1. QUICK RECAP: BASIC FORMULA (FOR PROPORTION)

For prevalence/proportion studies (very common in orthopaedics):

n=Z2⋅p⋅qd2n = \frac{Z^2 \cdot p \cdot q}{d^2}n=d2Z2⋅p⋅q​

Where:

• n = sample size

• Z = Z-score for confidence level (usually 1.96 for 95%)

• p = expected prevalence (proportion)

• q = 1 − p

• d = margin of error (precision)

If p is unknown → use 0.5 (50%)
If population 10,000 → apply finite population correction (FPC)
If non-response expected → divide by (1 − non-response rate)

2. CASE 1: POPULATION > 10,000, UNKNOWN PREVALENCE

Case Study 1

Research question:
“What is the prevalence of chronic low back pain among adults attending the orthopaedic clinic at a county referral hospital?”

• The hospital sees about 30,000 adult patients per year → treat as >10,000 (large population).

• You don’t know the prevalence of chronic low back pain → use p = 0.5 (50%) to be safe.

Given:

• Confidence level = 95% → Z = 1.96

• p = 0.5

• q = 1 − 0.5 = 0.5

• d = 0.05

Step-by-step calculation (more than 3 steps)

Step 1: Write the formula

n=Z2⋅p⋅qd2n = \frac{Z^2 \cdot p \cdot q}{d^2}n=d2Z2⋅p⋅q​

Step 2: Substitute the values

n=(1.96)2⋅0.5⋅0.5(0.05)2n = \frac{(1.96)^2 \cdot 0.5 \cdot 0.5}{(0.05)^2}n=(0.05)2(1.96)2⋅0.5⋅0.5​

Step 3: Calculate Z²

1.962=3.84161.96^2 = 3.84161.962=3.8416

Step 4: Multiply p × q

$$0.5\times 0.5=0.25$$

Step 5: Multiply the numerator

$$3.8416\times 0.25=0.9604$$

Step 6: Calculate d²

(0.05)2=0.0025(0.05)^2 = 0.0025(0.05)2=0.0025

Step 7: Divide numerator by denominator

n=0.96040.0025=384.16n = \frac{0.9604}{0.0025} = 384.16n=0.00250.9604​=384.16

Step 8: Round up

$$n\approx 385\text{ participants}$$

Because the population is large (>10,000), NO finite correction is needed.
Next we will later see how to adjust this for non-response.

3. CASE 2: POPULATION > 10,000, KNOWN PREVALENCE

Case Study 2

Research question:
“What is the prevalence of post-operative wound infection following ORIF (open reduction internal fixation) at a regional orthopaedic centre?”

Suppose previous records show wound infection rate = 10% (p = 0.10) among ORIF patients.

Given:

• Population (yearly ORIF patients) ≈ 12,000 → >10,000

• p = 0.10

• q = 1 − 0.10 = 0.90

• d = 0.03 (you want ±3% precision)

• Z = 1.96

Step-by-step calculation

Step 1: Formula

n=Z2⋅p⋅qd2n = \frac{Z^2 \cdot p \cdot q}{d^2}n=d2Z2⋅p⋅q​

Step 2: Substitute values

n=(1.96)2⋅0.10⋅0.90(0.03)2n = \frac{(1.96)^2 \cdot 0.10 \cdot 0.90}{(0.03)^2}n=(0.03)2(1.96)2⋅0.10⋅0.90​

Step 3: Compute Z²

1.962=3.84161.96^2 = 3.84161.962=3.8416

Step 4: Compute p × q

$$0.10\times 0.90=0.09$$

Step 5: Multiply for numerator

$$3.8416\times 0.09=0.345744$$

Step 6: Compute d²

0.032=0.00090.03^2 = 0.00090.032=0.0009

Step 7: Divide numerator by denominator

n=0.3457440.0009=384.16n = \frac{0.345744}{0.0009} = 384.16n=0.00090.345744​=384.16

Step 8: Round up

$$n\approx 385\text{ ORIF patients}$$

Again, no finite population correction because population is large (> 10,000).

4. CASE 3: POPULATION 10,000, KNOWN PREVALENCE (USE FPC)

Now we bring in the finite population correction (FPC).

Case Study 3

Research question:
“What is the prevalence of knee osteoarthritis among patients aged ≥50 years attending your rural orthopaedic clinic in one year?”

• Total number of patients aged ≥50 years in that year = N = 2,400

• Previous small audit suggests OA prevalence = 40% (p = 0.40)

Given:

• N = 2,400 (10,000 → we MUST correct)

• p = 0.40

• q = 0.60

• d = 0.05

• Z = 1.96

Step-by-step calculation – Part 1: Initial n (as if population is large)

Step 1: Formula for n

n=Z2⋅p⋅qd2n = \frac{Z^2 \cdot p \cdot q}{d^2}n=d2Z2⋅p⋅q​

Step 2: Substitute

n=(1.96)2⋅0.40⋅0.60(0.05)2n = \frac{(1.96)^2 \cdot 0.40 \cdot 0.60}{(0.05)^2}n=(0.05)2(1.96)2⋅0.40⋅0.60​

Step 3: Compute Z²

1.962=3.84161.96^2 = 3.84161.962=3.8416

Step 4: Compute p × q

$$0.40\times 0.60=0.24$$

Step 5: Multiply numerator

$$3.8416\times 0.24=0.921984$$

Step 6: Compute d²

0.052=0.00250.05^2 = 0.00250.052=0.0025

Step 7: Calculate n

n=0.9219840.0025=368.79n = \frac{0.921984}{0.0025} = 368.79n=0.00250.921984​=368.79

Round → n ≈ 369 (initial)

Step-by-step calculation – Part 2: Apply Finite Population Correction (FPC)

nadj=n1+n−1Nn_{adj} = \frac{n}{1 + \frac{n - 1}{N}}nadj​=1+Nn−1​n​

Step 1: Substitute n and N

nadj=3691+369−12400n_{adj} = \frac{369}{1 + \frac{369 - 1}{2400}}nadj​=1+2400369−1​369​

Step 2: Simplify numerator in the fraction

$$369-1=368$$ 3682400=0.15333‾≈0.1533\frac{368}{2400} = 0.1533\overline{3} \approx 0.15332400368​=0.15333≈0.1533

Step 3: Add 1 in the denominator

$$1+0.1533=1.1533$$

Step 4: Divide n by this value

nadj=3691.1533≈319.9n_{adj} = \frac{369}{1.1533} \approx 319.9nadj​=1.1533369​≈319.9

Step 5: Round up

Final corrected sample size:

nadj≈320 patientsn_{adj} \approx 320 \text{ patients}nadj​≈320 patients

So, because the population is only 2,400, you don’t need 369; 320 is enough.

5. CASE 4: POPULATION 10,000, UNKNOWN PREVALENCE

Case Study 4

Research question:
“What is the prevalence of chronic non-union in long bone fractures seen over 1 year in your district hospital?”

You don’t know the prevalence.

• Total fracture patients in 1 year: N = 1,000

• Unknown prevalence → use p = 0.5

• Z = 1.96

• d = 0.05

Part 1: Calculate initial n assuming large population

Step 1: Formula

n=Z2⋅p⋅qd2n = \frac{Z^2 \cdot p \cdot q}{d^2}n=d2Z2⋅p⋅q​

Step 2: Substitute values

n=(1.96)2⋅0.5⋅0.5(0.05)2n = \frac{(1.96)^2 \cdot 0.5 \cdot 0.5}{(0.05)^2}n=(0.05)2(1.96)2⋅0.5⋅0.5​

Step 3: Compute Z²

1.962=3.84161.96^2 = 3.84161.962=3.8416

Step 4: Compute p × q

$$0.5\times 0.5=0.25$$

Step 5: Numerator

$$3.8416\times 0.25=0.9604$$

Step 6: d²

0.052=0.00250.05^2 = 0.00250.052=0.0025

Step 7: n

n=0.96040.0025=384.16n = \frac{0.9604}{0.0025} = 384.16n=0.00250.9604​=384.16

So initial n ≈ 385

Part 2: Apply FPC (N = 1,000)

nadj=n1+n−1Nn_{adj} = \frac{n}{1 + \frac{n - 1}{N}}nadj​=1+Nn−1​n​

Step 1: Substitute

nadj=3851+385−11000n_{adj} = \frac{385}{1 + \frac{385 - 1}{1000}}nadj​=1+1000385−1​385​

Step 2: Simplify

$$385-1=384$$ 3841000=0.384\frac{384}{1000} = 0.3841000384​=0.384

Step 3: Add 1

$$1+0.384=1.384$$

Step 4: Divide

nadj=3851.384≈278.1n_{adj} = \frac{385}{1.384} \approx 278.1nadj​=1.384385​≈278.1

Step 5: Round up

nadj≈279 patientsn_{adj} \approx 279 \text{ patients}nadj​≈279 patients

So you only need 279 out of 1,000 fracture patients to estimate non-union prevalence.

6. CASE 5: UNKNOWN TOTAL POPULATION (TREAT AS VERY LARGE)

Sometimes you don’t know the exact population size:

• Example: You study community prevalence of back pain in a town, but you don’t have accurate census data for adults.

In such cases, we treat population as very large and do NOT apply FPC.

Case Study 5

“Prevalence of chronic low back pain among boda-boda riders in a large city.”

• No accurate count of all riders → population treated as infinite/very large.

• Suppose previous study estimates p = 0.25 (25%).

Given:

• p = 0.25 → q = 0.75

• d = 0.05

• Z = 1.96

Step 1: Formula

n=Z2⋅p⋅qd2n = \frac{Z^2 \cdot p \cdot q}{d^2}n=d2Z2⋅p⋅q​

Step 2: Substitute

n=(1.96)2⋅0.25⋅0.75(0.05)2n = \frac{(1.96)^2 \cdot 0.25 \cdot 0.75}{(0.05)^2}n=(0.05)2(1.96)2⋅0.25⋅0.75​

Step 3: Z²

1.962=3.84161.96^2 = 3.84161.962=3.8416

Step 4: p × q

$$0.25\times 0.75=0.1875$$

Step 5: Numerator

$$3.8416\times 0.1875=0.7203$$

Step 6: d²

0.052=0.00250.05^2 = 0.00250.052=0.0025

Step 7: n

n=0.72030.0025=288.12n = \frac{0.7203}{0.0025} = 288.12n=0.00250.7203​=288.12

Step 8: Round up → 289 riders

No FPC because population is treated as very large/unknown.

7. ADJUSTMENT FOR NON-RESPONSE (ALL CASES)

In real orthopaedic research, some patients will:

• refuse consent

• fail to return for follow-up

• not complete questionnaires

We adjust using:

nfinal=nadj1−non-response raten_{final} = \frac{n_{adj}}{1 - \text{non-response rate}}nfinal​=1−non-response ratenadj​​

If non-response rate is 10% → use 0.90 in denominator.
If 20% → 0.80, etc.

Example Using Case Study 3 (Knee OA, N = 2400, nₐ𝚍ⱼ = 320)

Assume 10% non-response.

Step 1: Write adjustment formula

nfinal=nadj1−NRn_{final} = \frac{n_{adj}}{1 - NR}nfinal​=1−NRnadj​​

Step 2: Substitute

nfinal=3201−0.10n_{final} = \frac{320}{1 - 0.10}nfinal​=1−0.10320​

Step 3: Calculate denominator

$$1-0.10=0.90$$

Step 4: Divide

nfinal=3200.90=355.56n_{final} = \frac{320}{0.90} = 355.56nfinal​=0.90320​=355.56

Step 5: Round up

nfinal≈356 patientsn_{final} \approx 356 \text{ patients}nfinal​≈356 patients

So you plan to recruit 356 patients, expecting that about 10% may not respond or complete the study.

8. SHORT NARRATIVE SUMMARY FOR STUDENTS

1. If population is very large (>10,000) or unknown
   → No finite correction
   → Use basic formula for n.

2. If population is small (10,000)
   → First compute n (as if large).
   → Then apply FPC formula to reduce n.

3. If prevalence is known from previous data
   → Use that p (e.g., 0.10, 0.25, 0.40).

4. If prevalence is unknown
   → Use p = 0.5 (gives maximum, safest n).

5. Always adjust for non-response
   → Divide by (1 − non-response rate).

6. Always round sample size UP, not down.