Calculating the Perfect Sample Size for your research

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How to calculate the sample size to estimate a mean or a proportion: the formula n = (zσ/E)², what to do without σ, examples and exercises with answers.

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Introduction

The process of choosing the individuals who will make up a SAMPLE is called SAMPLING. The researcher seeks to generalize the conclusions obtained from the SAMPLE, extending them to the entire POPULATION from which it was drawn.

A sample never represents a population perfectly. In other words, using a sample means accepting a margin of error, which we will call SAMPLING ERROR: the difference between a sample result and the true population result. These errors come from random fluctuations from sample to sample.

Non-sampling errors, on the other hand, occur when:

  • data are collected, recorded or analysed incorrectly;
  • a defective instrument is used for measurements;
  • a questionnaire or form has questions worded in a biased way.

To avoid non-sampling errors, check our lesson on samplinglink outside website.

We cannot avoid SAMPLING ERROR, but we can limit its size by choosing a sample of adequate size. SAMPLING ERROR and SAMPLE SIZE move in opposite directions: the larger the sample, the smaller the error, and vice versa. But not in the same proportion: the error falls with the square root of n, so to cut the error in half you need to quadruple the sample.

A survey of 400 people has a margin of error of 4 percentage points. Keeping the same confidence, how many people would be needed for a margin of 2 points?

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Answer: 1,600 — the margin of error is proportional to 1/√n. To divide the margin by 2, you must multiply n by 2² = 4. Doubling the sample (800) would only reduce the margin to about 2.8 points.

Determining the sample size

Determining the size of a sample is a very important problem because unnecessarily large samples lead to wasted time and money, and samples that are too small can lead to unreliable results.

In many cases it is possible to determine the minimum size of a sample to estimate a parameter, such as the POPULATION MEAN (µ).

But how do we calculate the sample size (n) from values that are already known? A confidence interval for the mean, with lower limit a and upper limit b, is:

a = x̄ − z · σ/√n

b = x̄ + z · σ/√n

where x̄ is the sample mean, σ the population standard deviation and z the standard normal value for the chosen confidence level (1.645 for 90%, 1.96 for 95%, 2.576 for 99%). The distance from the sample mean to each end of the interval is the margin of error (E):

E = z · σ/√n

Solving for n:

√n = z · σ/E

n = (z · σ/E)²

Since n must be a whole number and the margin cannot exceed E, the result is always rounded up. Therefore, to know the sample size, you need to:

  • choose the margin of error E;
  • choose the confidence level (and, with it, z);
  • know the population standard deviation (σ).

Examples

An economist wants to estimate the average income in the first year of work of a law graduate. How many income values must be collected if he wants to be 95% confident that the sample mean is less than R$ 500.00 (Brazilian reais) from the true population mean? Assume it is known from a previous study that for these incomes σ = R$ 6,250.00.

σ = 6,250; E = 500; 95% confidence, that is, z = 1.96.

n = (1.96 × 6,250/500)² = 24.5² = 600.25, which we round up: 601.

We must therefore obtain a sample of at least 601 first-year incomes, randomly selected, from law graduates. With it we will be 95% confident that the sample mean x̄ differs by less than R$ 500.00 from the true population mean µ.

Using the data from the previous example, use a larger margin of error, R$ 1,000.00, and determine the required sample size.

Since the margin of error doubled and it appears squared in the denominator, the sample size is divided by 2² = 4: n = (1.96 × 6,250/1,000)² = 150.06, that is, 151. Notice that more precise surveys may require many more resources.

A factory wants to estimate the average weight of a product with 95% confidence and a margin of error of 2 g. The known standard deviation is 20 g. What is the minimum sample size?

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Answer: 385 — n = (1.96 × 20/2)² = 19.6² = 384.16, and sample size is always rounded up. With 384 the margin would be slightly above 2 g.

What if σ is unknown?

The sample size formula requires a value for the population standard deviation σ. If it is unknown, we can use a preliminary value obtained in one of these ways:

  • Use the approximation σ ≈ range/4 (the range is the largest possible value minus the smallest).
  • Run a pilot study: with a first collection of at least 31 randomly selected values, calculate the sample standard deviation s and use it in place of σ. This value can be refined as more data are collected.
  • When what you want to estimate is a proportion p, use the standard deviation of a yes/no (Bernoulli) answer: σ = √(p · (1 − p)), with a prior estimate of p.
  • If there is no prior estimate of the proportion, use p = 0.5 (50%), the value that maximizes p · (1 − p). Then σ = √0.25 = 0.5, and the sample is large enough whatever the true proportion is.

For proportions, the formula then becomes n = z² · p · (1 − p)/E².

Examples

A social worker wants to know the sample size (n) needed to determine the proportion of the population served by a Health Unit in the municipality of Cariacica (Brazil). No previous survey was done, so the proportion is unknown. She wants to be 90% confident that the maximum estimation error (E) is ±5 percentage points (0.05). How many people need to be interviewed?

90% confidence, that is, z = 1.645;

p = 0.5 because it is unknown, so p · (1 − p) = 0.25 and σ = 0.5;

E = 0.05.

n = (1.645 × 0.5/0.05)² = 16.45² = 270.6, which we round up: 271.

We must therefore interview 271 people to estimate the proportion of Cariacica's population served by the Health Unit.

With no prior estimate, how many people must be interviewed to estimate a proportion with 95% confidence and a margin of error of 3 percentage points?

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Answer: 1,068 — with p = 0.5, n = 1.96² × 0.25/0.03² = 1,067.1, rounded up. Whoever gets 267 used 0.25 in place of σ: 0.25 is the variance p(1 − p), and the standard deviation is its square root, 0.5.

Why, without prior information, is p = 0.5 used to calculate the sample size?

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Answer: Because p = 0.5 gives the largest value of p(1 − p) and therefore the safest sample — 0.5 × 0.5 = 0.25 is the maximum. If the true proportion is different, the real margin will be smaller than planned, never larger.

Exercises

1) A researcher wants to estimate the proportion of rats that develop a certain type of tumour when exposed to radiation. He wants his estimate not to deviate from the true proportion by more than 0.02, with a probability of at least 90%. How many animals does he need to examine?

2) Using the data from the previous exercise, how could the sample size be reduced using the additional information that, in general, this type of radiation does not affect more than 20% of rats?

3) Before an election, a party wants to estimate the proportion of voters who favour its candidate. Determine the sample size needed for the estimation error to be at most 0.01, with a probability of 80%.

4) If, in the survey from the previous exercise, 55% of the voters interviewed favoured the candidate, build a 95% confidence interval for the proportion of the candidate's voters.

5) A scientist wants to estimate the proportion p of individuals with a certain disease in a region. He wants the probability that his estimate does not deviate from the true value of p by more than 0.02 to be at least 95%. What should the sample size be? Another scientist discovers that the disease is related to the concentration of substance A in the blood, and that every individual with a concentration of A below 1.488 mg/cm³ is considered ill. The concentration of A is known to be normally distributed with a standard deviation of 0.4 mg/cm³ and a mean greater than 2.0 mg/cm³. Can this information be used by the first scientist to reduce the sample size? What is the ideal size?

6) An opinion research institute wanted to assess what TV viewers in a region think of a certain sports commentator. It interviewed 380 viewers, randomly selected, and found that 180 wanted the commentator removed from TV. Determine a 90% confidence interval for p, the proportion of viewers in favour of the removal.

Answer key

1) 1,692 (z = 1.645 and p = 0.5: n = 1.645² × 0.25/0.02² = 1,691.3). With the table z rounded to 1.64, the result is 1,681.

2) 1,083 (using p = 0.2: n = 1.645² × 0.2 × 0.8/0.02² = 1,082.4). With z = 1.64, it gives 1,076.

3) 4,106 (z = 1.282 and p = 0.5: n = 1.282² × 0.25/0.01² = 4,105.9). With z = 1.28, it gives 4,096.

4) 95% CI from 53.5% to 56.5% (0.55 ± 1.96 × √(0.55 × 0.45/4,096) = 0.55 ± 0.015).

5) Without the information: 2,401. With it: the disease affects at most P(Z below (1.488 − 2.0)/0.4 = −1.28) ≈ 0.10, so using p = 0.10: n = 1.96² × 0.1 × 0.9/0.02² = 864.4, that is, 865.

6) p̂ = 180/380 = 0.474; 90% CI from 43.2% to 51.6% (0.474 ± 1.645 × √(0.474 × 0.526/380) = 0.474 ± 0.042).

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