I am not clear why np
Hint to use Poisson is the average no. of defect in the question: Average of one (1) flaw per square floot N = 3 because 3 fts is the sample size, p = 1 since 1 flaw per square foot in question, then np = 3 x 1 = 3
Probability of accept is only when p=0 defect, therefore chance to have P(0) when np = 3, which is 0.05 (from table) Because anyting above 0 is fail, we calculate chance of reject P(0) which is 1 -0.05 = 0.95
Hint to use Poisson is the average no. of defect in the question: Average of one (1) flaw per square floot N = 3 because 3 fts is the sample size, p = 1 since 1 flaw per square foot in question, then np = 3 x 1 = 3
My understanding: Even a process in statistical control, there will be a small amount of failure, evidence in the DPPM table for all different level of sigma level, therefore we cannot say all products meet specification even if the process is within statistical control. But the criteria to judge a process in statistic control is base on the 7 rules for the plots with the samples, and sample data fall within the statistical control limits.
Becaise the samples are being mixed with the previous batches when all parts are dumped into a central bin, the samples that is cause by special or assignable have mixed with the large population lower chance be picked during sampling.
what does "each individual plot
point" mean? Sample size =?
should be
"expect"?
When we do p chart, we are plotting no. of defects per lot. And the lot size can vary for each data point. Sometime , we may want to see the charasteristic for each individual lot, therefore we will do the control limit for each lot (individual plot point)
Estimate of process
standard deviation!!!
the formula on page X-58 specifies sampling standard deviation (S) should be used? Compare to ex.10.6
Because the question is asking the proability of the next sample plot, which is the probability of sample of 3, therefore we use sample std deviation instead of population std dev.
Why use sampling standard deviation? Formula
S=σ/(n
1/2) can be used?
My understanding: Even a process in statistical control, there will be a small amount of failure, evidence in the DPPM table for all different level of sigma level, therefore we cannot say all products meet specification even if the process is within statistical control. But the criteria to judge a process in statistic control is base on the 7 rules for the plots with the samples, and sample data fall within the statistical control limits.
When we do p chart, we are plotting no. of defects per lot. And the lot size can vary for each data point. Sometime , we may want to see the charasteristic for each individual lot, therefore we will do the control limit
Sample standard
deviation!!!
58 specifies sampling standard deviation (S) should be used? Compare to ex.10.6
Because the question is asking the proability of the next sample plot, which is the probability of sample of 3, therefore we use sample std deviation instead of population std dev.
Why just use UCL?
How anout LCL side?
Why use sampling standard deviation? Formula
) can be used?
Sample standard
deviation!!!
My understanding: Even a process in statistical control, there will be a small amount of failure, evidence in the DPPM table for all different level of sigma level, therefore we cannot say all products meet specification even if the process is within statistical control. But the criteria to judge a process in statistic control is base on the 7 rules for the plots with the samples, and sample data fall within the statistical control limits.
From the solution, it seems like 6 come from = 1 for overall freedom, 1 +1 +1 for (n-1) degree of freedom for 3 factors with 2 levels, 2 (n-1) degree of freedom for the factor with 3 levels And it says D.F equals no. of trails required
Expected value : For A defective 35x40/100 = 14 For A Good 65x40/100 = 26 For B defective 35x60/100 = 21
How to know?
Confidence interval is different from Confidence, the higher the confidence, the smaller the interval because it means the range of mean don't spread out that much. From the formula, as n increase, the +/- Z * sigma/sq root of n is smaller, thus the spread of the interval is closer (decrease)
My understanding is any factor could be treated as block due to its availability, therefore it has no direct relationship whether it has more levels than 2.
I have the same wrong answer, I don't know why.
I will pick this answer because it say fractional - factorial
From the solution, it seems like 6 come from = 1 for overall freedom, 1 +1 +1 for (n-1) degree of freedom for 3 factors with 2 levels, 2 (n-1) degree of freedom for the factor with 3 levels
Confidence interval is different from Confidence, the higher the confidence, the smaller the interval because it means the range of mean don't spread out that much. From the formula, as n increase, the +/- Z * sigma/sq root of n is smaller, thus the spread of the interval is closer (decrease)
