Interquartile Range: 86, 98, 120, 122, 138, 176, 234, 257,

[Monkeyseat]

For these values find the interquartile range:

86, 98, 120, 122, 138, 176, 176, 234, 257, 290, 298, 342, 354, 401

My working:

(n+1)/4 = 3.75th value (lower quartile)

3((n+1)/4) = 11.25th value (upper quartile)

How do I find the 3.75th value and 7.25th value to get the IQR? I got the median if it helps by doing (n+1)/2 = 15/2 = 7.5th value. This means it's (176+234)/2 which is 205. I thought the lower quartile might be 121.5 (0.75 of the way between 3rd and 4th value) and the upper quartile 309 (0.25 of the way between 11th and 12th value) but that's not the case according to the answers (says the IQR is 176).

I want to do it the "(n+1)/4" way ideally not by splitting it up like this and finding 2 medians:

86, 98, 120, [122], 138, 176, 176

and

234, 257, 290, [298], 342, 354, 401

I also know it works if you do n/2 then when you find that add 1 and divide by 2 to find the lower quartile value but I've been told this (n+1)/4 is the way to do it but I can't get the right answer. Baisically can anyone show me how to do it the (n+1)/4 way? Can it be done? If so how do I find the 3.75th value and 7.25th value to get the IQR?

Many thanks.

 

[royhaas]

Re: Interquartile Range

A quartile is not necessarily unique, especially for discrete data. In this case rounding off the ranks or using the two medians will give the same results. If the sample size is divisible by 2 then using the two medians is the easiest. The main problem with your interpolation scheme is that the sample size is rather small and it assumes that the data is uniform, i.e. equally spaced. The safe bet is to use the median approach, since it is less sensitive to outliers in the upper and lower range of the data.

 

[Monkeyseat]

So (n+1)/4 won't work for this? What if the data was larger but still even? I didn't know what you meant "necessarily unique, especially for discrete data" or the bit about outliers - I don't know how that affects it. Because it is all equally spaced you can't find .25 or .75? But why can you find .5 for the median when it's even and it gives the right answer...?

As I mentioned I know it works if you do n/2 then when you find that add 1 and divide by 2 to find the lower quartile value (i.e. 4th) but you can't times it by 3 to get the upper quartile value - you have to split the original full data in two and do it again to find the upper quartile which just gives the value above the median. It's baisically like finding 2 medians. That probably didn't make sense lol.

Sorry for asking lots of questions, I just did it the 2 median way but would like to understand why/if the (n+1)/4 method can't be used. Thanks.

 

[Monkeyseat]

The teacher today briefly told us that if the value is .75 here you find 3/4 of the distance between the 2 numbers but how could that work?

If it is the 3.75th value for example and numbers 3 and 4 are 120 and 122 how would that give 122 which is the lower quartile? I've tried it a load of times and finding 3/4 of the distance does not work - can anyone clear it up?

Thanks.