questions - cqe test (part02) - ak

8/10/2019 Questions - CQE Test (Part02) - AK

1/33

Just logic so that overall cost not increased with the cycle time reduction.


2/33


3/33


4/33

I am not clear why npHint to use Poisson is the average no. of defect in the question: Average of one (1) flaw per squareN = 3 because 3 fts is the sample size, p = 1 since 1 flaw per square foot in question, then np = 3 x 1 =Probability of accept is only when p=0 defect, therefore chance to have P(0) when np = 3, which is 0.05Because anyting above 0 is fail, we calculate chance of reject P(0) which is 1 -0.05 = 0.95


5/33

floot3(from table)


6/33


7/33

My understanding: Even a process in statistical control, there will be a small amount of failure, evidenceven if the process is within statistical control. But the criteria to judge a process in statistic control is b

Becaise the samples are being mixed with the previous batches when all parts are dumped into a centrduring sampling.

what does "each individual plotpoint" mean? Sample size =?

should be"expect"?


8/33

When we do p chart, we are plotting no. of defects per lot. And the lot size can vary for each data point.for each lot (individual plot point)

Estimate of processstandard deviation!!!

the formula on page X-58 specifies samplindeviation (S) should be used? Compare to


9/33

Because the question is asking the proability of the next sample plot, which is the probability of sample

Why use sampling stS= /( n1/2 ) can be use


10/33

Same as above, because the question is asking for the next data plot, which is also a sample size. The


11/33

in the DPPM table for all different level of sigma level, therefore we cannot say all products meet specse on the 7 rules for the plots with the samples, and sample data fall within the statistical control limits.

al bin, the samples that is cause by special or assignable have mixed with the large population lower ch


12/33

. Sometime , we may want to see the charasteristic for each individual lot, therefore we will do the contr

Sample standard

g standardex.10.6


13/33

of 3, therefore we use sample std deviation instead of population std dev.

Why just use UCL?How anout LCL side?

andard deviation? Formulad?

deviation!!!S= /( n1/2 )


14/33

refore we have to use sample std dev for calculation.


15/33

ification .

ance be picked


16/33

l limit


17/33


18/33


19/33

From the solution, it seems like 6 come from = 1 for overall freedom, 1 +1 +1 for (n-1) degree of freedo And it says D.F equals no. of trails required


20/33

Expected value : For A defective 35x40/100 = 14For A Good 65x40/100 = 26For B defective 35x60/100 = 21

How to know?

?


21/33

Confidence interval is different from Confidence, the higher the confidence, the smaller the interval becFrom the formula, as n increase, the +/- Z * sigma/sq root of n is smaller, thus the spread of the interval

My understanding is any factor could be treated as block due to its availability, therefore it has no direct


22/33

I have the same wrong answer, I don't know why.

I will pick this answer because it say fractional - factorial

This is mentioned in the first paragraph on Pg 50. Just follow, I don't know why.


23/33

Don't know why, just remember and follow


24/33

m for 3 factors with 2 levels, 2 (n-1) degree of freedom for the factor with 3 levels

21+2 2=6


25/33


26/33

ause it means the range of mean don't spread out that much. l is closer (decrease)

relationship whether it has more levels than 2.


27/33

Example 11.23 on page#XI-52 A B C?


28/33


29/33

?


30/33


31/33


32/33


33/33

questions - cqe test (part02) - ak

Documents