One-Sided T Confidence Bounds for a Mean

Why Use a One-Sided Bound?

Sometimes an estimate is needed in only one direction. A clinic might want an upper limit for the mean waiting time, or a manufacturer might need a lower limit for the mean strength of a material. A two-sided interval gives limits in both directions; a one-sided confidence bound focuses the available confidence on the direction relevant to the question.

A one-sided t confidence bound uses the sample mean and sample standard deviation to estimate a population mean when the population standard deviation is unknown. The choice between an upper and a lower bound is determined by the question being asked, not by which direction happens to make the result look more favorable.

Definition: A lower one-sided t confidence bound for a population mean $\mu$ is a value $L$ such that the confidence set extends upward from $L$, written $[L,\infty)$. An upper one-sided t confidence bound is a value $U$ such that the confidence set extends downward to $U$, written $(-\infty,U]$. The confidence level describes the long-run proportion of bounds constructed by the stated method that contain $\mu$.

The Model, Conditions, and Formulas

Let $X_1,\ldots,X_n$ be a random sample of quantitative observations from a population with mean $\mu$. The population standard deviation $\sigma$ is unknown, so the standard error is estimated by $s/\sqrt{n}$. For a one-sample t bound, observations should be independent; when sampling without replacement, the sample should be no more than 10% of the population. For a small sample, the population distribution should be approximately normal, with no strong skewness or extreme outliers. With a larger sample, moderate departures from normality are generally less consequential, but severe skewness or outliers still deserve attention.

If the observations are independent and come from a normal population, the statistic $(\bar{x}-\mu)/(s/\sqrt{n})$ follows a t distribution with $df=n-1$. This result is exact under that model. When the population is not exactly normal, a t procedure is an approximation whose reliability depends on the sample size and the data’s shape. The sampling and independence conditions, the 10% condition, and checks for normality and outliers are essential parts of deciding whether the procedure is appropriate.

Write $t_{\alpha,df}$ for the positive t critical value with upper-tail probability $\alpha$: if $T$ has a t distribution with $df$ degrees of freedom, then $P(T>t_{\alpha,df})=\alpha$. A one-sided confidence level of $C$ corresponds to $\alpha=1-C$. The same positive critical value is used for an upper or lower bound; the sign in the formula changes with the direction of the bound.

Formulas: For a one-sided confidence level $1-\alpha$, with $df=n-1$, $$ L=\bar{x}-t_{\alpha,df}\frac{s}{\sqrt{n}} \qquad\text{and}\qquad U=\bar{x}+t_{\alpha,df}\frac{s}{\sqrt{n}}. $$ The lower confidence set is $[L,\infty)$; the upper confidence set is $(-\infty,U]$. Each bound uses the full tail probability $\alpha$, not $\alpha/2$.

Why the One-Tail Critical Value Works

The formulas follow from the sampling distribution of the t statistic. The derivation also explains why a one-sided bound uses a critical value different from the one used for a two-sided interval at the same confidence level.

Start with the t statistic.
Under the one-sample t model, $T=(\bar{x}-\mu)/(s/\sqrt{n})$ has a t distribution with $df=n-1$.

Use the upper-tail quantile.
By definition, $P(T\le t_{\alpha,df})=1-\alpha$, because $P(T>t_{\alpha,df})=\alpha$.

Rearrange for a lower bound.
$T\le t_{\alpha,df}$ is equivalent to $\bar{x}-\mu\le t_{\alpha,df}s/\sqrt{n}$, which is equivalent to $\mu\ge\bar{x}-t_{\alpha,df}s/\sqrt{n}$. Thus the random lower bound is at or below $\mu$ with probability $1-\alpha$.

Rearrange for an upper bound.
$T\ge-t_{\alpha,df}$ has probability $1-\alpha$, by the symmetry of the t distribution. This inequality is equivalent to $\mu\le\bar{x}+t_{\alpha,df}s/\sqrt{n}$. Thus the random upper bound is at or above $\mu$ with probability $1-\alpha$.

The probability statements in this justification describe repeated samples: the endpoints vary from sample to sample, while $\mu$ is fixed. The method does not assign a probability to the fixed parameter after the data have been observed.

Worked Examples

Worked Example: Finding an Upper Bound for Mean Waiting Time

Suppose a hypothetical clinic randomly samples $n=12$ patient visits and records waiting time in minutes. The sample mean is $\bar{x}=5.6$ minutes and the sample standard deviation is $s=1.2$ minutes. Assume the observations are independent, the sample is less than 10% of the clinic’s visits in the sampling period, and inspection shows no extreme outliers or strong skewness. Find a 95% upper confidence bound for the population mean waiting time.

A 95% bound has $\alpha=0.05$. The degrees of freedom are $df=12-1=11$, and the upper-tail critical value is $t_{0.05,11}\approx1.796$. The conditions stated in the question support using the one-sample t procedure.

First calculate the estimated standard error and then the upper endpoint:

SE=\frac{s}{\sqrt{n}}=\frac{1.2}{\sqrt{12}}\approx0.3464\text{ minutes}, \qquad U=\bar{x}+t_{0.05,11}SE

U=5.6+1.796\left(\frac{1.2}{\sqrt{12}}\right) \approx5.6+1.796(0.3464) \approx5.6+0.6221 =6.2221\text{ minutes}.

The 95% upper confidence bound is approximately 6.22 minutes. We are 95% confident that the population mean waiting time is no greater than about 6.22 minutes. This is an upper bound on the mean, not a claim that 95% of individual visits have waiting times below 6.22 minutes.

Worked Example: Finding a Lower Bound for Mean Recovery Time

In a hypothetical setting, a random sample of $20$ patients has a mean recovery time of $\bar{x}=42.3$ hours and a sample standard deviation of $s=5.0$ hours. Suppose the observations are independent, the sample is less than 10% of the relevant patient population, and the sample distribution has no marked skewness or extreme outliers. Find a 90% lower confidence bound for the population mean recovery time.

For a 90% one-sided bound, $\alpha=0.10$. There are $df=20-1=19$ degrees of freedom, and $t_{0.10,19}\approx1.3277$. The estimated standard error is

SE=\frac{5.0}{\sqrt{20}}\approx1.1180\text{ hours}.

Use the lower-bound formula, subtracting the margin of error from the sample mean:

L=42.3-1.3277\left(\frac{5.0}{\sqrt{20}}\right) \approx42.3-1.4844 =40.8156\text{ hours}.

The 90% lower confidence bound is approximately 40.82 hours. We are 90% confident that the population mean recovery time is at least about 40.82 hours. The confidence set extends upward from this value; the bound does not provide a finite upper endpoint.

Worked Example: Matching a High Confidence Level to One Tail

A hypothetical materials laboratory takes an independent random sample of $9$ test pieces. Their measured strength has sample mean $\bar{x}=18.4$ units and sample standard deviation $s=2.1$ units. Suppose the sample is less than 10% of the production population and the strength distribution is approximately normal without extreme outliers. Find a 99% upper confidence bound for the population mean strength.

The desired confidence level is 99%, so $\alpha=0.01$. With $df=9-1=8$, the upper-tail critical value is $t_{0.01,8}\approx2.896$. The estimated standard error is $2.1/\sqrt{9}=0.7$ units. Therefore,

U=18.4+2.896(0.7) =18.4+2.0272 =20.4272\text{ units}.

The 99% upper confidence bound is about 20.43 units. The one-sided procedure gives a single upper endpoint because the question asks for a limit above the population mean. A greater confidence level requires a larger critical value, so this upper endpoint is farther above the sample mean than it would be at a lower confidence level.

Choosing the Direction and Interpreting the Result

Choose an upper bound when the practical question concerns how large $\mu$ could plausibly be, such as whether the mean duration is below a specified maximum. Choose a lower bound when the question concerns how small $\mu$ could plausibly be, such as whether the mean strength exceeds a required minimum. The direction should follow the question before looking at the sample result.

A confidence level of $1-\alpha$ places all of the allowed error probability in one tail. For example, a 95% one-sided bound uses $\alpha=0.05$ and the critical value $t_{0.05,df}$. A 95% two-sided interval instead leaves a total of 0.05 outside its two endpoints, with 0.025 in each tail when constructed symmetrically. Thus its critical value is $t_{0.025,df}$, which is larger. The two procedures answer different questions and should not be confused.

A one-sided bound can also be more informative when only one direction matters. It should not be used merely to make an endpoint appear closer to the estimate. If both unusually low and unusually high values would be important, a two-sided interval is appropriate. If a threshold is being assessed, make sure that the bound’s direction matches the claim: an upper bound below a maximum is relevant evidence that the mean is below that maximum, while a lower bound above a minimum is relevant evidence that the mean is above that minimum.

Common Mistake

Wrong reasoning: “For a 95% upper bound, the total error probability is 0.05, so I should divide it between two tails and use $t_{0.025,df}$.” This treats the requested bound as though it had two endpoints.

Correction: A one-sided bound has one finite endpoint and one unbounded direction. Its full error probability is in one tail, so use $t_{0.05,df}$ for a 95% bound. Dividing $\alpha$ by two would use a more conservative critical value than the specified one-sided procedure. Also check the sign: subtract the margin for a lower bound and add it for an upper bound.

Key takeaway: For a one-sided t confidence bound, choose the direction from the question, use $df=n-1$, and take $t_{\alpha,df}$ with $\alpha=1-\text{confidence level}$. Subtract the margin for a lower bound; add it for an upper bound. State the result as a bound on the population mean in its original units.

Check Your Understanding

Answer each question using the definitions and formulas above.

A 90% upper bound is requested. What is $\alpha$, and which upper-tail t critical value should be used?
A random sample has $n=16$. What are the degrees of freedom for a one-sample t bound?
In a lower one-sided bound, why is the margin of error subtracted from $\bar{x}$ rather than added?
A researcher uses $t_{0.025,df}$ for a 95% one-sided bound. Identify the error in the tail choice.
In context, what does it mean to say that a 95% upper confidence bound for a population mean is 12.4 minutes?