Question

Piston rings for an automotive engine are produced by a forging process. We wish to monitor the inside diameter of the rings manufactured by this process, using an x̄ and an s control chart. Samples of size 8 are to be taken at regular intervals, and the sample means and standard deviations are computed and plotted on the charts in time order. The target values for the inside diameter are a mean of μ = 75 mm and a standard deviation of σ = 0.38 mm. A table containing control chart constants is reproduced below.

sample size nlc4 ド |DE 2.606 2.276 2.088 1.964 0.9515 0.0291.874 9594 0.1131.806 965 0.179 1.751 0.7979 8862 9213 94 0.96930.2321.70 10 0.9727 0.276 1.669

Q1: The center line for the three sigma x̄ control chart would be

• 75.03
• 75.01
• 75.09
• 75

Q2: The upper control limit for the x̄ control chart would be

• 75
• 75.201525432638
• 75.403050865276
• 76.14

Q3: The lower control limit for the x̄ control chart would be

• 74.596949134724
• 75
• 73.86
• 74.798474567362

Q4: The center line for the three sigma s chart would be

• 0.38
• 0.19
• 0.3667
• 0.76

Q5: The lower control limit for the three sigma s chart would be

• 0
• 0.76
• 0.06802
• 0.38

Q6: The upper control limit for the three sigma s chart would be

• 100
• 0.66538
• 0.38
• 0.19

Q7:

Suppose we use a "run of nine" rule, that is, we declare the process out of control if nine consecutive points are all above the center line of the x̄ or if nine consecutive points are all below the center line of the x̄ chart. If the process is in statistical control, then with any particular collection of nine consecutive points, the probability that we declare the process out of control using this rule is (approximately):

• 7. I like the number 7, so I figure the probability is 7. After all, the 6 is afraid of the 7 because the 7 ate 9.
• 0.5^9. That is because we assume the distribution is approximately symmetric--about half the elements are above the center line and half are below. Furthermore, each element is independent. So we can figure the probability of one above x̄ is 0.5. Two above x̄ is 0.5*0.5 = 0.5^2. Three above is 0.5^3. For k above, the probability is 0.5^k. So for 9 above, it would be 0.5^9, or about 0.002. It is the same probability for a run of 9 below the x̄. It is roughly the same probability of observing a value outside the control limits.

I need help figuring out how to answer these questions. I don't just want the answers, I would like to know how to solve for them, as I am feeling completely lost. Please!!

Answer 1

The control limits for the $\bar{X}$chart are

CL=$\bar{X}$

UCL=$\bar{X}$+A₂$\bar{S}$

LCL=$\bar{X}$-A2$\bar{S}$

A₂=0.373 (From table of Control Chart constants)

Hence

Q1. Ans.:CL=$\bar{X}$=75

Q2.Ans:UCL=$\bar{X}$+A₂$\bar{S}$=75+(0.373x0.38)=75.14

Q3. Ans. LCL=$\bar{X}$-A2$\bar{S}$=75-(0.373x.38)=74.85

The control limits for s chart are:

CL=S

UCL=B₄S

LCL=B3S

B4=1.815 and B3=0.185 (From table of Control Chart constants)

Q4. Ans: CL=S=0.38

Q5.Ans:LCL=B3S=0.185x0.38=0.073

Q6: Ans:UCL=B₄S=1.815x0.38=0.6897