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Error Propagation

Error Propagation. Uncertainty reflects the knowledge that a measured value is related to the mean. Probable error is the range from the mean with a 50% certainty. For a normal distribution:. Uncertainty. The coefficient of variation (CV) is the ratio of the standard deviation to the mean.

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Error Propagation

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  1. Error Propagation

  2. Uncertainty reflects the knowledge that a measured value is related to the mean. Probable error is the range from the mean with a 50% certainty. For a normal distribution: Uncertainty

  3. The coefficient of variation (CV) is the ratio of the standard deviation to the mean. Fractional standard deviation The CV for the Poisson distribution is fixed by the mean. For Poisson or normal distributions the measured count n has a fixed estimate for s. Significance is 0.683 Coefficient of Variation

  4. Measurements often involve multiple variables. Assume two independent variables x, y. N measurements of xi, yi Assume small differences from means. The cross term (covariance) vanishes for large N. First order expansion: The variance in Q is: Multiple Uncertainties

  5. Counting experiments are expected to have Poisson statistics. Assume precise time measurement Gross count rate includes all counts in a time interval. Number of counts ng Time tg Count Rate Error

  6. Typical Problem Two technicians measure a sample with a 35% efficiency counter, Julie gets 19 counts in 1 min and Phil gets 1148 counts in 60 min. What is the confidence that the activity is the stated value of 42.0 min-1? Answer From the given value, the expected count rate rg = 14.7 min-1. The difference in Julie’s rate from expected is 4.3 cpm. By chance 27% of time The difference in Phil’s rate from expected is 4.4 cpm. By chance 10-19 of time Significant Difference

  7. Experiments typically have a background counting rate. Net count rate subtracts background This involves two variables with errors. Use standard deviation of gross and background rates Net Count Rates

  8. Typical Problem A counter with 28% efficiency measures a background of 2561 counts in 90 min. A 10 min measure of a sample is gets 1426 counts. How long must a sample be measured to get the activity within 5% with 95% confidence? Answer The activity will have the same significance as the count rate. Scales with efficiency Find the count rates. rg = 142.6 min-1 rb = 28.5 min-1 rn = 114 min-1 0.05 rn = 5.7 min-1 95% is 1.96s, snr = 2.91 min-1. Background Subtraction

  9. The time to measure can be optimized. Based on background and gross rates. A fixed total time can be partitioned into optimum segments. Variance can be used instead of standard deviation Optimum Counting

  10. Short Lives • A rapidly decaying does not obey Poisson statistics. • Still Bernoulli process • Use lt >> 1 • For large t, s 0 • The equations can be adapted for a fixed efficiency. • At long times s limited by efficiency

  11. Many measurements only seek to determine if any activity is present above background. The minimum significant measurement determines the conditions to assert that there is activity above background. Failure is false positive; type I error. The minimum detectable measurement determines the conditions to assert that there is no activity. Failure is false negative; type II error. False Measurements

  12. The type I and type II errors are due to statistical fluctuations. A type I error is set by a rate r1 Measurements in curve area a give a false positive. A type II error is set by a rate r2 Measurements in curve area b give a false negative. Tails of the Curve Pn(rn) a b r1 r2 rn 0

  13. The minimum significant measurement, r1, depends on the tolerance for error. Set the probability a. Area in tail of normal distribution equal to a Equate to number of standard deviations ka Type I Error for tg = tb,

  14. A type I error is a false positive. Can’t be made if a real signal is present. With no signal, the only measurement is background. For nb >> ka2, For well measured background, nb2 = B, Typical Problem A background reading at 29% efficiency is 410 counts in 10 minutes. The maximum risk of false positive is 5%. How accurate is the approximation? Answer From 5%, ka = 1.65. To compare, nb = 410. Times are equal, r1 = 4.9 with the full formula or 4.7 with the approximation. Background Error

  15. The minimum detectable measurement also depends on risk. Willingness to risk a miss Set the probability b. Area in tail of normal distribution equal to b Equate to number of standard deviations kb The result is related to the minimum significant measurement. Type II Error

  16. As with type I, there are approximations based on the knowledge of the background. With a large background count, nb >> ka2. The best case fixes the background, nb2 = B. Background Counts

  17. Counts are often made at a variety values for some variable. Time, position, angle, mass, frequency, energy, etc. Noise refers to the fluctuations in the counts from point to point in the variable. Statistical tests apply to one or more bins. Noise

  18. An autocorrelation function tests for a periodic signal in the presence of noise. The autocorrelation of the combination is based on mean values compared at two points. Cross terms are zero for long times. Autocorrelation

  19. Correlation Measure • Autocorrelated noise should peak strongly at 0. • Sinusoidal signals would show periodic peaks of auto correlation. • “This sample autocorrelation plot shows that the time series is not random, but rather has a high degree of autocorrelation between adjacent and near-adjacent observations.” nist.gov

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