Transformations

[NickL]

Hey guys,

The only concept I can't seem to grasp in further is linearising non-linear graphs.
How do I know when to use which transformation?

For an example, Q10 of Neap 2006 Exam 1.
The graph showed a scatterplot which was increasing exponentially.
How do I know to use x^2 as opposed to 1/y or logy when linearising such a graph?

[Stick]

Would you mind posting up the question?

[NickL]

Umm I don't have an electronic copy of this exam..

Question 10
The relationship between the two variables, y and x, as shown in the scatterplot below, is non-linear.


Which one of the following transformations, by itself, is most likely to linearise this data?
A) 1/x
B) 1/y
C) x^2
D) logx
E) logy

I know you can't really help without the actual graph but it's pretty much a scatterplot with all 6 points lying on what looks like an exponential graph or the right half of a x^2 +1 graph

[Stick]

Did the data points lie on a grid in which you could actually plug in the values into your calculator?

[NickL]

No. The only labels on the graph were the x and y axes. It's literally just a positive x and positive y graph with 6 dots on it

[Stick]

Alright, so this is what I was fearing it would turn out to be. Before I go into a very long explanation, I have one more question to ask to confirm my suspicions - was the correct answer x^2?

[NickL]

Yes it was x^2. I'd really like to know a general rule when applying transformations.
I have the circle of transformations (essentials p190) although it doesn't really specify which ones are more correct for different situations.

[Stick]

If you don't understand, remember that you will never be asked such a question on the real exam. The most similar type of question asked in any of the papers I've completed involve choosing and applying any appropriate transformation and possibly explaining why the data could/couldn't be linearised (sometimes, data is simply 'non-linear' and cannot be transformed - the performance of athletes is a good example). NEAP, Kilbaha and iTute exams should be avoided, because they are often too difficult compared to the actual standard and sometimes their solutions are incorrect.

Put simply, if the question stated the word 'exponentially' in it, then x^2 is the only appropriate exponential transformation. 1/y is also an exponential transformation, but as the diagram showed direct variation, it is incorrect. Perhaps if you do Methods you'll understand it a lot better. Otherwise, the question is fairly bogus.

[NickL]

The question did not state the word 'exponentially' in it, I am indeed a methods student and it is the only way I could describe the graph.
What do you mean by direct variation? I just want to know when to apply x^2 as opposed to 1/y, for instance.

[Yendall]

You have to realise what the transformations are doing.

is compressing the y values relative to the smaller data values

is compressing large y values relative to the smaller data values, to a greater extent than (values of y less than 1, become greater than 1, great becomes less etc.)

Spreads out the high x values relative to the smaller values.

I have the circle of transformations (essentials p190) although it doesn't really specify which ones are more correct for different situations.

The only way of figuring this out, I guess, is too see which transformation gives the best linear model.
It does not matter if you have to "guess", in essence, the points on the scatterplot because each transformations Coefficient of Determination is going to be different. So if you punch in your estimates on your calculator and note down all values for each transformation, you'll get your answer. It is tedious, but there isn't a way around it (to my knowledge). You could also create a residual plot in order to evaluate the linearity. Realistically, it won't take you more than a couple of minutes to do this, and it's the best way to get an accurate answer.

[Stick]

The question did not state the word 'exponentially' in it, I am indeed a methods student and it is the only way I could describe the graph.
What do you mean by direct variation? I just want to know when to apply x^2 as opposed to 1/y, for instance.

Direct variation means that as one variable increases, so does the other (and this occurs in this particular example). As you would know with a reciprocal graph, this does not occur.

It's a pretty vague explanation for a pretty poor question. Don't fret over it.

[astone788]

is compressing... to a greater extent than

True. As you can see in the green box, 1/Y ressembles a bow. WHEREAS log y is more straight. Hence, 1/Y is used to straighten the bow because of its ability to compress "to a greater extent." So as a general rule of thumb, if your line needs major straightening, use the 1/Y or 1/X transformatios when you have no time to test all of them 

note that for the Y2 transformation to apply, the scatterplot should peak or bottom around y = 0
note that for the X2 transformation to apply, the scatterplot should peak or bottom around x = 0


source: http://www.cambridge.edu.au/education/.../pageproofs_7_655904.pdf

[Yendall]

I recently came across a question in 2012 iTute exam, and thought it was appropriate that shared the solution with you guys:

Taking into consideration the nature of the increasing trend, you can eliminate transformations:

•                           

So we are left with:




We can see in the scatterplot that the x-values need compressing (but not to the extreme like ), rather then the y-values being spread, as identified by the two transformation diagrams. It depends on the scatterplot as to which is the best, and can definitely be determined via inspection. You just need to look closer at the scatterplot at the direction of the values.

[Stick]

As I said before, this is really getting into the nitty-gritty of the course and on all the VCAA multiple choice exams I've done they make sure the answer is very clear so there is no confusion. However, I guess it is definitely possible that it could appear on the exam so I'm glad this has been shared.

[Yendall]

I agree Stick, however it is good to have a clear understanding of the solutions to these types of questions. You never know what VCAA will throw at you. The more you know, the better you will go