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

1. Hilaire Ananda Perera
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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
NN
N
LCT
lim..



where N
SC = 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

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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

  
UL
P 
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

  
L
P 
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)”

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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,
22
r
T













,
22,
22
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

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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 t
j = Time of censorship and c = Number of censored items

(b) If failures are not replaced

T =



r
i
c
j
ji
tcrntt
11
*
)(




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, m
L1(0)
by setting r = 0 in the equation given before (See Page 3)

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Lower One-Sided Confidence Limits on MTBF (m) & Reliability (R)


mL1 =
2
22
2
T
r

;



Probability (m  m L1 ) = Confidence Level ................................... [1]



R
L1 =
e
t
m
L

1



Probability (R  R
L1 ) = 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 Perio
d
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

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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
22
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
0253 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
2326 2 1 2
.f
Can Use
EXCEL
Can Use
EXCEL

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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

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Example

Twenty units are subjected to a reliability test of 500 Hours duration. During this test five
failures occur at t
1 = 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 T 5 = 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) R
L1 (t = 2 hr) =
968
21

ee
L
m
t
= 0.997936
P [R(t = 2 Hr)  0.997936] = 0.90