Renu, 

Great question. I agree with most of the above responses, but sometimes I am unsatisfied with simply a mean and SD because my audience does not understand the nuances to the data. I suggest using the Potential for Conflict Index which you can calculate and read about on the links below. I like it because you can represent attitudes to several survey items and represent several populations in and easy to understand visual representation. 

Cheers!

Loren

http://warnercnr.colostate.edu/~jerryv/PCI2/

http://warnercnr.colostate.edu/~jerryv/PCI2/references/Manfredo_etal_(2003)_The_potential_for_conflict_index.pdf

Dear Researcher

The best measure of Central tendency is Mean because it involves all the score.

The options or response category for your scale may be

Agree / Neutral / Disagree i.e. A/B/C

scoring for Agree 3, neutral -2 disagree 1

for example - Norm for 10 items

maximum score will be 30 and the minimum score 10

Theoretical norm

a score below 20 indicates = Unfavourable or low level

21 to 30 - Favourable and High.

Mr. Suresh, examples of categories could be A- Efficiency  B-Teamwork C- Communication. Assume that a group of people has to grade the person using a Likert scale on each of these three categories. Finally one needs a consolidated score for the person being evaluated. I thought that taking arithmetic mean does not make sense in Likert scales as the intervals are not equidistant. In my opinion median would make more sense. 

Please visit the site for more information.

http://maduraipsychometrician.blogspot.in/2014/09/mean-median-and-mode.html

http://maduraipsychometrician.blogspot.in/2013/01/when-to-use-statistics_5310.html

Good question! But the answer is: it depends of what you are measuring... If numerical values are possible to replace the categorical data, you should see the distribuition. Remember that the mean is a central tendency that is strongly influenced by outliers... So, 20% trimmed mean or median may be more apropriated. However, Likert scales are sometimes treated as continuous data, and sometimes as categorical data.... Indeed, I think that "A, B, C" is not a Likert scale... is it possible to replace to " 1, 2, 3" values? Do they have the same distance, like ordinal scale?

Most people just go ahead and use mean (and standard deviation) with likert type scales, even though one could raise technical questions about whether they really constitute proper interval scales.

If you have likert type ratings on three separate categories (A,B,C), you can combine them into a single score only if you maintain that all three categories are measuring the same underlying variable. For example, three items on a personality scale might all be intended to measure the same trait (such as Extraversion). To some extent, each item measures something unique - but they all share something in common as well.

But if the three categories are really not overlapping, you can't do this. If, say, you had a rating for Extraversion, one for Neuroticism, and one for Openness, you couldn't add the three together and call it "Personality."

The easy way to tell whether the three can be combined is to correlate them. If they correlate positively with one another (at whatever level you pre-select), then they can be combined. A similar approach is to calculate the internal consistency (Cronbach's alpha or KR-20) of the 3-item scale and examine the corrected item-whole correlations. If the internal consistency (reliability) is high enough, they may be combined.

The mean is usually just fine. The mean works badly if the variables have extreme skew or extreme values. That tends to be unlikely with Likert-type scales.

However, the interesting question is how to model this kind of data. Ideally one would use an ordinal model (precisely because the variable is bounded - and thus generally can't have extreme skew or extreme values). Sometimes it won't matter (e.g., if most responses are central and few extreme then methods that assume normal distributions of residuals work rather well ... except perhaps for prediction).

As a Likert scale is minimally ordinal, you can always use the median. The advantage of the mean is that it contains more information. But at least you must have a look at the distribution, you can also transform your Likert values into standardized values and than use the mean without problems.

It depends on the data distribution. If the distribution is approximately normal (or at least not skewed), using the mean is feasible - if, of course, the scores can be considered discretizations of a continuous variable. 

If the scale is purely ordinal (e.g., hierarchy level in the armed forces, or education level), using the mean is wrong and should be avoided.

Likewise, if the measurement could in theory be a discretization of an underlying continuous variable, but the distributon is highly skewed, the median is probably a better measure of central tendency. 

As Philippe Golay suggested above, the total score of your scale (computed as the sum of individual items' scores) could be normally distributed even if the single items are highly skewed, so it would be better to check that distribution as well and decide accordingly. 

Lots of answers! My two cents:

A mean on an individual item, or aggregating across dimensions using an average, assumes interval scaling (as do Pearson correlations, which are used in alpha; the problems with alpha are a different post). So does standard deviation, for that matter. Median and interquartile range do not, but averaging them across dimensions will usually be unhelpful.

While you can often "get away with" treating ordinal responses as interval (particularly under conditions of minimal skew and larger numbers of response categories), the answer I'd suggest as technically correct is an ordinal IRT model or CFA. You don't have the sample size for those though, especially if you wanted to use all response categories, simply because some response categories are going to be minimally represented, if at all, in the data.

In your situation, if skew is minimal, mean is probably the best bet. If skew is high, dichotomizing might be better (e.g., floor and anything-but-floor) -- the mean of a set of binary variables is meaningful.

Pat

Renu, 

Great question. I agree with most of the above responses, but sometimes I am unsatisfied with simply a mean and SD because my audience does not understand the nuances to the data. I suggest using the Potential for Conflict Index which you can calculate and read about on the links below. I like it because you can represent attitudes to several survey items and represent several populations in and easy to understand visual representation. 

Cheers!

Loren

http://warnercnr.colostate.edu/~jerryv/PCI2/

http://warnercnr.colostate.edu/~jerryv/PCI2/references/Manfredo_etal_(2003)_The_potential_for_conflict_index.pdf

One sophisticated way to proceed consists in combining median and mean. You first consider the median, then you operate a recode Lo thru median = 0 (below median) Median thru hi = 1 (above median). You can than work with the mean of the dichotimized variable. The value of the mean equals the rate of cases with 1. e.g. if with n=10 you have twice 0 and 8 times 1, your mean will be ((0*2)+(1*8))/10=0.8 which says that 80% have the value 1.

@ Jasmine Loo: I do not support that in any case you could use mean and standard deviation with a Likert scale. This actually presupposes that scale requirements are met.

Charles, that's a good point. Any dichotomization scheme will have that benefit. The considerations of where to dichotomize would be substantively meaningful breakpoints and, secondarily, relatively balanced categories.

 Loren,  thanks for the links.

Charles and Patrick,  I find the idea of implementing a dichotomization scheme pretty interesting. Will look into it.

@ Renu: The main criterion is the distribution.

Quite simply if you are assuming data is interval and it is normally distributed - use the mean/SD etc. If cronbach's alpha shows reliability you can then go on to do a range of parametric statistics.

For ordinal data, the best measure of central tendency should be the mean. The mean of the responses to a particular question can be compared to the hypothesised mean of the Likert scale used. For a 5-point Likert scale, for example, the hypothesised mean is the mid-point value of 3. The significant agreement or otherwise with the notion being tested is determined by comparing the mean score with 3 ( see Coakes and Steed, 2001). This implies that any result significantly different from this uncommitted or unsure value of 3 iss assumed to be either positive or negative to the notion being tested (Pullin and Haidar, 2003).

References

Coakes, S.J. and Steed, L.G. (2001): SPSS: analysis without anguish. John Wiley & Sons, Milton

Pullin, L. and Haidar, A. (2003): Managerial values in local government – Victoria,

Australia, The International Journal of Public Sector Management, 16(4), 286-302

Means and other measures of central tendency have the problem that they reduce the data too far, potentially. A graphic representation of the data is more informative.

One way of doing this is to use a stacked bar chart, but to centre the bars so that positive responses are to the right of a vertical line and negative ones to the left.

I did this graph with -slideplot-, a Stata command written by Nick Cox: http://econpapers.repec.org/software/bocbocode/s432801.htm

It is possible to calculate the mean for a likert scale by adding all items in each construct and dividing that total by the number of constructs in a construct. See Spss manual 

Hello

And what about the Mode ?.

Think about trhat

@ Carlos: as the median does not represent a problem, Likert scales being ordinal, there is no need do go back to the mode, which would apply to a nominal scale, which is the strict minimum.

In some cases the Scale has been used as a continuous measure in which case the Mean can be calculated. The assumption here is that the intervals between points are the same. Others also argue that Likert Scales are nominal measurements in which case one can not estimate Mean scores. However, I will use the former argument to support the Mean as a best measure of Central Tendency for this Scale.    

I personally believe it depends on objective

I am agree with the answer added by Dr. Ronan Micheal Conroy ,, Best regards ,, Jawad Ali

LIKERT SCALE: (1,2,3,4,5) or (1,2,3,4,5,6,7) or even (1,2,3,4,5,6,7,8,9,10).... may be checked for central tendency by running Monte Carlo simulation to determine the size of N repetition and then determine the mean, SD of the randomly generated data for Nth repetitions.

These values: (1,2,3,4,5) or (1,2,3,4,5,6,7) or even (1,2,3,4,5,6,7,8,9,10), are used as the source for the simulation. use the simulated data for N repetition under Monte Carlo to verify CLT.

LINK:

http://web.mst.edu/~dux/repository/me360/ch8.pdf

http://www.phys.ubbcluj.ro/~zneda/edu/mc/mcshort.pdf

https://people.smp.uq.edu.au/DirkKroese/mccourse.pdf