# Central Tendency of 5-point Likert Scale

I have a Likert Scale (1-5) for rating of a product. So, the question is as follows:

Q: How would you are our platforms web experience?

1. Poor
2. Fair
3. Good
4. Very good
5. Excellent

I had 40 participants answering this, and I have been told that, a central tendency can be computed on this, mean and standard deviation. But, I have no idea. Using IBM SPSS, I get: EDIT: Ordinal Scale. My new calculation, • I cannot see what your question is. Can you edit your question to clarify just what you are trying to do? May 28, 2020 at 16:33
• I wanted to figure out, how to get mean and standard deviation from Likert Data. May 28, 2020 at 16:42

Likert scores taken ordinal categorical data. You have data as follows, I will put then into the data vector x, using R statistical software. Likert scores are unquestionably ordinal, so it is OK to compute quantiles. In particular, the median (50th percentile) is 4. So you might say that your data are 'centered' at the median 4. [Different statistical programs use use a variety of methods to compute quantiles, so these may be slightly different than you get in SPSS.]

score = 1:5
freq = c(1,3,15,17,4)
x = rep(score, freq)
quantile(x)
0%  25%  50%  75% 100%
1    3    4    4    5

table(x)
x
1  2  3  4  5
1  3 15 17  4


Likert scores assumed to be interval numerical data. If you are going to compute means and standard deviations from Likert data, you need to make the somewhat controversial assumption that these scores can be treated as interval numerical, so that addition and subtraction make sense. With this assumption, you have sample mean $$\bar X = 3.50$$ and sample standard deviation $$S_x = 0.88.$$

mean(x);  sd(x)
 3.5
 0.877058


To compute these summary statistics you use the 40 numbers as $$X_1, X_2, \dots, X_{50}.$$

x
 4 3 3 4 1 3 2 4 5 4 3 3 4 3 4 3 5 4 3 4
 4 3 3 4 3 2 4 4 5 5 4 2 4 4 3 4 4 3 3 3


The formulas are $$\bar X = \frac{1}{n} \sum_{i=1}^{40}X_i,\,$$ $$S_x^2 = \frac{1}{n-1}\sum_{i=1}^{40} (X_i - \bar X)^2,\,$$ and $$S_x = \sqrt{S_x^2}.$$

• Bruce, please see my latest EDIT. Thanks! May 28, 2020 at 8:35
• I see it, but don't understand what you may have done to get two different values for the standard deviation. May 28, 2020 at 8:43
• Bruce, I tried manually calculating with Excel, and did Variance = SUM((x-x̅)^2) / (n-1)  and Std. Variance^0.5 = 3.56 May 28, 2020 at 8:49
• Hi, hi, hi, I had made a miscalulation in the process, now I am getting 0.88 in Excel. May 28, 2020 at 9:27