $\sigma_i = \sigma$ for all variables. Is the change from $X_{hi}$ to $X_{h_i}$ just visual preference or does it have a different meaning? And there is a $H$ in the denominator because there are $H$ observations (each of which itself is a mean). Commented Feb 29 at 22:33
• For now, I've reverted the edits. If you think something can be clarified, I'm happy to do so, but I'm not convinced at this stage that you've understood what I wrote. Commented Feb 29 at 22:37

"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...