TF ErrorAna PropOfErr - Revision history

Foretony: /* Counting Experiment Example */

2017-11-03T20:05:06Z

Counting Experiment Example

2017-02-24T21:10:04Z

Weighted Mean

2016-02-26T19:56:31Z

Taylor Expansion

2016-02-19T19:00:11Z

Taylor Expansion

2015-02-27T00:49:00Z

Counting Experiment Example

2015-02-27T00:48:05Z

Counting Experiment Example

2015-02-27T00:47:07Z

Counting Experiment Example

2014-02-28T20:45:09Z

Example: Table Area

2014-02-28T20:44:25Z

Example: Table Area

2014-02-28T20:43:52Z

Example: Table Area

@@ Line 216: / Line 216: @@
 Now the combined probability that the first experiment measures the average <math>X_A</math> and the second <math>X_B</math> is given as the product of the two probabilities suth that
-:<math>P(x=X_A,X_B) \propto \frac{e^{-\frac{1}{2} \left ( \frac{X_A-X}{\sigma_A}\right )^2}}{\sigma_A} \frac{e^{-\frac{1}{2} \left ( \frac{X_B-X}{\sigma_B}\right )^2}}{\sigma_B} = \frac{e^{-\frac{1}{2}\left [  \left ( \frac{X_A-x}{\sigma_A}\right )^2+\left ( \frac{X_B-X}{\sigma_B}\right )^2\right ]}}{\sigma_A \sigma_B}\equiv \frac{e^{-\frac{1}{2}\left [  \chi^2\right ]}}{\sigma_A \sigma_B}</math>
+:<math>P(x=X_A,X_B) \propto \frac{e^{-\frac{1}{2} \left ( \frac{X_A-X}{\sigma_A}\right )^2}}{\sigma_A} \frac{e^{-\frac{1}{2} \left ( \frac{X_B-X}{\sigma_B}\right )^2}}{\sigma_B} = \frac{e^{-\frac{1}{2}\left [  \left ( \frac{X_A-X}{\sigma_A}\right )^2+\left ( \frac{X_B-X}{\sigma_B}\right )^2\right ]}}{\sigma_A \sigma_B}\equiv \frac{e^{-\frac{1}{2}\left [  \chi^2\right ]}}{\sigma_A \sigma_B}</math>
 where

@@ Line 20: / Line 20: @@
 The taylor expansion of a function with two variables<math> (x_1 , x_2)</math> about the average of the two variables<math> (\bar {x_1} , \bar{x_2} )</math> is given by
-<math>f(x, y)=f(\bar {x}, \bar{x})+(x-\bar {x}) \frac{\partial f}{\partial x}\bigg |_{(x = \bar {x}, y = \bar{y})} +(y-\bar{y}) \frac{\partial f}{\partial y}\bigg |_{(x = \bar {x}, y = \bar{x})}</math>
+<math>f(x, y)=f(\bar {x}, \bar{y})+(x-\bar {x}) \frac{\partial f}{\partial x}\bigg |_{(x = \bar {x}, y = \bar{y})} +(y-\bar{y}) \frac{\partial f}{\partial y}\bigg |_{(x = \bar {x}, y = \bar{y})}</math>
 or

@@ Line 13: / Line 13: @@
 For small values of x (x << 1) we can expand the function about 0 such that
-<math>\sqrt{1+x} = \left . \sqrt{1-0} \frac{1}{2}(1+x)^{-1/2}\right |_{x=0} \frac{x^1}{1!}+ \left . \frac{1}{2}\frac{-1}{2}(1+x)^{-3/2} \right |_{x=0} \frac{x^2}{2!}</math>
+<math>\sqrt{1+x} = \left . \sqrt{1-0}  + \frac{1}{2}(1+x)^{-1/2}\right |_{x=0} \frac{x^1}{1!}+ \left . \frac{1}{2}\frac{-1}{2}(1+x)^{-3/2} \right |_{x=0} \frac{x^2}{2!}</math>
 ;<math>=1 + \frac{x}{2} - \frac{x^2}{8}</math>

← Older revision		Revision as of 00:49, 27 February 2015
Line 111:		Line 111:


−	: <math>= 0.68 * 8 = 5</math>	+	: <math>= 0.68 * 8 = 5 \Rightarrow</math> expect at least 5 measurements to be within 1 <math>\sigma</math>

	Looks like we have 7/8 events within 1<math> \sigma</math> = 87.5%		Looks like we have 7/8 events within 1<math> \sigma</math> = 87.5%

← Older revision		Revision as of 00:48, 27 February 2015
Line 93:		Line 93:


−	If you approximate the Poisson distribution by a Gaussian then the probability any one measurement is within 1 <math>\sigma</math> of the mean is 68% = Probability that a measurement of a Gaussian variant will lie within 1 <math>\sigma</math> of the mean. For the Poisson distribution with a mean of 97 you would have 66% of the data occur within 1 <math>\sigma = \sqrt{97}= 9~~.8 < 10~~.8</math>.	+	If you approximate the Poisson distribution by a Gaussian then the probability any one measurement is within 1 <math>\sigma</math> of the mean is 68% = Probability that a measurement of a Gaussian variant will lie within 1 <math>\sigma</math> of the mean. For the Poisson distribution with a mean of 97 you would have 66% of the data occur within 1 <math>\sigma = \sqrt{97}= 9.8 </math>.

	<pre>		<pre>

← Older revision		Revision as of 00:47, 27 February 2015
Line 93:		Line 93:


−	If you approximate the Poisson distribution by a Gaussian then the probability any one measurement is within 1 <math>\sigma</math> of the mean is 68% = Probability that a measurement of a Gaussian variant will lie within 1 <math>\sigma</math> of the mean. For the Poisson distribution with a mean of 97 you would have 66% of the data occur within 1 <math>\sigma = \sqrt{97}</math>.	+	If you approximate the Poisson distribution by a Gaussian then the probability any one measurement is within 1 <math>\sigma</math> of the mean is 68% = Probability that a measurement of a Gaussian variant will lie within 1 <math>\sigma</math> of the mean. For the Poisson distribution with a mean of 97 you would have 66% of the data occur within 1 <math>\sigma = \sqrt{97}= 9.8 < 10.8</math>.

	<pre>		<pre>

← Older revision		Revision as of 20:45, 28 February 2014
Line 175:		Line 175:
	:<math>\sigma^2_A = \frac{\sum_{i=1}^{i=N} (A_i - \bar{A})^2}{N}</math>		:<math>\sigma^2_A = \frac{\sum_{i=1}^{i=N} (A_i - \bar{A})^2}{N}</math>

−	:<math>= \sum_{i=1}^{i=N}~~\frac{~~ \left [ (L-\bar{L}) \frac{\partial A}{\partial L} \bigg \|_{\bar L \bar W} + (W-\bar W) \frac{\partial A}{\partial W} \bigg \|_{\bar L \bar W} \right] ^2}{N}</math>	+	:<math>=\frac{ \sum_{i=1}^{i=N} \left [ (L-\bar{L}) \frac{\partial A}{\partial L} \bigg \|_{\bar L \bar W} + (W-\bar W) \frac{\partial A}{\partial W} \bigg \|_{\bar L \bar W} \right] ^2}{N}</math>

@@ Line 160: / Line 160: @@
 A quantity which is calculated using  quantities with known uncertainties will have an uncertainty based upon the uncertainty of the quantities used in the calculation.
-To determine the uncertainty in a quantity which is a function of other quantities, you can consider the dependence of these quantities in terms of a tayler expansion
+To determine the uncertainty in a quantity which is a function of other quantities, you can consider the dependence of these quantities in terms of a Tayler expansion
 Consider a calculation of a Table's Area

← Older revision		Revision as of 20:43, 28 February 2014
Line 175:		Line 175:
	:<math>\sigma^2_A = \frac{\sum_{i=1}^{i=N} (A_i - \bar{A})^2}{N}</math>		:<math>\sigma^2_A = \frac{\sum_{i=1}^{i=N} (A_i - \bar{A})^2}{N}</math>

−	:<math>= ~~\frac{~~\sum_{i=1}^{i=N} \left [ (L-\bar{L}) \frac{\partial A}{\partial L} \bigg \|_{\bar L \bar W} + (W-\bar W) \frac{\partial A}{\partial W} \bigg \|_{\bar L \bar WW} \right] ^2}{N}</math>	+	:<math>= \sum_{i=1}^{i=N}\frac{ \left [ (L-\bar{L}) \frac{\partial A}{\partial L} \bigg \|_{\bar L \bar W} + (W-\bar W) \frac{\partial A}{\partial W} \bigg \|_{\bar L \bar W} \right] ^2}{N}</math>