<p>Bryan's answer is just fine, I just wish to add some fundamentals on the median. A quick look on the <a href="https://en.wikipedia.org/wiki/Median" rel="nofollow noreferrer">wiki page</a> would've told you right away that medians of integers don't necessarily have to be integers when there are an even number of observations. Consider the data set [0 1]  - the median and mean are the same, namely 0.5. An uneven number of integers will always yield an integer median. </p>

<p>If you feel uncomfortable mentioning decimals of a Likert scale, consider rounding the outcomes to the nearest integer.  </p>

CC BY-SA 4.0
2019-07-26T19:15:57.407Z
+1 except for the last sentence...I would not recommend doing this, it wouldn't be a median anymore.
Bryan Krause
CC BY-SA 4.0
2019-07-26T19:16:32.057Z
@BryanKrause - rounding a number doesn't affect its meaning.
AliceD
CC BY-SA 4.0
2019-07-26T19:23:08.787Z
In the case of a Likert item, though, you'd change a median from 2.5 to 3. That's not a median anymore. They'd be reporting the median is one category when the median is actually between two categories, and the direction of choice would be arbitrary.
Bryan Krause
CC BY-SA 4.0
2019-07-26T20:52:58.410Z
@BryanKrause - agreed. It depends on the situation an the number of steps involved for sure. Obviously, I would simply report one decimal. But suppose I were to be reviewing a MS from another person, I wouldn't care too much if it were rounded up or down.
AliceD
CC BY-SA 4.0
2019-07-26T19:12:37.160Z
2019-07-26T19:12:37.160Z
AliceD
<p>You are using the wrong justification for using the median. The only reason you can even consider using a median or mean with Likert data is because there <em>is</em> meaning to values between Likert items. If you have "Agree" and "Strongly Agree" there is definitely the possibility of some value where someone agrees more strongly than agree but less strongly than strongly agree.</p>

<p>However, you lose a ton of information when you calculate the median of a Likert-like item, and it can be dangerous to do so when the shapes of the response distributions vary. Let's consider a case where you have 5 levels. These two count vectors give you the same median '3':</p>

<p>A = [5 5 50 20 20]</p>

<p>B = [20 20 50 5 5]</p>

<p>but the distributions are quite different and the median is not very informative about that difference.</p>

<p>You may consider using the "interpolated median" which treats a Likert item as indicating the interval between adjacent items, for example a "3" really means "somewhere on the interval [2.5 3.5]."</p>

<p>For the cases A and B above, the interpolated medians are:</p>

<p>m<sub>intA</sub> = 3.3</p>

<p>m<sub>intB</sub> = 2.7</p>

CC BY-SA 4.0
2019-07-26T15:27:38.397Z
2019-07-26T15:45:17.693Z
2019-07-26T15:45:17.693Z
Bryan Krause
<p>My colleague and I are working with non-parametric Likert data. </p>

<p>We have been taking the median, but just realised that we are still getting values of say 2.5 or 1.5. Our justification for using the median over the mean was that values which lie between our Likert data points do not have strong meaning. </p>

<p>We are unsure if we should run our analysis again, always taking the minimum value or...?</p>

<p>Thank you.</p>

<data><analysis>
Using the median for non parametric (Likert) data, when it gives you a decimal place
JoshuaESummers
# Using the median for non parametric (Likert) data, when it gives you a decimal place

(Asked by: username_0 on 2019-07-26)

My colleague and I are working with non-parametric Likert data.

We have been taking the median, but just realised that we are still getting values of say 2.5 or 1.5. Our justification for using the median over the mean was that values which lie between our Likert data points do not have strong meaning.

We are unsure if we should run our analysis again, always taking the minimum value or...?

Thank you.

**Tags:** data, analysis


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## 2 answers

### Answer by username_1 on 2019-07-26. Score: 1

Bryan's answer is just fine, I just wish to add some fundamentals on the median. A quick look on the [wiki page](https://en.wikipedia.org/wiki/Median) would've told you right away that medians of integers don't necessarily have to be integers when there are an even number of observations. Consider the data set [0 1] - the median and mean are the same, namely 0.5. An uneven number of integers will always yield an integer median.

If you feel uncomfortable mentioning decimals of a Likert scale, consider rounding the outcomes to the nearest integer.

> **Comments:**
> **username_2**: +1 except for the last sentence...I would not recommend doing this, it wouldn't be a median anymore.  
> **username_1**: @BryanKrause - rounding a number doesn't affect its meaning.  
> **username_2**: In the case of a Likert item, though, you'd change a median from 2.5 to 3. That's not a median anymore. They'd be reporting the median is one category when the median is actually between two categories, and the direction of choice would be arbitrary.  
> **username_1**: @BryanKrause - agreed. It depends on the situation an the number of steps involved for sure. Obviously, I would simply report one decimal. But suppose I were to be reviewing a MS from another person, I wouldn't care too much if it were rounded up or down.  

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### Answer by username_2 on 2019-07-26. Score: 1

You are using the wrong justification for using the median. The only reason you can even consider using a median or mean with Likert data is because there *is* meaning to values between Likert items. If you have "Agree" and "Strongly Agree" there is definitely the possibility of some value where someone agrees more strongly than agree but less strongly than strongly agree.

However, you lose a ton of information when you calculate the median of a Likert-like item, and it can be dangerous to do so when the shapes of the response distributions vary. Let's consider a case where you have 5 levels. These two count vectors give you the same median '3':

A = [5 5 50 20 20]

B = [20 20 50 5 5]

but the distributions are quite different and the median is not very informative about that difference.

You may consider using the "interpolated median" which treats a Likert item as indicating the interval between adjacent items, for example a "3" really means "somewhere on the interval [2.5 3.5]."

For the cases A and B above, the interpolated medians are:

mintA = 3.3

mintB = 2.7


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