Exact Student t Intervals For One Population Mean

Why Use a Student t Interval?

A confidence interval for a population mean gives a range of plausible values for that mean, with a stated long-run coverage probability. When the population variance is known, a normal critical value can be used. In many applications, however, the variance is unknown and must be estimated from the sample. That estimation adds uncertainty, which is accounted for by a Student t critical value.

This tutorial concerns a two-sided interval for one population mean. Its exact coverage depends on a specific model: the observations must be an independent random sample from a normal population. The procedure is aligned with the SOA VEE Mathematical Statistics syllabus.

Definition: Let $X_1,\ldots,X_n$ be an independent random sample from a normal population with mean $\mu$ and unknown variance $\sigma^2$. Define the sample mean and sample standard deviation by $$ \bar X=\frac{1}{n}\sum_{i=1}^{n}X_i, \qquad S=\sqrt{\frac{1}{n-1}\sum_{i=1}^{n}(X_i-\bar X)^2}. $$ For confidence level $1-\alpha$, the exact two-sided Student t confidence interval for $\mu$ is $$ \left(\bar X-t_{1-\alpha/2,n-1}\frac{S}{\sqrt n},\; \bar X+t_{1-\alpha/2,n-1}\frac{S}{\sqrt n}\right). $$ Here $t_{p,\nu}$ is the $p$-quantile of a Student t distribution with $\nu$ degrees of freedom.

Reading the Formula Correctly

The interval has the familiar form estimate plus or minus critical value times standard error. The point estimate is $\bar X$. Because $\sigma$ is unknown, the estimated standard error is $S/\sqrt n$, not $\sigma/\sqrt n$. The critical value is selected from a t distribution with $n-1$ degrees of freedom.

For a two-sided confidence level $1-\alpha$, the total probability left outside the central interval is $\alpha$. The two tails each contain probability $\alpha/2$, so the required upper quantile is $t_{1-\alpha/2,n-1}$. For example, a 90% interval has $\alpha=0.10$, uses the 0.95 quantile, and leaves 0.05 in each tail. A 95% interval uses the 0.975 quantile.

\text{margin of error} = t_{1-\alpha/2,n-1}\frac{S}{\sqrt n}, \qquad \text{interval}=\bar X\pm\text{margin of error}.

The confidence level describes the performance of the procedure over repeated samples: under the stated model, the method produces intervals that contain the fixed mean $\mu$ with probability $1-\alpha$. After a particular sample has been observed, the endpoints are fixed. It is not correct to interpret the interval as assigning probability $1-\alpha$ to the fixed parameter being inside those already calculated endpoints.

Why the t Interval Has This Form

For a normal sample, the standardized sample mean using the unknown population standard deviation has a standard normal distribution. Estimating that standard deviation with $S$ changes the standardized quantity: the resulting pivot has a Student t distribution with $n-1$ degrees of freedom. In precise form, under the model stated above,

T=\frac{\bar X-\mu}{S/\sqrt n}\sim t_{n-1}.

This pivot is useful because its distribution does not depend on the unknown $\mu$ or $\sigma$. The central t quantiles satisfy

P\left(-t_{1-\alpha/2,n-1} \leq \frac{\bar X-\mu}{S/\sqrt n} \leq t_{1-\alpha/2,n-1}\right)=1-\alpha.

To obtain an interval for $\mu$, multiply all parts of the inequality by the positive quantity $S/\sqrt n$, then rearrange to isolate $\mu$:

-t_{1-\alpha/2,n-1}\frac{S}{\sqrt n} \leq \bar X-\mu \leq t_{1-\alpha/2,n-1}\frac{S}{\sqrt n} $$ $$ \Longleftrightarrow \quad \bar X-t_{1-\alpha/2,n-1}\frac{S}{\sqrt n} \leq \mu \leq \bar X+t_{1-\alpha/2,n-1}\frac{S}{\sqrt n}.

This inversion explains why the interval is centered at $\bar X$ and why its margin of error is the t critical value multiplied by the estimated standard error. The exactness comes from the t pivot under the normal-sample model, not merely from the use of a t table.

Constructing an Interval from Raw Data

Worked Example: A 90% Interval for Processing Time

A hypothetical analyst records processing times, in minutes, for five independent claims from a process assumed to have a normal distribution. The observations are $10,12,14,16,18$. Construct a 90% confidence interval for the population mean processing time. Use $t_{0.95,4}=2.13185$ (approximately).

Method and model. The observations are an independent random sample from a normal population with unknown mean $\mu$ and variance $\sigma^2$. Use the one-sample t interval. The sample size is $n=5$, so the degrees of freedom are $n-1=4$. Because the interval is 90%, $\alpha=0.10$, and each tail has probability $0.05$. Thus the critical value is $t_{0.95,4}$.

Calculate the sample mean. $$ \bar X=\frac{10+12+14+16+18}{5}=\frac{70}{5}=14\text{ minutes}. $$

Calculate the sample standard deviation. The deviations from 14 are $-4,-2,0,2,4$, and their squared values sum to $16+4+0+4+16=40$. The sample variance uses denominator $n-1=4$: $$ S^2=\frac{40}{4}=10, \qquad S=\sqrt{10}\approx3.1623\text{ minutes}. $$

Calculate the standard error and margin. $$ \frac{S}{\sqrt n}=\frac{\sqrt{10}}{\sqrt5}=\sqrt2\approx1.4142, \qquad 2.13185\sqrt2\approx3.0149\text{ minutes}. $$

Form the endpoints. $$ 14-3.0149=10.9851, \qquad 14+3.0149=17.0149. $$ Rounded to two decimal places, the 90% confidence interval is (10.99, 17.01) minutes.

Check. The endpoints are equally far from the sample mean: $14-10.9851=3.0149$ and $17.0149-14=3.0149$. The interval is centered correctly and uses the 90% critical value, not the 95% critical value. A common mistake is to use $t_{0.975,4}\approx2.776$, which corresponds to a 95% two-sided interval, not a 90% interval.

Intervals from Summary Statistics

When the sample observations are not listed, the calculation can still be completed if the sample size, sample mean, and sample standard deviation are supplied. Be sure that the stated standard deviation is the sample standard deviation calculated with denominator $n-1$, as required by the t procedure.

Worked Example: Claim Processing Hours with Summary Statistics

In a hypothetical study, 12 independent processing times for claims are modeled as a normal sample with unknown mean and variance. The sample mean is 150 hours and the sample standard deviation is 18 hours. Construct a 95% confidence interval for the population mean. Use $t_{0.975,11}=2.201$.

Method and degrees of freedom. Since $\sigma$ is unknown and the model is normal, use the one-sample t interval. Here $n=12$, so the degrees of freedom are $12-1=11$. For a 95% interval, $\alpha=0.05$, giving $1-\alpha/2=0.975$.

Calculate the standard error. $$ \frac{S}{\sqrt n}=\frac{18}{\sqrt{12}} =\frac{18}{3.4641} \approx5.1962\text{ hours}. $$

Calculate the margin of error. $$ 2.201\left(\frac{18}{\sqrt{12}}\right) \approx2.201(5.1962) \approx11.4367\text{ hours}. $$

Calculate and report the endpoints. $$ 150-11.4367=138.5633, \qquad 150+11.4367=161.4367. $$ Rounded to two decimal places, the interval is (138.56, 161.44) hours.

Check. The midpoint is $(138.56+161.44)/2=150.00$ hours, and the unrounded margin is approximately 11.4367 hours. The common mistake avoided here is using the sample standard deviation, 18, as the margin itself; it must first be divided by $\sqrt{12}$, then multiplied by the t critical value.

How Confidence Level and Sample Size Affect the Interval

Holding the observed sample statistics fixed, increasing the confidence level increases the t critical value and therefore widens the interval. Increasing the sample size generally decreases the standard error $S/\sqrt n$, although the critical value also depends on the degrees of freedom. The direction and size of these changes should be understood through both parts of the margin, rather than by looking only at the critical value.

Worked Example: A 90% Interval for a Hypothetical Audit Measure

A hypothetical audit team obtains nine independent measurements from a normal population. The population variance is unknown; the sample mean is 42 units and the sample standard deviation is 6 units. Construct a 90% confidence interval, using $t_{0.95,8}=1.860$, and state its width.

Method and degrees of freedom. Use the one-sample t interval because the variance is unknown. The sample size is $n=9$, so the degrees of freedom are $8$. For a 90% interval, use the 0.95 quantile.

Calculate the standard error. $$ \frac{S}{\sqrt n}=\frac{6}{\sqrt9}=\frac{6}{3}=2\text{ units}. $$

Calculate the margin. $$ 1.860(2)=3.720\text{ units}. $$

Calculate the interval and width. $$ 42-3.720=38.280, \qquad 42+3.720=45.720. $$ The interval is (38.28, 45.72) units, rounded to two decimal places. Its width is $45.72-38.28=7.44$ units.

Check. The interval midpoint is $(38.28+45.72)/2=42.00$, and twice the margin is $2(3.72)=7.44$, matching the reported width. A higher confidence level would require a larger critical value and hence a wider interval for these same data.

Common Mistakes and Exam Traps

The most important checks are the tail probability, the degrees of freedom, and the standard error. Each one affects the endpoints, and confusing them can produce an interval that looks plausible but has the wrong coverage or width.

Using the wrong quantile. For a two-sided 90% interval, use $t_{0.95,n-1}$, not $t_{0.90,n-1}$ or $t_{0.975,n-1}$. The latter is used for a two-sided 95% interval.
Using $n$ degrees of freedom. The one-sample t pivot uses $n-1$ degrees of freedom. This is not a choice based on the number of observations minus one for convenience; it is the distribution of the pivot for this model.
Forgetting to divide by $\sqrt n$. The sample standard deviation estimates spread among individual observations. The standard error estimates spread of the sample mean and is $S/\sqrt n$.
Substituting a z critical value while claiming an exact t interval. The normal critical value does not account for estimating $\sigma$. Using it instead of the t critical value does not give the stated exact t procedure.
Calling the realized interval a probability statement about $\mu$. The mean is fixed; the random interval is what changes from sample to sample. The confidence level describes the repeated-sampling coverage of the procedure.

For example, in the 12-observation calculation above, using $1.96$ in place of $2.201$ would give a margin of approximately $1.96(5.1962)=10.1845$ hours and an interval of approximately $(139.82,160.18)$. That is not the requested exact 95% t interval. The proper calculation uses $t_{0.975,11}=2.201$, giving $(138.56,161.44)$ hours.

Key takeaway: For an independent normal sample with unknown variance, calculate $\bar X$ and $S$, use $S/\sqrt n$ as the estimated standard error, and use $t_{1-\alpha/2,n-1}$ for a two-sided $100(1-\alpha)\%$ interval. The exact coverage statement relies on the normal-sample model.

Check Your Understanding

For each problem, use the stated model and critical value. Round interval endpoints to two decimal places.

A hypothetical sample of 10 independent normal observations has sample mean 80 and sample standard deviation 5. Construct a 90% confidence interval using $t_{0.95,9}=1.833$. (Answer: $(77.10,\ 82.90)$)
A hypothetical normal sample has $n=4$, sample mean 25, and sample standard deviation 4. Construct a 95% confidence interval using $t_{0.975,3}=3.182$. (Answer: $(18.64,\ 31.36)$)
A hypothetical independent normal sample has $n=16$, sample mean 12.4, and sample standard deviation 2.4. Construct a 99% confidence interval using $t_{0.995,15}=2.947$. (Answer: $(10.63,\ 14.17)$)
Five independent measurements from a hypothetical normal population are $7,9,10,12,12$. Construct a 90% confidence interval for the population mean using $t_{0.95,4}=2.132$. (Answer: $(7.98,\ 12.02)$)

What You'll Learn