How Big a Sample Do I Need?
Copyright © 2007–2023 by Stan Brown, BrownMath.com
Copyright © 2007–2023 by Stan Brown, BrownMath.com
But before you perform the study, how can you decide how big a sample you need so that your confidence interval will have your desired margin of error or less?
The answer is that you take the formula for the margin of error, rearrange it algebraically to solve for the sample size, compute, and round up. This page shows the formulas for some common cases, with examples.
If you have a TI-83 or TI-84, you can skip the calculations in this Web page and download a program to do them. See MATH200A Program part 5.
For more advanced treatment of more cases:
If you know the standard deviation σ of the population, and you want to estimate the mean μ to within a given margin of error E in a 1−α confidence interval, here’s how to find the required sample size n:
E = z_{α/2} · σ/√n transforms to n = (z_{α/2} · σ/E)²
Example 1: You want to estimate the average hourly output of a machine to within ±1.5, with 90% confidence. Based on historical data, you have reason to believe that the standard deviation of the machine’s hourly output is 6.2. How large a sample do you need?
Solution: Note first that this is not a realistic situation. It’s pretty unlikely that you would know the standard deviation of a population but not know the mean of that population. However, statistics texts always begin with this case because it’s the simplest way to demonstrate the principles. You leave Perfectland and enter Realityville in the other cases. With that said—
Comments | Computation |
---|---|
It’s good practice to start any problem by writing down what you know and what you need, with symbols. | Given: E = 1.5,
σ = 6.2, 1−α = 0.90.
Wanted: sample size n |
The formula wants z_{α/2}. How do you compute it? Begin by finding α/2. | 1−α = 0.90 ⇒
α = 0.10 ⇒
α/2 = 0.05
Since α/2 = 0.05, z_{α/2} = z_{0.05} |
z_{rtail} is the
critical z, or the z score that divides the normal curve
leaving a right-hand tail with an area of rtail. You compute it
on your TI-83/84/89 as
invNorm(1− rtail) .
| z_{0.05} = invNorm(1−0.05) ≈ 1.6449 |
Now you have all the pieces. Don’t use the
rounded value of z_{α/2}, but use [2nd (-) makes ANS ] to keep
full precision.
| n = [ z_{α/2} × σ ÷ E ]² = (ANS×6.2÷1.5)² = 46.2227... → 47 |
Answer: Given a population standard deviation of 6.2 units per hour, if you have a sample size ≥47 the margin of error in a 90% confidence interval will be ≤1.5 units per hour.
Why do we round up? After computing 46.2227, why not report a sample size of 46? Well, the computation shows that a sample size of exactly 46.2227... would give a margin of error of exactly 1.5. If you go slightly lower, to 46, the margin of error will be slightly higher than 1.5. Since the sample size must be a whole number, 46 or 47, and your margin of error must not exceed 1.5, you have to choose the slightly higher number 47, which will give a margin of error slightly less than 1.5.
Note: Many basic statistics courses skip the material in this section and estimate sample sizes using a z distribution, so the material in this section might be an advanced extra for you. Check your course requirements.
This is the realistic case for estimating a population mean. Usually you don’t know the standard deviation of the population, so you have to use Student’s t distribution instead of the normal (z) distribution. You estimate the standard deviation of the population from the standard deviation of a sample obtained in a prior study or a small pilot study. Here is the formula for sample size:
E = t_{df,α/2} · s/√n transforms to n = t_{df,α/2} · s/E)²
There’s a certain element of Catch-22 in this formula for n. You don’t know n, so you don’t know the degrees of freedom df either and you can’t compute the critical t for the formula. How do you get around this?
Use what NIST/SEMATECH calls an iterative method. First compute the formula using z_{α/2} instead of t_{df,α/2}. Then, when you have a preliminary sample size determined by (ab)using z in this way, recompute the formula using that sample size minus 1 for df. The two numbers should not be very different, since t is generally not very different from z; but if they are, you can use the second number to compute t once again.
Caution: It may happen that the formula gives a sample size less than 30. But remember that your sample size must always be at least 30 unless you have good reason to believe that the underlying population is normally distributed.
See also: Sample Sizes Required in NIST/SEMATECH e-Handbook of Statistical Methods: scroll down to “More often we must compute the sample size with the population standard deviation being unknown”
Let’s illustrate the method using a modified form of the previous example.
Example 2: You want to estimate the average hourly output of a machine to within ±1.5, with 90% confidence. A small pilot study finds a sample standard deviation of the machine’s hourly output is 6.2. How large a sample do you need?
Solution: Use z instead of t to make a preliminary estimate, then recompute with t.
Comments | Computation |
---|---|
Marshal your data. | Given: E = 1.5,
s = 6.2, 1−α = 0.90.
Wanted: sample size n |
The formula wants t_{df,α/2}, but we approximate with z_{α/2}. Begin by finding α/2. | 1−α = 0.90 ⇒
α = 0.10 ⇒
α/2 = 0.05
Since α/2 = 0.05, z_{α/2} = z_{0.05} |
z_{0.05} is the critical z-score that divides the normal distribution such that the area of the right-hand tail is 0.05, and therefore the area of the left-hand tail is 1−0.05. | z_{0.05} = invNorm(1−0.05) ≈ 1.6449 |
Now you have all the pieces you need for the preliminary
sample size. Don’t use the
rounded value of z_{α/2}, but use [2nd (-) makes ANS ] to keep
full precision.
| n = [ z_{α/2} × s ÷ E ]² = (ANS×6.2÷1.5)² = 46.2227... → 47 |
Your preliminary sample size is 47, and next you use that to
compute t. df = n−1 = 46, so you need
t_{46,0.05}. On the TI-84 you can use the invT
function.
| t_{46,0.05} = 1.67866 |
Now recompute the formula using the t value. Remember, always round sample size up. | n = [ t_{df,α/2} × s ÷ E ]² = (ANS×6.2÷1.5)² = 48.142... → 49 |
Answer: Given a sample standard deviation of 6.2 units per hour, if you have a sample size ≥49 the margin of error in a 90% confidence interval will be ≤1.5 units per hour.
Remark: The sample size of 49 is a bit larger than the Case 0 sample size of 47. This makes sense. When you don’t know the standard deviation of the population, you have to use the t distribution. Student’s t is more spread out than z, so the confidence intervals are a bit wider, so you have to use a larger sample to keep the confidence interval to the same width.
For binomial data with true proportion p, the population standard deviation is σ = √p(1−p). Even though you don’t know p, the value p̂(1−p̂) from your sample — or your prior estimate, if you don’t have a sample — will be close to the true value p(1−p) in the population, because the product p(1−p) doesn’t vary much as p varies.
Replacing p with p̂ in a formula may seem like cheating, but n this case it’s not, because p(1−p) varies a lot less than p on its own. For instance, suppose the true population proportion is 45% but your estimate is 35%. The true p(1−p) is 0.45×0.55 = 0.2475, and your estimate is 0.35×0.65 = 0.2275. The difference between 0.2475 and 0.2275 is a lot less than the difference between 0.45 and 0.35.
Therefore you can use a z function, and the formulas are the same as Case 0 with √p̂(1−p̂) substituted for σ:
E = z_{α/2} · √p̂(1−p̂)/n transforms to n = (z_{α/2}/E)² · p̂(1−p̂)
If you don’t have a sample or any credible estimate, use p̂ = (1−p̂) = 0.5. This is the conservative procedure because the product p̂(1−p̂) takes its highest value when p̂ = 0.5. The conservative procedure may give you a sample size larger than necessary, but you can be sure your sample won’t be too small, forcing you to throw out your survey and start over.
Caution: The sample must not exceed 10% of the population. Another way to look at that is that 10 times sample size must be less than or equal to population size.
Example 3: What percent of the voters would vote for your candidate if the election were held today? You want 95% confidence in your answer, with a margin of error no more than 3.5%. Last month’s poll showed your candidate had 42% support. How many voters do you need to survey?
Comments | Computation |
---|---|
Marshal your data. Caution! 3.5% is 0.035, not 0.35. | Given: 1−α = 0.95,
E = 0.035, p̂ = 0.42
Wanted: sample size n |
To find z_{α/2}, first find α/2. | 1−α = 0.95 ⇒ α = 0.05 ⇒ α/2 = 0.025 |
z_{α/2} = z_{0.025}, the critical z for a right-hand tail area of 0.025. That’s invNorm(1−.025). | |
Divide by E and square. You’re going to
chain calculations so that you don’t have to re-enter
any of your intermediate numbers. Press [/ ], and notice how
the calculator responds Ans to let you know it’s
using the previous answer. Enter .035 for E
and press [ENTER ]; that gives you the result of the fraction. Press
[x² ] [ENTER ] to square it.
| |
The last link in the chain is multiplying by p̂ and then by (1−p̂). Your result is 764. Remember, always round sample size up, regardless of the decimal part. |
Answer: To find a 95% CI with a margin of error no more than ±3.5 percentage points, where the true population proportion is around 42%, you must survey at least 764 people.
10×764 = 7640; presumably the electorate is larger than that.
Example 4: Suppose you’re planning your first poll, and you have no idea of your candidate’s level of support. How big a sample would you need to be sure of a margin of error no more than 3.5% in a 95% CI?
Solution: Compute z_{α/2} = 1.9600 as in the previous example. But this time use p̂ = 0.5 since you have no estimate for p.
n = ( z_{α/2} ÷ E )² p̂ (1−p̂) = (invNorm(1-.025)/.035)² × 0.42 × (1−0.42) = 783.971... → 784
Answer: To find an 95% CI with a margin of error no more than ±3.5 percentage points, where you have no idea of the true population proportion, you must survey at least 784 people.
When you’re comparing two population proportions, it’s perfectly legitimate to have different-sized samples. The formula for margin of error, below left, is just an extension of the formula for one population proportion.
But when you’re planning sample size, you can’t solve one equation for two variables n_{1} and n_{2}. (If you had a reason to choose some particular value for one of them, you could solve for the other one.) You can solve for sample size if you decide to use the same size for both samples.
transforms to
n = n_{1} =
n_{2} =
[p̂_{1}(1−p̂_{1}) + p̂_{2}(1−p̂_{2})] · (z_{α/2}/E)²
For the reasons given above, if you have any prior estimates for the population proportions p_{1} and p_{2} you should use them; otherwise use 0.5.
Example 5: You’d like to know how your candidate’s support differs between men and women. You know that overall support is 42%. How many of each sex must you survey to answer the question with 90% confidence and a margin of error no more than 3%?
Comments | Computation |
---|---|
Marshal your data. (Caution! 3% is 0.03 not 0.3.) | Given: 1−α = 0.90, E = 0.03
Wanted: sample size n=n_{1}=n_{2} |
Do you have an estimate of p_{1} and p_{2}? Yes, since the overall support is 42% you expect that men’s and women’s support is not too different from that. (You do expect p_{1} and p_{2} are somewhat different, or you wouldn’t be doing the survey. But remember from one population proportion that p(1−p) doesn’t vary much when p varies.) | Prior: p̂_{1} = 0.42 and
1−p̂_{1} = 0.58
p̂_{2} = 0.42 and 1−p̂_{2} = 0.58 |
Compute z_{α/2} in the usual way. | 1−α = 0.90 ⇒
α = 0.10 ⇒ α/2 = 0.05
z_{0.05} = invNorm(1−0.05) ≈ 1.6449 |
Finish, using the unrounded value of z. Always remember to round sample sizes up. | [p̂_{1}(1−p̂_{1}) + p̂_{2}(1−p̂_{2})] [z_{α/2}÷E]² = (0.42×0.58+0.42×0.58)×(ANS÷0.03)² = 1464.60... → 1465 |
Answer: To find a 90% CI for the difference in your candidate’s support between men and women, with margin of error no more than 3%, you must survey at least 1465 men and at least 1465 women.
Remark: You might wonder why the samples must be so large. After all, to estimate one population proportion to ±3% in a 90% CI, with prior estimate p̂ = 42%, a sample of 752 is enough. (Check it!) Why do you need over 2900 people in two groups for the same margin of error?
The answer is that it’s not the same margin of error. If you surveyed 752 men and 752 women you’d have confidence intervals of ±3% for each, but that’s an overall margin of error of ±6% — think that the true proportion might be near the bottom of one group’s interval and near the top of the other group’s, or vice versa. (It’s not quite that simple, but that’s the basic idea.) To bring down that margin of error, you have to increase the sample size.
Example 6: Let’s modify the previous example. Suppose you have reason to believe your candidate appeals more strongly to women, with a gap of about 10%? That means you estimate men’s support at 37% and women’s support at 47%. p̂_{1} = 0.37, 1−p̂_{1} = 0.63, p̂_{2} = 0.47, 1−p̂_{2} = 0.52. Your required sample size becomes
z_{0.05} = invNorm(1−0.05) ≈ 1.6449
[p̂_{1}(1−p̂_{1}) + p̂_{2}(1−p̂_{2})] [z_{α/2}÷E]² = (0.37×0.63+0.47×0.53)×(ANS÷0.03)² = 1449.57... → 1450
Answer: For a 90% CI with margin of error ≤3%, when you think one population’s proportion is 37% and the other’s is 47%, you need a sample of at least 1450 from each group.
For χ² tests, the requirements are that all of the expected counts must be ≥5. The expected count for each category is sample size n times the model proportion in that category, so to find the the necessary sample size you divide that minimum expected value of 5 by the smallest proportion or percentage in the model:
n = 5 / (the smallest proportion in your model)
If your model is expressed in ratios, such as 9:3:3:1, the smallest proportion is the smallest number in the model divided by the total in the model, in this case 1/(9+3+3+1) or 1/16. So the required sample size is 5 / (1/16) = 80.
Example 7: You expect that customers will choose coffee, tea, bottled water, and Snapple in the proportions of 65%, 15%, 15%, 5%. How large a random sample must you take to test this model?
Solution: Take the least likely category and divide 5 by that percentage:
n = 5 / 5% = 5 / 0.05 = 100.
Answer: You need a random sample of at least 100 to test this model. (As always, the minimum sample will give a significant result only if the null hypothesis is extremely wrong. If the model is only moderately wrong, a larger sample will probably be needed to reveal that.)
Example 8: For a kids’ play area, you ordered five cartons of blue plastic balls, two of green, and three each of red and yellow. But your assistant dumped them all together after taking delivery. How many balls must you select randomly to see if the proportions are right?
Solution: You must divide 5 by the proportion in the smallest category in the model, which is green:
smallest proportion = 2 / (5+2+3+3) = 2/13
5 / (2/13) = 32.5 → 33
Answer: To meet the requirements, you need a random sample of at least 33 balls.
Example 9: You believe that plain M&Ms are distributed in the proportions 24% blue, 13% brown, 16% green, 18% orange, 15% red, 14% yellow. How large a sample do you need to test this model?
Solution: The smallest proportion in the model is 13%, so compute 5/13% = 5/0.13 = 38.46... → 39. (Remember, sample sizes always round up.)
Answer: The sample must contain at least 39 M&Ms.
Updates and new info: https://BrownMath.com/stat/