Sampling Question - Random Sample of 100 pcs and 11 pcs. have been rejected

G

GRISHAG

#1
I received a batch of 2,300 pcs. I took random sample of 100pcs, and 11 pcs. have been rejected after my inspection. (I have no special customer requirements).
What will be my conclusion regarding the batch?
 
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ScottK

Not out of the crisis
Leader
Super Moderator
#3
Re: sampling question

what Jim says.
as to whether you reject or accept the lot based on that information must take in a lot of factors and may be more a business decision than a quality decision.
 
G

GRISHAG

#4
Thanks. I want to add additional parameter: confidence level 90%.
I tried to do some calculations, but I can't paste them because I use symbol of root. Here is the beginning multiplicate by root:
Pmax=0.11+Z0.90 x
under root is 0.11x0.89 devided by 100
 

Bill McNeese

Involved In Discussions
#5
For a large sample, (n*p>10), you can use the following for confidence levels:

LCL = p - z*sqrt(pq/n)

UCL = p + z*sqrt(pq/n)

where z is the z value for alpha/2 and q = 1 - p. For small samples, the equation is a little more complex.
 
G

GRISHAG

#6
Thank you.
Can you help with calculation for my case?
What will be conclusion regarding the batch?
 

Bill McNeese

Involved In Discussions
#7
The calculation is straight forward. p = .11, q = .89, n = 100 and z = 1.645 for 90%. You can use NORMSINV in Excel to find z. What conclusion would you reach about the batch?
 
G

GRISHAG

#8
Thank you.
Using NORM.S.INV I received z=1.282 (not 1.645 as you wrote).
My conclusion is that in batch of 23,000 there is at least 15% rejected parts. Am I right?
 

Bill McNeese

Involved In Discussions
#9
All you can get from this is an estimated range. You had 11 parts rejected out of a sample size of 100. You want a confidence of 90%, which means alpha = 10%. Since this is two sided, the z value is based on alpha/2. Z for 0.05 is 1.645. Do the math and the confidence limits range from 6.85% to 17.21%. So, that is the range for % rejects based on this one sample result and size.
 
G

GRISHAG

#10
Thank you very much. I thought that this is one tail test, this is a reason that I choose 1.283.
How do you came to conclusion regarding that probability to find defective is between 6.85% and 17.21%? Could you send me formula for calculation?
 
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