New To This - Help With Understanding Chi-Square and Residuals

At the risk of getting crushed by more knowledgeable folks on this forum...

In very simple terms, the Chi square test is a ratio of the amount of adjustment applied to a network (the a posteriori error) to the initial estimates that are given for each measurement (the a priori error). What it tells you is if your initial assumptions about your data are valid. Things that contribute to the initial error estimates include the accuracy of the instrument used, estimates of the possible centering error of the instrument and target, and in the case of a GPS baseline the error estimates provided by the baseline processor.

Residuals are the difference between an observed value and an adjusted value, or most probable value. As I understand it, and again to put it in very simple terms, let's assume we measure a distance ten times with a 100' steel chain of approximately 450'. Due to various factors, we are going to come up with variations in our ten distances. Let's assume that our shortest distance is 449.94 and our largest is 450.03. Least squares will test each value in that range and come up with a most probable value; the MPV will be the value that has the smallest residuals. If I remember correctly, the formula for this is the square root of the sum of the squares of the residuals divided by the number of observations minus one - hence the name Least Squares.

Feel free to batter me at your leisure fellow posters...

BTW, to truly understand least squares you need to have at least a working knowledge of calculus and especially linear algebra. My son started school as a Geomatics major and figured out not too far into it that a dual major in mathematics was going to be of great benefit.

Steward Souten, post: 427881, member: 12714 wrote:
Can someone explain what residuals are in terms of surveying?

Sure. Survey measurements generally are not error-free quantities. They are best thought of as estimates of the true value of some quantity such as an angle, distance, or height difference. While the sort of estimates that a skilled surveyor can produce are for some practical purposes what many people would consider to be exact determinations of what was measured, they never are truly so perfect and always contain errors of some magnitude.

For example, a distance of nominally 300 ft. measured with a tape may be in error by 0.03 ft. or more and the same distance measured with a good quality EDM may be in error by only 0.003 ft.

The presence of angular errors may be demonstrated by a number of methods, the easiest one to describe being the classical method of surveying closed polygons where the geometric test of comparing the sum of the interior angles of the figure to the exact theoretical value gives a means of detecting some types of errors in the work.

The typical method of adjustment involves making small corrections to measurements so that the geometric tests are met - interior angles sum to (n-2) x 180å¡, for example, where n is the number of sides. Those small corrections added to measured quantities are known as residuals. While the residual is typically treated as being a correction that is added to a measurement , it's worth noting that in some survey literature, the residuals are conceived of as being the errors contained in the measurements and are to be subtracted from the measurements. Star*Net deals with residuals as small corrections to be added to observations (measurements).

In good quality survey measurements, errors should tend to be random (as likely to be + as -) and to cluster around a mean value of 0.

The random errors commonly found in survey measurements are typically treated as having a pattern such that the likelihood of a random error of a particular magnitude drops off as the error gets larger. In mathematical terms, the errors in survey measurements are commonly treated as having a mound-shaped Gaussian distribution. that is described by a characteristic value known as the standard error.

The standard error is a value characteristic of a measuring process, such as measuring angles with a particular instrument to targets centered over ground marks, that the errors in approximately 68% of a sufficiently large series of measurements would not be expected to exceed.

In order to compare the magnitues of residuals in examining the outputs of an adjustment such as Star*Net performs, the residuals are standardized by dividing them by the standard errors that were expected of the process that produced the measurement.

Nice primer, guys!

I pretty much agree with everything that's been said, but have the urge to talk about it, too.

In your example, you gave the program some estimates of the standard errors. Those need to be realistic estimates of what your instruments and procedures produce, such that 68% of the time (one standard deviation) your results are within the range of what you gave the program.

The program found that after doing its best adjustment, your overall residuals (the amount that your measurements didn't fit the best estimate it made of what you were measuring) was 1.517 times the standard error (averaged as rms over all types of measurements).

Of course, it also breaks it down so you see that the angle measurements fit pretty close to as good as you told it (1.127), the distance measurements fit much better than you told it (0.131), and the zenith values were not nearly as good as you told it (2.292). So the zenith measurements were the culprit.

The chi-sq test failed because, with the amount of redundancy you have in your collection of measurements, it was reasonable for the residuals to come out no more than 1.210 times your standard errors, but not 1.517 times.

One question not addressed was the difference between a sample and the population. If a distance is measured 20 times, then we have a sample of 20 measurements drawn from the population made up of every single measurement of that distance that could be made. Our sample size is 20, but the population is infinite.

MathTeacher, post: 427972, member: 7674 wrote: One question not addressed was the difference between a sample and the population. If a distance is measured 20 times, then we have a sample of 20 measurements drawn from the population made up of every single measurement of that distance that could be made. Our sample size is 20, but the population is infinite.

There are two different approaches that may be used to estimate the standard errors of survey measurements. One is based upon prior evaluation by testing and the other treats the standard errors as unknown quantities to be worked out from analysis of the residuals of the adjustment.

The former method is efficient in that the standard errors used to weight measurements are derived from a sufficiently large sample size that they are well characterized, with small uncertainties in the values of the standard errors. The second method tends to be very inefficient where the standard errors are to be worked out from relatively few redundant measurements, which is equivalent to estimating the standard error of some measuring process from a small number of test measurements.

Some processes, such as angle and distance measurements, taking into account target and instrument centering, are much more easily characterized by testing beforehand. The manufacturers even specify standard errors that their equipment should not perform worse than under normal conditions.

The Chi-Square test is a "goodness of fit" test in lay terms. It tells you how well your weighting strategy fits with your actual data.

The chi-square test can be failed by exceeding the bounds on either the upper or lower bounds.

Exceeding the lower bounds may be serious. This may indicate that your weighting strategy is too conservative or it could fail because of a small data set with low redundancy.

Exceeding the upper limit is ALWAYS serious. This may indicate that your weighting strategy is too optimistic, there me be blunders in the data, or there may be uncorrected systematic errors in the data set.

Lee D, post: 427890, member: 7971 wrote: BTW, to truly understand least squares you need to have at least a working knowledge of calculus and especially linear algebra. My son started school as a Geomatics major and figured out not too far into it that a dual major in mathematics was going to be of great benefit.

I disagree. A good grasp of Statistics is whats important.

Jim in AZ, post: 427991, member: 249 wrote: I disagree. A good grasp of Statistics is whats important.

Statistics to interpret the results. Linear algebra to perform the adjustment.

Lee D, post: 427997, member: 7971 wrote: Statistics to interpret the results. Linear algebra to perform the adjustment.

I'll agree with that, but does anyone actually do this longhand?

Jim in AZ, post: 428020, member: 249 wrote: I'll agree with that, but does anyone actually do this longhand?

I know surveyors who longhand calculated compass rule adjustments, dmd areas, average end area volumes, geodetic coordinates, solar observations, and on and on (heck I used to do all those, some of them almost everyday),,,,,,,,,
but I've never heard of anyone who sits down and does least squares,,,,,,,,,
It's a black box operation.

Back to the original questions. Getting good statistical results from the given network design is likely to be difficult and might result from pure chance.

Perhaps adding a point or two would help.

Jim in AZ, post: 428020, member: 249 wrote: I'll agree with that, but does anyone actually do this longhand?

As a practical matter, no. I'm under the influence of my son who just completed two semesters of Measurement Science. For his final he had to devise an LSA for an example traverse and write a program to execute it.

MathTeacher, post: 428023, member: 7674 wrote: Getting good statistical results from the given network design is likely to be difficult

Why is that? I would think a good combination of angle and distance measurements would work fine.

Bill93, post: 428025, member: 87 wrote: Why is that? I would think a good combination of angle and distance measurements would work fine.

That's an awfully linear geometrical figure...

Jim in AZ, post: 428029, member: 249 wrote: That's an awfully linear geometrical figure

You're probably thinking of angle-only intersections to measure or stake a point, which would have a lot of leverage against you making the point very uncertain in distance from your setup points. But that's no problem if you are measuring distances too. Run an example and look at the ellipses.