Has it all really come to this -

> From: Kevin Marks <[EMAIL PROTECTED]>
> This is completely statistically illiterate.
> 
> Firstly, the average of a 1-5 rating is 1+2+3+4+5/5 = 3, NOT 3.5. For
> it to be 2.5 you'd need to use a zero to five scale.
> 
> Your sample is 35 of 500, so expressing averages to 11 decimal places
> is not techie, it's foolish. Assuming a normal ditribution, the
> standard estimate of error is 1/sqrt(n) = about 17%. So your mean is
> 3.25 +/- 0.5.
> 
> (You could calculate a proper standard deviation from your data and
> quote that; the point is that the difference between 3.25 and 3 is
> well within the margin of error)
> 
> 
> Your bins are not the same size (1-2 has 3 datapoints on, the others
> have 2.) In any case, binning on an arbitrary scale is not really
> helpful. You'd be better off giving quintiles (rank the scores in
> order, break them into even-sized numbers of users, and report the
> means of those).


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