The Will Rogers Phenomenon The so-called Will Rogers phenomenon describes a way to tweak statistics by raising the average in two (multi)sets when one element is moved between the two sets. As a simple example, consider the two sets

A = {1, 2, 3}
B = {4, 5, 6}


Their arithmetic means are  2 and 5, respectively. If we move the 4 to A:

A = {1, 2, 3, 4}
B = {5, 6}


Now the averages are 2.5 and 5.5, respectively, so both averages have been raised through a simple regrouping.

As another example, consider

A = {3, 4, 5, 6} --&gt; A = {3, 5, 6}
B = {2, 3, 4, 5} --&gt; B = {2, 3, 4, 4, 5}


On the other hand, it's not possible to raise both averages for the sets

A = {1, 5, 9}
B = {4, 5, 7, 8}


The Challenge

Given two lists of non-negative integers, determine whether it is possible to raise both averages by moving a single integer from one list to the other.

The average of an empty list is not defined, so if one of the lists contains only one element, this element cannot be moved.

You may write a program or function, taking input via STDIN (or closest alternative), command-line argument or function argument and outputting the result via STDOUT (or closest alternative), function return value or function (out) parameter.

Input may be taken in any convenient string or list format.

You must not assume that the elements in each list are unique, nor that they are sorted. You may assume that both lists contain at least one element.

Output should be truthy if both averages can be raised by moving a single integer and falsy otherwise.

This is code golf, so the shortest answer (in bytes) wins.

Test Cases

Truthy:

[1], [2, 3]
[1, 2, 3], [4, 5, 6]
[3, 4, 5, 6], [2, 3, 4, 5]
[6, 5, 9, 5, 6, 0], [6, 2, 0, 9, 5, 2]
[0, 4], [9, 1, 0, 2, 8, 0, 5, 5, 4, 9]


Falsy:

[1], [2]
[2, 4], [5]
[1, 5], [2, 3, 4, 5]
[2, 1, 2, 3, 1, 3], [5, 1, 6]
[4, 4, 5, 2, 4, 0], [9, 2, 10, 1, 9, 0]


Leaderboards

Here is a Stack Snippet to generate both a regular leaderboard and an overview of winners by language.

To make sure that your answer shows up, please start your answer with a headline, using the following Markdown template:

# Language Name, N bytes


where N is the size of your submission. If you improve your score, you can keep old scores in the headline, by striking them through. For instance:

# Ruby, &lt;s&gt;104&lt;/s&gt; &lt;s&gt;101&lt;/s&gt; 96 bytes




&lt;script&gt;site = 'meta.codegolf'; postID = 5314; isAnswer = true; QUESTION_ID = 53913&lt;/script&gt;&lt;script src='https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js'&gt;&lt;/script&gt;&lt;script&gt;jQuery(function(){var u='https://api.stackexchange.com/2.2/';if(isAnswer)u+='answers/'+postID+'?order=asc&amp;sort=creation&amp;site='+site+'&amp;filter=!GeEyUcJFJeRCD';else u+='questions/'+postID+'?order=asc&amp;sort=creation&amp;site='+site+'&amp;filter=!GeEyUcJFJO6t)';jQuery.get(u,function(b){function d(s){return jQuery('&lt;textarea&gt;').html(s).text()};function r(l){return new RegExp('&lt;pre class="snippet-code-'+l+'\\b[^&gt;]*&gt;&lt;code&gt;([\\s\\S]*?)&lt;/code&gt;&lt;/pre&gt;')};b=b.items[0].body;var j=r('js').exec(b),c=r('css').exec(b),h=r('html').exec(b);if(c!==null)jQuery('head').append(jQuery('&lt;style&gt;').text(d(c[1])));if (h!==null)jQuery('body').append(d(h[1]));if(j!==null)jQuery('body').append(jQuery('&lt;script&gt;').text(d(j[1])))})})&lt;/script&gt;
Pyth, 29 28 26 24 bytes
Thanks to @Jakube for saving me 3 bytes with .p and L.
Very simple, checks if any elements in list 2 are greater than mean of list 1 and less than mean of list 2, then repeats with list 1 and list 2 switched.
Lcsblbff&amp;&gt;YyhT&lt;YyeTeT.pQ

Prints an a non-empty list for truthy, and [] for falsey.
L                    Define y(b). Pyth has no builtin for mean
 c                   Float div
  sb                 Sum of b
  lb                 Length of b
f        .pQ         Filter all permutations of input
 f     eT            Filter the last element of the filter var
  &amp;                  Logical and
   &gt;Y                Inner filter var greater than
    y                Call the mean function we defined earlier
     hT              First element of outer filter var
   &lt;Y                Inner filter var less than
    y                Mean of
     eT              Last element of outer filternvar

Try it online here.
Test Suite.