Let's look at the some examples using correlation and regression analysis.
An example data set:
The Goal here is to find the best relation between, Y the dependent variable, and X- the independent variable.
X is the variable that would measure because Y is more difficult, and in some cases might be impossble to Measure.
Since we are measuring X - the role of measurement error will be come important. More on that later.
X Y 10.0 12.5 8.5 11.1 16.8 22.3 11.2 15.4 17.8 25.3 5.4 8.4 21.6 32.6 9.6 18.5 14.0 15.3 13.5 16.8
The correlation between X and Y is shown here:
Ypred = 1.39X + 0.03 ; dispersion = 2.53 ; r = 0.94
Let's calculate the residuals for each data point now.
X Y Y-pred Residual Significance 10.0 12.5 13.93 1.43 0.56 8.5 11.1 11.85 0.75 0.29 16.8 22.3 23.38 1.08 0.43 11.2 15.4 15.60 0.20 0.08 17.8 25.3 24.77 -0.52 -0.21 5.4 8.4 7.53 -0.86 -0.34 21.6 32.6 30.05 -2.54 -1.00 9.6 18.5 13.37 -5.12 -2.02 14.0 15.3 19.49 4.19 1.65 13.5 16.8 18.80 2.00 0.79
Try rejection analysis to improve the fit (mainly lower the scatter). Reject the most deviant point in the above.
That new relation is plotted here:
Ypred = 1.48X -1.76 ; dispersion = 1.96
This representation of the data is a more reliable and robust and allows Y to be more accurately estimated.
About measurement errors in X.
Suppose I have two relations involving different quantities but both use the same independent variable X.
Y1 = 1.5X + 1.5 ; with a dispersion of 0.5 units
Y2 = 6.0X + 2.5 ; with a dispersion of 0.3 units
Suppose that I can only make measurements of X which are accurate to 10%. This means that, despite a lower disperions, Y2 is less well determined than Y1!
Example: x = 10 +/- 1
Y1 = 1.5*10 +1.5 = 16.5
Y1 = 1.5*11 +1.5 = 18.0
So 10% uncertainty in X translates into +/- 1.5 unit uncertainty in Y.
For Y2 = 6.0*10 +2.5 = 62.5
For Y2 = 6.0*11 +2.5 = 68.5
So 10% uncertainty in X translates into +/- 6.0 unit uncertainty
So relations which have steep slopes require that X be measured very accurately.
About 75 years ago, Astronomers used the simple technique of correlation to discover the Universe was expanding. For nearby galaxies they measured a redshift and plotted that against the distance to the galaxy. Here is the data:
The line through the data is a "best fit" linear relationship which shows that there is a linear relationship between the the velocity at which a galaxy moves away from us and its distance. This linear relatinship is consistent with a model of uniform expansion for the Universe.
Binning this noisy data:
There are 12 points between x = 0 and 1 and 9 points between x = 1 and 2.
The means, dispersions, and errors in the mean for that binned data set are as follows:
X = 0.58 +/- 0.32 +/- .09
Y = 373 +/- 265 +/- 76
X = 1.74 +/- 0.30 +/- 0.10
Y = 744 +/- 206 +/- 72
Is there a significant difference between the means?
X = (1.74) - (0.58)/1.5(.10) = 7.7
Y = (744) - (373)/1.5(70) = 3.5
So yes, at higher values of X, the mean value of Y is significantly larger so a correlation exists.
So if you have enough data and you can bin in X, often the binned correlation will be obvious even in noisy data.
The PNI is a good example of this averaging process.
For the Bonneville Dam data:
What about Steelhead vs Chinook at Bonneville Dam. This is an exercise in data inspection and thinking. A superficial glance or a machine processing of the data will not reveal what may actually be in there.
Formally there is a very little correlation. The correlation coefficient, r, is 0.31. But look at the data closer to notice that its kind of odd.
There are 9 distinct occurences where the Steelhead Count is significantly above average (this corresponds to counts above 250,000). If we ignore those 9 points (years) out of the total of 57 years worth of data, the average Steelhead count is
The mean count for those 9 higher years is
Is the difference in these means significant?
One can therefore to conclude that something produces very high Steelhead Counts. Examining the data in time shows that the high Steelhead Counts occured in 1952--1953 and again in 1984-1989 and 1991-1992. High Steelhead count, however, does not mean high chinook count (nor does it correlate with anyother species)
For the whole data set, the weak correlation (r = 0.31) is shown below:
While a social scientist might argue that a correlation exists, you should be able to do better than that.
Okay, what about using just the chinook counts as a tracer of the entire salmon population. How well does that work? Here is the data:
Your eye sees a correlation and indeed r = 0.79 for this data set. Of course, some trend is expected since roughly 30--40% of the total Salmon Population is chinook; the question is, what is the dispersion in total salmon counts that results from using chinook as the tracer?
The formal fit is:
This means that chinook counts can be used to predict the total Salmon counts to an accuracy of 97,000. Since the Salmon count ranges from 500,000 to 1 million, that means an accuracy of 10-20%. This suggests that, if you are only interested in total Salmon, you can use chinook as a reliable tracer, provided that you don't require accuracy better than 20%.
The fit as applied to the data is shown here. In this case, r =0.79 and the fit is a good fit. There are no strongly abberant data points.
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