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Statistical Confidence Level

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The statistical Confidence Level (C.L.) is the probability that the corresponding confidence interval covers the true ( but unknown ) value of a population parameter. Such confidence interval is often used as a measure of uncertainty about estimates of population parameters

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Statistical Confidence Level

  1. 1. Hilaire Ananda Perera http://www.linkedin.com/in/hilaireperera 1 Statistical Confidence Level The statistical confidence Level (C.L.) is the probability that the corresponding confidence interval covers the true (but unknown) value of a population parameter. Such confidence interval is often used as a measure of uncertainty about estimates of population parameters. In other words, rather than express statistical estimates as point estimates, it is much more meaningful to express them as a range (or interval), with an associated probability (or confidence) that the true value lies within such an interval T SC N N N LC T lim..   where NSC = Number of samples whose confidence limits on mˆ contain true m NT = Total number of samples tested 8 7 6 5 4 3 2 1 0 Infinity mL1,8 True MTBF (m) Mean Time Between Failures, m mL1,7 mL1,6 mL1,5 mL1,4 mL1,3 mL1,2 mL1,1
  2. 2. Hilaire Ananda Perera http://www.linkedin.com/in/hilaireperera 2 Confidence Limits and Intervals For most practical applications, we should be interested in the accuracy of the point estimate and the confidence which we can attach to it. We know that statistical estimates are more likely to be closer to the true value as the sample size increases. Only the impossible situation of having an infinitely large number of samples to test could give us 100 percent confidence or certainty that a measured value of a parameter coincides with the true value. For any practical situation, therefore, we must establish confidence intervals or ranges of values between which we know, with a probability determined by the finite sample size, that the true value of the parameter lies. Confidence intervals around point estimates are defined in terms of a lower confidence limit, L, and an upper confidence limit, U. If, for example, we calculate the confidence limits for a probability of, say, 95 percent, this means that in repeated sampling, 95 percent of the calculated intervals will contain the true value of the reliability parameter. If we want to be 99 percent sure that the true value lies within certain limits for a given sample size, we must widen the interval or test a larger number of samples if we wish to maintain the same interval width. The problem, then, is reduced to one of either determining the interval within which the true parametric value lies with a given probability for a given sample size, or determining the sample size required to assure us with a specified probability that true parametric value lies within a specific interval. Thus, we would be able to make statements such as   ULP  ˆˆ  =  --------- two-sided confidence limit, or confidence interval where  is some unknown population parameter, L and U are estimators associated with a random sample and  is a probability value such as 0.99, 0.95, 0.90, etc. If, for instance,  = 0.95, we refer to the interval (L <  < U) for particular values of Lˆ and Uˆ as a 95% confidence interval. In this case we are willing to accept a 5% risk that our statement is not, in fact, true. Or we may also want to make statements such as   LP  ˆ =  ---------- one-sided confidence limit in which case we make statements like, “we are 90% confident that the true MTBF is greater than some lower confidence limit (or measured value)”
  3. 3. Hilaire Ananda Perera http://www.linkedin.com/in/hilaireperera 3 Confidence Limits – Exponential Distribution Two situations have to be considered for estimating confidence intervals: one in which the test is run until a pre assigned number of failures (r* ) occurs, and one in which the test is stopped after a pre assigned number of test hours (t* ) is accumulated. The formula for the confidence interval employs the 2 (Chi-Square) distribution. The general notation used is dp, 2  where p and d are two constants used to choose the correct value from the table. The quantity p is a function of the confidence coefficient; d known as the degrees of freedom is a function of the number of failures Equations for Confidence Limits on Mean Life Type of Confidence Limits Fixed Number of Failures, r* Fixed Truncation Time, t* One Sided (Lower Limit)           , 2, 2 2 r T              , 22, 2 2 r T  Two Sided (Upper and Lower Limits)                         r T r T 2, 2 1 2 , 2, 2 2 22                              r T r T 2, 2 1 2 , 22, 2 2 22     The terms used are identified as follows: n = Number of items placed on test at time t = 0 t* = Time at which the life test is terminated  = Mean life (or MTBF for the case of replacement or repair upon failure) r = Number of failures accumulated at time t* r* = Preassigned number of failures  = Acceptable risk of error 1- = Confidence Level T = Total test time Degrees of Freedom (DF) is a function of “r” and for the case One-Sided (Lower Limit) Fixed Truncation Time, DF = 2r + 2
  4. 4. Hilaire Ananda Perera http://www.linkedin.com/in/hilaireperera 4 Total test time “T” is computed as follows, depending on the type of test procedure. Replacement Tests (failure replaced or repaired) T = nt* Non-Replacement Tests T =   r i i trnt 1 * )( Censored Items (withdrawal or loss of items which have not failed) (a) If failures are replaced and censored items are not replaced T =   c j j tcnt 1 * )( Where tj = Time of censorship and c = Number of censored items (b) If failures are not replaced T =     r i c j ji tcrntt 1 1 * )( MTBF (m) Determination when No Failures are Observed in a Situation If in a situation no failures are observed, one might be tempted to say that  0 ˆ T r T m , which of course cannot be correct, because from a finite sample and a short test time, during which no failures occur, we cannot say this equipment has an infinite m because a failure could have occurred any time after the examination of this situation. One can however calculate the Lower One-Sided Confidence Limit on m with zero failures, mL1(0) by setting r = 0 in the equation given before (See Page 3)
  5. 5. Hilaire Ananda Perera http://www.linkedin.com/in/hilaireperera 5 Lower One-Sided Confidence Limits on MTBF (m) & Reliability (R) mL1 = 2 2 22 T r ;  Probability (m  mL1 ) = Confidence Level ................................... [1] RL1 = e t mL  1 Probability (R  RL1 ) = Confidence Level ................................. [2] If a Customer requires the meeting of a MTBF or Reliability number at a certain Confidence Level, perform the following steps: (a) Use field Failure/Time data if available to show whether conditions [1] or [2] can be satisfied (b) If no field Failure/Time data is available, use Handbook failure rate predictions. It could be assumed that Handbook ( like MILL-HDBK-217 ) failure rates are at 60% UCL. Perform the following calculations using Chi-Square Distribution to obtain the MTTF(MTBF) at any other Confidence Level. T / 0.982 = MTBF@60% LCL for zero failures ................... [3] T / 2.303 = MTBF@90% LCL for zero failures .................... [4] From [3]&[4] MTBF@90% LCL = 0.426 of MTBF@60% LCL Indicate whether conditions [1] or [2] can be satisfied UCL = Upper Confidence Limit LCL = Lower Confidence Limit  2 = Chi-Square Distribution at  Significance Level ( = 1 - Confidence Level) r = Number of Failures 2r + 2 = Degrees of Freedom T = Use Time Period t = Mission Duration Typical Statements MTBF of Line Replaceable Unit will be greater than 20,000 Hours with 90% Confidence Reliability of Line Replaceable Unit will be greater than 0.99990 with 90% Confidence for the Mission Duration of 2 Hours
  6. 6. Hilaire Ananda Perera http://www.linkedin.com/in/hilaireperera 6 How To Calculate MTBF or MTBUR at a Confidence Level Lower One-Sided Confidence Limit on MTBF or MTBUR for a useTime Period T is: MTBF = T a Where a = Number of Failures at required Confidence Level MTBUR = T b Where b = Number of Removals at required Confidence Level a or b =  2 2 2 2 : r  Where  2 = Chi-Square Distribution at  Significance Level ( = 1 - Confidence Level) r = Number of Failures or Removals observed f = Degrees of Freedom = 2r + 2 For Degrees Of Freedom f < 30 use following table or the Generic Chi-Sq. Table: Number Of Failures Or Removals Degrees Of Freedom a or b At Confidence Level 99% 95% 90% 60% 50% 0 1 2 3 4 5 2 4 6 8 10 12 4.605 6.639 8.406 10.045 11.605 13.109 2.996 4.744 6.296 7.754 9.154 10.513 2.303 3.890 5.323 6.681 7.994 9.275 0.982 2.096 3.180 4.250 5.311 6.365 0.693 1.679 2.674 3.672 4.671 5.670 For f>30 Use following equations: a or b (60% Confidence Level) =  1 4 0 253 2 1 2 .  f a or b (95% Confidence Level) =  1 4 1645 2 1 2 .  f a or b (99% Confidence Level) =  1 4 2 326 2 1 2 .  f Can Use EXCEL Can Use EXCEL
  7. 7. Hilaire Ananda Perera http://www.linkedin.com/in/hilaireperera 7 How to Determine 2 Values from EXCEL fx Function Category: Statistical Function Name: CHIINV OK Enter Degrees of Freedom (DF) and Significance  (Risk Probability) for the requirement
  8. 8. Hilaire Ananda Perera http://www.linkedin.com/in/hilaireperera 8 Example Twenty units are subjected to a reliability test of 500 Hours duration. During this test five failures occur at t1 = 110 Hr, t2 = 180 Hr, t3 = 300 Hr, t4 = 410 Hr, and t5 = 480 Hr. The failed units are not replaced. Do the following: (a) Determine the mean-time-between failures, mˆ (b) Determine the lower one-sided confidence limit on m at a confidence level of 90% (c) Write a probabilistic statement concerning the result obtained in Case-b (d) Using the results for Case-b find the 90% confidence limit on the associated reliability for a mission of 2 Hours, and write a probabilistic statement Solution This is a time terminated test because the last failure occurred at T5 = 480 Hrs, whereas the test’s duration was t* = 500 Hrs, n = 20, r = 5,  (Significance (Risk)) = 1 – C.L. = 0.10 (a) r trnt r T m r i i   1 * )( ˆ Hoursm 1796 5 8980 5 500)520(480410300180110 ˆ    (b) Hours xT m xr L 968 549.18 1796017960898022 12;10.0 2 252;10.0 2 22; 21     (c) P (m  968 Hr) = 0.90 (e) RL1 (t = 2 hr) = 968 2 1   ee L m t = 0.997936 P [R(t = 2 Hr)  0.997936] = 0.90

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