Chapter 13, Problem 29P

(a)

Summary Introduction

Interpretation:

The place of indicating control limits on a mean chart by using the sampling procedure that produced the data.

Concept Introduction:

Control limits are the lines which define whether a process is out of control or not. These are the horizontal line which comes above and below the central line.

Explanation of Solution

Given information:

The data is given as below.

Sample	Sample mean	Sample range
$1$	$350.85\text{ml}$	$0.6\text{ml}$
$2$	$\text{350}\text{.80}\text{ml}$	$\text{0}\text{.8 ml}$
$3$	$\text{351}\text{.20}\text{ml}$	$\text{0}\text{.5 ml}$
$4$	$\text{351}\text{.00 ml}$	$\text{1}\text{.0 ml}$
$5$	$\text{350}\text{.62 ml}$	$\text{0}\text{.8 ml}$
$6$	$\text{351}\text{.12 ml}$	$\text{0}\text{.7 ml}$
$7$	$\text{351}\text{.50 ml}$	$\text{0}\text{.8 ml}$

First, calculate the average of the sample mean.

  $\begin{array}{l}\stackrel{-}{x}=\frac{350.85+350.80+351.20+351.00+350.62+351.12+351.50}{7}\\ \stackrel{-}{x}=351.013\text{ml}\end{array}$

Then calculate the average of the range.

  $\begin{array}{l}\stackrel{-}{R}=\frac{0.6+0.8+0.5+1.0+0.8+0.7+0.8}{7}\\ \stackrel{-}{R}=0.743\text{ml}\end{array}$

Find the value of $A_2$ from the mean chart table $=0.48$

Then,

Calculate the upper control limit by using below formula.

  $\begin{array}{l}UCL=\stackrel{-}{x}+A_2*\stackrel{-}{R}\\ UCL=351.013+0.48*0.743\\ UCL=351.37\end{array}$

Next, calculate the lower control limit by using below formula.

  $\begin{array}{l}LCL=\stackrel{-}{x}-A_2*\stackrel{-}{R}\\ LCL=351.013-0.48*0.743\\ LCL=350.65\end{array}$

(b)

Summary Introduction

Interpretation:

The position to indicate control limits on range chart by using the same sampling procedure that produced the data.

Concept Introduction:

Control limits are the lines which define whether a process is out of control or not. These are the horizontal line which comes above and below the central line.

Explanation of Solution

Given information:

The data is given as below.

Sample	Sample mean	Sample range
$1$	$350.85\text{ml}$	$0.6\text{ml}$
$2$	$\text{350}\text{.80}\text{ml}$	$\text{0}\text{.8 ml}$
$3$	$\text{351}\text{.20}\text{ml}$	$\text{0}\text{.5 ml}$
$4$	$\text{351}\text{.00 ml}$	$\text{1}\text{.0 ml}$
$5$	$\text{350}\text{.62 ml}$	$\text{0}\text{.8 ml}$
$6$	$\text{351}\text{.12 ml}$	$\text{0}\text{.7 ml}$
$7$	$\text{351}\text{.50 ml}$	$\text{0}\text{.8 ml}$

Find the values of $D_3$ and $D_4$ from the range chart for a sample of 6.

  $D_3=2.00$

  $D_4=0$

Then calculate the average of the range.

  $\begin{array}{l}\stackrel{-}{R}=\frac{0.6+0.8+0.5+1.0+0.8+0.7+0.8}{7}\\ \stackrel{-}{R}=0.743\text{ml}\end{array}$

Now, derive the upper control limit and lower control limit for the rage charts as shown below.

  $\begin{array}{l}UCL=D_3*\stackrel{-}{R}\\ UCL=2.00*0.743\\ UCL=1.486\text{ml}\\ And\\ LCL=D_4*\stackrel{-}{R}\\ LCL=0*0.743\\ LCL=0\end{array}$

(c)

Summary Introduction

Interpretation:

Using the control limits in part a and part b, the sample is in control or not.

Concept Introduction:

Control limits are the lines which define whether a process is out of control or not. These are the horizontal line which comes above and below the central line.

Explanation of Solution

Given information:

The data is given as below.

New sample data are $350.82\text{ml, 350}\text{.07 ml, 350}\text{.10 ml, 350}\text{.50 ml, 349}\text{.75 ml and 349}\text{.90 ml}\text{.}$

First, calculate the average of sample mean by using below formula.

  $\begin{array}{l}\stackrel{-}{x}=\frac{350.82+350.07+350.10+350.50+349.75+349.90}{6}\\ \stackrel{-}{x}=350.19\text{ml}\end{array}$

Then, calculate the range by using below formula.

  $\begin{array}{l}R=uppersamplemean-lowersamplemean\\ R=350.82-349.75\\ R=1.07\text{ml}\end{array}$

In this, the upper control limit $=351.37\text{ml}$ and lower control limit $\text{=350}\text{.65 ml}$ . It is clear that the sample mean of new sample at $\text{350}\text{.19 ml}$ , which is outside the mean control limits.

Hence, the process is out of control.