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

G

GRISHAG

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?
 

ScottK

Not out of the crisis
Leader
Super Moderator
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

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
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

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

Bill McNeese

Involved In Discussions
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

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
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

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?
 
Top Bottom