Question

The smallest defect in a computer chip will render the entire chip worthless. Therefore, tight quality control measures must be established to monitor these chips. In the past, the percentage of defective for these chips has been 3%. The sample size is 100. Calculate the upper and lower control limits of a P-chart to monitor this process. Also determine whether the process is in control if a sample of 100 chips indicated 7 defectives.

Please show the work and how to get Z

Answer 1

For 3 sigma level control of 99.73% service level, Z = 3

p= 3% or 0.03

n = 100 (Sample size)

standard deviation (s) = sqrt(p*(1-p)/n) = sqrt(0.03*(1-0.03)/100) = 0.017

UCL = p + z*s = 0.03 + 3*0.017 = 0.081

LCL = p - z*s = 0.03 - 3*0.017 = - 0.021. As it cannot be negative, it is 0

So, UCL is 0.081 or 8.1%

So, process with 7 defectives will be in control