# Statistical Quality Control Questions and Answers – Time-Weighted – EWMA Control Chart – 2

This set of Statistical Quality Control Multiple Choice Questions & Answers (MCQs) focuses on “Time-Weighted – EWMA Control Chart – 2”.

1. The EWMA charts are constructed by plotting _______________ versus the sample number i or the time.
a) The sample observation, xi
b) The sample variance, σ2
c) The variance of variable zi i.e. $$\sigma_{z_i}^2$$
d) The variable zi
View Answer

Answer: d
Explanation: The EWMA charts are the plots for knowing the trends in data due to assignable causes. They are constructed by plotting the variable zi versus the sample number i or time.

2. The UCL of the EWMA charts is given by ______________
a) UCL=$$μ_0+Lσ\sqrt{[\frac{λ}{2-λ}\big\{1-(1-λ)^{2i}\big\}]}$$
b) UCL=$$μ_0-Lσ\sqrt{[\frac{λ}{2-λ}\big\{1-(1-λ)^{2i}\big\}]}$$
c) UCL=$$μ_0-Lσ\sqrt{[\frac{λ}{2-λ}\big\{1-(1+λ)^{2i}\big\}]}$$
d) UCL=$$μ_0+Lσ\sqrt{[\frac{λ}{2-λ}\big\{1-(1+λ)^{2i}\big\}]}$$
View Answer

Answer: a
Explanation: The upper control limit of the EWMA charts is L times the standard deviation of the variable zi added, to the variable zi starting value μ0. It is expressed by,
UCL=$$μ_0+Lσ\sqrt{[\frac{λ}{2-λ}\big\{1-(1-λ)^{2i}\big\}]}$$

3. What is the center line value of the EWMA charts?
a) 0
b) 1
c) x
d) μ0
View Answer

Answer: d
Explanation: The center line value of the EWMA charts is the target value, i.e. the mean without shift. It is expressed as,
CL=μ0
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4. What is the value of L=2.27 upper control limit for a EWMA chart which has value of λ=0.10 and the value of the standard deviation σ=1 and the value of μ0=10?
a) 10.41
b) 9.59
c) 10.27
d) 10.10
View Answer

Answer: c
Explanation: We know that,
UCL=$$μ_0+Lσ\sqrt{[\frac{λ}{2-λ}\big\{1-(1-λ)^{2i}\big\}]}$$
Putting the values, we get UCL=10.27.

5. What is the value of LCL of EWMA control charts?
a) LCL=$$μ_0+Lσ\sqrt{[\frac{λ}{2-λ}\big\{1-(1-λ)^{2i}\big\}]}$$
b) LCL=$$μ_0-Lσ\sqrt{[\frac{λ}{2-λ}\big\{1-(1-λ)^{2i}\big\}]}$$
c) LCL=$$μ_0-Lσ\sqrt{[\frac{λ}{2-λ}\big\{1-(1+λ)^{2i}\big\}]}$$
d) LCL=$$μ_0+Lσ\sqrt{[\frac{λ}{2-λ}\big\{1-(1+λ)^{2i}\big\}]}$$
View Answer

Answer: b
Explanation: We know that the LCL of the EWMA charts are just L times the standard deviation. So the LCL of the EWMA charts is given by,
LCL=$$μ_0-Lσ\sqrt{[\frac{λ}{2-λ}\big\{1-(1-λ)^{2i}\big\}]}$$
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6. When the EWMA charts are in operation for quite a long period, what will be the value of the UCL of the chart?
a) UCL=$$μ_0+Lσ\sqrt{\left\{\frac{λ}{2-λ}\right\}}$$
b) UCL=$$μ_0-3Lσ\sqrt{\left\{\frac{λ}{2-λ}\right\}}$$
c) UCL=$$μ_0+2Lσ\sqrt{\left\{\frac{λ}{2-λ}\right\}}$$
d) UCL=$$μ_0-Lσ\sqrt{\left\{\frac{λ}{2-λ}\right\}}$$
View Answer

Answer: a
Explanation: EWMA charts, when are in operation for a long period, the term, 1-(1-λ)2i becomes unity as i becomes large. So the UCL can be written as,
UCL=$$μ_0+Lσ\sqrt{\left\{\frac{λ}{2-λ}\right\}}$$

7. What will be the value of the UCL for a control chart, which has the value of λ=0.10, L=2.7, σ=1 and μ0=10? The value of i can be assumed quite large.
a) UCL=10.67
b) UCL=10.62
c) UCL=9.38
d) UCL=9.37
View Answer

Answer: b
Explanation: We know that when the value of i is high, we calculate UCL by the following expression.
UCL=$$μ_0+Lσ\sqrt{\left\{\frac{λ}{2-λ}\right\}}$$
Putting the values, we get UCL=10.62.
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8. What is the LCL value when the value of i becomes high in the EWMA charts?
a) LCL=$$μ_0-6Lσ\sqrt{\left\{\frac{λ}{2-λ}\right\}}$$
b) LCL=$$μ_0+3Lσ\sqrt{\left\{\frac{λ}{2-λ}\right\}}$$
c) LCL=$$μ_0+Lσ\sqrt{\left\{\frac{λ}{2-λ}\right\}}$$
d) LCL=$$μ_0-Lσ\sqrt{\left\{\frac{λ}{2-λ}\right\}}$$
View Answer

Answer: d
Explanation: We know that the difference between the UCL and the CL is the same as the difference between the CL and the LCL. So we can write the expression for LCL when the process is running for quite a long period as,
LCL=$$μ_0-Lσ\sqrt{\left\{\frac{λ}{2-λ}\right\}}$$

9. After some time, i in the equation of LCL becomes high. What will be the value of λ when this occurs? You may use, LCL=9.38, the standard deviation= 1 and the target mean = 10.
a) 0.1
b) 2.8
c) 3.0
d) 6.0
View Answer

Answer: a
Explanation: We know that,
LCL=$$μ_0-Lσ\sqrt{\left\{\frac{λ}{2-λ}\right\}}$$
Here, we consider the high value of i. When we put the given values, we get, λ=0.1.
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10. Which of the parameters are chosen wisely in the case of EWMA charts to get a desired ARL?
a) Only L
b) Only σ
c) Both, L and λ
d) Only λ
View Answer

Answer: c
Explanation: We know that in the parameters used for constructing EWMA charts, only L and λ are the design parameters. So we choose them wisely to obtain desired ARL.

11. When the value of λ is small, and if the value of EWMA is on one side of the CL and the shift of mean occurs in the opposite direction, EWMA takes several periods to react to that shift. This is called _____
a) Weight effect
b) Stationary effect
c) Inertia effect
d) Tendency effect
View Answer

Answer: c
Explanation: The staying of EWMA on a side till a very heavy shift occurs in the process mean; is called the inertia effect of the EWMA charts. This occurs only when the value of λ is very small.

12. How is the effectiveness of EWMA charts by inertia effect?
a) Increase
b) Decrease
c) No change
d) Can either decrease or increase
View Answer

Answer: b
Explanation: As inertial effects increase the time for EWMA to react to a process mean shift, it ultimately is influencing the effectiveness of EWMA charts. This decreases the effectiveness of EWMA.

13. Which of these values of λ gives the weights given to current and previous observations matching as closely as possible the weights when these observations give by a Shewhart chart with the WE rules?
a) 0.7
b) 0.9
c) 0.2
d) 0.4
View Answer

Answer: d
Explanation: The value of λ is recommended by Hunter to be equal to 0.4 to get the weights given to current and previous observations, matching closely to the weights when these observations give by a Shewhart chart, with Western Electric rules.

14. The control limits of EWMA charts do not depend on the sample number when the sample number becomes quite large.
a) True
b) False
View Answer

Answer: a
Explanation: We know that the sample number is denoted by i in the control limit expression of EWMA charts. As i becomes large, it eliminates its term from the control limit expressions and they don’t depend on i. They become,
Control limits of EWMA = $$μ_0±Lσ\sqrt{\frac{λ}{2-λ}}$$

15. ARL of the process, when EWMA is used, does not depend on σ.
a) True
b) False
View Answer

Answer: b
Explanation: The design parameters of the EWMA charts are the multiple of sigma (L) and the value of λ. It is possible to choose only these, to give desired ARL performance for the EWMA charts.

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