Confidence-Interval Width, Margin of Error, and Sample Size
A confidence interval reports both an estimate and its sampling uncertainty. Its width depends primarily on three quantities:
- The critical value determined by the confidence level
- The standard deviation or estimated variability
- The sample size
Understanding these relationships makes it possible to plan sample sizes and evaluate reported margins of error.
General Form of a Confidence Interval
Many confidence intervals have the form\[ \boxed{ \text{estimate} \pm z^*\times SE(\text{estimate}) }. \]
Here:
- \(z^*\) is the critical value.
- \(SE\) is the standard error.
- The lower endpoint is the estimate minus \(z^*SE\).
- The upper endpoint is the estimate plus \(z^*SE\).
The quantity\[ \boxed{ ME=z^*SE } \]
is called the margin of error.
Therefore, the confidence interval is\[ \boxed{ \text{estimate}\pm ME }. \]
Margin of Error Versus Interval Width
The margin of error is the distance from the estimate to either endpoint.
The full interval width is twice the margin of error:\[ \boxed{ W=2ME }. \]
Since\[ ME=z^*SE, \]
the width is\[ \boxed{ W=2z^*SE }. \]
For example, consider the interval\[ (54.9\%,61.1\%). \]
Its midpoint is\[ \frac{54.9+61.1}{2}=58.0\%. \]
Its margin of error is\[ 61.1-58.0=3.1 \]
percentage points.
Its full width is\[ 61.1-54.9=6.2 \]
percentage points.
Thus, it can be reported as\[ 58.0\%\pm3.1\text{ percentage points}. \]
The Square-Root Law
For a sample mean based on independent observations,\[ SE(\bar{X}) = \frac{\sigma}{\sqrt{n}}. \]
Therefore, the margin of error is\[ \boxed{ ME = z^* \frac{\sigma}{\sqrt{n}} }. \]
The full width is\[ \boxed{ W = 2z^* \frac{\sigma}{\sqrt{n}} }. \]
This shows that confidence-interval width decreases at the rate\[ \frac{1}{\sqrt{n}}. \]
Increasing the sample size makes the interval narrower, but the improvement has diminishing returns.
How Sample Size Affects Width
Suppose the original sample size is \(n\), producing width\[ W_n = 2z^*\frac{\sigma}{\sqrt{n}}. \]
If the sample size is multiplied by a factor \(k\), the new width is\[ W_{kn} = 2z^*\frac{\sigma}{\sqrt{kn}}. \]
Since\[ \sqrt{kn} = \sqrt{k}\sqrt{n}, \]
we have\[ W_{kn} = \frac{W_n}{\sqrt{k}}. \]
Therefore:
| Change in sample size | Change in interval width |
|---|---|
| Multiply \(n\) by 4 | Divide width by 2 |
| Multiply \(n\) by 9 | Divide width by 3 |
| Multiply \(n\) by 16 | Divide width by 4 |
| Multiply \(n\) by 25 | Divide width by 5 |
| Multiply \(n\) by 100 | Divide width by 10 |
To cut the confidence-interval width in half, the sample size must be quadrupled.
To reduce the width to one-tenth of its original value, the sample size must be multiplied by 100.
Solving Directly for Sample Size
Suppose the desired margin of error is \(m\). Starting with\[ m = z^*\frac{\sigma}{\sqrt{n}}, \]
solve for \(n\):\[ \sqrt{n} = \frac{z^*\sigma}{m}. \]
Therefore,\[ \boxed{ n = \left( \frac{z^*\sigma}{m} \right)^2 }. \]
If this calculation produces a non-integer value, round upward. Rounding down would produce a margin of error larger than the target.
Example for a Population Mean
Suppose a measurement has standard deviation\[ \sigma=12 \]
and we want a 95% confidence interval with margin of error no greater than\[ m=2. \]
Using\[ z^*=1.96, \]
the required sample size is\[ n = \left( \frac{1.96(12)}{2} \right)^2. \]
Therefore,\[ n = (11.76)^2 \approx138.30. \]
Round upward:\[ \boxed{n=139}. \]
If \(\sigma\) is unknown during planning, it must be estimated using prior data, a preliminary sample, or a conservative assumption.
Confidence Level and the Critical Value
The critical value \(z^*\) is determined by the confidence level.
Common values include:
| Confidence level | \(z^*\) |
|---|---|
| 80% | 1.282 |
| 90% | 1.645 |
| 95% | 1.960 |
| 98% | 2.326 |
| 99% | 2.576 |
For fixed data and sample size:
- Higher confidence produces a larger \(z^*\).
- A larger \(z^*\) produces a larger margin of error.
- Therefore, higher confidence produces a wider interval.
This creates a tradeoff between confidence and precision.
The Confidence–Precision Tradeoff
Suppose the standard error is\[ SE=0.02. \]
An 80% confidence interval has margin of error\[ ME=1.282(0.02)=0.0256. \]
A 95% confidence interval has margin of error\[ ME=1.96(0.02)=0.0392. \]
A 99% confidence interval has margin of error\[ ME=2.576(0.02)=0.0515. \]
| Confidence level | Margin of error |
|---|---|
| 80% | 2.56 percentage points |
| 95% | 3.92 percentage points |
| 99% | 5.15 percentage points |
A narrower 80% interval is more precise but has a lower long-run coverage rate. A wider 99% interval is less precise but has a higher coverage rate.
It is inappropriate to choose a lower confidence level after seeing the data merely to obtain a desired conclusion. The confidence level should be selected according to the application and, ideally, before analyzing the results.
Confidence Intervals for Proportions
For a population proportion \(p\), the sample proportion is \(\hat{p}\). Under independent sampling, its standard error is\[ SE(\hat{p}) = \sqrt{ \frac{p(1-p)}{n} }. \]
Because \(p\) is unknown, the standard error is commonly estimated by\[ \widehat{SE}(\hat{p}) = \sqrt{ \frac{\hat{p}(1-\hat{p})}{n} }. \]
An approximate 95% Wald interval is\[ \hat{p} \pm 1.96 \sqrt{ \frac{\hat{p}(1-\hat{p})}{n} }. \]
Its margin of error is\[ ME = 1.96 \sqrt{ \frac{\hat{p}(1-\hat{p})}{n} }. \]
The Maximum Variability of a Proportion
The quantity\[ p(1-p) \]
depends on the population proportion. It is largest when\[ p=0.5. \]
To verify this, write\[ p(1-p) = p-p^2. \]
This is a downward-opening parabola whose maximum occurs at \(p=0.5\). At that point,\[ p(1-p) = 0.5(0.5) = 0.25. \]
Therefore,\[ \sqrt{p(1-p)} \leq0.5. \]
The standard error of a sample proportion can consequently be bounded by\[ SE(\hat{p}) \leq \frac{0.5}{\sqrt{n}}. \]
The Quick 95% Rule
For a 95% interval,\[ z^*\approx2. \]
Using the maximum possible standard deviation of 0.5,\[ ME \approx 2\frac{0.5}{\sqrt{n}}. \]
The factors of 2 and 0.5 cancel:\[ \boxed{ ME\approx\frac{1}{\sqrt{n}} }. \]
Thus, a quick conservative 95% interval for a proportion is\[ \boxed{ \hat{p} \pm \frac{1}{\sqrt{n}} }. \]
This expression uses the proportion scale. To express it in percentage points, multiply by 100:\[ \boxed{ 100\hat{p} \pm \frac{100}{\sqrt{n}} \text{ percentage points} }. \]
Why the Rule Is Conservative
The exact 95% normal critical value is 1.96 rather than 2. At \(p=0.5\), the maximum normal-theory margin of error is\[ 1.96\frac{0.5}{\sqrt{n}} = \frac{0.98}{\sqrt{n}}. \]
The quick rule uses\[ \frac{1}{\sqrt{n}}, \]
which is slightly larger than\[ \frac{0.98}{\sqrt{n}}. \]
It is also conservative when the true proportion is far from 0.5 because\[ \sqrt{p(1-p)}<0.5. \]
Quick-Rule Examples
Sample Size of 400
\[ ME \approx \frac{1}{\sqrt{400}} = \frac{1}{20} = 0.05. \]
Thus, the approximate margin of error is\[ 5\text{ percentage points}. \]
Sample Size of 1,000
\[ ME \approx \frac{1}{\sqrt{1{,}000}} \approx 0.0316. \]
Thus, the approximate margin of error is\[ 3.16\text{ percentage points}. \]
Sample Size of 2,500
\[ ME \approx \frac{1}{\sqrt{2{,}500}} = \frac{1}{50} = 0.02. \]
Thus, the approximate margin of error is\[ 2\text{ percentage points}. \]
Sample Size of 10,000
\[ ME \approx \frac{1}{\sqrt{10{,}000}} = 0.01. \]
Thus, the approximate margin of error is\[ 1\text{ percentage point}. \]
Quick Sample-Size Planning for a Percentage
Using\[ ME\approx\frac{1}{\sqrt{n}}, \]
solve for \(n\):\[ n\approx\frac{1}{ME^2}. \]
If the desired margin of error is expressed as a proportion \(m\), then\[ \boxed{ n\approx\frac{1}{m^2} }. \]
For a desired margin of error of 3 percentage points,\[ m=0.03. \]
Therefore,\[ n \approx \frac{1}{0.03^2} = \frac{1}{0.0009} \approx1{,}111.11. \]
Round upward:\[ \boxed{n\approx1{,}112}. \]
For a desired margin of error of 1 percentage point,\[ m=0.01, \]
so\[ n\approx\frac{1}{0.01^2}=10{,}000. \]
More Exact Sample-Size Formula for a Proportion
If a planning value \(p_0\) is available, the required sample size for margin of error \(m\) is\[ \boxed{ n = \frac{ z^{*2}p_0(1-p_0) }{ m^2 } }. \]
When there is no reliable prior estimate, use\[ p_0=0.5 \]
because it produces the largest required sample size.
For a 95% confidence level and a margin of error of 3 percentage points,\[ n = \frac{ 1.96^2(0.5)(0.5) }{ 0.03^2 }. \]
Thus,\[ n = \frac{3.8416(0.25)}{0.0009} \approx1{,}067.11. \]
Round upward:\[ \boxed{n=1{,}068}. \]
The quick rule gave approximately 1,112 because it uses 2 rather than 1.96.
The Rule of Thumb Has Conditions
The shortcut\[ \hat{p}\pm\frac{1}{\sqrt{n}} \]
is only a rough 95% normal-approximation rule. It works best when:
- Sampling is approximately independent.
- The sample is representative.
- Expected success and failure counts are sufficiently large.
- The estimate is not extremely close to 0 or 1.
- The design resembles a simple random sample.
- Sampling weights and clustering do not substantially alter variance.
For small samples or extreme proportions, methods such as the Wilson score interval are generally preferable.
Finite Population Correction
If sampling occurs without replacement from a finite population of size \(N\), the standard error includes\[ \sqrt{ \frac{N-n}{N-1} }. \]
For a proportion,\[ SE(\hat{p}) = \sqrt{ \frac{p(1-p)}{n} } \sqrt{ \frac{N-n}{N-1} }. \]
If \(n\) is small relative to \(N\), this factor is nearly 1. When the sample represents a substantial fraction of the population, it can noticeably reduce the interval width.
Complex Survey Designs
Polls often use clustering, stratification, weighting, and adjustments for nonresponse. The actual standard error may therefore differ from the simple random-sample formula.
A design effect can be used:\[ DEFF = \frac{ \operatorname{Var}_{\text{design}}(\hat{p}) }{ \operatorname{Var}_{\text{SRS}}(\hat{p}) }. \]
The design-adjusted standard error is approximately\[ SE_{\text{design}} = SE_{\text{SRS}}\sqrt{DEFF}. \]
If\[ DEFF=1.5, \]
then the standard error is multiplied by\[ \sqrt{1.5}\approx1.225. \]
The interval becomes about 22.5% wider.
Larger Samples Do Not Remove Bias
Increasing \(n\) reduces random sampling uncertainty:\[ SE\propto\frac{1}{\sqrt{n}}. \]
It does not automatically reduce:
- Selection bias
- Undercoverage
- Nonresponse bias
- Measurement error
- Question-wording effects
- Incorrect weighting
- Data-processing mistakes
A huge biased sample can produce a very narrow confidence interval centered on the wrong value.
Confidence-interval width measures precision under the statistical model, not total accuracy.
Reporting the Confidence Level
If a report gives an estimate and margin of error without stating a confidence level, readers often assume that it refers to a 95% confidence interval.
However, this convention should not be relied upon. A complete report should state:
- The confidence level
- The estimate
- The margin of error or interval endpoints
- The sample size
- The sampling method
- The population being estimated
- Whether survey-design adjustments were used
For example:
The estimated approval rating was 58.0%, with a 95% confidence interval from 54.9% to 61.1%, based on a sample of 1,000 respondents.
Key Takeaway
A confidence interval has the general form\[ \text{estimate}\pm z^*SE. \]
Its margin of error is\[ ME=z^*SE, \]
and its full width is\[ W=2z^*SE. \]
Because\[ SE=\frac{\sigma}{\sqrt{n}}, \]
confidence intervals become narrower as sample size increases, but only according to the square-root law. Cutting the width in half requires four times the sample size.
For a population proportion, a quick conservative 95% margin-of-error rule is\[ \boxed{ ME\approx\frac{1}{\sqrt{n}} } \]
on the proportion scale, or\[ \boxed{ ME\approx\frac{100}{\sqrt{n}} } \]
in percentage points.
Narrower intervals can be obtained by increasing the sample size or lowering the confidence level. Lowering the confidence level, however, reduces long-run coverage, while increasing sample size cannot correct systematic bias or poor data collection.