Forest Error Analysis for the Physical Sciences

From New IAC Wiki
Revision as of 21:59, 5 December 2009 by Oborn (talk | contribs) (→‎Round Off)
Jump to navigation Jump to search

Class Admin

Forest_ErrorAnalysis_Syllabus

Homework

Homework is due at the beginning of class on the assigned day. If you have a documented excuse for your absence, then you will have 24 hours to hand in the homework after being released by your doctor.

Class Policies

http://wiki.iac.isu.edu/index.php/Forest_Class_Policies

Instructional Objectives

Course Catalog Description
Error Analysis for the Physics Sciences 3 credits. Lecture course with computation requirements. Topics include: Error propagation, Probability Distributions, Least Squares fit, multiple regression, goodnes of fit, covariance and correlations.

Prequisites:Math 360.

Course Description
The course assumes that the student has very limited experience with the UNIX environment and C/C++ programming. Homework problems involve modifying and compiling example programs written in C++.

Systematic and Random Uncertainties

Although the name of the class is "Error Analysis" for historical purposes, a more accurate description would be "Uncertainty Analysis". "Error" usually means a mistake is made while "Uncertainty" is a measure of how confident you are in a measurement.

Accuracy -vs- Precision

Accuracy
How close does an experiment come to the correct result
Precision
a measure of how exactly the result is determine. No reference is made to what the result means.

Reporting Uncertainties

Notation

X [math]\pm[/math] Y = X(Y)

Significant Figures and Round off

Significant figures

Most Significant digit
The leftmost non-zero digit is the most significant digit of a reported value
Least Significant digit
The least significant digit is identified using the following criteria
1.) If there is no decimal point, then the rightmost digit is the least significant digit.
2.)If there is a decimal point, then the rightmost digit is the least significant digit, even if it is a zero.

In other words, zero counts as a least significant digit only if it is after the decimal point. So when you report a measurement with a zero in such a posiiton you had better mean it.

The number of significant digits in a measurement are the number of digits which appear between the least and most significant digits.

examples:

Measurement most Sig. digit least Sig. Num. Sig. Dig. Sient. Not.
5 5 5 1 [math]5[/math]
5.0 5 0 2 [math]5.0 [/math]
50 5 0 1 [math]5 \times 10^{1}[/math]
50.1 5 1 3 [math]5.01 \times 10^{1}[/math]
0.005001 5 1 4 [math]5.001\times 10^{-3}[/math]
Note
The value of "50" above is ambiguous unless we use scientific notation in which case we know if the zero is significant or not.

Round Off

Measurements that are reported which are based on the calculation of more than one measured quantity must have the same number of significant digits as the quantity with the smallest number of significant digits.

To accomplish this you will need to round of the final measured value that is reported.

To round off a number you:

1.) Increment the least significant digit by one if the digit below it (in significance) is greater than 5.

2.) Do nothing the least significant digit by one if the digit below it (in significance) is less than 5.

Then truncate the remaining digits below the least significant digit.

What happens if the next significant digit below the least significant digit is exactly 5?

To avoid a systematic error involving round off you would ideally randomly decide to just truncate or increment. If your calculation is not on a computer with a random number generator, or you don't have one handy, then the typical technique is to increment if the number after the least significant digit is odd (or even) and truncate if it is even (or odd).

Statistical Distributions

Average and Variance

Average

The word "average" is used to describe a property of a probability distribution or a set of observations/measurements made in an experiment which gives an indication of a likely outcome of an experiment.

The symbol

[math]\mu[/math]

is usually used to represent the average.

Other notations are

[math]\bar{x}[/math]

Variance

The word "average" is used to describe a property of a probability distribution or a set of observations/measurements made in an experiment which gives an indication how the average value fluctuates.

A typical variable used to denot the variance is

[math]\sigma[/math]

Standard Deviation

The standard deviation is defined as the square root of the variance

S.D. = [math]\sqrt{\sigma}[/math]

Average for an unknown probability distribution (parent population)

The "Parent Population


If you are just given a list of number with no indication of the probability distribution that they were drawn from, then the average and variance may be calculate as shown below.

Arithmetic Mean and variance

If [math]n[/math] observables are mode in an experiment then the arithmetic mean of those observables is defined as

[math]\bar{x} = \frac{\sum_{i=1}^{i=n} x_i}{n}[/math]


The "unbiased" variance of the above sample is defined as

[math]s^2 = \frac{\sum_{i=1}^{i=n} (x_i - \bar{x})^2}{n-1}[/math]
If you were told that the average is [math]\bar{x}[/math] then you can calculate the

"true" variance of the above sample as

[math]\sigma^2 = \frac{\sum_{i=1}^{i=n} (x_i - \bar{x})^2}{n}[/math]

Weighted Mean and variance

If each observable ([math]x_i[/math]) is accompanied by an estimate of the uncertainty in that observable ([math]\delta x_i[/math]) then weighted mean is defined as

[math]\bar{x} = \frac{ \sum_{i=1}^{i=n} \frac{x_i}{\delta x_i}}{\sum_{i=1}^{i=n} \frac{1}{\delta x_i}}[/math]

The variance of the distribution is defined as

[math]\bar{x} = \sum_{i=1}^{i=n} \frac{1}{\delta x_i}[/math]

The average of a sample drawn from any probability distribution is defined in terms of the expectation value E(x) such that

The expectation value for a discrete probability distribution is given by

[math]E(x) = \sum_i x_i P(x_i)[/math]

The expectation value for a continuous probability distribution is calculated as

[math]E(x) = \int x dP(x)[/math]

variance

The variance of a sample draw from a probability distribution

Binomial

Binomial random variable describes experiments in which the outcome has only 2 possibilities. The two possible outcomes can be labeled as "success" or "failure". The probabilities may be defined as

p
the probability of a success

and

q
the probability of a failure.


If we let [math]X[/math] represent the number of successes after repeating the experiment [math]n[/math] times

Experiments with [math]n=1[/math] are also known as Bernoulli trails.

Then [math]X[/math] is the Binomial random variable with parameters [math]n[/math] and [math]p[/math].

The number of ways in which the [math]x[/math] successful outcomes can be organized in [math]n[/math] repeated trials is

[math]\frac{n !}{ \left [ (n-x) ! x !\right ]}[/math] where the [math] ![/math] denotes a factorial such that [math]5! = 5\times4\times3\times2\times1[/math].

The expression is known as the binomial coefficient and is represented as

[math]{n\choose x}=\frac{n!}{x!(n-x)!}[/math]


The probability of any one ordering of the success and failures is given by

[math]P( \mbox{experimental ordering}) = p^{x}q^{n-x}[/math]


This means the probability of getting exactly k successes after n trials is

[math]P(x=k) = {n\choose x}p^{x}q^{n-x} [/math]


It can be shown that the mean of the distribution is

[math]\bar{x} = n p[/math]

and the variance is

[math]\sigma^2 = npq[/math]

Examples

The number of times a coin toss is heads.

The probability of a coin landing with the head of the coin facing up is

[math]P = \frac{\mbox{number of desired outcomes}}{\mbox{number of possible outcomes}} = \frac{1}{2}[/math]

Suppose you toss a coin 4 times. Here are the possible outcomes


order Number Trial # # of Heads
1 2 3 4
1 t t t t 0
2 h t t t 1
3 t h t t 1
4 t t h t 1
5 t t t h 1
6 h h t t 2
7 h t h t 2
8 h t t h 2
9 t h h t 2
10 t h t h 2
11 t t h h 2
12 t h h h 3
13 h t h h 3
14 h h t h 3
15 h h h t 3
16 h h h h 4


The probability of order #1 happening is

P( order #1) = [math]\left ( \frac{1}{2} \right )^0\left ( \frac{1}{2} \right )^4 = \frac{1}{16}[/math]

P( order #2) = [math]\left ( \frac{1}{2} \right )^1\left ( \frac{1}{2} \right )^3 = \frac{1}{16}[/math]

The probability of observing the coin land on heads 3 times out of 4 trials is.

[math]P(x=3) = \frac{4}{16} = \frac{1}{4} = {n\choose x}p^{x}q^{n-x} = \frac{4 !}{ \left [ (4-3) ! 3 !\right ]} \left ( \frac{1}{2}\right )^{3}\left ( \frac{1}{2}\right )^{4-3} = \frac{24}{1 \times 6} \frac{1}{16} = \frac{1}{4}[/math]

Count number of times a 6 is observed when rolling a die

p=1/6

Expectation value :

The expected (average) value from a single roll of the dice is
[math]E({\rm Roll\ With\ 6\ Sided\ Die}) =\sum_i x_i P(x_i) = 1 \left ( \frac{1}{6} \right) + 2\left ( \frac{1}{6} \right)+ 3\left ( \frac{1}{6} \right)+ 4\left ( \frac{1}{6} \right)+ 5\left ( \frac{1}{6} \right)+ 6\left ( \frac{1}{6} \right)=\frac{1 + 2 + 3 + 4 + 5 + 6}{6} = 3.5[/math]

Poisson

[math]P(x) = \frac{\left ( \lambda s \right)^x e^{-\lambda s}}{x!}[/math]

where

[math]\lambda[/math] = probability for the occurrence of an event per unit interval [math]s[/math]


Homework Problem (Bevington pg 38)
Derive the Poisson distribution assuming a small sample size

1.) Assume that the average rate of an event is constant over a given time interval and that the events are randomly distributed over that time interval.

2.) The probability of NO events occuring over the time interval t is exponential such that

[math]P(0,t,\tau) = \exp^{-t/\tau}[/math]

where \tau is a constant of proportionality associated with the mean time

the change in the probability as a function of time is given by

[math]dP(0,t,\tau) = - P(0,t,\tau) \frac{dt}{\tau}[/math]

Gaussian

Lorentzian

Propagation of Uncertainties

Statistical inference

For this class we shall define a hypothesis test as a test used to

There are two schools of thought on this

frequentist statistical inference

Statistical inference is made using a null-hypothesis test; that is, ones that answer the question Assuming that the null hypothesis is true, what is the probability of observing a value for the test statistic that is at least as extreme as the value that was actually observed?


The relative frequency of occurrence of an event, in a number of repetitions of the experiment, is a measure of the probability of that event. Thus, if nt is the total number of trials and nx is the number of trials where the event x occurred, the probability P(x) of the event occurring will be approximated by the relative frequency as follows:

[math]P(x) \approx \frac{n_x}{n_t}.[/math]

Bayesian inference.

Statistical inference is made by using evidence or observations to update or to newly infer the probability that a hypothesis may be true. The name "Bayesian" comes from the frequent use of Bayes' theorem in the inference process.

Bayes gave a special case involving continuous probability distribution|continuous prior and posterior probability distributions and discrete probability distributions of data, but in its simplest setting involving only discrete distributions, Bayes' theorem relates the conditional probability|conditional and marginal probability|marginal probabilities of events A and B, where B has a non-vanishing probability:

[math]P(A|B) = \frac{P(B | A)\, P(A)}{P(B)}\,\! [/math].

Each term in Bayes' theorem has a conventional name:

  • P(A) is the prior probability or marginal probability of A. It is "prior" in the sense that it does not take into account any information about B.
  • P(A|B) is the conditional probability of A, given B. It is also called the posterior probability because it is derived from or depends upon the specified value of B.
  • P(B|A) is the conditional probability of B given A.
  • P(B) is the prior or marginal probability of B, and acts as a normalizing constant.

Bayes' theorem in this form gives a mathematical representation of how the conditional probabability of event A given B is related to the converse conditional probabablity of B given A.

Example

Suppose there is a school having 60% boys and 40% girls as students. The female students wear trousers or skirts in equal numbers; the boys all wear trousers. An observer sees a (random) student from a distance; all the observer can see is that this student is wearing trousers. What is the probability this student is a girl? The correct answer can be computed using Bayes' theorem. The event A is that the student observed is a girl, and the event B is that the student observed is wearing trousers. To compute P(A|B), we first need to know: P(A), or the probability that the student is a girl regardless of any other information. Since the observers sees a random student, meaning that all students have the same probability of being observed, and the fraction of girls among the students is 40%, this probability equals 0.4. P(B|A), or the probability of the student wearing trousers given that the student is a girl. As they are as likely to wear skirts as trousers, this is 0.5. P(B), or the probability of a (randomly selected) student wearing trousers regardless of any other information. Since P(B) = P(B|A)P(A) + P(B|A')P(A'), this is 0.5×0.4 + 1×0.6 = 0.8. Given all this information, the probability of the observer having spotted a girl given that the observed student is wearing trousers can be computed by substituting these values in the formula:

[math]P(A|B) = \frac{P(B|A) P(A)}{P(B)} = \frac{0.5 \times 0.4}{0.8} = 0.25.[/math]

Another, essentially equivalent way of obtaining the same result is as follows. Assume, for concreteness, that there are 100 students, 60 boys and 40 girls. Among these, 60 boys and 20 girls wear trousers. All together there are 80 trouser-wearers, of which 20 are girls. Therefore the chance that a random trouser-wearer is a girl equals 20/80 = 0.25. Put in terms of Bayes´ theorem, the probability of a student being a girl is 40/100, the probability that any given girl will wear trousers is 1/2. The product of two is 20/100, but you know the student is wearing trousers, so you remove the 20 non trouser wearing students and receive a probability of (20/100)/(80/100), or 20/80.


Chi-Square

comparing experiment with theory/function

Comparing 2 experiments

P-value

Root fundtion to evaluate meaning of Chi-square


PDG => Rather, the p-value is the probability, under the assumption of a hypothesis H , of obtaining data at least as incompatible with H as the data actually observed.

From http://en.wikipedia.org/wiki/P-value and http://en.wikipedia.org/wiki/Statistical_significance

the p-value is the frequency or probability with which the observed event would occur, if the null hypothesis were true. If the obtained p-value is smaller than the significance level, then the null hypothesis is rejected.

In some fields, for example nuclear and particle physics, it is common to express statistical significance in units of "σ" (sigma), the standard deviation of a Gaussian distribution. A statistical significance of "[math]n\sigma[/math]" can be converted into a value of α via use of the error function:

[math]\alpha = 1 - \operatorname{erf}(n/\sqrt{2}) \, [/math]

The use of σ is motivated by the ubiquitous emergence of the Gaussian distribution in measurement uncertainties. For example, if a theory predicts a parameter to have a value of, say, 100, and one measures the parameter to be 109 ± 3, then one might report the measurement as a "3σ deviation" from the theoretical prediction. In terms of α, this statement is equivalent to saying that "assuming the theory is true, the likelihood of obtaining the experimental result by coincidence is 0.27%" (since 1 − erf(3/√2) = 0.0027).


What is a p-value?

A p-value is a measure of how much evidence we have against the null hypothesis. The null hypothesis, traditionally represented by the symbol H0, represents the hypothesis of no change or no effect.

The smaller the p-value, the more evidence we have against H0. It is also a measure of how likely we are to get a certain sample result or a result “more extreme,” assuming H0 is true. The type of hypothesis (right tailed, left tailed or two tailed) will determine what “more extreme” means.

Much research involves making a hypothesis and then collecting data to test that hypothesis. In particular, researchers will set up a null hypothesis, a hypothesis that presumes no change or no effect of a treatment. Then these researchers will collect data and measure the consistency of this data with the null hypothesis.

The p-value measures consistency by calculating the probability of observing the results from your sample of data or a sample with results more extreme, assuming the null hypothesis is true. The smaller the p-value, the greater the inconsistency.

Traditionally, researchers will reject a hypothesis if the p-value is less than 0.05. Sometimes, though, researchers will use a stricter cut-off (e.g., 0.01) or a more liberal cut-off (e.g., 0.10). The general rule is that a small p-value is evidence against the null hypothesis while a large p-value means little or no evidence against the null hypothesis. Please note that little or no evidence against the null hypothesis is not the same as a lot of evidence for the null hypothesis.

It is easiest to understand the p-value in a data set that is already at an extreme. Suppose that a drug company alleges that only 50% of all patients who take a certain drug will have an adverse event of some kind. You believe that the adverse event rate is much higher. In a sample of 12 patients, all twelve have an adverse event.

The data supports your belief because it is inconsistent with the assumption of a 50% adverse event rate. It would be like flipping a coin 12 times and getting heads each time.

The p-value, the probability of getting a sample result of 12 adverse events in 12 patients assuming that the adverse event rate is 50%, is a measure of this inconsistency. The p-value, 0.000244, is small enough that we would reject the hypothesis that the adverse event rate was only 50%.

A large p-value should not automatically be construed as evidence in support of the null hypothesis. Perhaps the failure to reject the null hypothesis was caused by an inadequate sample size. When you see a large p-value in a research study, you should also look for one of two things:

a power calculation that confirms that the sample size in that study was adequate for detecting a clinically relevant difference; and/or a confidence interval that lies entirely within the range of clinical indifference. You should also be cautious about a small p-value, but for different reasons. In some situations, the sample size is so large that even differences that are trivial from a medical perspective can still achieve statistical significance.

As a statistician, I am not in a good position to advise you on whether a difference is trivial or not. As a medical expert, you need to balance the cost and side effects of a treatment against the benefits that the therapy provides.

The authors of the research paper should inform you what size difference is clinically relevant and what sized difference is trivial. But if they don't, you should. Ask yourself how much of a difference would be large enough to cause you to change your practice. Then compare this to the confidence interval in the research paper. If both limits of the confidence interval are smaller than a clinically relevant difference, then you should not change your practice, no matter what the p-value tells you.

You should not interpret the p-value as the probability that the null hypothesis is true. Such an interpretation is problematic because a hypothesis is not a random event that can have a probability.

Bayesian statistics provides an alternative framework that allows you to assign probabilities to hypotheses and to modify these probabilities on the basis of the data that you collect.

Example

A large number of p-values appear in a publication

Consultation Patterns and Provision of Contraception in General Practice Before Teenage Pregnancy: Case-Control Study. Churchill D, Allen J, Pringle M, Hippisley-Cox J, Ebdon D, Macpherson M, Bradley S. British Medical Journal 2000: 321(7259); 486-9. [Abstract] [Full text] [PDF]

by Churchill et al 2000. This was a study of consultation practices among teenagers who become pregnant. The researchers selected 240 patients (cases) with a recorded conception before the age of 20. Three controls were selected for each case and were matched on age and practice.

The not too surprising finding is that the cases were more likely to have consulted certain health professionals in the year before conception and were more likely to request contraceptive protection. This demonstrates that teenagers are not reluctant to seek advice about contraception.

For example, 91% of the cases (219/240) sought the advice of a general practitioner in the year before conception compared to 82% of the controls (586/719) during a similar time frame. This is a large difference. The odds ratio is 2.37. The p-value is 0.001, which indicates that this ratio is statistically significantly different from 1.0. The 95% confidence interval for the odds ratio is 1.45 to 3.86.

In contrast, 23% of the cases (56/240) sought advice from a practice nurse while 24% of the controls (170/719) sought advice. This is a small difference and the odds ratio is 0.98. The p-value is 0.905, which indicates that this odds ratio does not differ significantly from 1. As with any negative finding, you should be concerned about whether the result is due to an inadequate sample size. The confidence interval, however, is 0.69 to 1.39. This indicates that the research study had a good amount of precision and that the sample size was reasonable.


root [3] TMath::Prob(1.31,11)

Double_t Prob(Double_t chi2, Int_t ndf)
 Computation of the probability for a certain Chi-squared (chi2)
 and number of degrees of freedom (ndf).

 Calculations are based on the incomplete gamma function P(a,x),
 where a=ndf/2 and x=chi2/2.

 P(a,x) represents the probability that the observed Chi-squared
 for a correct model should be less than the value chi2.

 The returned probability corresponds to 1-P(a,x),
 which denotes the probability that an observed Chi-squared exceeds
 the value chi2 by chance, even for a correct model.

--- NvE 14-nov-1998 UU-SAP Utrecht

t-test

Komolgorov test

References

1.) "Data Reduction and Error Analysis for the Physical Sciences", Philip R. Bevington, ISBN-10: 0079112439, ISBN-13: 9780079112439

CPP programs for Bevington

Bevington programs

2.)An Introduction to Error Analysis, John R. Taylor ISBN 978-0-935702-75-0