# Probability: Sampling to find defects

#### Joerice

##### New Member
Hi, I'm new to the site.

We discovered that a process is resulting in a defect rate of 14% for some batches.
Each batch of product contains 2,500 units.

I want to sample each batch to determine if the problem exists in that batch or not, to a 95% confidence level. I need to calculate the sample size required that will give me a 95% chance of detecting the problem.

I have performed a calculation using probability where we sample without replacement and based on this it gives me a sample size of 20 per batch, which shows we should be 95% confident of identifying a defective unit. Data below.

I also considered solving this problem using the Binomial distribution with Minitab, but this seems to give me a different answer.

Is this a sound approach or can you suggest an alternative?

Unit No. good Total post sample Prob. Good unit Prob. next pump good
1 2150 2500 0.86
2 2149 2499 0.860 0.740
3 2148 2498 0.860 0.636
4 2147 2497 0.860 0.547
5 2146 2496 0.860 0.470
6 2145 2495 0.860 0.404
7 2144 2494 0.860 0.347
8 2143 2493 0.860 0.299
9 2142 2492 0.860 0.257
10 2141 2491 0.859 0.221
11 2140 2490 0.859 0.190
12 2139 2489 0.859 0.163
13 2138 2488 0.859 0.140
14 2137 2487 0.859 0.120
15 2136 2486 0.859 0.103
16 2135 2485 0.859 0.089
17 2134 2484 0.859 0.076
18 2133 2483 0.859 0.066
19 2132 2482 0.859 0.056
20 2131 2481 0.859 0.048
21 2130 2480 0.859 0.042

Thank you for reading,

Joe

#### vinux

##### Dark Knight
Hi, I'm new to the site.
Welcome to Talkstats forum:wave:.

I also considered solving this problem using the Binomial distribution with Minitab, but this seems to give me a different answer.
As per Binomial also the answer is same, the sample size as 20. What was the 'different answer' are talking here?

#### Joerice

##### New Member
Vinux,
firstly thank so much you for your reply, I'm really pleased that this is the correct sample size, I was starting to doubt myself!

Using the Binomial in Minitab I get the following:

"Welcome to Minitab, press F1 for help.

Cumulative Distribution Function

Binomial with n = 20 and p = 0.86

x P(*X*≤*x*)
1 0.00000
2 0.00000
3 0.00000
4 0.00000
5 0.00000
6 0.00000
7 0.00000
8 0.00000
9 0.00002
10 0.00014
11 0.00080
12 0.00384
13 0.01534
14 0.05067
15 0.13748
16 0.30412
17 0.54498
18 0.79157
19 0.95103
20 1.00000"

Which suggests a sample size of 19. Perhaps I am not using the correct function in minitab.

Any suggestions would be welcome.

Joe

#### vinux

##### Dark Knight
Vinux,
firstly thank so much you for your reply, I'm really pleased that this is the correct sample size, I was starting to doubt myself!

Using the Binomial in Minitab I get the following:

"Welcome to Minitab, press F1 for help.

Cumulative Distribution Function

Binomial with n = 20 and p = 0.86

x P(*X*≤*x*)
1 0.00000
2 0.00000
3 0.00000
4 0.00000
5 0.00000
6 0.00000
7 0.00000
8 0.00000
9 0.00002
10 0.00014
11 0.00080
12 0.00384
13 0.01534
14 0.05067
15 0.13748
16 0.30412
17 0.54498
18 0.79157
19 0.95103
20 1.00000"

Which suggests a sample size of 19. Perhaps I am not using the correct function in minitab.

Any suggestions would be welcome.

Joe
This is not the way to calculate the sample size.
Hint: In your calculation n is fixed (x is varying). You need to calculate the sample size required that will give me a 95% chance of detecting the problem. So, sample size is the variable.