# Confidence interval estimates for mean time between failures

Usually, in reliability studies in the industrial sector, the time to item failure is assumed to be exponentially distributed. Thus, the item failure rate becomes constant and, in turn, the mean time between failures (MTBF) is simply the reciprocal of the failure rate.

In testing a sample of parts/items with exponentially distributed times to failures, a point estimate of MTBF can be made; however, this figure only provides an incomplete picture because it fails to provide any surety of measurement. Nonetheless, it would probably be more realistic if we say, for example, that after testing a sample of items for t hours, n number of failures have occurred and the MTBF lies somewhere between certain lower and upper limits with certain confidence.

The confidence intervals on MTBF can be calculated by utilizing the y2 (chi-square) distribution. The usual notation used for obtaining chi-square values is as follows:

X2(p,df) (7.19)

where

df is the degrees of freedom.

p is a quantity function of the confidence coefficient.

The following list of symbols is used in subsequent associated formulas in this section [6, 7]:

9 is the acceptable error risk.

p is the mean life or MTBF.

C = 1 - 0 is the confidence level.

n is the number of items that were placed on test at zero time (i.e., t = 0).

m is the number of failures accumulated to time f, where f’denotes the life test termination time.

in' is the number of preassigned failures.

There are following two different cases to estimate confidence intervals:

• • Testing is terminated at a preassigned time t'.
• • Testing is terminated at a preassigned number of failures, in.

Thus, for the above two cases, to compute upper and lower limits of MTBF, the following formulas can be used [6, 7]:

• • Preassigned truncation time, t
• 2Y 2Y

Z2(|z2'" + 2)

(7.20)

• • Preassigned number of failures, in
• 2Y
• 2Y

%2(l-|,2m|

(7.21)

It is to be noted that the value of Y is determined by the test types: replacement test (i.e., the failed unit is replaced or repaired) and non-replacement test.

Thus, for the replacement test, we have

Y = nt'

• (7.22)
• (7.23)

Similarly, for the non-replacement test, we have

w

Y = (H-w)t*+^t;

/-1

where

t; is the jth failure time.

In the case of censored items/units (i.e., withdrawal or loss of unfailed items/units), the value of Y becomes as follows:

• • For replaced failed units/items but non-replacement of censored units/items
• (7.24)

where

c is the number of censored items/units.

t, is the ith, censorship time.

• For non-replaced failed and censored items/units

C HI

Y = (n-c-m)t' (725)

>=i /=i

Some tabulated values of %2 (p,df) are presented in Table 7.1.

Table 7.1 Values of chi-square distribution

 Degrees of freedom X2 value 2 0.1 0.21 4.6 5.99 4 0.71 1.06 7.77 9.44 6 1.63 2.2 10.64 12.59 8 2.73 3.49 13.36 15.5 10 3.94 4.86 15.98 18.3 12 5.22 6.3 18.54 21.02 14 6.57 7.79 21.06 23.68 16 7.96 9.31 23.54 26.29 18 9.39 10.86 25.98 28.86 20 10.85 12.44 28.41 31.41 Probability value: 0.95 0.9 0.1 0.05

Example 7.3

Assume that 20 identical electronic items were placed on test at time t = 0 and none of the failed items were replaced and the test was terminated after 200 hours. Three electronic items failed after 20,40, and 60 hours of operation. Calculate the electronic items' MTBF and its upper and lower limits with 90% confidence level.

By substituting the given data values into Equation (7.23), we get

Y = (20 - 3)(200)+(20 + 40 + 60) = 3520 hours

Thus, the electronic items' MTBF is given by

„ 3520

P = \$ = 1173.3 hours

By substituting the given and other values into Equation (7.20) and using Table 7.1, we obtain the following values of MTBF upper and lower limits:

Upper limit =

2(3520) X2 (0-95,6)

_ 2(3520)

1.63

= 4319.01 hours

T . 2(3520)

Lower limit = -r-,-------

X2 (0.05,8) 2(3520)

15.5 = 454.19 hours

Thus, we can state with 90% confidence that the electronic items' true MTBF will lie within 454.19 hours and 4319.01 hours or

454.19 <0 <4319.01-

Example 7.4

Assume that 16 electronic items were put on test at zero time and at the occurrence of eighth failure, the testing was stopped. The eighth failure occurred at 100 hours and all the failed items were replaced.

Calculate the MTBF of the electronic items and upper and lower limits on MTBF at 80% confidence level.

By substituting the given data values into Equation (7.22), we obtain

Y = (16)(100)=1600 hours

Thus, the electronic items' MTBF is given by

• 1600
• 8

= 200 hours

By substituting the given and other values into Equation (17.21) and using Table 7.1, we obtain the following values of MTBF upper and lower limits:

Upper limit =

2(1600)

X2 (0.90,16)

_ 2(1600)

9.31

= 343.7 hours

Lower limit =

2(1600)

X2 (0-1,16)

_ 3200

23.54

= 135.9 hours

Thus, at 80% confidence level, the electronic items' true MTBF will lie within 135.9 hours and 343.7 hours or 135.9 < 0 < 343.7.