PRIN.OF OPERATIONS MANAGEMENT-MYOMLAB
PRIN.OF OPERATIONS MANAGEMENT-MYOMLAB
11th Edition
ISBN: 9780135226742
Author: HEIZER
Publisher: PEARSON
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Chapter 6.S, Problem 13P

a)

Summary Introduction

To determine: Determine the UCL and LCL if the sample size is 100 and 3-sigma limits.

Introduction: Control charts used to determine whether the process is under control or not. Attributes and variables are the factors under the control charts.

a)

Expert Solution
Check Mark

Answer to Problem 13P

Hence, the UCL is 0.05145 and LCL is 0.

Explanation of Solution

Given information:

Defect rate is given as 1.5%.

Determine the UCL and LCL if the sample size is 100 and 3-sigma limits:

UCLp=p¯+z×σp

LCLp=p¯z×σp

Here,

z refers to the number of standard deviation for setting the limit. Here, it is 3-sigma limit.

σp is the standard deviation of the sampling distribution.

Calculate the value of the standard deviation of the sampling distribution σp :

σp=p¯×(1p¯)n

The given values of p¯=1.5% and the sample size n=100 and obtain the value of σp

σp=0.015×(0.985)100=0.014775100=0.00014775=0.012155

Hence, the standard deviation of the sampling distribution σp=0.012155

The given values of p¯=1.5% and z=3 and σp=0.012155 has been determine. Substitute the values in UCL formula to determine the value of UCL:

UCLp=p¯+z×σp=0.015+3×0.01215=0.05145

Hence, the upper control chart limit UCLp=0.05145

The given values of p¯=1.5% and z=3 and σp=0.012155 has been determine. Substitute the values in LCL formula to determine the value of LCL:

LCLp=p¯z×σp=0.0153×0.01215=0.02145=0

Hence, lower control chart limit cannot be negative therefore LCLp=0

b)

Summary Introduction

To determine: Determine the UCL and LCL if the sample size is 50 and 3-sigma limits.

Introduction: Control charts used to determine whether the process is under control or not. Attributes and variables are the factors under the control charts.

b)

Expert Solution
Check Mark

Answer to Problem 13P

Hence, the UCL is 0.06657 and LCL is 0.

Explanation of Solution

Given information:

Defect rate is given as 1.5%.

Determine the UCL and LCL if the sample size is 50 and 3-sigma limits:

UCLp=p¯+z×σp

LCLp=p¯z×σp

Here,

z refers to the number of standard deviation for setting the limit. Here, it is 3-sigma limit.

σp is the standard deviation of the sampling distribution.

Calculate the value of the standard deviation of the sampling distribution σp :

σp=p¯×(1p¯)n

The given values of p¯=1.5% and the sample size n=50 and obtain the value of σp

σp=0.015×(0.985)50=0.01477550=0.0002955=0.017190

Hence, the standard deviation of the sampling distribution σp=0.017190

The given values of p¯=1.5% and z=3 and σp=0.017190 has been determine. Substitute the values in UCL formula to determine the value of UCL:

UCLp=p¯+z×σp=0.015+3×0.017190=0.06657

Hence, the upper control chart limit UCLp=0.06657

The given values of p¯=1.5% and z=3 and σp=0.017190 has been determine. Substitute the values in LCL formula to determine the value of LCL:

LCLp=p¯z×σp=0.0153×0.017190=0.03657=0

Hence, lower control chart limit cannot be negative therefore LCLp=0

c)

Summary Introduction

To determine: Determine the UCL and LCL if the sample size is 100 and 2-sigma limits.

Introduction: Control charts used to determine whether the process is under control or not. Attributes and variables are the factors under the control charts.

c)

Expert Solution
Check Mark

Answer to Problem 13P

Hence, the UCL is 0.0393 and LCL is 0.

Explanation of Solution

Given information:

Defect rate is given as 1.5%.

Determine the UCL and LCL if the sample size is 100 and 2-sigma limits:

UCLp=p¯+z×σp

LCLp=p¯z×σp

Here,

z refers to the number of standard deviation for setting the limit. Here, it is 2-sigma limit.

σp is the standard deviation of the sampling distribution.

Calculate the value of the standard deviation of the sampling distribution σp :

σp=p¯×(1p¯)n

The given values of p¯=1.5% and the sample size n=100 and obtain the value of σp

σp=0.015×(0.985)100=0.014775100=0.00014775=0.012155

Hence, the standard deviation of the sampling distribution σp=0.012155

The given values of p¯=1.5% and z=2 and σp=0.012155 has been determine. Substitute the values in UCL formula to determine the value of UCL:

UCLp=p¯+z×σp=0.015+2×0.01215=0.0393

Hence, the upper control chart limit UCLp=0.0393

The given values of p¯=1.5% and z=2 and σp=0.012155 has been determine. Substitute the values in LCL formula to determine the value of LCL:

LCLp=p¯z×σp=0.0152×0.01215=0.0093=0

Hence, lower control chart limit cannot be negative therefore LCLp=0

d)

Summary Introduction

To determine: Determine the UCL and LCL if the sample size is 50 and 2-sigma limits.

Introduction: Control charts used to determine whether the process is under control or not. Attributes and variables are the factors under the control charts.

d)

Expert Solution
Check Mark

Answer to Problem 13P

Hence, the UCL is 0.04938 and LCL is 0.

Explanation of Solution

Given information:

Defect rate is given as 1.5%.

Determine the UCL and LCL if the sample size is 50 and 2-sigma limits:

UCLp=p¯+z×σp

LCLp=p¯z×σp

Here,

z refers to the number of standard deviation for setting the limit. Here, it is 2-sigma limit.

σp is the standard deviation of the sampling distribution.

Calculate the value of the standard deviation of the sampling distribution σp :

σp=p¯×(1p¯)n

The given values of p¯=1.5% and the sample size n=50 and obtain the value of σp

σp=0.015×(0.985)50=0.01477550=0.0002955=0.017190

Hence, the standard deviation of the sampling distribution σp=0.017190

The given values of p¯=1.5% and z=2 and σp=0.017190 has been determine. Substitute the values in UCL formula to determine the value of UCL:

UCLp=p¯+z×σp=0.015+2×0.017190=0.04938

Hence, the upper control chart limit UCLp=0.04938

The given values of p¯=1.5% and z=2 and σp=0.017190 has been determine. Substitute the values in LCL formula to determine the value of LCL:

LCLp=p¯z×σp=0.0153×0.017190=0.01938=0

Hence, lower control chart limit cannot be negative therefore LCLp=0

e)

Summary Introduction

To determine: What happen to σ^p if the sample size is larger.

Introduction: Control charts used to determine whether the process is under control or not. Attributes and variables are the factors under the control charts.

e)

Expert Solution
Check Mark

Explanation of Solution

Determine what happen to σ^p if the sample size is larger:

The standard deviation of the sampling distribution σp . reduces when the sample size increases. However, in the calculation the sample size is in the denominator.

f)

Summary Introduction

To determine: Why LCL cannot be less than 0.

Introduction: Control charts used to determine whether the process is under control or not. Attributes and variables are the factors under the control charts.

f)

Expert Solution
Check Mark

Explanation of Solution

Determine why LCL cannot be less than 0:

The lower control limit for defectives cannot be less than zero because it becomes negative. A negative lower control limit is meaningless.

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The defect rate for data entry of insurance claims hashistorically been about 1.5%.a) What are the upper and lower control chart limits if you wishto use a sample size of 100 and 3-sigma limits?b) What if the sample size used were 50, with 3s ?c) What if the sample size used were 100, with 2s ?d) What if the sample size used were 50, with 2s ?e) What happens to snp when the sample size is larger?f) Explain why the lower control limit cannot be less than 0.
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PRIN.OF OPERATIONS MANAGEMENT-MYOMLAB

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