J O U R N A L O F M A T E R I A L S S C I E N C E L E T T E R S 2 2, 2 0 0 3, 1771 – 1773
Determining the confidence intervals of Weibull parameters estimated using a more precise probability estimator J . A . G R I G G S ∗, Y U N L O N G Z H A N G Department of Biomaterials Science, Baylor College of Dentistry, Texas A&M University System Health Science Center, 3302 Gaston Avenue, Dallas, TX 75246, USA E-mail:
[email protected]
The two-parameter Weibull model [1] has been used to model the statistical variation in the failure stress of nominally identical ceramic components [2]. The probability that a given component will fracture at or below a tensile stress, σ , can be predicted as m σ Pf = 1 − exp − σ0
(1)
where m is the shape parameter, or Weibull modulus, and σ0 is the scale parameter, or characteristic strength. The distribution parameters, m and σ0 , can be estimated by fitting Equation 1 to a random sample of n specimens with failure stresses, σi , and cumulative probabilities of failure, Pi , where i is the rank of each specimen. Cumulative probability of failure is usually calculated by applying a discrete probability estimator of the form [3–5] i −α Pˆ i = n+β
(2)
where α and β are constants historically selected to ˆ minimize the bias of the sample Weibull modulus, m, ˆ so that m/m ≈ 1. Gong recently suggested the use of ˆ such that if the approa correction factor, km = m/m, priate value of km can be determined, then a corrected ˆ m , may serve sample Weibull modulus, mˆ unbiased = m/k as an unbiased estimate of m [6]. Gong showed that values of α = −0.999 and β = 1000 resulted in a low standard deviation for km and therefore a precise estimate of m. In another recent study, Gong described a method for determining the confidence interval for km as a function of the sample size, n, when α = −0.5 and β = 0 [7]. Since α = −0.999 and β = 1000 provide a more precise estimate of m, it is desirable to apply the method of confidence interval determination to this case. In addition, an estimate of the population characteristic strength, σ0 , is necessary to predict the probability of failure, so it is desirable to determine the confidence interval for an additional correction factor, kσ 0 = σˆ 0 /σ0 . These were the goals of the present study. The Monte Carlo method was used to generate random samples of 30 specimens each having a
Weibull distribution. Failure stress values, σi , were calculated as σi = σ0 ln
1 1 − Xi
1/m (3)
where X i was a random variable with uniform distribution between 0 and 1. Ten thousand samples were generated for each combination of population parameters, m and σ0 , listed in Table I. The failure stress values for each sample were ranked, and a cumulative probability of failure was assigned to each stress value according to Equation 2 with α = −0.999 and β = 1000. Least squares regression of ln ln(1 − Pi )−2 vs. ln σi was used to determine the Weibull parameˆ ters, mˆ and σˆ 0 , and the correction factors, km = m/m and kσ 0 = σˆ 0 /σ0 , for each sample. These results are summarized in Table I. It was previously reported that km can be described by a log normal distribution with little error when n ≥ 30 [7, 8]. This observation was verified in the present study, and kσ 0 was similarly distributed. km was independent of m and σ0 , and kσ 0 was independent of σ0 , so population parameters were limited to m = 5, 10, 15, or 20 and σ0 = 1 for convenience in the remainder of the present study. The above procedure was performed for each combination of population parameters using sample sizes of n = 10, 20, 30, 40, 50, and 100. The results are summarized in Figs 1–3. The mean values of ln km and m ln kσ 0 were dependent on sample size and were well
T A B L E I Correction factors, km and kσ 0 , to be used in estimating Weibull distribution parameters, m and σ0 , for a sample size of n = 30 m
σ0
ln km
ln k0
5 10 15 20 10 10
1 1 1 1 70 500
0.215 (0.166) 0.217 (0.164) 0.215 (0.164) 0.216 (0.164) 0.215 (0.163) 0.217 (0.165)
0.667 (0.109) 0.333 (0.054) 0.223 (0.036) 0.167 (0.027) 0.333 (0.054) 0.333 (0.054)
Mean (standard deviation).
∗ Author to whom all correspondence should be addressed. C 2003 Kluwer Academic Publishers 0261–8028
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fitted (R 2 > 0.99) by the following relations: Mm = −0.0008n − 0.0448 + 9.3573
Mσ 0
Figure 1 Mean value and standard deviation of ln km , to be used in estimating Weibull modulus, m, as a function of sample size.
1 n
1 −26.7284 2 (4) n 1 1 −0.0075n + 3.8638 − 9.7062 = m n 1 (5) + 15.5686 2 n
where Mm is the mean of the ln km values, Mσ 0 is the ˆ exp(Mm ). The mean of the ln kσ 0 values, and m ≈ m/ standard deviations of ln km and m ln kσ 0 were also dependent on sample size and were well fitted (R 2 > 0.99) by the following relations: 1 1 Sm = −0.0003n + 0.0936 + 2.5199 − 5.3329 2 n n (6) 1 1 −0.0015n + 0.3447 + 8.1178 Sσ 0 = m n 1 (7) −29.1186 2 n
Figure 2 Mean value of ln kσ 0 , to be used in estimating characteristic strength, σ0 , as a function of Weibull modulus and sample size.
where Sm is the standard deviation of the ln km values and Sσ 0 is the standard deviation of the ln kσ 0 values. Equations 4 and 6 are different than those originally suggested by Gong [7]. The equations suggested by Gong fit the data reasonably for sample sizes of n = 10–50, but are not acceptable for larger sample sizes. Equations 4–7 may be used in determining confidence intervals for the corrected estimates of Weibull modulus and characteristic strength. Since km and kσ 0 approximately follow a log normal distribution,
Figure 3 Standard deviation of ln kσ 0 , to be used in estimating characteristic strength, σ0 , as a function of Weibull modulus and sample size.
Figure 4 Ninety-five percent confidence interval for population Weibull modulus, m, as a function of sample size when m = 10 (Equation 10).
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where Z is the two-tailed test statistic for a normal distribution (1.96 for a 95% confidence interval). If one sample of n specimens is tested to estimate the population parameters, then the confidence intervals for m and σ0 may be calculated as follows: Sm m = mˆ exp − Mm ± Z √ n Sσ 0 σ0 = σˆ 0 exp − Mσ 0 ± Z √ n
(10) (11)
Fig. 4 illustrates the results of Equation 10 as a function of sample size when m = 10. Fig. 5 illustrates the results of Equation 11 as a function of sample size when m = 5, 10, or 20 and σ0 = 1. Figure 5 Ninety-five percent confidence interval for population characteristic strength, σ0 , as a function of Weibull modulus and sample size when σ0 = 1 (Equation 11).
References 1. W . W E I B U L L , J. Appl. Mech. 8 (1951) 293. 2. R . W . D A V I D G E , Phil. Trans. R. Soc. Lond. A 310 (1983) 113. 3. J . D . S U L L I V A N and P . H . L A U Z O N , J. Mater. Sci. Lett. 5
confidence intervals for the correction factors are as follows:
4. 5. 6. 7. 8.
(1986) 1245.
Sm km = exp Mm ± Z √ n Sσ 0 kσ 0 = exp Mσ 0 ± Z √ n
(8) (9)
A . K H A L I L I and K . K R O M P , ibid. 26 (1991) 6741. B . B E R G M A N , ibid. 3 (1984) 689. J . G O N G , ibid. 19 (2000) 827.
Idem., ibid. 18 (1999) 1405. ´ D E Z - S AE ´ Z and C . N A V A R R O , E . B A R B E R O , J . F E R N AN ibid. 20 (2001) 847.
Received 9 June and accepted 9 July 2003
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