One-Sided Z Confidence Bounds For A Mean

Why Use a One-Sided Confidence Bound?

A two-sided confidence interval gives a lower and an upper endpoint. Some questions instead require a bound in just one direction: for example, a hypothetical benefits analyst may need evidence that an average processing time exceeds a target, or a hypothetical insurer may want an upper confidence bound for a mean claim amount. In either case, putting confidence in the direction that answers the question is the main task.

A one-sided confidence bound does not estimate both ends of a plausible range. It gives a procedure that covers the population mean from below or from above with a specified long-run frequency. This tutorial treats the mean when the population variance is known, using a standard normal pivot.

Model and assumptions: Let $X_1,\ldots,X_n$ be independent observations from a population with mean $\mu$ and known variance $\sigma^2$, where $\sigma>0$. If the population is normal, then $\bar{X}\sim N(\mu,\sigma^2/n)$ exactly. If the population is not normal, a normal approximation for $\bar{X}$ may be appropriate for sufficiently large $n$, under the conditions for the central limit theorem. The Z bounds below are exact in the normal model and approximate when justified by a large-sample normal approximation.

Write $\bar{x}$ for the observed sample mean and define the standard error $SE=\sigma/\sqrt{n}$. Let $\Phi$ be the standard normal cumulative distribution function, and let $z_c=\Phi^{-1}(c)$, so that $P(Z\leq z_c)=c$ for $Z\sim N(0,1)$. Here $c$ is the one-sided confidence level and $1-c$ is the tail probability.

Definition: A lower one-sided confidence bound $L$ for $\mu$ at confidence level $c$ is constructed so that $P_\mu(L\leq\mu)=c$ under the model. Its confidence set is $[L,\infty)$. An upper one-sided confidence bound $U$ is constructed so that $P_\mu(\mu\leq U)=c$; its confidence set is $(-\infty,U]$. The probability statement describes the repeated-sampling coverage of the procedure, not a probability distribution for the fixed parameter after the data are observed.

Choosing the Quantile and the Direction

For a lower bound, the procedure must leave probability $c$ in the lower part of the standard normal distribution. For an upper bound, it must leave probability $c$ in the upper part. Symmetry of the standard normal distribution makes the resulting formulas look similar, but the sign of the margin is different.

L=\bar{x}-z_c\frac{\sigma}{\sqrt{n}} \qquad\text{and}\qquad U=\bar{x}+z_c\frac{\sigma}{\sqrt{n}}.

For the confidence levels commonly used in applications, $c>0.5$, so $z_c>0$. The lower bound is then below $\bar{x}$, and the upper bound is above $\bar{x}$. More generally, the formulas are valid for any $0<c<1$, but $z_c$ is negative when $c<0.5$; the endpoints then need not lie on the intuitively expected sides of $\bar{x}$. The coverage statements, rather than an endpoint-side check, are the definitive test.

Key result: For a one-sided confidence level $c$, use $z_c=\Phi^{-1}(c)$ for either bound. Subtract $z_cSE$ for a lower bound and add $z_cSE$ for an upper bound. Do not substitute the two-sided critical value $z_{(1+c)/2}$.

For example, at $c=0.95$, the one-sided critical value is $z_{0.95}\approx1.6449$. The critical value for a two-sided 95% interval is $z_{0.975}\approx1.9600$, because that interval allocates 2.5% to each tail. Using the latter for a one-sided bound produces a bound with more than 95% coverage under the stated model.

Why the Formulas Give the Required Coverage

The standardized sample mean is the pivot $Z=(\bar{X}-\mu)/SE$, which has a standard normal distribution under the normal model. The bounds follow by translating an event involving this pivot into an event involving $\mu$.

Start with the pivot.
Under the model, $Z=(\bar{X}-\mu)/SE\sim N(0,1)$, so $P(Z\leq z_c)=c$.

Rearrange for a lower bound.
$Z\leq z_c$ is equivalent to $\bar{X}-z_cSE\leq\mu$. Therefore $P(\bar{X}-z_cSE\leq\mu)=c$, and the observed lower bound is $L=\bar{x}-z_cSE$.

Use symmetry for an upper bound.
Since $P(Z\geq-z_c)=c$, the event $Z\geq-z_c$ is equivalent to $\mu\leq\bar{X}+z_cSE$. Thus $P(\mu\leq\bar{X}+z_cSE)=c$, giving $U=\bar{x}+z_cSE$.

This derivation also shows why the direction matters. For the lower bound, the event being covered is $\bar{X}-z_cSE\leq\mu$; for the upper bound, it is $\mu\leq\bar{X}+z_cSE$. The probability comes from the sampling distribution of $\bar{X}$, not from choosing a sign by visual preference.

Worked Calculations

Worked Example: A Lower Bound for Mean Processing Time

A hypothetical service center takes a simple random sample of $n=36$ completed cases. The observed mean processing time is $\bar{x}=74.60$ hours. Assume independent observations from a normal population with unknown mean $\mu$ and known standard deviation $\sigma=12$ hours. Construct a 95% lower confidence bound for $\mu$. There is no interest-rate or mortality assumption in this sampling problem.

Method. Use the one-sided Z lower-bound formula. The standard error is $SE=12/\sqrt{36}=12/6=2$ hours. For $c=0.95$, $z_c=z_{0.95}=1.6449$.

The margin is $1.6449(2)=3.2898$ hours. Therefore $L=74.60-3.2898=71.3102$ hours, or 71.31 hours to two decimal places. The confidence set is $[71.31,\infty)$ hours.

Check and common mistake. The bound is below the observed mean, as expected for $c>0.5$. Adding the margin would give $74.60+3.2898=77.8898$, which is the upper-bound calculation, not the requested lower bound. The result means the procedure has 95% coverage under the stated model; it does not mean there is a 95% probability that this fixed $\mu$ exceeds 71.31 after observing the sample.

Worked Example: An Upper Bound for Mean Claim Amount

A hypothetical portfolio analyst samples $n=64$ claims from a population modeled as normal with mean $\mu$ and known standard deviation $\sigma=\$16$. The sample mean is $\bar{x}=\$42.50$. Find a 90% upper confidence bound for the population mean claim amount. No interest rate or mortality assumption is relevant; the model is the normal sampling model just stated.

Method. The standard error is $SE=16/\sqrt{64}=16/8=\$2$. The required one-sided quantile is $z_{0.90}=1.2816$. The margin is $1.2816(\$2)=\$2.5632$.

For an upper bound, add the margin: $U=42.50+2.5632=\$45.0632$. Rounded to the nearest cent, the upper bound is \$45.06, and the confidence set is $(-\infty,\$45.06]$.

Check and common mistake. The endpoint is above the sample mean, which is expected because $c=0.90>0.5$. Subtracting the margin would give $\$39.94$ (to the nearest cent), the lower-bound direction. Also, $z_{0.95}$ is not the right quantile for a 90% one-sided bound; the correct value is $z_{0.90}$.

Worked Example: A Confidence Level Below One-Half

A small study has $n=25$ independent observations from a normal population with mean $\mu$ and known standard deviation $\sigma=10$. The observed mean is $\bar{x}=18.00$. Calculate a 40% lower confidence bound and explain whether it must be below the sample mean. There is no interest-rate or mortality assumption.

Method. Here $c=0.40$, so $z_{0.40}=-0.2533$. The standard error is $SE=10/\sqrt{25}=10/5=2$. The lower-bound formula remains $L=\bar{x}-z_cSE$: $L=18.00-(-0.2533)(2)=18.00+0.5066=18.5066$.

Rounded to two decimal places, the lower bound is 18.51. Although it is above $\bar{x}$, its lower-bound coverage statement still holds: $P(L\leq\mu)=P(Z\leq-0.2533)=0.40$.

Check and common mistake. It would be incorrect to claim that every lower bound must be at or below $\bar{x}$. That endpoint check is valid for $c\geq0.5$, when $z_c\geq0$, but not for all confidence levels. Here the requested confidence level is below one-half, so the quantile is negative and the formula correctly places $L$ above $\bar{x}$.

Choosing and Reporting a Bound

Begin by identifying which direction the question asks for. A lower bound supports a statement of the form “the mean is at least this value,” while an upper bound supports “the mean is at most this value.” Then calculate the standard error, obtain the quantile for the stated one-sided confidence level, and apply the matching sign. The following sequence is useful under exam conditions.

State the model.
Record $n$, $\bar{x}$, the known $\sigma$, and whether normality gives an exact result or a large-sample approximation.

Calculate the standard error.
Use $SE=\sigma/\sqrt{n}$, not $\sigma/n$ and not $\sigma^2/\sqrt{n}$.

Match the quantile to the confidence level.
For a one-sided level $c$, use $z_c$. The direction determines whether the margin is subtracted or added.

Report the endpoint and its meaning.
Name it as a lower or upper confidence bound, include the measurement units, and state the confidence level and model qualification.

A one-sided bound uses the full tail probability $1-c$ on its one relevant side. This differs from a two-sided interval at the same confidence level, which splits its total excluded probability between two tails. The formula here also requires a known population standard deviation. When $\sigma$ is not known, replacing it with the sample standard deviation does not preserve this Z procedure; that setting requires a different method.

Direction error: adding the margin gives the upper bound; subtracting gives the lower bound.
Critical-value error: use $z_c$ for a one-sided confidence level, not the two-sided quantile.
Standard-error error: divide $\sigma$ by $\sqrt{n}$, not by $n$.
Interpretation error: confidence is a long-run property of the procedure, not a probability assigned to a fixed mean after the data are observed.
Endpoint-check error: the usual “lower below the mean, upper above the mean” check requires $c\geq0.5$; for $c<0.5$, $z_c<0$.

Key takeaway: Under the known-variance Z model, compute $SE=\sigma/\sqrt{n}$, find $z_c=\Phi^{-1}(c)$, and use $L=\bar{x}-z_cSE$ or $U=\bar{x}+z_cSE$ according to the requested direction. Verify coverage and assumptions rather than relying only on where an endpoint falls relative to $\bar{x}$. This treatment is aligned with the SOA VEE Mathematical Statistics syllabus.

Check Your Understanding

For each problem, identify the requested direction, use the one-sided quantile, and check the interpretation against the model assumptions.

A normal population has known $\sigma=15$. A sample of $n=25$ has mean $\bar{x}=60$. Find a 90% lower confidence bound for $\mu$, rounded to two decimals. (Answer: 56.16)
For a normal population with known $\sigma=30$, a sample of $n=100$ has mean $\bar{x}=12.40$. Find a 95% upper confidence bound, rounded to two decimals. (Answer: 17.33)
A normal population has known $\sigma=14$; a sample of $n=49$ has mean $\bar{x}=105$. Find a 99% lower confidence bound, rounded to two decimals. (Answer: 100.35)
A normal population has known $\sigma=8$. For $n=64$, the observed sample mean is $20.50$. A report requests a 90% upper bound. Give the bound to two decimals and state why using $z_{0.95}$ would be a quantile mismatch. (Answer: 21.78; the correct one-sided quantile is $z_{0.90}$, not $z_{0.95}$)
A normal population has known $\sigma=5$. For $n=100$, $\bar{x}=30.00$, and the requested confidence level is $c=0.40$. Calculate the lower bound to two decimals and decide whether it must be below the observed mean. (Answer: 30.13; no, because $z_{0.40}<0$)

What You'll Learn