Learning Objectives
By the end of this chapter, you will be able to:
- Calculate required sample size for estimating a mean
- Calculate required sample size for estimating a proportion
- Understand the relationship between sample size and precision
- Handle situations where population variance is unknown
- Apply sample size formulas to practical problems
Why Determine Sample Size?
Before collecting data, we need to know how many observations are required to achieve:
- Desired precision (margin of error)
- Desired confidence level
- Cost-effectiveness (not too large or too small)
flowchart TD
A[Sample Size Planning]
A --> B[Too Small]
A --> C[Just Right]
A --> D[Too Large]
B --> B1["Imprecise estimates<br/>Wide confidence intervals"]
C --> C1["Good precision<br/>Cost-effective"]
D --> D1["Wasted resources<br/>Unnecessary cost"]
Sample Size for Estimating Mean (σ Known)
Formula
Starting from margin of error:
\[ME = z^* \cdot \frac{\sigma}{\sqrt{n}}\]Solving for n:
\[n = \left(\frac{z^* \cdot \sigma}{ME}\right)^2\]Where:
- $n$ = required sample size
- $z^*$ = z-value for confidence level
- $\sigma$ = population standard deviation
- $ME$ = desired margin of error
Step-by-Step Example 1: Sample Size for Mean
Problem: A researcher wants to estimate the average monthly expenditure of households. Historical data suggests σ = NPR 5,000. How large a sample is needed for a 95% confidence interval with margin of error NPR 800?
Solution:
Step 1: Identify given values
- σ = 5,000
- ME = 800
- Confidence = 95% → z* = 1.96
Step 2: Apply formula \(n = \left(\frac{z^* \cdot \sigma}{ME}\right)^2 = \left(\frac{1.96 \times 5000}{800}\right)^2\)
\[n = \left(\frac{9800}{800}\right)^2 = (12.25)^2 = 150.0625\]Step 3: Round up (always!) \(n = 151\)
Answer: A sample of at least 151 households is required.
Step-by-Step Example 2: Effect of Confidence Level
Problem: Using the same scenario, compare sample sizes needed for: (a) 90% confidence (b) 95% confidence (c) 99% confidence
Solution:
Using formula: $n = \left(\frac{z^* \times 5000}{800}\right)^2$
(a) 90% confidence (z* = 1.645): \(n = \left(\frac{1.645 \times 5000}{800}\right)^2 = (10.28)^2 = 105.7 \approx 106\)
(b) 95% confidence (z* = 1.96): \(n = \left(\frac{1.96 \times 5000}{800}\right)^2 = (12.25)^2 = 150.1 \approx 151\)
(c) 99% confidence (z* = 2.576): \(n = \left(\frac{2.576 \times 5000}{800}\right)^2 = (16.1)^2 = 259.2 \approx 260\)
| Confidence Level | Required n |
|---|---|
| 90% | 106 |
| 95% | 151 |
| 99% | 260 |
Conclusion: Higher confidence requires larger sample size.
What If σ is Unknown?
Options:
- Use pilot study: Take a small preliminary sample to estimate σ
- Use prior research: Use σ from similar studies
- Use range rule: Estimate σ ≈ Range/4
- Use conservative estimate: Use the largest reasonable σ
Example 3: Using Range Estimate
Problem: Salaries range from NPR 20,000 to NPR 80,000. Estimate σ and calculate sample size for 95% CI with ME = 5,000.
Solution:
\[\sigma \approx \frac{\text{Range}}{4} = \frac{80,000 - 20,000}{4} = \frac{60,000}{4} = 15,000\] \[n = \left(\frac{1.96 \times 15,000}{5,000}\right)^2 = (5.88)^2 = 34.6 \approx 35\]Answer: Approximately 35 observations needed.
Sample Size for Estimating Proportion
Formula
\[n = \frac{(z^*)^2 \cdot \hat{p}(1-\hat{p})}{ME^2}\]Or equivalently:
\[n = \hat{p}(1-\hat{p})\left(\frac{z^*}{ME}\right)^2\]Where:
- $\hat{p}$ = estimated proportion (from pilot study or prior knowledge)
- $ME$ = desired margin of error for proportion
Step-by-Step Example 4: Sample Size for Proportion
Problem: A researcher wants to estimate the proportion of households with internet access. A pilot study suggests about 40% have access. How many households should be surveyed for 95% confidence with margin of error 5%?
Solution:
Step 1: Identify given values
- $\hat{p}$ = 0.40
- ME = 0.05
- z* = 1.96
Step 2: Apply formula \(n = \frac{(1.96)^2 \times 0.40 \times 0.60}{(0.05)^2}\)
\[n = \frac{3.8416 \times 0.24}{0.0025}\] \[n = \frac{0.922}{0.0025} = 368.8 \approx 369\]Answer: At least 369 households should be surveyed.
Conservative Sample Size (When p is Unknown)
Maximum Variability
The product $p(1-p)$ is maximized when $p = 0.5$:
\[p(1-p) = 0.5 \times 0.5 = 0.25\]Conservative Formula
When p is completely unknown, use:
\[n = \frac{(z^*)^2 \times 0.25}{ME^2} = \frac{0.25 \times (z^*)^2}{ME^2}\]This gives the largest sample size needed regardless of actual p.
Step-by-Step Example 5: Conservative Approach
Problem: No prior information is available about a proportion. Calculate the sample size for 95% confidence with 3% margin of error.
Solution:
Using conservative formula with p = 0.5:
\[n = \frac{(1.96)^2 \times 0.25}{(0.03)^2}\] \[n = \frac{3.8416 \times 0.25}{0.0009}\] \[n = \frac{0.9604}{0.0009} = 1067.1 \approx 1068\]Answer: 1,068 observations needed.
Comparison: Effect of p on Sample Size
| Estimated p | p(1-p) | Sample Size (95%, ME=5%) |
|---|---|---|
| 0.10 | 0.09 | 139 |
| 0.20 | 0.16 | 246 |
| 0.30 | 0.21 | 323 |
| 0.40 | 0.24 | 369 |
| 0.50 | 0.25 | 385 |
Insight: Sample size is maximized at p = 0.50.
Sample Size Quick Reference
For Means
\[n = \left(\frac{z^* \cdot \sigma}{ME}\right)^2\]For Proportions
\[n = \frac{(z^*)^2 \cdot \hat{p}(1-\hat{p})}{ME^2}\]Conservative (Proportion)
\[n = \frac{0.25 \times (z^*)^2}{ME^2}\]Step-by-Step Example 6: Exam-Style Problem
Problem: A government department wants to estimate: (a) The average processing time (σ estimated at 12 minutes) with ME = 2 minutes at 99% confidence (b) The proportion of satisfied customers (previous estimate: 75%) with ME = 4% at 95% confidence
Calculate the required sample sizes.
Solution:
(a) Sample size for mean:
- σ = 12, ME = 2, z* = 2.576
(b) Sample size for proportion:
- $\hat{p}$ = 0.75, ME = 0.04, z* = 1.96
Answers:
- (a) 239 samples for processing time
- (b) 451 customers for satisfaction survey
Practical Considerations
1. Always Round UP
A sample of 150.1 requires 151 observations, not 150.
2. Account for Non-Response
If expected response rate is 80%: \(n_{\text{adjusted}} = \frac{n_{\text{required}}}{0.80}\)
3. Budget Constraints
Sometimes you can’t afford the calculated sample size. Options:
- Accept larger margin of error
- Accept lower confidence level
- Use stratified sampling for efficiency
4. Finite Population Correction
For finite population N: \(n_{\text{corrected}} = \frac{n}{1 + \frac{n-1}{N}}\)
Example 7: Adjusting for Non-Response
Problem: A survey requires 400 responses. If the expected response rate is 70%, how many should be contacted?
Solution:
\[n_{\text{contact}} = \frac{400}{0.70} = 571.4 \approx 572\]Answer: Contact 572 people.
Decision Flow for Sample Size
flowchart TD
A[What are you estimating?]
A --> B{Mean or Proportion?}
B -->|Mean| C{Is σ known?}
B -->|Proportion| D{Is p estimated?}
C -->|Yes| E["Use n = (z*σ/ME)²"]
C -->|No| F["Estimate σ from range<br/>or pilot study"]
D -->|Yes| G["Use n = z²p(1-p)/ME²"]
D -->|No| H["Use p = 0.5<br/>(conservative)"]
Practice Problems
Problem 1
How large a sample is needed to estimate population mean with:
- σ = 20
- ME = 3
- 95% confidence
Problem 2
What sample size is required to estimate a proportion with:
- Estimated p = 0.60
- ME = 0.04
- 99% confidence
Problem 3
If we want to halve the margin of error, by what factor must we increase the sample size?
Problem 4
A survey of voters requires ME = 3% at 95% confidence. No prior estimate of p is available. (a) Calculate conservative sample size (b) If p is estimated at 0.35, recalculate
Problem 5
A researcher has budget for only 200 samples. If σ = 15 and 95% confidence is required, what is the achievable margin of error?
Summary
| Scenario | Formula |
|---|---|
| Mean (σ known) | $n = \left(\frac{z^* \sigma}{ME}\right)^2$ |
| Proportion (p estimated) | $n = \frac{(z^*)^2 \hat{p}(1-\hat{p})}{ME^2}$ |
| Proportion (p unknown) | $n = \frac{0.25(z^*)^2}{ME^2}$ |
| Halve ME | Quadruple n |
| Double confidence | Increase n |
Unit 4 Complete!
You have completed Unit 4: Estimation. You now understand:
- Sampling distributions and standard errors
- Properties of good estimators
- How to construct confidence intervals
- How to determine sample size
In Unit 5, we will study Hypothesis Testing - the most extensive unit covering statistical tests for making decisions about population parameters.
