Solution

To specify the control limits, initially, we have to calculate each sample’s mean and standard deviation (${\overline{X}}_i$ and S_i). After that, we have to calculate the mean of the sample means and the mean of the sample standard deviations:

$\stackrel{̿}{X}=\frac{\sum _{i=1}^{20}{\overline{X}}_i}{20}=\frac{117.15}{20}=8.86$

$\overline{S}=\frac{\sum _{i=1}^{20}{S}_i}{20}=\frac{3.64}{20}=0.18$

These calculations are shown in detail in Table 22.E.4.

Table 22.E.4

Calculating $\stackrel{̿}{X}$ and $\overline{S}$

Sample	Measurements
	m1	m2	m3	m4	${\overline{X}}_i$	Si
1	8.62	8.12	8.44	8.33	8.38	0.21
2	8.10	8.27	8.65	8.48	8.38	0.24
3	8.64	8.59	8.87	8.96	8.77	0.18
4	9.01	8.87	9.22	9.15	9.06	0.16
5	9.14	9.08	8.87	8.74	8.96	0.19
6	9.24	9.41	9.37	9.52	9.39	0.12
7	9.05	9.12	8.87	8.69	8.93	0.19
8	8.56	8.74	8.45	8.72	8.62	0.14
9	8.45	8.14	8.24	8.53	8.34	0.18
10	8.48	8.97	8.64	8.45	8.64	0.24
11	8.89	8.69	9.03	9.15	8.94	0.20
12	9.24	9.34	9.09	9.41	9.27	0.14
13	9.25	9.42	9.36	9.65	9.42	0.17
14	8.75	9.21	8.83	8.42	8.80	0.32
15	8.47	8.68	8.49	8.68	8.58	0.12
16	9.01	9.24	9.36	9.48	9.27	0.20
17	9.22	8.76	8.94	8.86	8.95	0.20
18	9.14	9.54	9.36	9.51	9.39	0.18
19	8.5	8.97	8.65	8.72	8.71	0.20
20	8.27	8.34	8.47	8.41	8.37	0.09
				Sum	177.15	3.64
				Mean	8.86	0.18

From Table O in the Appendix, we can determine the values of constants A₃ = 1.628, B₃ = 0, and B₄ = 2.266 of the control limits for $\overline{X}$ and S, considering n = 4.

Therefore, the control limits for $\overline{X}$ by using Expression (22.23) can be written as:

$\mathit{UCL}=\stackrel{̿}{X}+A_3\overline{S}=8.8574+1.628\times 0.1822=9.1539$

$\mathit{CL}=\stackrel{̿}{X}=8.8574$

$\mathit{LCL}=\stackrel{̿}{X}-A_3\overline{S}=8.8574-1.628\times 0.1822=8.5608$

On the other hand, the control limits for S by using Expression (22.24) are:

$\mathit{UCL}=B_4\overline{S}=2.266\times 0.1822=0.4128$

$\mathit{CL}=\overline{S}=0.1822$

$\mathit{LCL}=B_3\overline{S}=0$

$\overline{X}$ and S control charts will be generated by using SPSS and are not available on Stata.

Solution by Using SPSS Software

The data in Example 22.2 are available in the file Example_22.2.sav and can be seen in Fig. 22.10.

Similar to Example 22.1, we must click on Analyze → Quality Control → Control Charts..., select, in Variables Charts, the option X-bar, R, s and define the variables we are interested in. However, in Charts, we must select the options X-bar using standard deviation and Display s chart, as shown in Fig. 22.11.

Figs. 22.12 and 22.13 show the control charts for $\overline{X}$ and S, respectively, which were obtained from SPSS software.

The S charts indicate that the process variation is in control. The same is not true in relation to the process mean. From the $\overline{X}$ control charts, we can see that there are several samples outside the control limits, which suggests that the process is not in statistical control, that is, the process mean has changed.

Solution

We have m = 25 samples and n = 60 observations for each sample.

The proportion of defective sheets for each sample is presented in Table 22.E.5. Therefore, the average proportion of defective steel sheets for these 25 samples ($\overline{p}$) is calculated as:

$\overline{p}=\frac{0.100+0.117+\dots +0.100}{25}=0.1187$

or:

$\overline{p}=\frac{\sum _{i=1}^m{D}_i}{\mathit{mn}}=\frac{178}{25\times 60}=0.1187$

By applying Expression (22.32) of the control limits to $\overline{p}$, we have:

$\mathit{UCL}=0.1187+3\sqrt{\frac{0.1187\bullet \left(0.8813\right)}{60}}=0.2439$

$\mathit{CL}=0.1187$

$\mathit{LCL}=max\left(0.118733-3\sqrt{\frac{0.1187\bullet \left(0.8813\right)}{60}},0\right)=0$

Solution by Using Stata Software

To construct the p chart on Stata, we have to use the following command:

pchart reject_var unit_var ssize_var [, pchart_options]

The data in Example 22.3 are available in the file Example_22.3.dta. The first variable, called sample, represents each one of the 25 samples being analyzed. The second column refers to the variable rejects, which computes the total number of defective parts for each sample. Finally, the variable ssize represents the number of steel sheets in each sample that, in this case, is 60. Hence, the command to be typed to construct the p chart on Stata is:

pchart rejects sample ssize

Fig. 22.14 shows the p chart obtained on Stata.

Solution by Using SPSS Software

The data in Example 22.3 are available in the file Example_22.3.sav. The first column represents the sample or subgroup under analysis. The total number of defective parts for each sample is listed in sequence order in the second column, called rejects, as shown in Fig. 22.15. The third column, called ssize, is optional, since in the cases in which the sample sizes are constant, its respective value can be defined on the SPSS screen, as shown in Fig. 22.18.

So, we must click on Analyze → Quality Control → Control Charts..., define the option p, np in Attribute Charts and the option Cases are subgroups in Data Organization, select the variables of interest, and click on p (Proportion nonconforming) in Chart, as shown in Figs. 22.16, 22.17, and 22.18. Also in Fig. 22.18, in Sample Size, we must select the option Constant and define the value 60. Another alternative is to select the option Variable and insert the variable ssize, as shown in Fig. 22.19. By clicking on OK, we obtain the p chart on SPSS, as shown in Fig. 22.20.

We can verify through Fig. 22.20 that the process is in statistical control, since there are no points outside the control limits.

Solution

We have n = 60 and $\overline{p}=0.1187$, as calculated in the previous example.

By applying Expression (22.34) of the control limits to $n\overline{p}$, we have:

$\mathit{UCL}=60\times 0.1187+3\sqrt{60\times 0.1187\left(0.8813\right)}=14.6350$

$\mathit{CL}=60\times 0.1187=7.1200$

$\mathit{LCL}=max\left(60\times 0.1187-3\sqrt{60\times 0.1187\left(0.8813\right)},0\right)=0$

Stata does not offer the np chart. Therefore, Example 22.4 will only be solved on SPSS.

Solution by Using SPSS

Open the file Example_22.3.sav. Constructing the np chart follows the same logic as the p chart. Once again, click on Analyze → Quality Control → Control Charts..., define the option p, np in Attribute Charts and the option Cases are subgroups in Data Organization. Once again, select the variables of interest, the sample size and, this time, click on the option np (Number of nonconforming) in Chart, as shown in Fig. 22.21. By clicking on OK, we obtain the np chart on SPSS, as shown in Fig. 22.22.

Through Fig. 22.22, we can confirm that the process is in statistical control, since there are no points outside the control limits.

Solution

For this example, each sample consists of a single inspection unit (n = 1). The average number of defects observed in the 30 samples is given by:

$\overline{\lambda }=\frac{129}{30}=4.3$

The control limits for $\overline{\lambda }$ from Expression (22.38) can be specified as:

$\mathit{UCL}=4.3+3\sqrt{4.3}=10.5209$

$\mathit{CL}=4.3$

$\mathit{LCL}=max\left(4.3-3\sqrt{4.3},0\right)=0$

Solution by Using Stata Software

To construct the c chart on Stata, we have to use the following command:

cchart defect_var unit_var [, cchart_options]

The data in Example 22.5 are available in the file Example_22.5.dta. The first variable, called sample, represents each one of the 30 samples being analyzed. The second column refers to the variable defects, which computes the total number of defects per coffee maker. Hence, the command to be typed to construct the c chart on Stata is:

cchart defects sample

Fig. 22.23 shows the c chart obtained on Stata.

Solution by Using SPSS

Open the file Example_22.5.sav. Constructing the c chart follows the same logic as the p and np charts. Once again, click on Analyze → Quality Control → Control Charts...; however, this time, define the option c, u in Attribute Charts and the option Cases are subgroups in Data Organization, as shown in Fig. 22.24. Once again, select the variables of interest, the sample size and click on the option c (Number of nonconformities) in Chart, as shown in Fig. 22.25. By clicking on OK, we obtain the c chart on SPSS, as shown in Fig. 22.26.

Through Fig. 22.25, we can see that it is not necessary to select the variable sample in Subgroups Labeled by, since the sample size is constant and was specified as 30 in Sample Size. Fig. 22.26 shows that there is one point above the upper control limit, indicating that the process is not in statistical control.

Solution

By using Expression (22.40), the average number of defects per scooter is:

$\overline{u}=\frac{6152}{5100}=1.206$

Since the samples are not the same size, the control limits must be calculated for each sample by using Expression (22.41), as shown in Table 22.E.8.

Table 22.E.8

Calculating the Control Limits for Example 22.6

Sample i	Sample Size (ni)	UCL = $\overline{u}+3\sqrt{\frac{\overline{u}}{n_i}}$	LCL = $\overline{u}-3\sqrt{\frac{\overline{u}}{n_i}}$
1	200	1.439	0.973
2	250	1.415	0.998
3	200	1.439	0.973
4	250	1.415	0.998
5	300	1.397	1.016
6	300	1.397	1.016
7	200	1.439	0.973
8	250	1.415	0.998
9	300	1.397	1.016
10	300	1.397	1.016
11	250	1.415	0.998
12	200	1.439	0.973
13	250	1.415	0.998
14	300	1.397	1.016
15	300	1.397	1.016
16	250	1.415	0.998
17	250	1.415	0.998
18	300	1.397	1.016
19	250	1.415	0.998
20	200	1.439	0.973

The u chart will be generated by using SPSS and it is not available on Stata.

Solution on SPSS

Open the file Example_22.6.sav. Constructing the u chart follows the same logic as the c chart. Once again, click on Analyze → Quality Control → Control Charts..., choose the option c, u in Attribute Charts one more time, and the option Cases are subgroups in Data Organization, as shown in Fig. 22.24. Select the variables of interest and click on the option u (Nonconformities per unit) in Chart, as shown in Fig. 22.27. Notice that, different from the c chart, we must select the option Variable in Sample Size and insert the variable sample_size. By clicking on OK, we obtain the u chart on SPSS, as shown in Fig. 22.28.

Fig. 22.28 suggests that the number of defects per scooter in sample 9 is below the lower control limit, while in samples 17, 18, and 20, the numbers are above the upper limits, indicating that the process is not in statistical control.

Solution

To construct the $\overline{X}$ and R control charts in Example 22.1, the mean of the sample means and the mean of the sample ranges were calculated:

$\stackrel{̿}{X}=14.5417$

$\overline{R}=0.9872$

Thus, to calculate the indexes, we will use $\stackrel{̿}{X}$ and $\overline{R}/d_2$ as estimators of μ and σ, respectively. Through Table O in the Appendix, we obtain the value of constant d₂ = 2.326 for n = 5.

1. a) LCL = 13 mm, UCL = 16 mm and T = 14.5.

Based on the data found in Table 22.E.1, we can see that there are no items outside the specifications limits.

Indexes C_p, C_pk, C_pm, and C_pmk can be calculated as:

$C_p=\frac{\mathit{UCL}-\mathit{LCL}}{6\sigma }=\frac{16-13}{6\left(\overline{R}/d_2\right)}=\frac{3}{6\times 0.9872/2.326}=1.178$

$C_{\mathit{pk}}=min\left(\frac{\mathit{UCL}-\mu }{3\sigma },\frac{\mu -\mathit{LCL}}{3\sigma }\right)=min\left(\frac{16-14.54}{3\times 0.9872/2.326},\frac{14.54-13}{3\times 0.9872/2.326}\right)=1.145$

$C_{\mathit{pm}}=\frac{\mathit{UCL}-\mathit{LCL}}{6\sqrt{{\sigma }^2+{\left(\mu -T\right)}^2}}=\frac{16-13}{6\sqrt{{\left(0.9872/2.326\right)}^2+{\left(14.54-14.5\right)}^2}}=1.172$

$C_{\mathit{pmk}}=min\left\{\frac{16-14.54}{3\sqrt{{\left(0.9872/2.326\right)}^2+{\left(14.54-14.5\right)}^2}},\frac{14.54-13}{3\sqrt{{\left(0.9872/2.326\right)}^2+{\left(14.54-14.5\right)}^2}}\right\}=1.140$

From the estimated mean of the process and the specification limits, we notice that the process is centered on the specifications mean. This can also be proven from the indexes calculated, since C_p ≅ C_pk and C_pm ≅ C_pmk. We can also see that C_p ≅ C_pm and C_pk ≅ C_pmk, since the process mean is much closer to the nominal value of the specification (μ ≅ T).

Since 1 ≤ C_p ≤ 1.33, the process is classified as acceptable. The same interpretation is valid for the other indexes.

1. b) LCL = 12 mm, UCL = 15 mm, and T = 13.5.

Based on the data found in Table 22.E.1, we can see that 22.4% of the items are outside the specifications limits (above the upper specification limit). The calculations of the indexes are:

$C_p=\frac{15-12}{6\left(\overline{R}/d_2\right)}=\frac{3}{6\times 0.9872/2.326}=1.178$

$C_{\mathit{pk}}=min\left(\frac{15-14.54}{3\times 0.9872/2.326},\frac{14.54-12}{3\times 0.9872/2.326}\right)=0.360$

$C_{\mathit{pm}}=\frac{15-12}{6\sqrt{{\left(0.9872/2.326\right)}^2+{\left(14.54-13.5\right)}^2}}=0.445$

$C_{\mathit{pmk}}=min\left\{\frac{15-14.54}{3\sqrt{{\left(0.9872/2.326\right)}^2+{\left(14.54-13.5\right)}^2}},\frac{14.54-12}{3\sqrt{{\left(0.9872/2.326\right)}^2+{\left(14.54-13.5\right)}^2}}\right\}=0.136$

Different from the previous case, the process is not centered on the specifications mean. This can also be proven from the calculated indexes, since C_pk < C_p and C_pmk < C_pm. Since the C_p index only considers the process variability, its value did not change, leading to incorrect interpretations.

We can also see that C_pm < C_p and C_pmk < C_pk, since the process mean differs from the nominal value of the specification (μ ≠ T).

The process is classified as incapable.

Solution Through SPSS Software

The data are available in the file Example_22.1.sav. Initially, we must follow the same steps for constructing the $\overline{X}$ and R control charts presented in Example 22.1. Once again, click on Analyze → Quality Control → Control Charts..., define the option X-bar, R, s in Variables Charts and the option Cases are subgroups in Data Organization, select the variables of interest and click on the options X-bar using range and Display R chart (optional) in Charts, as shown in Figs. 22.5, 22.6, and 22.7. On Statistics..., define the specification limits: Upper, Lower and Target, select the desired Process Capability Indices (CP, CpK, and CpM). In addition, select the option Estimate using R-bar in Capability Sigma, as shown in Figs. 22.29 and 22.30. For case (a), the following specification limits were used: Upper (16), Lower (13), and Target (14.5), in accordance with Fig. 22.29. On the other hand, for case (b), the specification limits are: Upper (15), Lower (12), and Target (13.5), as shown in Fig. 22.30. To conclude, click on Continue and OK. Notice that SPSS does not provide the C_pkm index. The results for each case can be seen in Figs. 22.31 and 22.32.

We will not present the calculation of the capability indexes on Stata for this example. Since Stata only uses the standard deviation S as an estimator of σ, it does not offer the option $\overline{R}/d_2$.

Solution

Based on Table 22.E.3, we can see that 7 of the 80 observations (8.75%) are outside the specifications limits.

To construct the $\overline{X}$ and S control charts in Example 22.2, the mean of the sample means and the mean of the sample standard deviation were calculated:

$\stackrel{̿}{X}=8.8574$

$\overline{S}=0.1822$

that will be the estimators of μ and σ, respectively, to calculate the indexes C_p, C_pk, C_pm, and C_pmk.

We have UCL = 8.82 + 0.66 = 9.48 and LCL = 8.82 - 0.66 = 8.16. Therefore, indexes C_p, C_pk, C_pm, and C_pmk can be calculated as:

$C_p=\frac{\mathit{UCL}-\mathit{LCL}}{6\sigma }=\frac{9.48-8.16}{6\times 0.18}=1.2077$

$C_{\mathit{pk}}=min\left(\frac{\mathit{UCL}-\mu }{3\sigma },\frac{\mu -\mathit{LCL}}{3\sigma }\right)=min\left(\frac{9.48-8.86}{3\times 0.18},\frac{8.86-8.16}{3\times 0.18}\right)=1.1394$

$C_{\mathit{pm}}=\frac{\mathit{UCL}-\mathit{LCL}}{6\sqrt{{\sigma }^2+{\left(\mu -T\right)}^2}}=\frac{9.48-8.16}{6\sqrt{{0.18}^2+{\left(8.86-8.82\right)}^2}}=1.1831$

$C_{\mathit{pmk}}=min\left\{\frac{\mathit{UCL}-\mu }{3\sqrt{{\sigma }^2+{\left(\mu -T\right)}^2}},\frac{\mu -\mathit{LCL}}{3\sqrt{{\sigma }^2+{\left(\mu -T\right)}^2}}\right\}=min\left\{\frac{9.48-8.86}{3\sqrt{{0.18}^2+{\left(8.86-8.82\right)}^2}},\frac{8.86-8.16}{3\sqrt{{0.18}^2+{\left(8.86-8.82\right)}^2}}\right\}=1.1161$

Indexes C_p and C_pk, in addition to indexes C_pm and C_pmk, are relatively near, indicating that the process mean is relatively close to the center of the specification limits. We can also see that C_p ≅ C_pm and C_pk ≅ C_pmk, since the process mean is much closer to the specification target value (μ ≅ T). Since 1 ≤ C_p ≤ 1.33, the process is classified as acceptable. The same interpretation is valid for the other indexes.

Example 22.8 will be solved on Stata, since it uses the standard deviation S as an estimator of σ to calculate the capability indexes. However, it only provides the calculation of indexes C_p and C_pk. Even though SPSS also offers the estimation option using the $\overline{S}$, its results are different when compared to the indexes above and to the results obtained on Stata.

Solution by Using Stata Software

To calculate indexes C_p and C_pk on Stata, we have to use the following command:

pciest mean sd, f(#) s(#)

where mean and sd are the estimates of μ and σ, respectively. On the other hand, f(#) and s(#) correspond to the lower and upper specification limits, respectively.

It is not necessary to open the file Example_22.2.dta to execute the command; however, we need to specify the function values. The command to be typed is, therefore:

pciest 8.857375 0.182157, f(8.16) s(9.48)

whose results can be seen in Fig. 22.33.