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Suppose I have a number N of independent ratings of a given item, where each rating is an integer between 1 and 7 (inclusive). For simplicity sake, let us assume the ratings are normally distributed, though the mean and stddev will be different for each item.

I'm looking for a single number which will reflect how much trust I should put into the currently calculated mean. As an example, if I have N=1000 where every single rating is of 4, this trust should be pretty close to 100%. If on the other hand I have only N=2 where one rating is 1 and the other is 7, my trust that the mean is 4 should be very low.

I realise that a compute-trust function such as the one I've described above will likely need further parameterisation beyond the set of ratings. If you prefer, I can reformulate compute-trust in terms of "with 95% confidence, what is the probability that the population mean will be found within a given interval around the sample mean?"

Any thoughts on how I can compute this trust in a manner that is statistically sound?

Edit: changed 'median' to 'mean'.

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the median and stddev will be different Are you actually thinking of the median or the mean (expected value)? –  leonbloy Jul 8 '11 at 13:59
    
Yes, I meant the "mean". Sorry, English is not my first language. –  Jon Smark Jul 8 '11 at 15:07
    
Normality assumption is unreasonable. Do not understand meaning of $\mu$ and $\sigma$ being different "for each one." But put as a confidence interval problem, sample mean should, except for very small samples, have a normal enough distribution, and usual techniques will work well enough. –  André Nicolas Jul 8 '11 at 18:40

1 Answer 1

Practically speaking, the estimate of the standard deviation seems like a reasonable measure of the "trust" you can put in the mean.

Statistically speaking, you could consider the lengths of (say, 95%) confidence intervals to measure that "trust". If the ratings are normally distributed, those lengths are proportional to the estimates of the standard deviation (see this section), so this would actually be the same measure.

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