Sample Size Calculator (Surveys and Margin of Error)

There are two ways to use this calculator. ① Enter the confidence level, proportion and margin of error to find the sample size you need. ② Enter the confidence level, proportion and sample size to find the margin of error. Enter the population size only if you know it.

Confidence level (%)

Expected proportion (%)

Population size N (leave blank if unknown or very large)

① Margin of error ±ε (%)

② Sample size n

Enter all values as percentages (for example, "95" for a 95% confidence level and "5" for a ±5% margin of error). Fill in only one of "margin of error" and "sample size", depending on what you want to find.

Result

Enter the confidence level and other values in the fields on the left and press "Calculate". The result will appear here.

[ CLOSE ]

What you can do on this page
What is this calculation used for?
How to Use
Formula
Symbols and terms
- Symbols
- Terms
Good to know before you start
How to calculate it in Excel
How to calculate it in Google Sheets
How to calculate it in Python
How to write it in LaTeX and other math languages (copy and paste)
How to have ChatGPT do the calculation
DataChef Features
Related Features
NumberChef Calculators List

What you can do on this page

"How many people do I need to survey for a 95% confidence level and a ±5% margin of error?" Get the sample size your survey needs on the spot
The other way around, "How accurate, plus or minus, is a survey of 1,000 people?" Find the margin of error on the same page
If you know the size of the population (500 employees, 1,000 members, and so on), the finite population correction narrows it down to the number you really need
A plain-language explanation of the formulas and copy-and-paste formulas for Excel, Google Sheets and Python are all on this page

This calculation assumes that respondents are chosen from the population at random, and that the survey measures a proportion (%), such as "yes" or "no" answers. Planning a survey that estimates a mean (such as average income) needs a different formula that uses the standard deviation.

What is this calculation used for?

Understanding the "±3 points" in polls and election coverage

National polls in the news usually survey about 1,000 to 2,000 randomly chosen people out of more than 200 million adults. At a 95% confidence level, a ±3% margin of error and a 50% proportion, the required sample size is \(1.96^2 \times 0.25 \div 0.03^2 \approx 1067.1\) (about 1,068 people). This formula answers the question, "Can about 1,000 people really tell us what hundreds of millions think?"
The other way around, you can see that if "45% versus 47% approval" is within a ±3-point margin of error, you cannot say for sure that there is a statistical difference.

Planning customer and employee satisfaction surveys (business)

"How many of our 1,000 members do we need to ask?" is a classic case for the finite population correction. At a 95% confidence level, a ±5% margin of error and a 50% proportion, the value before correction, 384.16, drops to \(384.16 \div (1 + 383.16 \div 1000) \approx 277.7\) (278 people after rounding up).
If you decide in advance how many responses are enough for the precision you need, before sending reminder after reminder to everyone, you can balance the cost and the precision of the survey with numbers.

Sample inspections in manufacturing (quality control)

This formula is also used to estimate the defect rate of mass-produced parts without inspecting every one. If past records suggest a defect rate of about 10%, a 90% confidence level and a ±4% margin of error give \(1.65^2 \times 0.1 \times 0.9 \div 0.04^2 \approx 153.1\), so a sample of about 154 parts is enough.
The example also shows that the farther the proportion is from 50%, the smaller the spread \(\hat{p}(1-\hat{p})\), and the fewer items you need.

Planning the number of participants in medical and epidemiological studies

In an epidemiological study of "the share of people in an area who have this condition", the study plan must state how many participants are needed. If earlier studies suggest a prevalence of about 20%, a 95% confidence level and a ±3% margin of error give \(1.96^2 \times 0.2 \times 0.8 \div 0.03^2 \approx 683.0\), about 683 people.
Real clinical research also uses more precise methods that fit the study design (such as power analysis). This formula is the basic form they start from.

TV ratings come from a sample of homes - how big is the error?

TV ratings are estimated from a panel of sample households, not from every home. As a simplified example (real ratings methods are more complex), take a panel of 2,700 households and a show watched in 10% of them. At a 95% confidence level, the margin of error is \(1.96 \times \sqrt{0.1 \times 0.9 \div 2700} \approx 0.0113\) (about ±1.1%).
You can then see that "the gap between a 9.5% and a 10.5% rating may be within the margin of error", and read the numbers in the news one step deeper.

Formula

Required sample size (when the population is very large)

Standard notation (the usual math form)

\(n_0\) \(=\) \(z^2\) \(\times\) \(\hat{p}\) \(\times\) \((1-\hat{p})\) \(\div\) \(\varepsilon^2\)

In words (symbols replaced with words)

⑤ \(n_0\): calculated value \(=\) ① z-score \(z\) squared \(\times\) ② \(\hat{p}\): expected proportion \(\times\) ③ the rest, \((1-\hat{p})\) \(\div\) ④ margin of error \(\varepsilon\) squared

The formula in words

① Take the square of the z-score \(z\) set by the confidence level

② multiply it by the \(\hat{p}\): expected proportion

③ and by the rest of the proportion, \((1-\hat{p})\)

④ divide by the square of the margin of error \(\varepsilon\)

⑤ and you get the calculated required sample size \(n_0\)

Quick example

With a 95% confidence level (z = 1.96), a proportion of 50% (0.5) and a margin of error of ±5% (0.05), the sample size you need is

\(n_0\): calculated value \(=\) 1.96 squared (3.8416) \(\times\) proportion (0.5) \(\times\) rest (1 − 0.5 = 0.5) \(\div\) 0.05 squared (0.0025)

\(3.8416 \times 0.5 \times 0.5 \div 0.0025 = 3.8416 \times 0.25 \div 0.0025 = 384.16\)

Key idea

Round the result up to get a number of people (384.16 → 385 people). If you rounded down, the margin of error would end up slightly larger than your target. If you have no idea of the proportion in advance, enter 50%. The spread \(\hat{p}(1-\hat{p})\) is largest when \(\hat{p}\) is 50%, so this gives the largest (safest) estimate of the number of people. The usual z-scores for each confidence level are 80% → 1.28, 85% → 1.44, 90% → 1.65, 95% → 1.96, 98% → 2.33 and 99% → 2.58 (the calculator on this page also uses these customary rounded values; statistics textbooks sometimes give 90% → 1.645 and 99% → 2.576).

Finite population correction for the sample size (when you know the population size \(N\))

Standard notation (the usual math form)

\(n\) \(=\) \(n_0\) \(\div\) \((\) \(1\) \(+\) \((n_0 - 1)\) \(\div\) \(N\) \()\)

In words (symbols replaced with words)

⑤ \(n\): corrected value \(=\) ④ \(n_0\): value before correction \(\div\) \((\) ③ \(1\) \(+\) ① value before correction minus 1, \((n_0 - 1)\) \(\div\) ② \(N\): population size \()\)

The formula in words

① Take the value before correction minus 1, \((n_0-1)\)

② divide it by the \(N\): population size

③ and add \(1\)

④ Divide the \(n_0\): value before correction by that number

⑤ and you get the corrected required sample size \(n\)

Quick example

If the value before correction is 384.16 (95% confidence level, 50% proportion, ±5% margin of error) and the population is 500 people, then

corrected value \(=\) value before correction (384.16) \(\div\) \((\) 1 \(+\) 384.16 − 1 = 383.16 \(\div\) population (500 people) \()\)

\(384.16 \div \left( 1 + \dfrac{383.16}{500} \right) = 384.16 \div 1.76632 \approx 217.49\)

Key idea

When the population is small, this correction cuts the number of people you need a lot (in this example, 385 people → 218 after rounding up). If you can study a large share of the population directly, the result is that much more certain. The divisor "\(1 + (n_0 - 1) \div N\)" shows how strong the correction is. If the population \(N\) is much larger than the calculated value \(n_0\), \((n_0 - 1) \div N\) is almost 0, so the divisor is almost 1 (dividing changes nothing, the same as no correction). The smaller the population, the larger the divisor gets above 1, and the fewer people you need. The "1" is the baseline of "divide by 1 (leave it as is) when there is no correction". On the other hand, if the population is much larger than the calculated value, the correction hardly changes anything, so you can leave "population size" in the calculator blank (no correction). As in the first formula, round up (217.49 → 218 people).

Margin of error (when the population is very large)

Standard notation (the usual math form)

\(\varepsilon\) \(=\) \(z\) \(\times\) \(\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}\)

In words (symbols replaced with words)

③ \(\varepsilon\): margin of error \(=\) ① \(z\): z-score \(\times\) ② square root of \(\dfrac{\text{spread of the proportion }\hat{p}(1-\hat{p})}{\text{sample size }n}\)

The formula in words

① Take the z-score \(z\) set by the confidence level

② multiply it by the square root of \(\dfrac{\text{spread of the proportion }\hat{p}(1-\hat{p})}{\text{sample size }n}\)

③ and you get the \(\varepsilon\): margin of error

Quick example

At a 95% confidence level (z = 1.96), for a survey where 60% of 100 people answered "yes", the margin of error is

\(\varepsilon\): margin of error \(=\) z-score (1.96) \(\times\) square root of "0.6 × 0.4 ÷ 100"

\(1.96 \times \sqrt{0.6 \times 0.4 \div 100} = 1.96 \times \sqrt{0.0024} \approx 1.96 \times 0.0490 \approx 0.0960\ \ (9.60\%)\)

Key idea

For \(\hat{p}\) in the spread \(\hat{p}(1-\hat{p})\), use the proportion you actually got in the survey (the sample proportion). The sample size \(n\) is in the denominator inside the square root, so to cut the error in half you need 4 times the sample, and to cut it to a third you need 9 times. "Adding more sample helps only slowly" is an important property of this formula.

Finite population correction for the margin of error (when you know the population size \(N\))

Standard notation (the usual math form)

\(\varepsilon\) \(=\) \(\varepsilon_0\) \(\times\) \(\sqrt{\dfrac{N-n}{N-1}}\)

In words (symbols replaced with words)

③ \(\varepsilon\): corrected margin of error \(=\) ① \(\varepsilon_0\): margin of error before correction \(\times\) ② square root of the correction factor \(\dfrac{\text{rest of the population }(N-n)}{\text{population size minus 1 }(N-1)}\)

The formula in words

① Take the margin of error before correction, \(\varepsilon_0\) (from the third formula)

② multiply it by the square root of the correction factor \(\dfrac{\text{rest of the population }(N-n)}{\text{population size minus 1 }(N-1)}\)

③ and you get the \(\varepsilon\): corrected margin of error

Quick example

If you survey 200 of a population of 1,000 people and the margin of error before correction is 6.35%, the corrected margin of error is

\(\varepsilon\): corrected margin of error \(=\) margin of error before correction (6.35%) \(\times\) square root of "800 ÷ 999"

\(6.35 \times \sqrt{\dfrac{1000 - 200}{1000 - 1}} = 6.35 \times \sqrt{0.8008} \approx 6.35 \times 0.8949 \approx 5.68\,(\%)\)

Key idea

If you study a large part of the population, the result is that much more certain. The correction factor shows this: as the sample size \(n\) gets close to the population size \(N\), it gets close to 0 (a census that covers everyone has zero error), and when \(N\) is very large it is almost 1 (the same as no correction). You can multiply the margin of error as a percentage (the correction factor is a number with no units that is 1 or less).

The required sample size is "z-score squared × spread of the proportion ÷ margin of error squared", rounded up. The margin of error is "z-score × √(spread ÷ sample size)". When you know the population size, the finite population correction lowers the number of people you need (and makes the error smaller).

Symbols and terms

Symbols

\(n\)	en	The sample size. For a survey, it is the number of people whose answers you collect; for an inspection, the number of items you check.
\(n_0\)	n sub zero	The calculated required sample size when the population is treated as very large. It is the value before the finite population correction.
\(z\)	zee	The z-score (a value on the standard normal distribution) for the confidence level. The customary values are 1.96 for 95% and 2.58 for 99%.
\(\hat{p}\)	p-hat	The proportion (such as the share of "yes" answers you want to know). To find the sample size, use your expected value; to find the margin of error, use the proportion you actually got in the survey (the sample proportion). Write it as a decimal from 0 to 1 (0.5 for 50%).
\(\varepsilon\)	epsilon	The margin of error, the gap you allow between the survey result and the true value. It is shown with ±, as in "±5%".
\(N\)	capital N	The population size, the number of people or items in the whole group you want to study (example - all 500 employees).

Terms

population	The whole group you really want to know about, such as "every voter in the country" or "all 500 employees".
sample	The part of the population that you pick out to actually study. The number of people or items in the sample is called the sample size.
random sampling	Choosing the sample so that everyone in the population has the same chance of being picked, like drawing names from a hat. The formulas on this page assume random sampling and do not work for surveys where only the people who want to answer do.
population proportion	The proportion you really want to know, such as the share of "yes" in the whole population. The proportion you get from a sample is called the sample proportion to tell them apart.
confidence level	How certain you can be: if you repeated the same survey many times, the percentage of times the result would land within the margin of error. 95% is used most often.
confidence interval	The range you get when you show a survey result with some width, as in "52% ± 3% (49% to 55%)". The margin of error ±ε on this page is that width.
margin of error	The gap you allow between the survey result and the true value. It is what news polls mean by "a margin of error of ±3 points".
finite population correction	A correction that lowers the required sample size or the margin of error when you know the population has a limited size. It reflects the fact that the more of the population you study, the more certain the result.

Good to know before you start

Here is what helps you use the calculation on this page with real understanding, not just by pressing the button.

Percents and ratios (Grade 6)	Being able to switch between percents and decimals, such as 50% = 0.5 and 5% = 0.05 Knowing that "whole × percent = part"
Square roots (Grade 8)	Knowing a square root as "the number that gives this number when squared", as in \(\sqrt{0.0025} = 0.05\) Having a feel for how square roots scale: when the number inside becomes a quarter, the square root becomes a half
Random sampling (Grade 7)	Knowing the difference between a census and a sample survey, and what "population" and "sample" mean Knowing that unless the sample is chosen at random, you cannot correctly estimate what the population looks like
The normal distribution and statistical inference (high school statistics / AP Statistics)	Knowing that sample proportions vary in a shape close to a normal distribution (the idea of the central limit theorem) Knowing the confidence interval idea that a 95% confidence level matches a z-score of 1.96

How to calculate it in Excel

Copy the whole table below and paste it into cell A1 in Excel. It works as is.

Table to find the required sample size (very large population)

z-score (1.96 for 95%)	1.96
Expected proportion (%)	50
Margin of error (%)	5
Calculated value n0	=B1^2(B2/100)(1-B2/100)/(B3/100)^2
Required sample size (rounded up)	=ROUNDUP(B4,0)

Table for the finite population correction of the sample size

z-score (1.96 for 95%)	1.96
Expected proportion (%)	50
Margin of error (%)	5
Population size N	500
Value before correction n0	=B1^2(B2/100)(1-B2/100)/(B3/100)^2
Required sample size (corrected, rounded up)	=ROUNDUP(B5/(1+(B5-1)/B4),0)

Table to find the margin of error (very large population)

z-score (1.96 for 95%)	1.96
Proportion from the survey (%)	60
Sample size n	100
Margin of error (%)	=B1SQRT(B2/100(1-B2/100)/B3)*100

Table for the finite population correction of the margin of error

z-score (1.96 for 95%)	1.96
Proportion from the survey (%)	30
Sample size n	200
Population size N	1000
Margin of error before correction (%)	=B1SQRT(B2/100(1-B2/100)/B3)*100
Corrected margin of error (%)	=B5*SQRT((B4-B3)/(B4-1))

After pasting, the upper cells (from B1) are your inputs and the last row shows the calculated result.
In the first table, B4 shows 384.16 and B5 shows 385. The answer in the second table is 218, in the third about 9.60 (%), and in the fourth about 5.68 (%).
"^" is a power (squared), "SQRT" is the square root, and "ROUNDUP" is the function for rounding up.
Note that the calculator on this page adds one more for safety even when the calculated value is exactly a whole number, so for such round values it can differ from Excel's ROUNDUP by 1.

How to calculate it in Google Sheets

Copy the whole table below and paste it into cell A1 in Google Sheets. It works as is.

Table to find the required sample size (very large population)

z-score (1.96 for 95%)	1.96
Expected proportion (%)	50
Margin of error (%)	5
Calculated value n0	=B1^2(B2/100)(1-B2/100)/(B3/100)^2
Required sample size (rounded up)	=ROUNDUP(B4,0)

Table for the finite population correction of the sample size

z-score (1.96 for 95%)	1.96
Expected proportion (%)	50
Margin of error (%)	5
Population size N	500
Value before correction n0	=B1^2(B2/100)(1-B2/100)/(B3/100)^2
Required sample size (corrected, rounded up)	=ROUNDUP(B5/(1+(B5-1)/B4),0)

Table to find the margin of error (very large population)

z-score (1.96 for 95%)	1.96
Proportion from the survey (%)	60
Sample size n	100
Margin of error (%)	=B1SQRT(B2/100(1-B2/100)/B3)*100

Table for the finite population correction of the margin of error

z-score (1.96 for 95%)	1.96
Proportion from the survey (%)	30
Sample size n	200
Population size N	1000
Margin of error before correction (%)	=B1SQRT(B2/100(1-B2/100)/B3)*100
Corrected margin of error (%)	=B5*SQRT((B4-B3)/(B4-1))

The same formulas as in Excel work as is. Copy the whole table, paste it into cell A1, and replace the numbers in column B with the values for your own survey.

How to calculate it in Python

import math

# Quick table from confidence level (%) to z-score (customary rounded values)
z_table = {80: 1.28, 85: 1.44, 90: 1.65, 95: 1.96, 98: 2.33, 99: 2.58}

# ---- Way 1: find the required sample size ----
confidence_level = 95    # confidence level (%)
population_ratio = 50    # expected proportion (%)
margin_of_error = 5      # margin of error (%)
population_size = None   # population size (None if unknown or very large)

z = z_table[confidence_level]
p = population_ratio / 100
e = margin_of_error / 100

raw_size = z ** 2 * p * (1 - p) / e ** 2   # value before correction
if population_size is not None:
    raw_size = raw_size / (1 + (raw_size - 1) / population_size)   # finite population correction
required_size = math.floor(raw_size) + 1   # smallest whole number above the calculated value (rounded up for safety)
print(f"Required sample size: {required_size}")

# ---- Way 2: find the margin of error ----
sample_size = 1000       # sample size actually surveyed
sample_ratio = 50 / 100  # proportion from the survey (0.5 for 50%)

error_ratio = z * math.sqrt(sample_ratio * (1 - sample_ratio) / sample_size)
print(f"Margin of error: ±{error_ratio * 100:.2f}%")

Runs with the standard library only. Way 1 calculates the required sample size (385 with the example values) and way 2 calculates the margin of error (±3.10% with the example values). Replace the numbers at the top with the conditions of your own survey and run it.

How to write it in LaTeX and other math languages (copy and paste)

Required sample size (when the population is very large)

n₀ = z² × p̂(1 − p̂) ÷ ε²

n_0 = \dfrac{z^2 \, \hat{p}(1-\hat{p})}{\varepsilon^2}

<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
  <mrow>
    <msub><mi>n</mi><mn>0</mn></msub>
    <mo>=</mo>
    <mfrac>
      <mrow>
        <msup><mi>z</mi><mn>2</mn></msup>
        <mover accent="true"><mi>p</mi><mo>^</mo></mover>
        <mo>(</mo><mn>1</mn><mo>&#x2212;</mo><mover accent="true"><mi>p</mi><mo>^</mo></mover><mo>)</mo>
      </mrow>
      <msup><mi>&#x3B5;</mi><mn>2</mn></msup>
    </mfrac>
  </mrow>
</math>

n_0 = (z^2 hat(p)(1 - hat(p)))/(epsilon^2)

z^2*p*(1 - p)/err^2

n0 := z^2*p*(1 - p)/err^2;

n0 = z^2*p*(1 - p)/err^2;

n_0 = (z^2 p̂(1 − p̂))/ε^2

Finite population correction for the sample size (when you know the population size \(N\))

n = n₀ ÷ (1 + (n₀ − 1) ÷ N)

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

<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
  <mrow>
    <mi>n</mi>
    <mo>=</mo>
    <mfrac>
      <msub><mi>n</mi><mn>0</mn></msub>
      <mrow>
        <mn>1</mn><mo>+</mo>
        <mfrac>
          <mrow><msub><mi>n</mi><mn>0</mn></msub><mo>&#x2212;</mo><mn>1</mn></mrow>
          <mi>N</mi>
        </mfrac>
      </mrow>
    </mfrac>
  </mrow>
</math>

n = n_0/(1 + (n_0 - 1)/N)

n0/(1 + (n0 - 1)/nPop)

n := n0/(1 + (n0 - 1)/N);

n = n0/(1 + (n0 - 1)/N);

n = n_0/(1 + (n_0 − 1)/N)

Margin of error (when the population is very large)

ε = z × √(p̂(1 − p̂) ÷ n)

\varepsilon = z \sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}

<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
  <mrow>
    <mi>&#x3B5;</mi>
    <mo>=</mo>
    <mi>z</mi>
    <msqrt>
      <mfrac>
        <mrow>
          <mover accent="true"><mi>p</mi><mo>^</mo></mover>
          <mo>(</mo><mn>1</mn><mo>&#x2212;</mo><mover accent="true"><mi>p</mi><mo>^</mo></mover><mo>)</mo>
        </mrow>
        <mi>n</mi>
      </mfrac>
    </msqrt>
  </mrow>
</math>

epsilon = z sqrt((hat(p)(1 - hat(p)))/n)

z*Sqrt[p*(1 - p)/n]

err := z*sqrt(p*(1 - p)/n);

err = z*sqrt(p*(1 - p)/n);

ε = z√((p̂(1 − p̂))/n)

Finite population correction for the margin of error (when you know the population size \(N\))

ε = ε₀ × √((N − n) ÷ (N − 1))

\varepsilon = \varepsilon_0 \sqrt{\dfrac{N-n}{N-1}}

<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
  <mrow>
    <mi>&#x3B5;</mi>
    <mo>=</mo>
    <msub><mi>&#x3B5;</mi><mn>0</mn></msub>
    <msqrt>
      <mfrac>
        <mrow><mi>N</mi><mo>&#x2212;</mo><mi>n</mi></mrow>
        <mrow><mi>N</mi><mo>&#x2212;</mo><mn>1</mn></mrow>
      </mfrac>
    </msqrt>
  </mrow>
</math>

epsilon = epsilon_0 sqrt((N - n)/(N - 1))

err0*Sqrt[(nPop - n)/(nPop - 1)]

err := err0*sqrt((N - n)/(N - 1));

err = err0*sqrt((N - n)/(N - 1));

ε = ε_0 √((N − n)/(N − 1))

How to have ChatGPT do the calculation

You are a calculation assistant for statistical surveys. Do the following calculation by actually running Python code, and base your answer only on the numbers from the execution result (do not answer by mental math or guessing).

A service with 1,000 members is running a satisfaction survey. Use a 95% confidence level (z-score 1.96), an expected proportion of 50%, and a margin of error of ±5%.
Find each of the following:
1. The required sample size when the population is treated as very large (the calculated value and the number of people after rounding up)
2. The required sample size with the finite population correction for a population of 1,000
3. The margin of error (with the finite population correction, in %) if only 200 people respond

Show the formulas you used and the numbers from the execution result.

How to Use

1

Enter your numbers

Type the numbers you want to calculate with into the input fields
2

Calculate

Press the "Calculate" button
3

Check the result

The result appears on the spot. The same page also explains the idea behind the calculation and the formula