# Mean calculation

I have a situation where I have to calculate a mean in a group of N=80. But 7 participants have a score of the measured variable <3. It's a medial measure and I don't have the exact numeric value for the measure; I just know it's lower than 3.

How can I calculate the mean in this situation? I can't eliminate the participants because the variable is very important for the study.

• Do you mean "less than 3" or "greater than 3" (which is what ">3" means)? What is the range of the other values? Is there a natural upper bound for the value of the variable itself? What does "medial" mean?
– whuber
Jan 23, 2017 at 15:54
• I mean less then 3, it's a typo, the values are fom 0-3. Its vitamin D, and the values for the most participants are from 10-40. Jan 23, 2017 at 16:04
• Are you saying these are measurements of concentrations? If so, that's important to know, because it gives a lot of information about what "<3" might mean. Although it's not unusual to see a large gap between a reported limit of detection (such as 3) and the smallest quantified value (such as 10), this, too, is useful information (because in some cases it suggests one might want to reinterpret "<3" as "<10").
– whuber
Jan 23, 2017 at 16:06
• What the data interpretation is is methods dependent. For example, in a (Vitamin D) Schillings test, incomplete urine collection can result in low recovered percentage of administered marker. So, what test are you doing?
– Carl
Jan 23, 2017 at 16:40
• Why do you "have to calculate a mean"? Are you sure a median wouldn't do? Or just plotting the data. What is special about a mean in your context? Jan 23, 2017 at 18:40

You could also try a range of plausible values for the unknown cases. Sort of a worst and best case scenario. So try 0, 1, 2, and 3, and see if any of those substantially alter the mean.

If they do, you can discuss what the mean is under the best and worst case scenarios. If they do not, you can include that in your discussion as a justification.

• This is a good idea. Even with the minimal information provided you can go much further: since $7$ values are recorded as "$\lt 3$" and cannot be negative, the range of uncertainty in the total of those values is from $7\times 0 =0$ to $7\times 3 = 21$. Since all other values lie between $10$ and $40$, their sum lies between $72\times10+40=760$ and $10+72\times40=2890$. The total of all $80$ will be some interval of width $21-0$ ranging from $[760,781]$ up to $[2890,2911]$. Dividing by $80$ shows the mean is an interval of width $21/80\approx0.3$ ranging from $[9.5,9.8]$ to $[36.1,36.4]$.
– whuber
Jan 23, 2017 at 17:01

You can't calculate an exact mean if those values are unknown. You could replace them with a realistic value e.g 3 or 1.5, and then calculate the mean. If you do this you should add a caveat to your results to explain what you did.

• I thought to to that, but I'm not sure is this a valid method, is it usual to do it this way? Jan 23, 2017 at 16:05
• Although this method is commonly employed by non-statisticians, it is not considered valid in the chemometrics literature. It can be valid in some circumstances, but its validity ought to be checked each time.
– whuber
Jan 23, 2017 at 16:08
• It does depend on the context, and your aims. If you just want to find a rough estimate of the mean, then this method is fine. I would not recommend using it in a scientific publication, where it could mislead people who are skim-reading the results into thinking it was the true mean.
– rw2
Jan 23, 2017 at 19:20