How to check Likert scale summaries for plausibility
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Suppose you are reading a paper that uses Likert scale responses. The paper reports the mean, standard deviation, and number of responses. If we are — for some reason — suspicious of a paper, we might ask, “Are these summary statistics possible for this number of responses, for this Likert scale?” Someone asked me this recently, so I wrote some simple code to help check. In this blog post, I outline how the code works.Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.
Suppose we are reading a paper that uses a 5-category Likert scale response (0-4) and they report that for 100 responses, the mean response was .5, and the standard deviation of the responses was 5. I have intentionally chosen these numbers to be impossible: for the mean to be .5, the responses have to be on average near the bottom of the scale. But if the responses are near the bottom of the scale, the standard deviation also has to be very low, because there is a bound at 0. The standard deviation of 5 is much to large for the mean.
Another possible inconsistency arises due to the discreteness of Likert scales. For 10 Likert responses on a 3 point scale (0-2), the mean must be a multiple of .1. This, in turn, imposes a complicated constraint on the standard deviation.
Checking whether a response pattern is possible may be simple for low N, but it gets complex as N increases. Suppose we are wondering, for a 6-item Likert scale (0-5), whether N=94, M=.83, and SD=1.21 are possible summary statistics. There are several ways we could go about this using R.
The code for both is available here:Â https://gist.github.com/richarddmorey/787d7b7547a736a49c3f
Option 1: Brute force
In the brute force method, we create all possible response patterns (defined as the number of responses in each response category) and then check them, finding the ones closest to our desired response pattern. The code I linked above has two functions: count.ns, which counts the total number of response patterns for a given Likert scale and N. For the summaries above, this will be
> count.ns(N, nlev)
[1] 71523144
or about 72 million response patterns. Brute force isn’t pretty, but this job is possible in a minute or two on a modern PC. A note: on some older Windows computers this may exhaust your memory.
The function get.ns will compute all possible response patterns and put them in a matrix. This may take a bit of time (on my Macbook Pro it takes about a minute and a half):
> x = get.ns( 94, nlev=6)
> x[1:10,]
   [,1] [,2] [,3] [,4] [,5] [,6]
 [1,]   0   0   0   0   0  94
 [2,]   0   0   0   0   1  93
 [3,]   0   0   0   0   2  92
 [4,]   0   0   0   0   3  91
 [5,]   0   0   0   0   4  90
 [6,]   0   0   0   0   5  89
 [7,]   0   0   0   0   6  88
 [8,]   0   0   0   0   7  87
 [9,]   0   0   0   0   8  86
[10,] Â Â 0 Â Â 0 Â Â 0 Â Â 0 Â Â 9 Â 85
As you can see, x now contains the possible response patterns for N=94 responses, with K=6 Likert response categories. Above I show the first 10 patterns.
All that remains now is to compute the mean and standard deviation of each response pattern and compare it to the target. We can use the sum of the squared deviations from the target mean and standard deviation to sort the possible solutions. Note that if we wanted both deviations to be less than .005 (to account for rounding to the nearest .01) then we would want solutions that are no greater than 2 times .005^2 =Â .00005 in their summed squared error.
The code linked above does all the sorting and places the solutions in an object called res. After running the code, the first 10 rows of res are:
| > res[1:10,] |
   resp0 resp1 resp2 resp3 resp4 resp5    mean sum x^2 std. dev.   sum error
 [1,]   50   30   0   9   4   1 0.8297872   200  1.206063 1.554821e-05
 [2,]   54   22   0   17   0   1 0.8297872   200  1.206063 1.554821e-05
 [3,]   56   18   1   18   1   0 0.8297872   200  1.206063 1.554821e-05
 [4,]   57   15   4   17   1   0 0.8297872   200  1.206063 1.554821e-05
 [5,]   59   3   27   0   4   1 0.8297872   200  1.206063 1.554821e-05
 [6,]   59   10   7   18   0   0 0.8297872   200  1.206063 1.554821e-05
 [7,]   60   1   27   2   3   1 0.8297872   200  1.206063 1.554821e-05
 [8,]   60   7   10   17   0   0 0.8297872   200  1.206063 1.554821e-05
 [9,]   41   46   1   0   0   6 0.8297872   200  1.206063 1.554821e-05
[10,] Â Â 43 Â Â 42 Â Â 2 Â Â 1 Â Â 1 Â Â 5 0.8297872 Â Â 200 Â 1.206063 1.554821e-05
The first six columns contain the possible response pattern; the next three columns contain the summary statistics; the final column contains the sum of the error. We wished to match M=.83 and SD=1.21, and there are many solutions that yield these summary statistics. There is no reason to be suspicious of these summaries.
Option 2: Linear Inverse Models
The brute force solution above does the job, but it is slow and memory intensive. If we had a 7-item Likert scale, the number of possible response patterns would be about 1.2 billion; add more, and you can see that the amount of time and memory for the brute force method becomes prohibitive. We can actually use an approximate method — linear inverse models — to get close.
The idea of linear inverse models is that we can try to minimize a quadratic function, subject to some constraints. In our case, suppose we would like to find proportions of responses that would yield as summary statistics as close as possible to our given summary statistics. We have some reasonable constraints:
- Proportions must sum to 1 (equality constraint).
- Our summary statistics should be as close as possible to our target summaries (approximate equality constraint)
- Proportions must be between 0 and 1 (inequality constraint).
The linSolve package in R allows us to sample approximate solutions for our response proportions according to these constraints. I will not go into detail here about how to define these constraints for the linSolve package; see my code linked above and the linSolve manual. I will, however, show you the output of limSolve‘s xsample function, which we use to sample possible solutions:
| > xs <- limSolve::xsample(A = A, B = B, E = E, F = F, G = G, H = H, sdB = 1) |
| > xs$X[1:10,] |
      [,1]    [,2]    [,3]    [,4]    [,5]    [,6]
 [1,] 0.5231726 0.2862201 0.12002470 0.02459116 0.00000000 0.04599141
 [2,] 0.5231726 0.2862201 0.12002470 0.02459116 0.00000000 0.04599141
 [3,] 0.5231726 0.2862201 0.12002470 0.02459116 0.00000000 0.04599141
 [4,] 0.5231726 0.2862201 0.12002470 0.02459116 0.00000000 0.04599141
 [5,] 0.5231726 0.2862201 0.12002470 0.02459116 0.00000000 0.04599141
 [6,] 0.5085929 0.3226566 0.08151242 0.01938088 0.03202715 0.03582995
 [7,] 0.5085929 0.3226566 0.08151242 0.01938088 0.03202715 0.03582995
 [8,] 0.5085929 0.3226566 0.08151242 0.01938088 0.03202715 0.03582995
 [9,] 0.5860701 0.2541897 0.04073821 0.03497458 0.01822376 0.06580361
[10,] 0.5860701 0.2541897 0.04073821 0.03497458 0.01822376 0.06580361
The possible solutions for the proportions of responses in each response category are in each row of the xs$XÂ element of the output. We need to multiply these by N=94 to see the response patterns themselves:
| > xs$X[1:10,]*94 |
     [,1]   [,2]    [,3]   [,4]   [,5]   [,6]
 [1,] 49.17823 26.90469 11.282322 2.311569 0.000000 4.323192
 [2,] 49.17823 26.90469 11.282322 2.311569 0.000000 4.323192
 [3,] 49.17823 26.90469 11.282322 2.311569 0.000000 4.323192
 [4,] 49.17823 26.90469 11.282322 2.311569 0.000000 4.323192
 [5,] 49.17823 26.90469 11.282322 2.311569 0.000000 4.323192
 [6,] 47.80774 30.32972  7.662167 1.821802 3.010553 3.368016
 [7,] 47.80774 30.32972  7.662167 1.821802 3.010553 3.368016
 [8,] 47.80774 30.32972  7.662167 1.821802 3.010553 3.368016
 [9,] 55.09059 23.89383  3.829392 3.287611 1.713034 6.185539
[10,] 55.09059 23.89383 Â 3.829392 3.287611 1.713034 6.185539
There are several things to notice here. First, there are some duplicates, which is to be expected from the sampling method. Second, these solutions are not integers. Since response patterns must be integers, we would have to round these and make sure they sum to N=94 before testing them to see if their summaries are acceptably close.Â
Take the first solution: we might round this to obtain
49 27 11 2 0 4
However, this only sums to N=93, so we need to add 1 somewhere. Suppose we add it to the fourth category, to obtain
49 27 11 3 0 4
The mean response for this response pattern is almost exactly .83, our target mean. The standard deviation is 1.20, which is only .01 away from our target standard deviation.Â
Finally, note the similarity of these solutions to the ones we obtained by brute force. The linear inverse model method has gotten us in the neighborhood of the good solutions without the inelegance, time, and memory hogging of the brute force method. However, unlike the brute force method, it is not constrained to integer solutions, so we need to do some post processing.Â
Conclusion
Both the brute force and linear inverse model solution yield the same answer: for a 6-item Likert scale (0-5), N=94, M=.83, and SD=1.21 are not problematic summary statistics. Which of these methods one uses depends largely on the situation, but one could even combine them for a more efficient search. As a reminder, the complete code can be found here: Â https://gist.github.com/richarddmorey/787d7b7547a736a49c3f.
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