Sample Size Calculation for Research Studies

"How many respondents do I need?" is one of the first questions every thesis student asks, and "Justify your sample size" is one of the most common comments examiners write. A sample that is too small cannot detect real effects; one that is too large wastes time and money. This guide shows the main formulas, works through them by hand and explains how to write the justification.

What Drives Sample Size

Every sample size calculation balances a few ingredients:

• Confidence level (usually 95%, giving \( z = 1.96 \)): how sure you want to be.
• Margin of error \( e \) (often ±5%): how precise your estimate must be.
• Variability: the expected proportion \( p \) for categorical outcomes, or standard deviation \( \sigma \) for continuous ones.
• Significance level \( \alpha \) (usually .05) and power \( 1 - \beta \) (usually .80) for hypothesis tests.
• Effect size: the smallest difference or relationship worth detecting.
• Population size \( N \), when the population is small and known.

Two broad situations arise. Descriptive surveys estimate a proportion or mean with a given precision. Analytical studies test a hypothesis, so they need a power analysis.

Worked Example 1: Cochran's Formula for a Large Population

A researcher wants to estimate the proportion of university students in Punjab who use AI tools for assignments, at 95% confidence with a ±5% margin of error. No prior estimate exists, so she uses \( p = 0.5 \), which gives the largest (safest) sample.

$$ n_0 = \frac{z^2 p(1-p)}{e^2} $$

Step 1: Substitute.

$$ n_0 = \frac{1.96^2 \times 0.5 \times 0.5}{0.05^2} = \frac{3.8416 \times 0.25}{0.0025} = \frac{0.9604}{0.0025} = 384.16 $$

Step 2: Round up. Always round sample sizes up, so \( n_0 = 385 \).

This explains why "384" or "385" appears in so many theses: it is the 95%, ±5%, \( p = 0.5 \) answer for a large population.

If a pilot study suggested that about 30% of students use AI tools, then \( p = 0.3 \) and

$$ n_0 = \frac{3.8416 \times 0.3 \times 0.7}{0.0025} = \frac{0.80674}{0.0025} = 322.7 $$

so 323 students would be enough.

Worked Example 2: Adjusting for a Small, Known Population

Now suppose the study is limited to one college with \( N = 2000 \) students. The finite population correction reduces the required sample:

$$ n = \frac{n_0}{1 + \dfrac{n_0 - 1}{N}} $$

Step 1: Substitute \( n_0 = 384.16 \).

$$ n = \frac{384.16}{1 + \dfrac{383.16}{2000}} = \frac{384.16}{1 + 0.19158} = \frac{384.16}{1.19158} = 322.4 $$

Step 2: Round up to 323 students.

Yamane's Simplified Formula

Many theses in Pakistan and South Asia use Taro Yamane's 1967 formula, which assumes 95% confidence and \( p = 0.5 \):

$$ n = \frac{N}{1 + Ne^2} = \frac{2000}{1 + 2000(0.05)^2} = \frac{2000}{1 + 5} = \frac{2000}{6} = 333.3 $$

Rounded up, this gives 334. It is slightly more conservative than Cochran's corrected value. The Krejcie and Morgan (1970) table, also widely cited, gives 322 for a population of 2000, consistent with Cochran's approach.

Worked Example 3: Power Analysis for Comparing Two Means

A researcher plans an experiment comparing a new teaching method with the standard method. Past data suggest test scores have a standard deviation of \( \sigma = 10 \) marks, and a difference of \( \Delta = 5 \) marks would be educationally important. She wants \( \alpha = .05 \) (two-tailed) and 80% power.

The normal-approximation formula for each group is

$$ n = \frac{2(z_{1-\alpha/2} + z_{1-\beta})^2\sigma^2}{\Delta^2} $$

Step 1: z values. \( z_{0.975} = 1.960 \) and \( z_{0.80} = 0.842 \), so their sum is 2.802 and its square is 7.849.

Step 2: Substitute.

$$ n = \frac{2 \times 7.849 \times 100}{25} = \frac{1569.8}{25} = 62.8 $$

Step 3: Round up to 63 students per group. The exact calculation based on the t-distribution, as done by G*Power, gives 64 per group, because the t-distribution has slightly heavier tails. The standardised effect size here is Cohen's \( d = 5/10 = 0.5 \), a "medium" effect.

Step 4: Allow for dropout. If about 10% of participants may drop out, divide by 0.9:

$$ \frac{64}{0.9} = 71.1 $$

so she should recruit 72 per group, or 144 in total.

Worked Example 4: Sample Size for a Correlation

To detect a correlation of \( r = 0.30 \) with \( \alpha = .05 \) (two-tailed) and 80% power, use Fisher's z transformation:

$$ C = \frac{1}{2}\ln\left(\frac{1+r}{1-r}\right) = \frac{1}{2}\ln\left(\frac{1.3}{0.7}\right) = \frac{1}{2}\ln(1.857) = 0.3095 $$

$$ n = \left(\frac{z_{1-\alpha/2} + z_{1-\beta}}{C}\right)^2 + 3 = \left(\frac{2.802}{0.3095}\right)^2 + 3 = 81.9 + 3 = 84.9 $$

Rounded up, 85 participants are needed.

Rules of Thumb for Multivariate Analyses

Formulas are not always available for complex models, so researchers also cite these guidelines:

• Multiple regression: \( N \ge 50 + 8k \) for testing the overall model, where \( k \) is the number of predictors (Green, 1991). With 5 predictors, that is \( 50 + 40 = 90 \).
• Factor analysis: at least 5 to 10 respondents per item, and ideally 200 or more.
• SEM: 200 or more is a common minimum; complex models need more. PLS-SEM users often apply the "10 times rule" or, better, a power analysis.
• Qualitative interviews: sample size is guided by data saturation, often reached somewhere between about 12 and 30 interviews, depending on the topic.

Software for Sample Size

• G*Power (free) handles t-tests, ANOVA, correlation, regression and chi-square.
• R: the pwr package.
• Stata: the power command.
• Excel: easy to program Cochran's and Yamane's formulas.
• Python:

python from statsmodels.stats.power import TTestIndPower TTestIndPower().solve_power(effect_size=0.5, alpha=0.05, power=0.80) # about 64 per group

How to Justify It in Your Methodology Chapter

A strong justification names the formula, every input, the source of the inputs, the result and any adjustment:

An a priori power analysis in G*Power 3.1 indicated that 64 participants per group were required to detect a medium effect (d = 0.5) with α = .05 and power = .80 in a two-tailed independent t-test. Allowing for 10% attrition, 72 participants per group were recruited.

Did You Know?

• The sample size needed for a ±5% margin barely changes once the population passes about 20,000; a national survey of millions needs roughly the same 385 as a large city.
• Halving the margin of error from 5% to 2.5% roughly quadruples the required sample, because \( e \) appears squared in the denominator.
• Jacob Cohen, who popularised small, medium and large effect sizes, warned that many published studies had less than a 50% chance of detecting medium effects.

Key Takeaways

• Descriptive surveys use precision formulas such as Cochran's; hypothesis tests need power analysis.
• With 95% confidence, ±5% margin and \( p = 0.5 \), a large population needs 385 respondents.
• Apply the finite population correction or Yamane's formula for small known populations.
• Always round up and inflate for expected non-response or dropout.
• Report the formula, inputs, software and final number in your methodology chapter.

Unsure how many respondents your thesis needs, or how to justify the number to your supervisor? Amathify's experts and PhD scholars can calculate and write up your sample size step by step; reach us via WhatsApp or the Ask a Question form.