Interval Width, Units, and Precision

Why Width and Units Matter

A confidence interval gives a range of parameter values consistent with a procedure and the observed data. Its endpoints describe the range; its width describes how far apart those endpoints are. Width is often used as a practical measure of an interval’s precision: all else being comparable, a narrower interval reports a more concentrated range of estimates.

“All else being comparable” is essential. A width has the same units as the parameter, and a comparison is most informative when the intervals estimate the same parameter under comparable confidence levels and valid assumptions. A narrow interval is not automatically better if its procedure has inadequate coverage, uses an invalid model, or estimates a different quantity.

Definition: If a confidence interval has lower endpoint $L$ and upper endpoint $U$, its total width is $U-L$. If the interval is centered at an estimate $\hat{\theta}$, its half-width (also called the margin of error in this setting) is the distance from the center to either endpoint. For a symmetric interval, $L=\hat{\theta}-h$, $U=\hat{\theta}+h$, and the total width is $2h$.

The width is not the upper endpoint, the lower endpoint, or the distance from the estimate to one endpoint. For a symmetric interval, that last distance is only half the width. Keeping these quantities distinct prevents a common factor-of-two error.

Deriving the Width of a Symmetric Interval

Suppose $X_1,\ldots,X_n$ are independent observations from a normal population with mean $\mu$ and known standard deviation $\sigma$. The sample mean is $\bar X$. The standardized statistic $\frac{\bar X-\mu}{\sigma/\sqrt{n}}$ has a standard normal distribution. For a two-sided confidence level $1-\alpha$, let $z_{1-\alpha/2}$ denote the standard normal quantile with cumulative probability $1-\alpha/2$. Inverting the central probability statement gives the interval below.

\bar X-z_{1-\alpha/2}\frac{\sigma}{\sqrt{n}} \leq \mu \leq \bar X+z_{1-\alpha/2}\frac{\sigma}{\sqrt{n}}.

To find its width, subtract the lower endpoint from the upper endpoint. The two copies of $\bar X$ cancel, while the two positive margin terms add. Thus the center of this symmetric interval affects its location, but not its width.

\begin{aligned} W &=\left(\bar X+z_{1-\alpha/2}\frac{\sigma}{\sqrt{n}}\right) -\left(\bar X-z_{1-\alpha/2}\frac{\sigma}{\sqrt{n}}\right)\\ &=2z_{1-\alpha/2}\frac{\sigma}{\sqrt{n}}. \end{aligned}

This expression also explains the units. The quantile $z_{1-\alpha/2}$ is dimensionless, while $\sigma/\sqrt n$ has the same units as the observations and the population mean. Multiplication by a dimensionless number leaves those units unchanged. Therefore, if $\mu$ measures dollars, the width is in dollars; if it measures hours, the width is in hours.

Theorem: For the normal-mean interval with known $\sigma$, the width is $2z_{1-\alpha/2}\sigma/\sqrt n$. It does not depend on the observed value of $\bar X$. At fixed $n$ and $\sigma$, a higher two-sided confidence level uses a larger critical quantile and produces a wider interval.

The condition matters: the formula just derived assumes independent observations from a normal population and a known population standard deviation. It is not the correct formula when $\sigma$ is unknown and replaced by the sample standard deviation. Under independent normal sampling with unknown $\sigma$, the corresponding exact interval uses a $t$ critical value and the sample standard deviation $s$:

\bar X\ \pm\ t_{1-\alpha/2,\,n-1}\frac{s}{\sqrt n}, \qquad W=2t_{1-\alpha/2,\,n-1}\frac{s}{\sqrt n}.

Here $t_{1-\alpha/2,\,n-1}$ is the $t$-distribution quantile with $n-1$ degrees of freedom. This interval requires an independent sample from a normal population; using the $t$ formula does not remove that assumption for an exact small-sample interval.

Worked Example: Comparing Two Confidence Levels

A hypothetical calibration study measures a component’s operating time in hours. An independent normal sample of $n=64$ components has sample mean $\bar X=48.5$ hours. The population standard deviation is known to be $\sigma=12$ hours. Calculate the 90% and 95% confidence-interval widths and interpret the difference. Use $z_{0.95}=1.645$ and $z_{0.975}=1.960$. The method is the normal $z$ interval for a mean with known standard deviation.

Step 1: Check the standard error. The assumed model is independent normal observations with known $\sigma=12$ hours. Therefore,

\frac{\sigma}{\sqrt n} =\frac{12}{\sqrt{64}} =\frac{12}{8} =1.5\text{ hours}.

Step 2: Calculate the 90% interval. Its half-width is $1.645(1.5)=2.4675$ hours, so its total width is $2(2.4675)=4.935$ hours. The endpoints are $48.5-2.4675=46.0325$ and $48.5+2.4675=50.9675$ hours.

Step 3: Calculate the 95% interval. Its half-width is $1.960(1.5)=2.94$ hours, so its total width is $2(2.94)=5.88$ hours. The endpoints are $48.5-2.94=45.56$ and $48.5+2.94=51.44$ hours.

Step 4: Compare and check. The 95% interval is wider by $5.88-4.935=0.945$ hours. Both intervals are centered at 48.5 hours, and subtracting their endpoints gives the stated widths. Reported to three decimal places, the widths are 4.935 and 5.880 hours. The higher confidence level produces the wider interval here because the standard error and method are fixed and the 95% critical value is larger. A common mistake is to report 2.94 hours as the 95% width; it is the half-width, so the total width is twice as large.

Comparing Precision Responsibly

A comparison of widths is most useful when it holds constant the target parameter, its units, and the confidence level, and when each procedure’s assumptions are reasonable. For intervals with the same confidence level and method, differences in sample size or observed variability can explain differences in width. A larger standard error generally makes a symmetric interval wider; a larger critical value does too.

Width describes the spread between the endpoints of the interval actually calculated. It does not tell you whether that interval contains the true parameter: the observed endpoints alone cannot answer that question. Nor does a narrower realized interval by itself prove that a procedure is superior. A procedure that is too narrow because its assumptions or stated coverage are not valid is not a more precise solution to the same inferential problem.

The estimate’s numerical location is also distinct from its precision. For a symmetric interval, changing $\bar X$ while keeping the standard error, confidence level, and method fixed shifts both endpoints equally and leaves the width unchanged. A value near the center of the parameter space is not inherently more precise than another value; precision is reflected by the interval’s spread.

Worked Example: Width for an Unknown Standard Deviation

A hypothetical laboratory records the processing time, in minutes, for a random independent sample of $n=9$ batches from a normal population. The sample mean is $\bar X=20.4$ minutes and the sample standard deviation is $s=3$ minutes. Find the 95% confidence interval and its width. Use $t_{0.975,8}=2.306$. The method is the one-sample $t$ interval; the population standard deviation is unknown.

Step 1: Confirm the model and degrees of freedom. The calculation assumes an independent sample from a normal population. Since $n=9$, the degrees of freedom are $n-1=8$. The standard error is

\frac{s}{\sqrt n} =\frac{3}{\sqrt 9} =\frac{3}{3} =1\text{ minute}.

Step 2: Find the half-width. The 95% critical value is 2.306, so the half-width is $2.306(1)=2.306$ minutes.

Step 3: Find the endpoints and total width. The lower endpoint is $20.4-2.306=18.094$ minutes, and the upper endpoint is $20.4+2.306=22.706$ minutes. Thus the width is $22.706-18.094=4.612$ minutes, also equal to $2(2.306)=4.612$ minutes. The interval is centered at 20.4 minutes, as a check. Reported to three decimal places, it is $[18.094,22.706]$ minutes with width 4.612 minutes.

A common mistake is to use a standard normal critical value merely because the target is a mean. With unknown population standard deviation and this small normal sample, the specified exact method uses $t_{0.975,8}$, not $z_{0.975}$.

Units and Fair Comparisons

The units must travel with the width. If a parameter is a mean annual payment measured in dollars, a width of 8,000 means a range 8,000 dollars wide. If the same payments and interval are expressed in thousands of dollars, the numerical width is 8, not 8,000, but the interval represents the same amount of uncertainty. The number changes because the measurement unit changed; the underlying interval has not become more or less precise.

When comparing two procedures, first check whether they address the same parameter. A width of 2 years cannot be ranked against a width of $2 as “smaller” in a meaningful precision comparison. Even for the same parameter and units, check that the confidence levels are comparable. A lower confidence level can produce a narrower interval simply because it is designed to cover the parameter less often in repeated sampling.

A practical comparison sequence is:

Identify the target and units.
Confirm that both intervals estimate the same parameter, such as the same population mean, and express both widths in the same units.

Check the confidence level and method.
Different confidence levels or assumptions can account for width differences, so do not attribute the difference solely to precision without noting them.

Calculate total width consistently.
For each interval use upper endpoint minus lower endpoint. Do not compare one total width with another interval’s half-width.

State the limited conclusion.
Say which realized interval is narrower and by how much, while retaining the assumptions and coverage context.

Worked Example: Comparing Precision for the Same Mean

Two hypothetical independent normal samples estimate the same population mean annual repair cost, with costs recorded in dollars. In each procedure the population standard deviation is known to be $\sigma=\$12{,}000$, and each interval has 95% confidence, using $z_{0.975}=1.96$. Procedure A has $n_A=36$ and $\bar X_A=\$103{,}200$; Procedure B has $n_B=64$ and $\bar X_B=\$102{,}400$. Find both interval widths and compare their realized precision. The method for both is the known-$\sigma$ normal interval.

Step 1: Calculate Procedure A’s standard error and width.

\operatorname{SE}_A=\frac{12{,}000}{\sqrt{36}}=\frac{12{,}000}{6}=\$2{,}000, \qquad W_A=2(1.96)(2{,}000)=\$7{,}840.

Its half-width is $1.96(2{,}000)=\$3{,}920$, so the interval is $[\$99{,}280,\$107{,}120]$. Subtraction confirms the width: $107{,}120-99{,}280=7{,}840$ dollars.

Step 2: Calculate Procedure B’s standard error and width.

\operatorname{SE}_B=\frac{12{,}000}{\sqrt{64}}=\frac{12{,}000}{8}=\$1{,}500, \qquad W_B=2(1.96)(1{,}500)=\$5{,}880.

Its half-width is $1.96(1{,}500)=\$2{,}940$, so the interval is $[\$99{,}460,\$105{,}340]$. Subtraction confirms $105{,}340-99{,}460=5{,}880$ dollars.

Step 3: Compare. Procedure B’s realized interval is narrower by $7{,}840-5{,}880=\$1{,}960$. The width ratio is $5{,}880/7{,}840=0.75$, so B’s width is 25% less than A’s. Since the target, units, known standard deviation, method, and confidence level are shared, this width comparison is interpretable as a comparison of the realized interval precision. The sample means differ, but that changes the interval locations, not their widths. A common mistake is to conclude that the interval with the smaller center is more precise; precision is compared through width, not through the estimated mean’s value.

Common Exam Traps

Confusing margin of error and total width. For $[\hat{\theta}-h,\hat{\theta}+h]$, the margin is $h$ and the total width is $2h$.
Dropping the units. A width is measured in the units of the estimated parameter. Always include them in the calculation and interpretation.
Claiming higher confidence means greater precision. For a fixed method and data quality in the symmetric intervals considered here, increasing confidence increases the critical value and widens the interval. It does not make the interval narrower.
Comparing unrelated intervals by their raw numbers. Widths measured in different units, or for different parameters, do not make a meaningful direct precision ranking.
Treating narrowness as proof of validity. Precision comparisons are useful only in the context of a valid procedure and its coverage level.

Key takeaway: Compute total width as upper endpoint minus lower endpoint, retain the parameter’s units, and compare widths only with the target, confidence level, and procedure in view. A narrower interval indicates a tighter realized range—not, by itself, a more trustworthy procedure.

Check Your Understanding

For each problem, calculate total width where requested and state the relevant comparison or interpretation.

A known-$\sigma$ normal interval for a mean has $n=25$, $\sigma=5$ kilograms, and a 95% confidence level with $z_{0.975}=1.96$. Find its total width. (Answer: 3.920 kilograms.)
A symmetric interval is $[14.2,18.8]$ hours. Find its total width and half-width, and identify the units. (Answer: total width 4.6 hours; half-width 2.3 hours.)
An independent normal sample of $n=16$ has $s=4$ millimeters. An exact 90% $t$ interval for the mean uses $t_{0.95,15}=1.753$. Find its total width. (Answer: $2(1.753)(4/\sqrt{16})=3.506$ millimeters.)
Two 95% known-$\sigma$ intervals estimate the same mean in dollars, with $\sigma=\$8{,}000$. Procedure A uses $n=16$; Procedure B uses $n=64$. With $z_{0.975}=1.96$, find both widths and say which is narrower. (Answer: A is $\$7{,}840$; B is $\$3{,}920$; B is narrower by $\$3{,}920$.)
A reported 90% interval for a mean payment is $[\$2{,}350,\$2{,}950]$, while a 95% interval for the same mean is $[\$2{,}200,\$3{,}100]$. Compare their widths and give one qualification needed before calling one procedure more precise. (Answer: widths are $\$600$ and $\$900$, respectively; the 90% interval is narrower, but the confidence levels differ, so width alone does not establish a fair procedure-precision comparison.)

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