1. Statistics for goods income tests with zero response
Zero response
If n tested items of the LOT show a zero response, i.e. zero non-conforming items
are found, a probability exists that non-conforming items are still in the LOT. For
n/M ≤ 0.05, i.e. large LOT size compared to the sample size, the binomial
distribution can be used to calculate the upper limit of non-conforming items for a
given confidence level C: [1]
P(k) =
n
k
pk
(1 − p)n−k
(1)
with n the sample size, k the number of non-conforming items and p the fraction of
non-conforming items in a LOT. For k = 0:
P(0) = (1 − p)n
. (2)
0.5 0.6 0.7 0.8 0.9 1.0
0.00.10.20.30.4
n = 11
Confidence level
Fractionofnon−conformingitems
Figure 1: Confidence level versus upper fraction of non-conforming items for n =
11.
Title: Zero response
Revision: 1
Effective: 2015-10-25
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ID-Number:
Author: Dr. Peter Drechsler
Developed: 2015-10-25
Phone: +49 711 46 92 73 40
Mobile: +49 173 2462720
peter.drechsler@qtec-group.com
qtec group R
qtec services GmbH
Humboldtstr. 30
70771 Leinfelden-Echterdingen
Germany

2. Statistics for goods income tests with zero response
Table 1: Maximal fraction of non-conforming items in a LOT in percent as a function
of samples n taken and zero non-compliance items k are found for
confidence intervals between 75% and 99%.
Samples Confidence limit
n 0.75 0.90 0.95 0.975 0.99
1 75.00 90.00 95.00 97.50 99.00
3 37.00 53.58 63.16 70.76 78.46
5 24.21 36.90 45.07 52.18 60.19
7 17.97 28.03 34.82 40.96 48.21
9 14.28 22.57 28.31 33.63 40.05
11 11.84 18.89 23.84 28.49 34.21
17 7.83 12.67 16.16 19.51 23.73
21 6.39 10.38 13.29 16.11 19.69
25 5.39 8.80 11.29 13.72 16.82
50 2.73 4.50 5.82 7.11 8.80
100 1.38 2.28 2.95 3.62 4.50
150 0.92 1.52 1.98 2.43 3.02
200 0.69 1.14 1.49 1.83 2.28
250 0.55 0.92 1.19 1.46 1.83
300 0.46 0.76 0.99 1.22 1.52
400 0.35 0.57 0.75 0.92 1.14
500 0.28 0.46 0.60 0.74 0.92
1000 0.14 0.23 0.30 0.37 0.46
and a upper fraction of non-conforming items 1 − C. This leads to
Cn
+ C − 1 = 0. (3)
Solved for n:
n =
ln(1 − C)
ln(C)
. (4)
One can ask for the largest p the makes P(0) reasonably small. Therefore, we set
P(0) = 1 − C, where C is the confidence level for the upper limit of non-conforming
items pu:
1 − C = (1 − pu)n
(5)
or
pu = 1 − n
√
1 − C (6)
Title: Zero response
Revision: 1
Effective: 2015-10-25
Page 2 of 3
ID-Number:
Author: Dr. Peter Drechsler
Developed: 2015-10-25
Phone: +49 711 46 92 73 40
Mobile: +49 173 2462720
peter.drechsler@qtec-group.com
qtec group R
qtec services GmbH
Humboldtstr. 30
70771 Leinfelden-Echterdingen
Germany

3. Statistics for goods income tests with zero response
Table 2: Sample size n for a given confidence level C.
C/% pu n
99.99 0.0001 92099
99.90 0.0010 6904
99.5 0.0050 1057
99 0.0100 458
97 0.0300 115
95 0.0500 58
93 0.0700 37
90 0.1000 22
70 0.3000 4
50 0.5000 1
Sometimes it’s very convienient to calculate the sample size n for a given
confidence level C
References
[1] ASTM E 2334-09. Standard practice for setting an upper confidence bound for
a fraction or number of non-conforming items, or a rate of occurence for
non-conformities, using attribute data, when there is a zero response in the
sample, 2009.
Title: Zero response
Revision: 1
Effective: 2015-10-25
Page 3 of 3
ID-Number:
Author: Dr. Peter Drechsler
Developed: 2015-10-25
Phone: +49 711 46 92 73 40
Mobile: +49 173 2462720
peter.drechsler@qtec-group.com
qtec group R
qtec services GmbH
Humboldtstr. 30
70771 Leinfelden-Echterdingen
Germany