Probability and Statistics for Engineering and the Sciences
Probability and Statistics for Engineering and the Sciences
9th Edition
ISBN: 9781305251809
Author: Jay L. Devore
Publisher: Cengage Learning
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Chapter 16, Problem 41SE

Observations on shear strength for 26 subgroups of test spot welds, each consisting of six welds, yield Σ x ¯ i = 10 , 980 , Σ s i = 402 , and Σ r i = 1074 . Calculate control limits for any relevant control charts.

Expert Solution & Answer
Check Mark
To determine

Find the control limits for any relevant control chart.

Answer to Problem 41SE

The limits are (402.42,442.20).

The control chart is,

LCL=402.42CL=422.31UCL=442.20

Explanation of Solution

Given info:

The data related to the shear strength for 26 subgroups where each subgroup consists of six welds. It is also given that x¯i=10,980, si=402 and ri=1,074.

Calculation:

One can consider X¯ chart based on the sample standard deviations.

Control limits for X¯ chart based on the sample standard deviations are defined as,

LCL=x¯¯3s¯ann and UCL=x¯¯+3s¯ann,

where x¯¯=i=1kx¯ik, xi be the mean of ith sample, s¯=i=1ksik, si be the sample standard deviation of ith sample and an=2Γ(n2)n1Γ(n12).

The random variable X defines the shear strength.

There are 26 subgroups where each subgroup consists of six welds. Thus, n=6 and k=26.

Hence,

x¯¯=126i=126x¯i=10,98026=422.31

Using the given information the sample mean of s is defined as,

s¯=126i=126si=40226=15.4615

According to book, the value of an for n=6 is, a6=0.952.

Now, the control limits are,

x¯¯±3s¯ann=422.31±3(15.4615(0.952)6)=422.31±(3)(15.46152.332)=422.31±(3)(6.630)=422.31±19.891

=(402.42,422.20)

Thus, the control limits are (402.42,442.20).

Therefore, the control chart is,

LCL=402.42CL=422.31UCL=442.20

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Chapter 16 Solutions

Probability and Statistics for Engineering and the Sciences

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