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September 15[edit]

Box and Whisker Plot[edit]

Ok I understand the general idea behind this graph method. The very top hash represents the maximum value in the sample, the fatter part of each line represents the data between the 1st and 3rd quartile (interquartile range I assume?), The bolder line within the fat box is the sample median, and the bottom hash represents the lowest value in the set. Does that all sound right? OK...

So, in experiment # 3 above How is it that the lowest value also represents the firs quartile? Or is it possible for that to happen?

I am currently looking a published article where the fat part of the box starts at 0 and ends at 3 with no median line and no bottom range line, but the fat box does end at 3 and the top of the range is indicated normally. So does this mean that the lowest value in the data set, the median, and the first quartile are all 0?

Much thanks in advance. Let me know if you need clarification

Mrdeath5493 (talk) 00:13, 15 September 2009 (UTC)[reply]

Lowest value can also be first quartile as in 1 1 2 2 3 3 4, for example. And the median can be the same as the first quartile, and the lowest value, eg. 1 1 1 1 5 5 5. Can you link to the published article? Without seeing it, I would imagine that the bottom range line is obscured by the box because the lowest value is also the median, as in your example of #3 above. RupertMillard (Talk) 00:31, 15 September 2009 (UTC)[reply]

I think you answered my question well enough. I'm not sure exactly which article the graph is from, but I bet it isn't licensed under GNU or whatever. It was patient reported data measuring pain from 0-10 after receiving 30 minutes of a IV pain reliever. So, I bet The low value, 1st quartile, and median were all 0 since they were all on pain meds. That makes sense :) Thank you for the help. Mrdeath5493 (talk) 00:46, 15 September 2009 (UTC)[reply]

I don't know if the first and last bar mean what you said they mean. I am in a stat class right now and my book calls your description a "skeletal box plot". The skeletal box plot has only minimum, first quartile, median, third quartile, and max. The thing is, your pictures have things other than that. Notice there are also a few little circles. That matches a "box plot" in my book, and also in the article box plot. The difference is the top and bottom lines are no longer max and min. They are upper adjusted value and lower adjusted value (that could be wrong terms, but it is UAV and LAV... wiki article does not have those terms). But, the definition of UAV and LAV I remember, and it agrees with wiki article. You look at Q_1 - 1.5 IQR. LAV is the smallest measurement that is bigger than that number. Then you consider Q_3 + 1.5 IQR. UAV is the biggest measurement which is smaller than that. Then, anything that does not fall in that range, from Q_1 - 1.5 IQR to Q_3 + 1.5 IQR is called an outlier. And, those 3 little circles represent outliers. See the Wikipedia article for more detail. StatisticsMan (talk) 01:13, 15 September 2009 (UTC)[reply]

You wrote:

The very top hash represents the maximum value in the sample,

That is incorrect. You see some outlying values in some of these plots—beyond the top or bottom hash marks. You'd better pay attention to those.

You glaringly omitted the sample sizes. Think about where the quartile boundaries are in a sample of only 3. That explains how the minimum and first quartile can coincide. Michael Hardy (talk) 20:30, 15 September 2009 (UTC)[reply]

I'm not sure what the sample size was, it wasn't indicated by my professor. I was looking at this plot out of the context of the study. I never took stats, but am now basically assumed to have mastered it in a class I'm currently taking. While I guess it is possible that there were only 3 people in the data set I was looking at, I'm pretty sure there were just a lot of 0's in the data set I was referring to. Since negative values were impossible and since the median was zero, this necessitated that the min value and first quartile would be as well. I have learned that the extra dots (outliers) on the graph even come in two flavors, hollow and filled, referring to the degree of outlierness they possess. Again, thank you for your help, even as my quest for knowledge was "glaringly" misguided. —Preceding unsigned comment added by 67.167.150.120 (talk) 03:37, 16 September 2009 (UTC)[reply]

Brain Teaser[edit]

Hey. I recently ran into a brain teaser that I havn't been quite able to figure out. Maybe someone here could help? I read this is the place to ask knowledge questions. OK, here goes:
There are 5 values, whose sum is 60 and the sum of whose squares is 765. I can't quite get past this part though. Help? Thanx Lucretion (talk) 02:15, 15 September 2009 (UTC)[reply]

I assume you want them integer. But how many are odd, and how many are even, of your 5 numbers if the sum is even and the sum of squares is odd? --pma (talk) 03:38, 15 September 2009 (UTC)[reply]

This is the intersection of a sphere with a hyperplane, so it's a sphere of smaller dimension. Maybe more later.......... Michael Hardy (talk) 20:33, 15 September 2009 (UTC)[reply]

A quick program search shows that there aren't any solutions to this problems for integers in the range -27 thru +27. FWIW, the same program rapidly finds a solution for sum = 60 with sum of squares = 766. -- SGBailey (talk) 19:32, 16 September 2009 (UTC)[reply]

As pma has hinted, you don't need a program search to know that there are no integer solutions. -- Meni Rosenfeld (talk) 19:44, 16 September 2009 (UTC)[reply]

Zeros of Derivatives[edit]

Okay, here it goes. My deficiencies in analysis have forced to me to ask for help from all of you again. I am doing this proof in numerical analysis regarding some error analysis and this is the information given to me. The domain given is [-1,1] and G(x) is a continuous given function. We know that G(-1)=G(1)=0 and also there is another third real distinct zero of G(x), denoted t so G(t)=0 for some t in (-1,1) and we have no idea what t actually is. In addition, we also know that $G'(0)=G''(0)=0$ as well. Using all this, how can I deduce that $G''''(x)$ , the fourth derivative, has a zero in (-1,1)? I have a proof but it is ugly and complicated. I hate it and there are a billion subcases. I am sure that there is a nice concise proof. Thanks!-Looking for Wisdom and Insight! (talk) 02:37, 15 September 2009 (UTC)[reply]

The way to do these questions is always the same: apply Rolle's theorem a lot of times. It tells you (under some hypotheses...) that between every two zeros of f, there's a zero of f'. Use this, and the other zeros you were given, enough and the result follows ~~with no subcases~~.Silverfish70 (talk) 09:02, 15 September 2009 (UTC)[reply]

These simple notions should help to make it short. Let f(x) be a smooth function on (a,b) and a<x₀<b.

Say as usual that x₀ is a zero of f(x) with multiplicity m (at least) iff f vanishes at x₀ together with its derivatives of all orders up to m-1 included. In what follows, we count zeros with multiplicity.
Fact: if f has n zeros in the interval I (an interval of any kind: open, closed..), then f' has at least n-1 zeros in I (then also, of course, f^(j) has at least n-j zeros in I &c).
Fact: (Generalized Rolle's theorem). If f(a)=f(b), and f(x)≠0 in (a,b), then f' has an odd number of zeros in (a,b), or infinitely many.
I assume your G is four times differentiable. Case I : G(0)=0. Then, 0 is a zero of G of multiplicity 3; with 1 and -1 this makes at least a multiplicity 5, so G⁽⁴⁾ has at least a zero. Case II : G(0)≠0, so e.g. t>0. Then by the Rolle's thm G' has a zero in (t,1), and, by the generalization, an odd number of zeros in [-1,t] (precisely, in the largest interval I around 0 where G(x)≠0) , thus at least 3, as 0 is a zero of G' of multiplicity at least 2. Hence G' has at least 4 zeros, and again G⁽⁴⁾ has a zero.

Note: since you wanted the zero of G⁽⁴⁾ to be in (-1,1), some small details are to be added. Finally, a great reference for these elementary yet subtle facts is Pólya - Szegő : "Problems & Theorems in Analysis".--pma (talk) 16:07, 15 September 2009 (UTC)[reply]