Student t Quantiles and Tail Probabilities

Why t Quantiles and Tail Areas Need Care

A Student t calculation can be short and still go wrong if the probability is read from the wrong tail, the degrees of freedom do not match the model, or a negative cutoff is handled as if it were positive. These distinctions matter whenever a t probability or critical value is used. The key is to translate the question into a statement about the cumulative distribution function, then read or calculate the requested area consistently.

This tutorial focuses on probability and quantile calculations for a Student t variable with specified degrees of freedom. It assumes the standard properties of the t distribution, including its symmetry, and does not derive when a sample statistic has a t distribution.

Distribution, Cumulative Probabilities, and Quantiles

Definition: A random variable $T$ has a Student t distribution with $\nu$ degrees of freedom, written $T\sim t_\nu$, if its density is $$ f_\nu(t)=\frac{\Gamma((\nu+1)/2)}{\sqrt{\nu\pi}\,\Gamma(\nu/2)} \left(1+\frac{t^2}{\nu}\right)^{-(\nu+1)/2},\qquad -\infty<t<\infty, $$ where $\nu>0$ and $\Gamma$ is the gamma function. Its cumulative distribution function is $F_\nu(t)=P(T\le t)$. The $p$-quantile $t_{\nu,p}$ is the value satisfying $P(T\le t_{\nu,p})=p$, for $0<p<1$.

For the degrees of freedom commonly used in sampling problems, t tables often give an upper-tail critical value instead of a left-tail quantile. We will denote that value by $c_{\nu,\alpha}$, defined by $P(T>c_{\nu,\alpha})=\alpha$. Because the distribution is continuous, this is equivalently $F_\nu(c_{\nu,\alpha})=1-\alpha$, so $c_{\nu,\alpha}=t_{\nu,1-\alpha}$. Always check the table's heading: some tables index the area in one tail, while others index the cumulative probability or the total area in two tails.

Theorem: The Student t density is symmetric about zero. Consequently, $F_\nu(-x)=1-F_\nu(x)$, $t_{\nu,p}=-t_{\nu,1-p}$, and $P(T<-x)=P(T>x)$ for $x>0$.

The distribution depends on its degrees of freedom. A critical value for $t_7$, for example, is not interchangeable with a critical value for $t_{15}$. The numerical values in the examples below are t-table or distribution-quantile values, rounded to four decimal places. The stated probabilities are rounded to four decimal places as well.

A Short Justification for Symmetry

The density formula contains $t$ only through $t^2$, so $f_\nu(t)=f_\nu(-t)$. For $x>0$, changing the integration variable from $t$ to $-u$ shows that the area to the left of $-x$ equals the area to the right of $x$:

P(T<-x)=\int_{-\infty}^{-x} f_\nu(t)\,dt =\int_x^\infty f_\nu(-u)\,du =\int_x^\infty f_\nu(u)\,du =P(T>x).

Since the total area under the density is 1 and there is no point mass at a cutoff, $P(T\le x)=1-P(T>x)$. These facts give the stated CDF and quantile relationships. They let us calculate many negative-cutoff probabilities using a positive upper-tail entry in a table.

A Reliable Calculation Process

First identify the degrees of freedom, then express the requested event as a left-tail probability, upper-tail probability, or interval. For an interval $(a,b)$, use the CDF at both endpoints. For a negative cutoff, use symmetry rather than treating its table entry as if it were a positive cutoff.

Record the model and degrees of freedom.
Write $T\sim t_\nu$, and use the specified $\nu$ throughout the calculation.

Translate the event.
For example, $P(T>x)$ is an upper tail, while $P(T<x)$ is a left tail. Split a union of disjoint tails into separate probabilities.

Match the table convention.
If a table gives upper-tail areas, use $c_{\nu,\alpha}$ directly for $P(T>c_{\nu,\alpha})=\alpha$. If it gives cumulative probabilities, use $t_{\nu,p}$ for $P(T\le t_{\nu,p})=p$.

Apply symmetry and check the result.
Use $P(T<-x)=P(T>x)$ for $x>0$, and verify that the requested area is between 0 and 1 and has the expected size.

Worked Probability and Quantile Calculations

Worked Example: An Interval With Unequal Endpoints

A hypothetical measurement score $T$ is modeled as $t_9$. Find $P(-1.8331<T<2.2622)$, using the t quantiles supplied here: $t_{9,0.95}=1.8331$ and $t_{9,0.975}=2.2622$. No interest or mortality assumption applies; the probability model is $T\sim t_9$.

Method. For a continuous variable, the probability between two cutoffs is the CDF at the upper cutoff minus the CDF at the lower cutoff: $P(a<T<b)=F_9(b)-F_9(a)$.

The upper endpoint is the 0.975 quantile, so $F_9(2.2622)=0.975$. The positive value $1.8331$ is the 0.95 quantile. By symmetry, $-1.8331$ is the 0.05 quantile, so $F_9(-1.8331)=0.05$. Therefore,

$$ P(-1.8331<T<2.2622)=0.975-0.05=0.925. $$

The requested probability is 0.9250 to four decimal places. The two endpoints do not have equal magnitudes, so this is not a symmetric central interval. As a check, the interval excludes 0.05 in the lower tail and 0.025 in the upper tail; the remaining area is $1-0.05-0.025=0.925$. The common mistake is to subtract the upper-tail area at the lower endpoint from the CDF at the upper endpoint without first converting conventions.

Worked Example: Combining a Left Tail and a Right Tail

A hypothetical process produces a standardized score $T\sim t_7$. Find the probability that $T<-2.3646$ or $T>1.4149$. Use $t_{7,0.975}=2.3646$ and $t_{7,0.90}=1.4149$. No interest or mortality assumption applies; the distribution and degrees of freedom are $t_7$.

Method. The two events are disjoint, so their probabilities add. First, $P(T>1.4149)=1-0.90=0.10$, because $1.4149$ is the 0.90 quantile. Next, symmetry gives $P(T<-2.3646)=P(T>2.3646)$. Since $2.3646$ is the 0.975 quantile, its upper-tail area is $1-0.975=0.025$.

P(T<-2.3646\ \text{or}\ T>1.4149) =0.025+0.10=0.125.

The probability is 0.1250 to four decimal places. The answer is plausible: the event includes a 2.5% left tail and a 10% right tail. A common mistake is to assign probability 0.975 to the left-tail event below $-2.3646$; that confuses the positive cutoff's cumulative probability with the negative cutoff's cumulative probability. The correct left-tail area is 0.025.

Worked Example: Finding a Central Interval and Transforming It

A hypothetical index is $Y=4+1.5T$, where $T\sim t_{15}$. Find the central 98% interval for $Y$. Use the supplied upper-tail critical value $c_{15,0.01}=2.6025$, so $P(T>2.6025)=0.01$. No interest or mortality assumption applies; the model for $T$ is $t_{15}$.

Method. A central 98% interval leaves $1-0.98=0.02$ outside the interval. By symmetry, each tail has area $0.02/2=0.01$. Thus the central interval for $T$ is $(-2.6025,2.6025)$.

The transformation $Y=4+1.5T$ has positive multiplier 1.5, so it preserves the order of the endpoints. Transform each endpoint:

4+1.5(-2.6025)=4-3.90375=0.09625, \qquad 4+1.5(2.6025)=4+3.90375=7.90375.

Rounding only the final endpoints to four decimal places gives the central interval (0.0963, 7.9038). Its probability is 0.98 because the original $T$ interval has 0.01 in each tail, and the increasing transformation carries that interval to the stated interval for $Y$. The common mistake is to use the 0.98 quantile as the positive endpoint; a central 98% interval requires the 0.99 quantile because only 1% is above its positive endpoint.

Common Table and Tail Errors

The notation printed in a table is not universal. Before using a number, check whether the heading describes a cumulative probability $P(T\le t)$, an upper-tail probability $P(T>t)$, or a total two-tail probability $P(|T|>t)$. For example, a central 95% interval has 2.5% in each tail. Its positive endpoint is $t_{\nu,0.975}$, equivalently the upper-tail critical value $c_{\nu,0.025}$. It is not the 0.95 quantile if the table's probabilities are cumulative.

A particularly common wrong calculation is to say that $P(T<-c_{\nu,0.05})=0.95$, because $c_{\nu,0.05}$ has an upper-tail area of 0.05. The correction uses symmetry: $P(T<-c_{\nu,0.05})=P(T>c_{\nu,0.05})=0.05$. The 0.95 area lies to the left of the positive cutoff, not to the left of its negative.

Use the specified degrees of freedom. A t critical value changes with $\nu$; do not borrow a value from a nearby row without a stated approximation.
Keep the tail convention visible. Write the probability represented by a table value before substituting it.
For intervals, subtract CDF values. $P(a<T<b)=F_\nu(b)-F_\nu(a)$ whenever $a<b$.
For central intervals, split the excluded probability equally. A central probability $1-\alpha$ leaves $\alpha/2$ in each tail.
Do not round intermediate probabilities. Keep the table values and arithmetic unrounded until the final reported probability.

Key takeaway: A t probability is determined by the degrees of freedom, the cutoff, and which tail or interval is requested. State all three, identify the table convention, and then use the CDF or symmetry to convert the table entry to the desired area.

Check Your Understanding

For each question, use the stated t quantile or critical value. Probabilities should be reported to four decimal places.

Let $T\sim t_6$, with $t_{6,0.95}=1.9432$. Find $P(T>1.9432)$. (Answer: 0.0500)
Let $T\sim t_{10}$, with $t_{10,0.975}=2.2281$. Find $P(T<-2.2281)$. (Answer: 0.0250)
Let $T\sim t_8$, with $t_{8,0.95}=1.8595$ and $t_{8,0.975}=2.3060$. Find $P(-1.8595<T<2.3060)$. (Answer: 0.9250)
For $T\sim t_{12}$, the 0.95 quantile is $1.7823$. Find the central 90% interval for $T$. (Answer: $(-1.7823,1.7823)$)
Let $T\sim t_5$, with $t_{5,0.95}=2.0150$ and $t_{5,0.975}=2.5706$. Find $P(T<-2.0150\ \text{or}\ T>2.5706)$. (Answer: 0.0750)

What You'll Learn