I have survey data in which the answer choices were "categorical" (0, <15%, 15-30%, 30-45%, 45-60%, 60-75%, 75-90%, >90%). In retrospect, this should have been a free response question, but I'm trying to report a single number (mean and confidence interval) that represents all of the responses. Is there a statistically-defensible (citable) way to do this? For example, the median response is 15-30%, but I'd like to report an actual mean response with hopefully a narrower "confidence interval" than 15-30%...

Unfortunately, the width of the categories is not consistent, although I could pool 0 and <15% and throw out >90%, potentially, and each group with be 15% across. If I then treat 0-15% as 1, 15-30% as 2...75-90% as 6 and take the mean, that gives me say 2.43. Would the mean would be 2.43*15 = 36%?

  • $\begingroup$ 0 <15% 15-30% 30-45% 45-60% 60-75% 75-90% >90% 114 721 860 390 153 61 17 3 $\endgroup$
    – user59450
    Oct 28, 2014 at 2:29
  • $\begingroup$ You can give upper and lower bounds on the sample mean readily enough (assuming >90% is still bounded above by 100%). One question: with 15-30% vs 30-45%, if I have exactly 30% which group should it go in? $\endgroup$
    – Glen_b
    Oct 29, 2014 at 6:49

1 Answer 1


I guess this would divide experienced users of statistics. Some would just observe that the stable door is open and the horse is long since away. But there are some things you can do with extreme caution.

What can do is plot your data so you have several points on the cumulative distribution, the fractions less than or equal to 0, 15, 30, 45, 60, 75, 90 and 100 (the last being 1 by definition). (I set aside ambiguities in your presentation of classes: what exactly was presented to people and why are 30, 45, 60 and 75 possible values of two classes?)

Given this plot, all rising curves that pass through these points are consistent with the data. That gives a lot of scope for different estimates of the mean. The simplest method is to interpolate linearly within classes, so that you use midpoints 0, 7.5, 22.5, 37.5, 52.5, 67.5, 82.5 and 95. But just because that's simple doesn't mean it's a good idea. Or you could try something based on a guess at the form of the distribution or interpolation using cubic splines or piecewise cubic Hermite interpolation followed by numeric integration. The fact that the median is somewhere between 15 and 30% suggests a right skewed distribution, so that the weight within each interval is more likely to be towards the lower end of each interval. If so, linear interpolation will overestimate the mean.

Getting a mean is hard enough; methods for getting a confidence interval could be imagined, but smack of unjustified precision. You can express uncertainty as an interval between an estimate based on the assumption that all values are at the lower end of each interval and one that all are at the upper end of each interval.

I see no point to your suggested method. Pooling categories and throwing out one category because it's narrower than the others are not defensible. The consequences are discarding information (when you have so little to start with!) and indeed bias.

  • $\begingroup$ Many thanks! It sounds like the midpoints method is going to meet some resistance with statistical reviewers. I added the actual data above. Assuming I go with the first method you suggested (cumulative distribution), is there a way to estimate some sort of uncertainty interval? I did the far right and far left of each and got 15.8% to 29.9%, which isn't much of an improvement over 15-30%... $\endgroup$
    – user59450
    Oct 28, 2014 at 2:32
  • $\begingroup$ I just plotted the cumulative distributions and used a 6th order polynomial, and it looks great to me. Can you think of any previous examples of published work on surveys that have gone this route or biostatistics papers that I can cite? $\endgroup$
    – user59450
    Oct 28, 2014 at 3:02
  • $\begingroup$ I think you should use cubic splines to interpolate as @Nick Cox has indicated, rather than a single high order polynomial (there is no reason to believe a 6th order polynomial is an appropriate distribution whereas cubic splines are at least in common usage) $\endgroup$
    – tristan
    Oct 28, 2014 at 7:39
  • $\begingroup$ I agree with @Tristan. If there's one technique likely to make statistical reviewers explode it's a high order polynomial. $\endgroup$
    – Nick Cox
    Oct 28, 2014 at 9:40

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