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How to use this manual: In the lab, copy the blank working table at the start of the Calculation into your record book and fill it as you compute the subgroup statistics or sampling metrics. The Calculation section shows the completed table with the arithmetic and control limits, and the Result states whether the process is in control (or the sampling plan's properties).
Control-chart constants used (subgroup size \(n = 5\)): \(A_2 = 0.577,\; D_3 = 0,\; D_4 = 2.114,\; A_3 = 1.427,\; B_3 = 0,\; B_4 = 2.089,\; d_2 = 2.326\).

List of Practical Experiments (Official Syllabus)

  1. Construction of Mean (\(\bar X\)) and Range (\(R\)) charts.
  2. Construction of Mean (\(\bar X\)) and Standard Deviation (\(S\)) charts.
  3. Construction of \(p\) chart for fixed sample size.
  4. Construction of \(p\) chart for variable sample size.
  5. Construction of \(np\) chart.
  6. Construction of \(c\) chart.
  7. Construction of \(u\) chart.
  8. Single sampling plan — OC curve, risks, AOQ, AOQL, ATI.
  9. Determination of a single sampling plan (lot-quality and average-quality approaches).

Phase 1 — Variable Control Charts

Experiment 1 — Mean (\(\bar X\)) and Range (\(R\)) Charts

1. Problem

Ten subgroups of size \(n = 5\) of a quality characteristic are recorded (standards unspecified). Construct the \(\bar X\) and \(R\) charts and decide whether the process is in control.

SubgroupObservations
11214111310
21316121115
3101213119
41413171216
5111014128
61513121416
71215111317
8109121113
91314121511
101213111410

2. Aim

To construct the \(\bar X\) and \(R\) control charts and assess statistical control of the process mean and dispersion.

3. Formula

\[ \text{UCL}_{\bar X} = \bar{\bar X} + A_2\bar R,\quad \text{LCL}_{\bar X} = \bar{\bar X} - A_2\bar R; \qquad \text{UCL}_R = D_4\bar R,\quad \text{LCL}_R = D_3\bar R \]

Applying it:

  1. For each subgroup compute \(\bar X_j\) and \(R_j = \max - \min\).
  2. Compute \(\bar{\bar X} = \frac{1}{k}\sum \bar X_j\) and \(\bar R = \frac{1}{k}\sum R_j\).
  3. Set the limits using \(A_2, D_3, D_4\); plot both charts.

4. Calculation

Blank working table:

Subgroup\(\bar X_j\)\(R_j\)
1–10
Mean\(\bar{\bar X} =\)\(\bar R =\)
Subgroup12345678910
\(\bar X_j\)12.013.411.014.411.014.013.611.013.012.0
\(R_j\)4545646444

\(\bar{\bar X} = 12.54,\; \bar R = 4.6\).

\(\bar X\) chart: UCL \(= 12.54 + 0.577(4.6) = 15.19\), LCL \(= 12.54 - 0.577(4.6) = 9.89\).

\(R\) chart: UCL \(= 2.114(4.6) = 9.72\), LCL \(= 0\).

5. Result

All \(\bar X_j\) lie within (9.89, 15.19) and all \(R_j\) within (0, 9.72), so the process is in statistical control with respect to both mean and dispersion.

Experiment 2 — Mean (\(\bar X\)) and Standard Deviation (\(S\)) Charts

1. Problem

For the same ten subgroups (\(n = 5\)), construct the \(\bar X\) and \(S\) charts and compare with the \(\bar X\)–\(R\) result.

2. Aim

To construct \(\bar X\) and \(S\) charts (standards unspecified) using \(A_3, B_3, B_4\).

3. Formula

\[ \text{UCL}_{\bar X} = \bar{\bar X} + A_3\bar S,\quad \text{LCL}_{\bar X} = \bar{\bar X} - A_3\bar S; \qquad \text{UCL}_S = B_4\bar S,\quad \text{LCL}_S = B_3\bar S \]

Applying it:

  1. For each subgroup compute the sample standard deviation \(S_j\).
  2. Compute \(\bar S = \frac{1}{k}\sum S_j\).
  3. Set the limits with \(A_3, B_3, B_4\).

4. Calculation

Blank working table:

Subgroup\(\bar X_j\)\(S_j\)
1–10
Mean\(\bar{\bar X} =\)\(\bar S =\)
Subgroup12345678910
\(S_j\)1.582.071.582.072.241.582.411.581.581.58

\(\bar{\bar X} = 12.54,\; \bar S = 1.828\).

\(\bar X\) chart: UCL \(= 12.54 + 1.427(1.828) = 15.15\), LCL \(= 12.54 - 1.427(1.828) = 9.93\).

\(S\) chart: UCL \(= 2.089(1.828) = 3.82\), LCL \(= 0\).

5. Result

All points lie within the limits, so the process is in control. The \(\bar X\) limits (9.93, 15.15) agree closely with the \(\bar X\)–\(R\) limits (9.89, 15.19), confirming both methods.

Phase 2 — Attribute Control Charts

Experiment 3 — \(p\) Chart (Fixed Sample Size)

1. Problem

Ten samples of \(n = 100\) items each are inspected; the numbers of defectives are 4, 3, 5, 2, 6, 4, 3, 5, 4, 4. Construct the \(p\) chart.

2. Aim

To monitor the fraction defective with a \(p\) chart when the sample size is constant.

3. Formula

\[ \bar p = \frac{\sum d_j}{kn}, \qquad \text{UCL/LCL}_p = \bar p \pm 3\sqrt{\frac{\bar p(1-\bar p)}{n}} \]

Applying it:

  1. Compute \(\bar p = \sum d_j /(kn)\).
  2. Set 3-sigma limits about \(\bar p\).

4. Calculation

Blank working table:

QuantityValue
\(\sum d_j\)
\(\bar p\)
UCL / LCL
QuantityValue
\(\sum d_j\)40
\(\bar p = 40/1000\)0.04
UCL \(= 0.04 + 3\sqrt{0.04(0.96)/100}\)0.0988
LCL \(= 0.04 - 0.0588\)0 (set to 0)

5. Result

\(\bar p = 0.04\); limits (0, 0.0988). All ten fractions defective lie within the limits, so the process is in control.

Experiment 4 — \(p\) Chart (Variable Sample Size)

1. Problem

Five samples of sizes 100, 120, 90, 110, 100 have 4, 6, 3, 5, 4 defectives. Construct the \(p\) chart with individual limits.

2. Aim

To monitor the fraction defective when the sample size varies, using subgroup-specific limits.

3. Formula

\[ \bar p = \frac{\sum d_j}{\sum n_j}, \qquad \text{UCL/LCL}_j = \bar p \pm 3\sqrt{\frac{\bar p(1-\bar p)}{n_j}} \]

Applying it:

  1. Compute the pooled \(\bar p = \sum d_j / \sum n_j\).
  2. For each subgroup compute limits using its own \(n_j\).

4. Calculation

Blank working table:

\(n_j\)\(d_j\)\(p_j\)UCL\(_j\)LCL\(_j\)
1004
1206
903
1105
1004

\(\bar p = 22/520 = 0.0423\).

\(n_j\)\(d_j\)\(p_j\)UCL\(_j\)LCL\(_j\)
10040.0400.10270
12060.0500.09740
9030.0330.10600
11050.0450.09990
10040.0400.10270

5. Result

\(\bar p = 0.0423\); every \(p_j\) lies below its (subgroup-specific) UCL, so the process is in control. Smaller samples get wider limits.

Experiment 5 — \(np\) Chart

1. Problem

For the ten samples of \(n = 100\) in Experiment 3 (defectives 4, 3, 5, 2, 6, 4, 3, 5, 4, 4), construct the \(np\) chart.

2. Aim

To monitor the number of defectives directly when the sample size is constant.

3. Formula

\[ \text{UCL/LCL}_{np} = n\bar p \pm 3\sqrt{n\bar p(1-\bar p)} \]

Applying it:

  1. Use \(\bar p = 0.04\) from Experiment 3; \(n\bar p = 4\).
  2. Set 3-sigma limits about \(n\bar p\); plot \(d_j\) directly.

4. Calculation

Blank working table:

QuantityValue
\(n\bar p\)
UCL / LCL
QuantityValue
\(n\bar p = 100(0.04)\)4
UCL \(= 4 + 3\sqrt{4(0.96)}\)9.88
LCL \(= 4 - 5.88\)0 (set to 0)

5. Result

\(n\bar p = 4\); limits (0, 9.88). All counts of defectives lie within the limits ⇒ process in control (equivalent to the \(p\) chart of Experiment 3).

Experiment 6 — \(c\) Chart

1. Problem

The numbers of defects in ten inspection units of constant size are 5, 3, 4, 6, 2, 5, 4, 3, 7, 1. Construct the \(c\) chart.

2. Aim

To monitor the number of defects per inspection unit using a \(c\) chart (Poisson model).

3. Formula

\[ \text{UCL/LCL}_c = \bar c \pm 3\sqrt{\bar c} \]

Applying it:

  1. Compute \(\bar c = \frac{1}{k}\sum c_j\).
  2. Set limits \(\bar c \pm 3\sqrt{\bar c}\).

4. Calculation

Blank working table:

QuantityValue
\(\bar c\)
UCL / LCL
QuantityValue
\(\bar c = 40/10\)4
UCL \(= 4 + 3\sqrt{4}\)10
LCL \(= 4 - 6\)0 (set to 0)

5. Result

\(\bar c = 4\); limits (0, 10). All defect counts lie within the limits, so the process is in control.

Experiment 7 — \(u\) Chart

1. Problem

Five inspection lots of sizes 1, 1.5, 2, 1, 2.5 units have 5, 6, 10, 4, 12 defects. Construct the \(u\) chart (defects per unit) with individual limits.

2. Aim

To monitor defects per unit when the inspection-unit size varies.

3. Formula

\[ \bar u = \frac{\sum c_j}{\sum n_j}, \qquad \text{UCL/LCL}_j = \bar u \pm 3\sqrt{\frac{\bar u}{n_j}} \]

Applying it:

  1. Compute \(\bar u = \sum c_j / \sum n_j\).
  2. For each lot compute limits using its own \(n_j\).

4. Calculation

Blank working table:

\(n_j\)\(c_j\)\(u_j\)UCL\(_j\)LCL\(_j\)
15
1.56
210
14
2.512

\(\bar u = 37/8 = 4.625\).

\(n_j\)\(c_j\)\(u_j\)UCL\(_j\)LCL\(_j\)
155.0011.080
1.564.009.890
2105.009.190.06
144.0011.080
2.5124.808.710.55

5. Result

\(\bar u = 4.625\); every \(u_j\) lies within its individual limits, so the process is in control.

Phase 3 — Acceptance Sampling

Experiment 8 — Single Sampling Plan (OC Curve, Risks, AOQ, AOQL, ATI)

1. Problem

For the single sampling plan \(n = 50,\; c = 1\) with lot size \(N = 1000\), construct the OC curve and compute the AOQ, AOQL, ATI, producer's risk (at AQL = 0.01) and consumer's risk (at LTPD = 0.08).

2. Aim

To evaluate a single sampling plan: probability of acceptance, average outgoing quality (and its limit) and average total inspection.

3. Formula

\[ P_a = \sum_{k=0}^{c}\frac{e^{-np}(np)^{k}}{k!}, \quad \text{AOQ} = P_a\,p\,\frac{N-n}{N}, \quad \text{ATI} = n + (1-P_a)(N-n) \]

Applying it:

  1. Compute \(P_a = P(X \le c)\) with \(X \sim\) Poisson(\(np\)) for a range of \(p\).
  2. AOQ \(= P_a\, p\,(N-n)/N\); the maximum over \(p\) is the AOQL.
  3. ATI \(= n + (1 - P_a)(N - n)\).
  4. Producer's risk \(= 1 - P_a\) at AQL; consumer's risk \(= P_a\) at LTPD.

4. Calculation

Blank working table:

\(p\)\(np\)\(P_a\)AOQATI
0.01
0.02
0.03
0.05
0.08
0.10
\(p\)\(np\)\(P_a\)AOQATI
0.010.50.90980.00864135.7
0.021.00.73580.01398301.0
0.031.50.55780.01590470.1
0.052.50.28730.01365727.1
0.084.00.09160.00696913.0
0.105.00.04040.00384961.6

The OC curve falls from \(P_a = 0.91\) to \(0.04\) as \(p\) rises. AOQ peaks at AOQL = 0.0160 (near \(p = 0.0325\)). Producer's risk \(= 1 - 0.9098 = 0.090\) at AQL = 0.01; consumer's risk \(= 0.0916\) at LTPD = 0.08.

5. Result

The plan \((50, 1)\) has AOQL ≈ 0.016, producer's risk ≈ 9 % and consumer's risk ≈ 9 %. ATI rises from 136 (good lots) toward 962 (bad lots), reflecting more 100 % screening as quality worsens.

Experiment 9 — Determination of a Single Sampling Plan

1. Problem

(a) Lot-quality approach: design a plan for AQL = 0.02 (\(\alpha = 0.05\)) and LTPD = 0.13 (\(\beta = 0.10\)). (b) Average-quality approach: design a plan for a specified AOQL ≈ 0.012 with lot size \(N = 1000\).

2. Aim

To determine \((n, c)\) meeting stated risk requirements (lot-quality) or a target AOQL (average-quality), using the Poisson unity-value table.

3. Formula

\[ R = \frac{\text{LTPD}}{\text{AQL}}, \qquad n = \frac{np_1}{\text{AQL}} \quad\text{(with } P_a = 1-\alpha \text{ at } np_1\text{)} \]

Applying it:

  1. Lot-quality: form \(R = \text{LTPD}/\text{AQL}\); from the table pick \(c\) whose ratio \(np_2/np_1\) (at \(P_a = 0.10\) and \(0.95\)) is closest to \(R\); then \(n = np_1/\text{AQL}\).
  2. Average-quality: for the target AOQL and \(N\), read the Dodge–Romig \((n, c)\); verify the AOQL by maximising AOQ.

4. Calculation

Blank working table:

ApproachKey quantityPlan \((n, c)\)
(a) Lot-quality\(R =\)
(b) Average-qualityAOQL =

(a) \(R = 0.13/0.02 = 6.5\). The Poisson table gives \(c = 2\) (ratio 6.51), with \(np_1 = 0.818\) at \(P_a = 0.95\). Then \(n = 0.818/0.02 = 40.9 \approx 41\); checking with LTPD, \(np_2 = 5.322\Rightarrow n = 5.322/0.13 = 40.9\) — consistent. Verify: \(P_a(0.02) = 0.95\), \(P_a(0.13) = 0.10\).

(b) For AOQL ≈ 0.012 and \(N = 1000\), the Dodge–Romig table gives \((n = 100, c = 2)\); maximising AOQ for this plan confirms AOQL = 0.0123.

ApproachKey quantityPlan \((n, c)\)
(a) Lot-quality\(R = 6.5\)\((41, 2)\)
(b) Average-qualityAOQL = 0.0123\((100, 2)\)

5. Result

(a) The plan \((n = 41,\; c = 2)\) meets both risks exactly (\(\alpha = 0.05\) at AQL, \(\beta = 0.10\) at LTPD). (b) The plan \((n = 100,\; c = 2)\) delivers AOQL ≈ 0.012 for \(N = 1000\).

Lab Record Format (to be followed for every experiment)

  1. 1. Problem — the data/plan and what is to be constructed or determined.
  2. 2. Aim — the chart or sampling metric the experiment produces.
  3. 3. Formula — the formula, then the numbered steps that apply it.
  4. 4. Calculation — the filled table with statistics and control limits.
  5. 5. Result — the control decision or plan properties, with interpretation.
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