I'm calculating the average rating from a selection of people where they rate something out of 5 stars. The mean is $4.73$ with a standard deviation of $0.55$. The breakdown is roughly:

 5* |  4* |  3* |  2* |  1*
76% | 20% |  3% |  0% |  1%
  1. Am I correct in saying that the margin of error for the mean is $0.55$?
  2. Can the margin of error be communicated as $4.73 \pm 0.55$?
  3. Is calculating the margin of error relevant for an average of 5 star ratings? (Surely $\pm 0.55$ on a mean value of $4.73$ could mean that the error bounds take it beyond 5 out-of-five 5, for instance.)
  4. If I'm wrong, how would you calculate the margin of error for this type of data? What is the value of calculating the margin of error for this type of data?

You seem to be confusing standard deviation and standard error. If I understand correctly, you're looking for a measure of precision of the mean. See this section of the Wikipedia Article on Standard Deviation, standard error is what you are looking for. $\sigma_{mean} \approx \frac{1}{\sqrt{N}}s$ is an estimator for the standard error of the mean.

Here is another explanation, which also goes into detail on how to calculate the margin of error, starting from the sample standard deviation, then calculating the standard error, then calculating the Margin of error from that given a confidence level. Mean + 1SE or margin of error can still lie outside the bounds, though.

Mean + 1*SD outside the bounds is a common occurence for skewed distributions and no problem at all. Here's great explanation by Glen_b on why and how they come up.

  • $\begingroup$ Thank you David, I'll have a read through what you've referenced. $\endgroup$ – David Ingledow Mar 23 '18 at 14:41

When I report on very skewed data such as rating scores (e.g., products or student evaluations), I would report the following: 
    •  the mean & standard deviation;
    •  the median & the % of the sample at or above (or below) the median;
    •  the cumulative % of the top two rankings (or bottom two rankings); and
    •  (possibly) the margin of error;
where I would decide on the direction (above or below) based on the direction of the skew of the data.  For the sample data you”ve provided, I would report it as:  “The responses were distributed with a mean (SD) of 4.73(0.55) and a median of 5; 76% of the respondents were at the median or above, and 96% of respondents were within one response unit of the median.”

If you do want to calculate the margin of error (my preference in 98%...as opposed to the more conventional 95%), you could use this formula for a reasonable approximation: $$ME = \frac{7}{3}*\frac{SD}{\sqrt{n}}$$ Hope this helps.  Happy to elaborate on anything as need be.

  • $\begingroup$ That's very valuable, thanks Gregg! One question about the margin of error: where does the 7/3 value come from and what value is 'n'? The example data I'm playing with comes from 576 responses. $\endgroup$ – David Ingledow Mar 23 '18 at 14:39
  • $\begingroup$ 1. 7/3 is actually my revised empirical rule (very briefly, the critical value for a 98% confidence interval is $z_{\alpha/2}=\pm2.33$...essentially, 7/3...happy to provide reference, if needed). 2. n would be the sample size, so $n=576$ for your data. $\endgroup$ – Gregg H Mar 23 '18 at 15:19
  • $\begingroup$ @GreggH Please do provide that reference. $\endgroup$ – Jim Mar 23 '18 at 16:57
  • $\begingroup$ Harbaugh, A. G. (2018, January). Clarifying and Reimagining the Empirical Rule: An Introduction to the By-Thirds Rule. Paper presented at 2018 Joint Mathematics Meetings, the joint annual meeting of the Mathematical Association of America and the American Mathematical Society, San Diego, CA. $\endgroup$ – Gregg H Mar 23 '18 at 17:54

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