20.3 Attribute Control Charts: Defectives (p, np) and Defects (c, u)

Attribute control charts evaluate qualitative characteristics that are counted rather than measured on a continuous dimensional scale. Attribute data arises when products are classified into discrete binary states (e.g., conforming vs. nonconforming, pass vs. fail, functional vs. inoperable) or when surface blemishes, solder bridges, or packaging flaws are tallied across an inspection unit. While attribute charts require larger sample sizes than variable charts to achieve equivalent statistical power, they offer tremendous administrative advantages: inspection is often rapid, non-destructive, inexpensive (e.g., go/no-go plug gauges or automated vision sensors), and capable of summarizing complex multi-defect assemblies into a single metric.

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1. Classification of Attribute Data: Defectives vs. Defects

The single most critical conceptual hurdle on the FE exam is distinguishing between nonconforming units (defectives) and nonconformities (defects):

                     Attribute Data Taxonomy
                               │
        ┌──────────────────────┴──────────────────────┐
        ▼                                             ▼
  DEFECTIVE ITEM                                DEFECT COUNT
(Nonconforming Unit)                           (Nonconformity)
─────────────────────                         ─────────────────
• The ENTIRE unit is rejected                 • Specific flaw, scratch, or blemish
• Binary: Conforming vs Nonconforming         • Multiple flaws can reside on 1 unit
• Statistical Model: BINOMIAL                 • Statistical Model: POISSON
• Sample Size: n items inspected              • Sample Size: Inspection area / volume
        │                                             │
   ┌────┴────┐                                   ┌────┴────┐
   ▼         ▼                                   ▼         ▼
p-Chart   np-Chart                            c-Chart   u-Chart
Fraction  Count                               Count     Rate
Defective Defective                           Defects   Defects/Unit
(n varies (n CONSTANT)                        (Unit     (Unit
or fixed)                                     CONSTANT) varies)

Defectives (Nonconforming Units) — Binomial Distribution

An item is either conforming (acceptable) or nonconforming (defective). A smartphone either powers on or it does not; a sealed hydraulic fitting either leaks or holds pressure. Regardless of whether a part has 1 flaw or 10 flaws, it counts as exactly one defective item.

• Governing Distribution: Binomial distribution with parameters $n$ (sample size) and $p$ (probability of nonconformance).
• Applicable Charts: $p$-chart (fraction nonconforming) and $np$-chart (number of nonconforming items).

Defects (Nonconformities) — Poisson Distribution

A defect is a specific physical imperfection or departure from a quality standard. An automobile door panel may have two paint pinholes, one scratch, and one dent: that is four defects on a single unit. A circuit board may have five cold solder joints but still pass functional testing.

• Governing Distribution: Poisson distribution with parameter $\lambda$ (average count of independent, rare events per inspection area of opportunity).
• Applicable Charts: $c$-chart (count of defects per constant unit) and $u$-chart (defects per unit for variable inspection sizes).

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2. The $p$-Chart (Fraction Nonconforming)

The $p$-chart monitors the proportion (or fraction) of nonconforming items in a sample. It is the most versatile attribute chart because it accommodates both constant and variable subgroup sample sizes $n_i$.

Fundamental Formulas

For $m$ subgroups, where subgroup $i$ contains $n_i$ inspected items and $d_i$ nonconforming items:

• Fraction Nonconforming for Subgroup $i$: $$p_i = \frac{d_i}{n_i}$$
• Center Line ($\bar{p}$): The overall weighted historical average fraction nonconforming: $$\bar{p} = \frac{\sum_{i=1}^m d_i}{\sum_{i=1}^m n_i} = \frac{\text{Total Defective Items Inspected}}{\text{Total Items Inspected}}$$

CRITICAL EXAM TRAP: If sample sizes $n_i$ vary across subgroups, $\bar{p}$ must never be calculated as the simple arithmetic average of the subgroup proportions ($\frac{1}{m}\sum p_i$). Doing so treats small batches with equal weight to large batches, introducing mathematical bias!

Control Limits for Constant Sample Size $n$

Under the Binomial distribution, the variance of sample proportion $p$ is $\sigma_p^2 = \frac{p(1-p)}{n}$. Using historical $\bar{p}$:

$$CL_p = \bar{p}$$ $$UCL_p = \bar{p} + 3 \sqrt{\frac{\bar{p}(1 - \bar{p})}{n}}$$ $$LCL_p = \max\left(0, \; \bar{p} - 3 \sqrt{\frac{\bar{p}(1 - \bar{p})}{n}}\right)$$

Mandatory Truncation Rule: If the mathematical calculation yields $LCL_p < 0$, the Lower Control Limit is set to $LCL = 0$. A negative proportion of defectives is physically impossible!

Handling Variable Sample Sizes on a $p$-Chart

When subgroup size $n_i$ varies from batch to batch:

1. Variable Control Limits (Exact Method): Calculate separate control limits for each individual subgroup $i$ based on its specific $n_i$: $$UCL_i = \bar{p} + 3 \sqrt{\frac{\bar{p}(1 - \bar{p})}{n_i}}, \qquad LCL_i = \max\left(0, \; \bar{p} - 3 \sqrt{\frac{\bar{p}(1 - \bar{p})}{n_i}}\right)$$ Notice: Larger sample sizes produce narrower control bands; smaller sample sizes produce wider control bands.
2. Average Sample Size Method ($\bar{n}$): If individual sample sizes vary by no more than $\pm 25\%$ from the average sample size $\bar{n} = \frac{1}{m}\sum n_i$, a single set of constant control limits may be computed using $\bar{n}$.

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3. The $np$-Chart (Number of Nonconforming Items)

The $np$-chart monitors the raw count of defective items ($d_i$) rather than the fraction $p_i$.

The Mandatory Operational Rule for $np$-Charts

STRICT CONSTRAINT: The $np$-chart requires a strictly CONSTANT sample size $n$ across all subgroups ($n_1 = n_2 = \dots = n$). If sample size varies, the $np$-chart cannot be used!

Fundamental Formulas

For constant sample size $n$ and baseline fraction nonconforming $\bar{p}$:

• Center Line ($CL_{np}$): $$CL_{np} = n\bar{p}$$
• Control Limits: $$UCL_{np} = n\bar{p} + 3 \sqrt{n\bar{p}(1 - \bar{p})}$$ $$LCL_{np} = \max\left(0, \; n\bar{p} - 3 \sqrt{n\bar{p}(1 - \bar{p})}\right)$$

Why Use an $np$-Chart over a $p$-Chart?

On factory production floors, line operators prefer the $np$-chart because it eliminates division. If $n = 100$ stamping parts are pulled, the operator simply counts the defective parts (e.g., 4 parts) and plots the integer 4 directly on the chart without calculating $4/100 = 0.04$.

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4. The $c$-Chart (Defects per Constant Inspection Unit)

The $c$-chart monitors the count of nonconformities (defects) observed in a single, standardized inspection unit of fixed size.

The Mandatory Operational Rule for $c$-Charts

STRICT CONSTRAINT: The area of opportunity (inspection unit size) must be strictly CONSTANT across all samples (e.g., exactly 1 refrigerator, exactly $50\text{ m}^2$ of fabric, exactly 1 complex circuit board).

Fundamental Formulas

Under the Poisson distribution, the mean equals the variance ($\lambda = \sigma^2 = c$). Therefore, the standard deviation of defect count is $\sigma_c = \sqrt{c}$. For $m$ inspected units, where unit $i$ exhibits $c_i$ defects:

• Center Line ($\bar{c}$): $$CL_c = \bar{c} = \frac{1}{m} \sum_{i=1}^m c_i$$
• Control Limits: $$UCL_c = \bar{c} + 3 \sqrt{\bar{c}}$$ $$LCL_c = \max\left(0, \; \bar{c} - 3 \sqrt{\bar{c}}\right)$$

If $\bar{c} - 3\sqrt{\bar{c}} < 0$, truncate $LCL_c$ to 0.

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5. The $u$-Chart (Defects per Unit for Variable Inspection Sizes)

When the inspection unit size varies from sample to sample (e.g., inspecting 5 aircraft wings on Monday, 3 on Tuesday; or inspecting continuous rolls of sheet steel where roll lengths vary), the $c$-chart is invalid. The engineer must use a $u$-chart, which monitors the average defect rate per standardized inspection unit.

Fundamental Formulas

For subgroup $i$ with $n_i$ inspection units and $c_i$ total defects:

• Defects per Unit for Subgroup $i$: $$u_i = \frac{c_i}{n_i}$$
• Center Line ($\bar{u}$): The overall weighted defect rate: $$\bar{u} = \frac{\sum_{i=1}^m c_i}{\sum_{i=1}^m n_i} = \frac{\text{Total Defects Counted}}{\text{Total Inspection Units Examined}}$$
• Control Limits for Subgroup $i$ (Variable Limits): $$CL_u = \bar{u}$$ $$UCL_{u, i} = \bar{u} + 3 \sqrt{\frac{\bar{u}}{n_i}}$$ $$LCL_{u, i} = \max\left(0, \; \bar{u} - 3 \sqrt{\frac{\bar{u}}{n_i}}\right)$$

If sample size is constant ($n_i = n$), the control limits remain fixed across all subgroups.

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6. Master Control Chart Selection Guide

The following decision logic guarantees selecting the correct chart on the FE exam:

                    Master Chart Selection Decision Tree

                        What type of data do you have?
                                      │
               ┌──────────────────────┴──────────────────────┐
               ▼                                             ▼
       VARIABLE (Continuous)                        ATTRIBUTE (Discrete)
        (mm, kg, MPa, cP)                         (Good/Bad, Flaw Counts)
               │                                             │
       What is sample size n?                    What are you counting?
         ┌─────┼─────┐                         ┌─────────────┴─────────────┐
         ▼     ▼     ▼                         ▼                           ▼
       n = 1  2<=n<=9 n>=10               DEFECTIVES                    DEFECTS
         │     │     │                 (Items Rejected)              (Flaw Counts)
       I-MR   X-R   X-S                        │                           │
                                         Is n constant?              Is area constant?
                                           ┌───┴───┐                   ┌───┴───┐
                                           ▼       ▼                   ▼       ▼
                                         CONST   VARIES              CONST   VARIES
                                           │       │                   │       │
                                          np       p                   c       u
                                         (or p)                       (or u)

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7. Step-by-Step Worked Engineering Calculations

Worked Example 20.3.1: $p$-Chart and $np$-Chart with Lower Limit Truncation

Problem Statement: An automated optical inspection (AOI) station evaluates surface-mount PCB assemblies for component misalignment. A quality engineer inspects $m = 20$ production lots, each consisting of a constant sample size of $n = 250$ boards. Across all 20 lots, a total of 150 defective boards are identified.

1. Calculate the center line $\bar{p}$ for the $p$-chart.
2. Compute the 3-sigma Upper and Lower Control Limits for the $p$-chart.
3. Compute the Center Line and 3-sigma Control Limits if the data were plotted on an $np$-chart.
4. If Lot 12 exhibits 18 defective boards, evaluate whether this lot represents an out-of-control condition on both charts.

Solution:

Step 1: Compute Center Line for $p$-Chart

• Total defective boards: $\sum d_i = 150$
• Total boards inspected: $\sum n_i = 20 \times 250 = 5,000$

$$\bar{p} = \frac{150}{5,000} = 0.030 \quad (3.0\% \text{ nonconforming})$$

Step 2: Calculate $p$-Chart Control Limits

• Standard error of sample proportion: $$\sigma_p = \sqrt{\frac{\bar{p}(1 - \bar{p})}{n}} = \sqrt{\frac{0.030(1 - 0.030)}{250}} = \sqrt{\frac{0.030(0.970)}{250}} = \sqrt{\frac{0.0291}{250}} = \sqrt{0.0001164} \approx 0.010789$$
• 3-sigma margin: $3\sigma_p = 3(0.010789) = 0.032367 \approx 0.0324$
• Control Limits: $$CL_p = \bar{p} = 0.0300$$ $$UCL_p = \bar{p} + 3\sigma_p = 0.0300 + 0.0324 = 0.0624 \quad (6.24\%)$$ $$LCL_p = \bar{p} - 3\sigma_p = 0.0300 - 0.0324 = -0.0024$$
• Applying the Mandatory Truncation Rule: Because $LCL_p < 0$, set $LCL_p = 0.000$.

Step 3: Calculate $np$-Chart Limits Because sample size is constant ($n = 250$):

• Center Line: $CL_{np} = n\bar{p} = 250(0.030) = 7.50$ defectives
• Standard deviation: $\sigma_{np} = \sqrt{n\bar{p}(1 - \bar{p})} = \sqrt{250(0.030)(0.970)} = \sqrt{7.275} \approx 2.6972$
• 3-sigma margin: $3\sigma_{np} = 3(2.6972) = 8.0916$
• Control Limits: $$UCL_{np} = 7.50 + 8.0916 = 15.59 \text{ defectives}$$ $$LCL_{np} = \max(0, \; 7.50 - 8.0916) = \max(0, -0.59) = 0.00 \text{ defectives}$$

Step 4: Evaluate Lot 12

• For Lot 12: $d_{12} = 18$ defectives.
  • On the $np$-chart: $d_{12} = 18 > UCL_{np} (15.59)$. Out of control!
  • On the $p$-chart: $p_{12} = \frac{18}{250} = 0.0720$. Since $p_{12} = 0.0720 > UCL_p (0.0624)$. Out of control!
• Conclusion: Both charts provide identical statistical conclusions. Lot 12 indicates a special-cause disturbance (e.g., feeder misalignment or solder paste stencil blockage) that must be investigated.

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Worked Example 20.3.2: $c$-Chart vs. $u$-Chart with Variable Inspection Units

Problem Statement: An industrial coating facility inspects newly electroplated steel panels for surface pitting defects. Over five consecutive shifts, inspectors record the total number of pitting defects found across varying numbers of standard panels inspected:

1. Explain why a $c$-chart is mathematically invalid for this operational scenario.
2. Calculate the center line $\bar{u}$ for the appropriate $u$-chart.
3. Compute the specific Upper and Lower Control Limits for Shift 3 ($n_3 = 8$ panels) and Shift 4 ($n_4 = 20$ panels).
4. Evaluate whether Shift 3 ($c_3 = 20$ defects) represents an out-of-control condition.

Solution:

Step 1: Chart Justification The data records nonconformities (defects), which follow a Poisson model. However, the sample size varies from shift to shift ($n_i$ ranges from 8 to 20 panels). A $c$-chart strictly requires a constant area of opportunity ($n_i = \text{constant}$). Plotting raw counts when sample size varies would cause large batches to trigger false alarms simply due to higher inspection volume. Therefore, a $u$-chart (defects per unit) must be used.

Step 2: Calculate Center Line $\bar{u}$

• Total defects counted: $\sum c_i = 18 + 24 + 20 + 32 + 16 = 110$ defects
• Total panels inspected: $\sum n_i = 10 + 15 + 8 + 20 + 12 = 65$ panels

$$\bar{u} = \frac{110}{65} = 1.6923 \approx 1.692 \text{ defects per panel}$$

Step 3: Calculate Control Limits for Shifts 3 and 4 For Shift 3 ($n_3 = 8$ panels):

• Standard error: $\sigma_{u, 3} = \sqrt{\frac{\bar{u}}{n_3}} = \sqrt{\frac{1.6923}{8}} = \sqrt{0.21154} \approx 0.4599$
• Limits: $$UCL_{u, 3} = 1.6923 + 3(0.4599) = 1.6923 + 1.3798 = 3.072 \text{ defects/panel}$$ $$LCL_{u, 3} = \max(0, \; 1.6923 - 1.3798) = 0.313 \text{ defects/panel}$$

For Shift 4 ($n_4 = 20$ panels):

• Standard error: $\sigma_{u, 4} = \sqrt{\frac{\bar{u}}{n_4}} = \sqrt{\frac{1.6923}{20}} = \sqrt{0.08462} \approx 0.2909$
• Limits: $$UCL_{u, 4} = 1.6923 + 3(0.2909) = 1.6923 + 0.8727 = 2.565 \text{ defects/panel}$$ $$LCL_{u, 4} = \max(0, \; 1.6923 - 0.8727) = 0.820 \text{ defects/panel}$$

(Notice: For $n = 20$, the control limits are significantly tighter than for $n = 8$).

Step 4: Evaluate Shift 3

• For Shift 3, the observed defect rate per panel is: $$u_3 = \frac{c_3}{n_3} = \frac{20}{8} = 2.500 \text{ defects per panel}$$
• Comparing to Shift 3 control limits: $$LCL_{u, 3} (0.313) \le u_3 (2.500) \le UCL_{u, 3} (3.072)$$
• Conclusion: Shift 3 is in statistical control. Even though 20 defects appeared high in raw count, when standardized against the small sample size ($n = 8$), the rate of 2.500 defects/panel remains within 3-sigma random variation.

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8. NCEES Reference Handbook Tips & Realistic Exam Traps

• Defect vs. Defective Trap: Always ask: Can an item have more than one of these flaws and still be one item? If yes, it is a defect ($c$ or $u$ chart). If the item is simply rejected as bad or accepted as good, it is a defective ($p$ or $np$ chart).
• Mandatory LCL Truncation: Whenever an attribute control chart formula generates a negative lower control limit, you must truncate it to zero ($LCL = 0$). NCEES exam questions frequently include the raw negative number as an attractive distractor!
• Weighted Average for $\bar{p}$ and $\bar{u}$: If an exam problem presents varying sample sizes, never take the simple average of the percentages. You must sum all defectives (or defects) and divide by the sum of all inspected units: $$\bar{p} = \frac{\sum d_i}{\sum n_i}, \qquad \bar{u} = \frac{\sum c_i}{\sum n_i}$$
• When $n$ Varies, $np$ and $c$ are Banned: If subgroup sizes fluctuate, you cannot use an $np$-chart or a $c$-chart. You must select a $p$-chart (with variable limits or $\bar{n}$) or a $u$-chart.