Chapter 16, Problem 41SE

Observations on shear strength for 26 subgroups of test spot welds, each consisting of six welds, yield $\Sigma {\bar{x}}_i=10,980,\Sigma s_i=402,\text{and}\Sigma r_i=1074$. Calculate control limits for any relevant control charts.

To determine

Find the control limits for any relevant control chart.

Answer to Problem 41SE

The limits are $\left(402.42,442.20\right)$.

The control chart is,

$\begin{array}{c}\text{LCL}=402.42\\ \text{CL}=422.31\\ \text{UCL}=442.20\end{array}$

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 ${\displaystyle \sum {\overline{x}}_i=10,980}$, ${\displaystyle \sum s_i=402}$ and ${\displaystyle \sum r_i=1,074}$.

Calculation:

One can consider $\overline{X}$ chart based on the sample standard deviations.

Control limits for $\overline{X}$ chart based on the sample standard deviations are defined as,

$\text{LCL}=\overline{\overline{x}}-3\frac{\overline{s}}{a_n\sqrt{n}}$ and $\text{UCL}=\overline{\overline{x}}+3\frac{\overline{s}}{a_n\sqrt{n}}$,

where $\overline{\overline{x}}=\frac{{\displaystyle \sum _{i=1}^k{\overline{x}}_i}}{k}$, $x_i$ be the mean of i^th sample, $\overline{s}=\frac{{\displaystyle \sum _{i=1}^k{s}_i}}{k}$, $s_i$ be the sample standard deviation of i^th sample and ${\text{a}}_n=\frac{\sqrt{2}\Gamma \left(\frac{n}{2}\right)}{\sqrt{n-1}\Gamma \left(\frac{n-1}{2}\right)}$.

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,

$\begin{array}{c}\overline{\overline{x}}=\frac{1}{26}{\displaystyle \sum _{i=1}^{26}{\overline{x}}_i}\\ =\frac{10,980}{26}\\ =422.31\end{array}$

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

$\begin{array}{c}\overline{s}=\frac{1}{26}{\displaystyle \sum _{i=1}^{26}{s}_i}\\ =\frac{402}{26}\\ =15.4615\end{array}$

According to book, the value of $a_n$ for $n=6$ is, $a_6=0.952$.

Now, the control limits are,

$\begin{array}{c}\overline{\overline{x}}\pm 3\frac{\overline{s}}{a_n\sqrt{n}}=422.31\pm 3\left(\frac{15.4615}{\left(0.952\right)\sqrt{6}}\right)\\ =422.31\pm \left(3\right)\left(\frac{15.4615}{2.332}\right)\\ =422.31\pm \left(3\right)\left(6.630\right)\\ =422.31\pm 19.891\end{array}$

$=\left(402.42,422.20\right)$

Thus, the control limits are $\left(402.42,442.20\right)$.

Therefore, the control chart is,

$\begin{array}{c}\text{LCL}=402.42\\ \text{CL}=422.31\\ \text{UCL}=442.20\end{array}$