# how to rank items measured with a likert scale?

I have a 70 or so items all measured with a 5 point likert scale. What I'd like to do is take their mean value and rank the items according to their means but with respect to statistical significance in terms of differences in means. So items that do not have statistically significant means get the same rank.

Item - Likert Mean - Rank
A - 5 - 1
B - 4 - 2
C - 3.1 - 3
D - 3.099 - 3
E - 2 - 4
F - 1 - 5


How would I go about doing this?

• Of what are you taking a mean? The 70 likert items? How large is your sample of means? – Placidia Jun 3 '13 at 1:28
• sample is about 200. the mean is the mean value of the item measured with a likert scale. – incognito2 Jun 3 '13 at 1:45
• What do you do about the potential problem of not being able to distinguish say C from D nor D from E, but C is significantly different from E. What do you do then? – Glen_b -Reinstate Monica Jun 3 '13 at 3:05

Since you are averaging the ranks of 70 items, I think we can safely view your response as a continuous variable, and probably normal. If the population mean for this situation is $\mu$, then each observation has model $x_i=\mu + \epsilon_i$ and the only difference between two observations $x_1$ and $x_2$ is independent random error, with variance $2 \sigma^2$. With respect to the model, there is no meaningful difference between $x_1$ and $x_2$. Statistically, we would regard the difference between them to be significant only if $x_1-x_2$ is inconsistent with a normal distribution having mean 0 and variance $2 \sigma^2$.
However this is not the kind of solution you wanted. Your probably wanted an answer where the number of distinct ranks is almost equal to the number of items. In other words, you are assuming a model of $x_i=\mu_i+\epsilon_i$, where occasionally, $\mu_i=\mu_j$. But that model has way too many parameters. You can't distinguish the $\mu$'s from the $\epsilon$'s. Any sensible measure of dispersion will ensure that most subjects are statistically indistinguishable from the others -- and the number of distinct ranks (i.e. clusters) will be small.