Statistics using dependant variables with different units

[xMoorez]

Hi guys,

 

I'm currently trying to do statistics to compare the variance of results in two different methods.

However, my results are in totally different units. So I think I need a way of scaling my data so that statistical analysis can take place.

 

The two methods are very different ways of measuring something, and the units cannot be converted to match. Is there a way of scaling, or perhaps a method concerning ratio?

 

The data sets are BOTH dependant variables and here are a few of the equivalent data points:

 

183 - 7.8

173 - 7.7

173 - 7.7

175 - 7.6

166 - 7.4

174 - 7.4

 

Any help would be GREATLY appreciated!

 

Thanks

 

A x

[studiot]

 

The two methods are very different ways of measuring something, and the units cannot be converted to match. Is there a way of scaling, or perhaps a method concerning ratio?

 

Please explain your aim in more detail.

 

You have said that the results are two different methods of measuring some property and that they cannot be interconverted.

 

Since they cannot be interconverted what do you want us to do?

 

 

 

I can think of a situation like this with hardness measurements where there is no direct correlation between the Mohs, Vickers and Brinell scales and the units have differenct physical dimensions.

 

Is your situation like this?

[mathematic]

I don't know if this is what you need. Get the average for each set, and divide the members of each set by its average

[MonDie]

I only took one statistics class, but I imagine that slope at inflection point could vary—a sort of center skew—even after both data sets are converted into standard deviations. For example, suppose one variable increases exponentially as the other increases linearly. Perhaps as an initial test, plot them as x and y, then see whether it's curved.

Edited May 14, 2015 by MonDie

[imatfaal]

Normalize the data?

 

What form of normalization will depend on the data

[studiot]

I don't see how any of the suggestions would help with my hardness example.

 

The OP was careful to specify that

 

The two methods are very different ways of measuring something

[MonDie]

Imatfaal, here's an easy to calculate, symmetric example.

 

With the frequency of each value going from 1 to 4 back to 1.

Data set one: -3, -2, -1, 0 1, 2, 3; [math]\bar{x}[/math]=0, σ=1.58

Data set two: -9, -4, -1, 0, 1, 4, 9; [math]\bar{x}[/math]=0, σ=3.81

 

Variance Calculations
(0x4 + 1x6 + 4x4 + 9x2) / 16 = (6+16+18)/16 = 40/16 = 2.5

(0x4 + 1x6 + 16x4 + 81x2) / 16 = (6+64+162)/16 = 232/16 = 14.5

 

Variance to Standard Deviation

2.5^0.5 = 1.58

14.5^0.5 = 3.81

 

After normalization

Data set one: -1.90, -1.27, -0.63, 0.00, 0.63, 1.27, 1.90,

Data set two: -2.36, -1.05, -0.26, 0.00, 0.26, 1.05, 2.36,

 

They don't look identical to me. Set two has a more compact center to compensate for the larger residuals of its outliers. In a best-fit curve, this difference should be apparent as a less steep inflection point. Kurtosis perhaps, but that's assuming the skewedness is greatest at the edges, at the outliers, and least in the center, at the mean, as it is here for convenience.

Edited May 14, 2015 by MonDie