Historically, the average proportion of defective bars has been 0.015. Samples will be of 100 bars each. Construct a p-chart using $$z = 3$$ . Suppose a sample had 0.054 defectives. What would be the best action to take in this scenario?
A) Nothing; it appears to be random variation.
B) Look for assignable causes.
C) Change z to 2 and take another sample.
D) Change z to 4 and continue sampling.

Question

Quizplus · Accepted Answer

Given a z-value of 3, the control limits for the p-chart are calculated using the average proportion of defects (p) and the standard deviation of the proportion (sqrt[p(1-p)/n]), where n is the sample size. The upper control limit (UCL) is p + z*sqrt[p(1-p)/n], and the lower control limit (LCL) is p - z*sqrt[p(1-p)/n]. With p = 0.015, n = 100, and z = 3, the UCL is significantly lower than 0.054, indicating that the sample with 0.054 defectives is outside the control limits. This suggests that the process is not in control, and it is not due to random variation. Therefore, the best action is to look for assignable causes of the variation.

Historically, the Average Proportion of Defective Bars Has Been 0 z=3z = 3z=3

Historically, the Average Proportion of Defective Bars Has Been 0 $z = 3$