Testing "closeness of data" I have data from two different sets (each with fields named Mobile and Landline). There are about 5 elements from each field in each set. They go something like so:  
Set A:  
Mobile: 45 61 78 65 71  
Landline: 56 71 75 98 12

Set B:  
Mobile: 67 71 75 64 23  
Landline: 9 34 56 21 22

I need to prove/disprove that the figures in the Mobile field are closer to the figures in the Landline field in Set A as against Set B. How do I go about doing that? Would a distance measure between the two suffice? If so, what distance measure?
P.S: There may not be enough data in the two sets, but that is all I have at the moment. 
Edit 1: Here is what I have tried in the interim. I made a single vector out of the Mobile and Landline scores for Set A and did the same for Set B. I then compute mean and variance for the two vectors. In case the variance/standard deviation for the vector from Set A is lesser than that of Set B, I can say that the Mobile scores in Set A are "closer" to Landline scores in Set A than Set B. Is this approach correct? Why or why not?
 A: I think a good distance measure here would be the Mahalanobis distance.
In the one-dimensional case, the measure simplifies to
$$
\text{distance} = \frac{|\mu_M - \mu_L|}{\sigma_{M+L}}
$$
Here, $\mu_M$ is the mean of the mobile data, $\mu_L$ is the mean of the landline data, and $\sigma_{M+L}$ is the standard deviation of the mobile and landline data combined together.
Running the numbers through from the question, I get
$$
\text{distance}_{\text{Set A}} = \frac{|64.0 - 62.4|}{22.86} = 0.07
$$
$$
\text{distance}_{\text{Set B}} = \frac{|60.0 - 28.4|}{24.81} = 1.27
$$
Basically, this measure is using the spread of the data to normalize the distance between the means.
A: There are classes of functions known as f-divergences that measure the difference between two probability distributions.  That sort of seems like what you are trying to do, but note that these functions are not guaranteed to have the properties of a distance metric (they might not even be symmetrical!).  The Kullback-Leibler divergence is a popular example of an f-divergence.
This might actually be overkill for what you are trying to do, and you might want something a bit more intuitive instead.
EDIT:  Upon closer examination of your question, it looks like it probably makes little sense to interpret your particular data in terms of probability distributions, so f-divergences are probably not want you want in this case.
A: My proposal
For each set calculate:


*

*|min(Mobile)-min(Landline)|

*|max(Mobile)-max(Landline)|

*|mean(Mobile)-mean(Landline)| (alternative: median)


This approach measures only how close the intervals are to each other. 
Then you have one vector for each set. Now the difference of sets can be calculated by using e.g. Euclidean Distance.
Regarding Edit1
Yes, this is certainly a way to go. Two remarks:


*

*Maybe instead of performing a "is-smaller"-test you may want to calculate the euclidean difference between the variances of both sets. This way you have a measure of the increase of quality

*Note that by using the variance of the union of Mobile and Landline you are not only measuring how close Mobile and Landline are to each other but also how the variances of Mobile and Landline are respectively. 


The second point marks the difference to my proposal. If Mobile and Landline contains data in the interval [0,100] (i.e. 0 and 100 are occuring too), my proposal will lead to the vector(0,0,0) which represents a perfect match, however, the variance of Mobile and Landline respectively is way too high, hence the matching is trivially correct but useless. 
One the other hand, if the variance in Mobile and/or Landline is fixed, I suggest to use my proposal to focus entirely on the closeness of the intervals.
General hint: If you create a metric, always check how to game it. Then ask yourself whether the imagined situation can actually happen or not.
A: What is the significance of the numbers ? If they represent some kind of quantity that admits a distance (i.e it makes sense to say that 54 is "closer" to 55 than to 56), then you could use the earthmover distance. 
