Chapter 19, Problem 16SE

Samples of size 5 provided the following 20 sample means for a production process that is believed to be in control.

95.72	95.24	95.18
95.44	95.46	95.32
95.40	95.44	95.08
95.50	95.80	95.22
95.56	95.22	95.04
95.72	94.82	95.46
95.60	95.78

1. a. Based on these data, what is an estimate of the mean when the process is in control?
2. b. Assume that the process standard deviation is σ = .50. Develop the $\bar{x}$ control chart for this production process. Assume that the mean of the process is the estimate developed in part (a).
3. c. Do any of the 20 sample means indicate that the process was out of control?

To determine

Find the estimate of the mean when the process is in control.

Answer to Problem 16SE

The estimate of the mean for production process that is in control is 95.4.

Explanation of Solution

Calculation:

The 20 sample means for a production process are provided.

The estimate of the mean for production process is given below:

$\begin{array}{c}\mu =\frac{{\displaystyle \sum \overline{x}}}{n}\\ =\frac{\left(\begin{array}{l}95.72+95.44+95.40+95.50+95.56+95.72+95.60+95.24+95.46+95.44+\\ 95.80+95.22+94.82+95.78+95.18+95.32+95.08+95.22+95.04+95.46\end{array}\right)}{20}\\ =\frac{1,908}{20}\\ =95.4\end{array}$

Thus, the estimate of the mean for production process that is in control is 95.4.

To determine

Construct a $\overline{x}$ control chart for the production process.

Answer to Problem 16SE

Output obtained using MINITAB software is given below:

Explanation of Solution

Calculation:

Upper control limits for $\overline{x}$chart:

$\begin{array}{c}\text{UCL}=\mu +3\frac{\sigma }{\sqrt{n}}\\ =95.4+3\frac{0.50}{\sqrt{5}}\\ =95.4+0.6696\\ =96.07\end{array}$

Lower control limits for $\overline{x}$chart:

$\begin{array}{c}\text{LCL}=\mu -3\frac{\sigma }{\sqrt{n}}\\ =95.4-3\frac{0.50}{\sqrt{5}}\\ =95.4-0.6696\\ =94.73\end{array}$

Software procedure:

Step-by-step software procedure to construct the $\overline{x}$ chart using MINITAB software:

• Choose Graph > Time Series Plot or Stat > Time Series > Time Series Plot.
• Choose Simple, and then click OK.
• In Series, select Mean.
• In Show reference lines at Y values, enter 96.07, 95.4, and 94.73.
• Click OK.

Thus, the control limits for $\overline{x}$ chart are $\text{UCL}=96.07\text{and}\text{LCL}=94.73$.

To determine

Explain whether any of the 20 sample means indicate that the process is out of control or not.

Explanation of Solution

From the $\overline{x}$ chart, it is observed that all the sample means lie within the UCL and LCL.

Thus, the 20 sample means do not indicate that the process was out of control.