# Statistical Quality Control Questions and Answers – Control Charting Techniques – Statistical Process Control for Short Production Runs – 6

«
»

This set of Statistical Quality Control Objective Questions & Answers focuses on “Control Charting Techniques – Statistical Process Control for Short Production Runs – 6”.

1. Which of the error probability does the acceptance control chart take into account?
a) Type I
b) Type II
c) Neither of the type I not the type II
d) Both, type I and type II

Explanation: The acceptance control charts were first designed by Freund. These charts take both, the probability of type I error and the probability of type II error into account.

2. How many approaches can be used to develop the acceptance control charts?
a) 2
b) 1
c) 4
d) 5

Explanation: The acceptance control charts can be designed by using two approaches. First uses the specified n, and a process fraction nonconforming γ that we would like to reject with probability 1-β. Another uses the values of n, δ and α.

3. What is the UCL of the acceptance control chart when the sample size n, the fraction nonconforming γ are specified?
a) $$UCL=USL-\left(Z_γ+\frac{Z_β}{\sqrt{n}}\right)$$
b) $$UCL=USL+\left(Z_γ-\frac{Z_β}{\sqrt{n}}\right)$$
c) $$UCL=USL+\left(Z_γ+\frac{Z_β}{\sqrt{n}}\right)$$
d) $$UCL=USL-\left(Z_γ-\frac{Z_β}{\sqrt{n}}\right)$$

Explanation: The Upper control limit of the acceptance control chart, when sample size n, and the fraction nonconforming for 1-β probability rejection are specified, is expressed as,
$$UCL=USL-\left(Z_γ+\frac{Z_β}{\sqrt{n}}\right)$$

4. What is the value of the LCL of the acceptance control chart when sample size n and the fraction nonconforming for 1-β probability of rejection are specified?
a) $$LCL=LSL-\left(Z_γ+\frac{Z_β}{\sqrt{n}}\right)$$
b) $$LCL=LSL-\left(Z_γ-\frac{Z_β}{\sqrt{n}}\right)$$
c) $$LCL=LSL+\left(Z_γ-\frac{Z_β}{\sqrt{n}}\right)$$
d) $$LCL=LSL+\left(Z_γ+\frac{Z_β}{\sqrt{n}}\right)$$

Explanation: The LCL value for the acceptance control chart uses the value of lower specification limit instead of the USL. The LCL value, when n, and γ are specified to reject with 1-β probability are specified, is
$$LCL=LSL+\left(Z_γ+\frac{Z_β}{\sqrt{n}}\right)$$

5. When n, γ, and β are specified, the values of control limits are between ________
a) μL And μU
b) LSL And μU
c) μL And USL
d) μL And LSL

Explanation: We know that the control limits of an acceptance control chart are,
$$LCL=LSL+\left(Z_γ+\frac{Z_β}{\sqrt{n}}\right)$$; $$UCL=USL-\left(Z_γ+\frac{Z_β}{\sqrt{n}}\right)$$
By the above equations, we get the control limit values between μL and μU.

6. When the values of n, δ, and α, are specified, the LCL of the acceptance control charts, lie in between ________
a) μL And USL
b) μL And LSL
c) USL And μU
d) LSL And μU

Explanation: When the values of n, β, and γ are specified, the control limits for an acceptance control chart lie between μL and μU. But, when the values of n, δ, and α, are specified, the LCL lies in between μL and LSL.

7. When the values of δ and α are specified instead of the β and γ, what is the position of UCL of the acceptance control charts?
a) Between USL and μU
b) Between LSL and μU
c) Between μL and USL
d) Between μL and LSL

Explanation: The value of UCL lies between μL and μU when the sample size n, β and γ are specified, but when the values of δ and α are specified instead of β and γ, the UCL of acceptance control charts lies between the USL and μv.

8. What is the expression for n which gives desired values of α, β, γ, and δ for any acceptance control chart?
a) $$n=\left(\frac{Z_α-Z_β}{Z_δ+Z_γ}\right)^2$$
b) $$n=\left(\frac{Z_α+Z_β}{Z_δ+Z_γ}\right)^2$$
c) $$n=\left(\frac{Z_α-Z_β}{Z_δ-Z_γ}\right)^2$$
d) $$n=\left(\frac{Z_α+Z_β}{Z_δ-Z_γ}\right)^2$$

Explanation: It is possible to choose a sample size for an acceptance control chart so that the specified values of α, β, γ, and δ are obtained. The required sample size will be,
$$n=\left(\frac{Z_α+Z_β}{Z_δ+Z_γ}\right)^2$$

9. if δ= 0.01, α=0.00135, γ=0.05, and β=0.2, what will be the sample size for the desired values of these?
a) n=38
b) n=2
c) n=32
d) n=45

Explanation: We know that,
$$n=\left(\frac{Z_α+Z_β}{Z_δ+Z_γ}\right)^2$$
Putting the required values, we get,
$$n=\left(\frac{Z_{0.00135}+Z_{0.2}}{Z_{0.01}+Z_{0.05}}\right)^2$$=31.43≅32.

10. Modified control chart design is not based upon ____________
a) Type I error probability
b) Sample size n
c) Type II error probability
d) Process fraction nonconforming

Explanation: The modified control chart design includes the development of the control limits for it, which includes only the sample size n, the process fraction nonconforming δ, and the type I error probability.

11. 2-sigma limits are recommended for the modified control charts. Which of these is not one of the reasons for this?
a) The spread must be low
b) The limits must be greater than the spread
c) Gives smaller β-risk
d) Gives higher β-risk

Explanation: The 2-sigma limits are recommended for the modified control charts, because the spread of the process is quite low than the limits, plus it also gives smaller β-risk.

12. Low type I error probability is obtained when 2-sigma limits are used for modified control chart.
a) True
b) False

Explanation: The use of 2-sigma limits gives the smaller β-risk, which is an indicator of low type II error probability. Thus the α-risk or the type I error probability is increased.

13. It is impossible to get desired values of α, β, γ, and δ in the acceptance control charts.
a) True
b) False

Explanation: It is possible to choose such a value for the sample size n for getting desired values of α, β, γ, and δ in the acceptance control charts. This may be done by using the following expression,
$$n=\left(\frac{Z_α+Z_β}{Z_δ+Z_γ}\right)^2$$