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Propagating Uncertainty & The Chi-square Test Physics 270 – Experimental Physics

Propagating Uncertainty & The Chi-square Test

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Propagating Uncertainty & The Chi-square Test. Physics 270 – Experimental Physics. Propagation of Errors. Let say we are given a functional relationship between several measured variables Q ( x , y, … ) What is the uncertainty in Q if the uncertainties in x and y are known? - PowerPoint PPT Presentation

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Page 1: Propagating Uncertainty & The Chi-square Test

Propagating Uncertainty

& The Chi-square TestPhysics 270 – Experimental Physics

Page 2: Propagating Uncertainty & The Chi-square Test

Let say we are given a functional relationship between several measured variables Q(x, y, …)

What is the uncertainty in Q if the uncertainties in x and y are known?Example: Power in an electric circuit is

P = I2RLet I = 1.0 ± 0.1 A

R =10.0 ± 1.0 .Determine the power and its uncertainty, assuming

I and R are uncorrelated.

Propagation of Errors

x y and x yx y

Page 3: Propagating Uncertainty & The Chi-square Test

We can use the variance in Q(x,y) as a function of the variances in x and y to find the fractional uncertainties associated with each variable in the function. 

yQ

xQ

yQ

xQ

xyyxQ 22

22

22

If x and y are uncorrelated, then this is zero.

The derivation of this equation is in the lab manual.

Page 4: Propagating Uncertainty & The Chi-square Test

z = xy

Multiplication

2222

and

yxf xy

xyfy

xf

2

2

2

2

yxfyxf

Power in an electric circuit is P = I2R.Let I = 1.0 ± 0.1 A and R = 10.0 ± 0.5 .Determine the power and its variance using propagation of errors, assuming I and R are uncorrelated.

2 22 2 2 2 2 2 2 2

1 10

222 2 2 2

(2 ) ( )

(0.1) (2)(1.0 A)(10.0 ) (0.5 ) (1.0 A) 4.25 watts

2.06 W

P I R I RI R

P

P P IR II R

P = I2R = 10.0 ± 2.0 W

Page 5: Propagating Uncertainty & The Chi-square Test

Z = x + y

Example: I = I1 + I2

I1 = 4.0 ± 0.1 A I2 = 6.0 ± 0.2 A

Addition

2 2

1 and 1

f x y

f fx y

yxfyxf

22

A

A

2 22 21 2 0.1 0.2 0.22 A

4.0 A 6.0 A 0.2 A 10.0 0.2 AI

I

0.22 0.022 2.2 %10.0

I

I

Page 6: Propagating Uncertainty & The Chi-square Test

Addition and Subtraction

2 2 2 2 2 2z

2.0 0.2 cm 3.0 0.6 cm 4.52 0.02 cm2.0 3.0 4.52 cm = 0.5 cm

(0.2 cm) (0.6 cm) (0.02 cm) 0.63 cm

0.5 0.6 cm

x y w

z x y wx y wz

z

Page 7: Propagating Uncertainty & The Chi-square Test

Exponents

22

2

n p

yxz

z x y

pnz x y

Page 8: Propagating Uncertainty & The Chi-square Test

C = 2πRThen, ΔC = 2π ΔR

For example, if R = 3.1 ± 0.2 cm, thenC = 2π(3.0 cm) = 19.478 cm = 19.5 cmΔC = 2π (0.2 cm) = 1.257 cm

C = 19.5 ± 1.3 cm

Exact Numbers

Page 9: Propagating Uncertainty & The Chi-square Test

Unphysical situations can arise if we use the propagation of errors results blindly!

Example: Suppose we measure the volume of a cylinder: V = 2πRL.

Let R = exactly 1.0 cm, and L = 1.0 ± 0.5 cm.Using propagation of errors:

σV = 2πRσL = π/2 cm3

V = π ± π/2 cm3

If the error on V (σV) is to be interpreted in the Gaussian sense

☞ finite probability (≈ 3%) that the volume (V) is < 0 since V is only 2σ away from than 0!

☞ Clearly this is unphysical! ☞ Care must be taken in interpreting the meaning of σV.

Page 10: Propagating Uncertainty & The Chi-square Test

quick estimation of uncertainty for a function

Upper Bound / Lower Bound

q = 38.0° ± 0.6° Evaluate sinf q

max

min

sin(38.6 ) 0.62388

sin(37.4 ) 0.60738

f

f

0.62388+0.60738 0.61562

0.62388 0.60738 0.00825 = 0.0082

0.616 0.008

f

f

f

Page 11: Propagating Uncertainty & The Chi-square Test

method for determining whether experimental findings validate a theoretical prediction

Observed values are those that the researcher obtains empirically through direct observation. The theoretical or expected values are developed on the basis of an established theory or a working “hypothesis.”

Chi-Square Test for Goodness of Fit

Page 12: Propagating Uncertainty & The Chi-square Test

For example, we might expect that if we flip a coin 200 times, that we would tally 100 heads and 100 tails.

In checking this statistical expectation, we might find only 92 heads and 108 tails. Should we reject this coin as being fair? Should we just attribute the difference between expected and observed frequencies to random fluctuation?

Chi-Square Test

Page 13: Propagating Uncertainty & The Chi-square Test

Chi-Square Test

Face Frequency1 422 553 384 575 646 44

Consider a second example: let’s suppose that we have an unbiased, six-sided die. We roll this die 300 times and tally the number of times each side appears: Ideally, we might expect every side to appear 50 times. What should we conclude from these results? Is the die biased?

Page 14: Propagating Uncertainty & The Chi-square Test

The null hypothesis is a statement that is assumed true. The use of the chi-squared distribution is hypothesis testing follows this process:(1) a null hypothesis H0 is stated,(2) a test statistic is calculated, the observed value

of the test statistic is compared to a critical value, and

(3) a decision is made whether or not to reject the null hypothesis.

Null Hypothesis

Page 15: Propagating Uncertainty & The Chi-square Test

The null hypothesis rejected only when the data has a degree of statistical confidence that the null hypothesis is false, when the level of confidence exceeds a pre-determined level, usually 95 %, that causes a rejection of the null hypothesis.

If experimental observations indicate that the null hypothesis should be rejected, it means either that the hypothesis is indeed false or the measured data gave an improbable result indicating that the hypothesis is false, when it is really true.

Null Hypothesis

Page 16: Propagating Uncertainty & The Chi-square Test

Chi-squared Calculations

k

i i

ii

EEO

1

22 Face O E (OE)2 / E

Heads 92 100 0.64Tails 108 100 0.64

Totals 200 200 2 =1.28

Table I: The Coin Toss Example

Face O E (OE)2 / E1 42 50 1.282 55 50 0.503 38 50 2.884 57 50 0.985 64 50 3.926 44 50 0.72

Totals 300 300 2 =10.28

Table II: The 6-sided Die Example

Page 17: Propagating Uncertainty & The Chi-square Test

The chi-square distribution is tabulated and available in most texts on statistics. To use the table, one must know how many degrees of freedom df are associated with the number of categories in the sample data.

This is because there is a familyof chi-square distribution, each afunction of the number of degreesof freedom.

The number of degrees of freedom is typically equal to k 1.

Coin Example: df = 1 Die Example: df = 5

Degrees of Freedom

Page 18: Propagating Uncertainty & The Chi-square Test

A chi-square table lists the chi-squared distribution in terms of df and in terms of the level of confidence, = 1 p.

Levels of Confidence

df = 0.10 = 0.05 = 0.025 df = 0.10 = 0.05 = 0.025

1 2.706 3.841 5.024 11 17.275 19.675 21.920

2 4.605 5.991 7.378 12 18.549 21.026 23.337

3 6.251 7.815 9.348 13 19.812 22.362 24.736

4 7.779 9.488 11.143 14 21.064 23.685 26.119

5 9.236 11.071 12.833 15 22.307 24.996 27.488

6 10.645 12.592 14.449 16 23.542 26.296 28.845

7 12.017 14.067 16.013 17 24.769 27.587 30.191

8 13.362 15.507 17.535 18 25.989 28.869 31.526

9 14.684 16.919 19.023 19 27.204 30.144 32.852

10 15.987 18.307 20.483 20 28.412 31.410 34.170

Page 19: Propagating Uncertainty & The Chi-square Test

The standard practice in the world of statistics is to use a p = 95 % level of confidence in the hypothesis decision making. Thus, if the value of chi-squared that is calculated indicates a value of 1-p that is less than or equal to 0.05, then the null hypothesis should be rejected. In the coin-flip example, you can toss a coin and get 14 heads out of twenty flips and find chi-squared = 0.0577. This would indicate that such an observation can happen by chance and the coin can be considered a fair coin. Such a finding would be described by statisticians as “not statistically significant at the 5 % level.”

Levels of Confidence

Page 20: Propagating Uncertainty & The Chi-square Test

If one found 15 heads out of 20 tosses, then chi-squared would be somewhat less than 0.05 and the coin would be considered biased. This would be described as “statistically significant at the 5 % level.”

The significance level of the test is not determined by the p value. It is pre-determined by the experimenter. You can choose a 90 % level, a 95 % level, a 99 % level, etc.

Levels of Confidence

Page 21: Propagating Uncertainty & The Chi-square Test

If I asked you to tell me the average income of all of the employees of State Farm Insurance Company, I am asking for the value of the parameter , the mean income for the entire population of State Farm employees.

Since the number of employees is very large and spread around the United States, you would be forced to take a sample of that population. You would have to try and collect data from many job levels and from people from around the country. You would probably send out a survey and get substantial return of the forms to provide an acceptable sample. Then, you could report an average salary for your sample, which we’ll call . If I then ask, what is the average salary for employees of State Farm, you would tell me the value you got from your sample . You would be making a point estimate.

Estimation

x

x

x

Page 22: Propagating Uncertainty & The Chi-square Test

A public opinion poll may report that the results have a margin of error of ± 3%, which means that readers can be 95% confident (not 68% confident) that the reported results are accurate within 3 percentage points.

In physics, the same average result would be reported with an uncertainty of ± 1.5% to indicate the 68% confidence interval.

Margin of Error