Is the third quartile in 13 data the tenth value or the mean of the tenth and eleventh value?

Question

If you have 13 data values written in order and you choose the median as the 7th value, this leaves 6 values on either side of the median. The median of these 6 values is the mean of the two central values. This should be the third quartile.

But the third quartile is the 75th percentile. $13*0.75=9.75$, this rounds up to 10. The 75th percentile and hence the third quartile is the 10th value.

Should you be getting the same result from these two methods?

Is my approach correct for these two methods?

So what is the third quartile and 75th percentile for 13 orders values? Is it the 10th value or is it the mean of the 10th and 11th value?

Answer 1

It is the $10^{\mbox{th}}$ value (call it $x_{10}$) since that is the only number such that

1) At least $75\%$ of the sample are at most $x_{10}$

and

2) At least $25\%$ of the sample are at least $x_{10}$.

(breaking ties in an appropriate way...)

I believe the double-median approach is problematic in your case because you ignore a value ($x_7$), and you cannot afford that when computing percentiles. However, you don't ignore a value if the sample size is even in the first place, and then the $75^{\mbox{th}}$ percentile is actually equal to the double-median result.

Answer 2

I don't think there is a strict universal rule to compute percentiles of a discrete sequence. The R function quantile() for example let the user choose between 3 different formulas for discrete samples and 6 formulas for continuous samples. By default it seems to give the 10th value in your case.

EDIT: @Mau314 I like the rule from your answer because it almost always give a unique value ${q}_{{\alpha}} = {x}_{\left(\left\lceil {\alpha} N\right\rceil \right)}$. Indeed ${\alpha} n \leqslant k = \left\lceil {\alpha} N\right\rceil \leqslant {\alpha} N+1$, which means that at least a proportion ${\alpha}$ of the sample are at most ${x}_{\left(k\right)}$ and at least a proportion $\left(1-{\alpha}\right)$ of the sample are at least ${x}_{\left(k\right)}$. The only case where there are two candidates is when ${\alpha} N \in \mathbb{N}$ in which case $k = \left\lfloor {\alpha} N\right\rfloor $ and $k = \left\lceil {\alpha} N\right\rceil $ both satisfy the property.

However, depending on the rule chosen in $R$'s function quantile(), the result seems to be different, here is an example with the 3 first formulas.

> x <- c(2,5,7,8,12,17)
> z <- seq(from = 0, to = 1, by = 0.05)
> quantile(x, probs=z, type=1)
0%   5%  10%  15%  20%  25%  30%  35%  40%  45%  50%  55% 
2    2    2    2    5    5    5    7    7    7    7    8 
60%  65%  70%  75%  80%  85%  90%  95% 100% 
8    8   12   12   12   17   17   17   17 
> quantile(x, probs=z, type=2)
0%   5%  10%  15%  20%  25%  30%  35%  40%  45%  50%  55% 
2.0  2.0  2.0  2.0  5.0  5.0  5.0  7.0  7.0  7.0  7.5  8.0 
60%  65%  70%  75%  80%  85%  90%  95% 100% 
8.0  8.0 12.0 12.0 12.0 17.0 17.0 17.0 17.0 
> quantile(x, probs=z, type=3)
0%   5%  10%  15%  20%  25%  30%  35%  40%  45%  50%  55% 
2    2    2    2    2    5    5    5    5    7    7    7 
60%  65%  70%  75%  80%  85%  90%  95% 100% 
8    8    8    8   12   12   12   17   17 

You can see for example that the percentile for $55 \mathrm{\%}$ is the 4th value for the 2 first rules but it is the 3rd value for rule number 3. As ${6}\times{0.55} = 3.3$, your rule would always give the 4th value.