Math Core

In Unit 6 you built confidence intervals and tests for proportions using $z$ critical values and the standard normal curve. Means are different in one important way: you almost never know the population standard deviation $\sigma$ , so you have to estimate it from the sample. That extra estimate adds extra uncertainty, and the $t$ -distributions are the tool that accounts for it.

The problem with standardizing a sample mean

From Unit 5, the sampling distribution of $\bar{x}$ has mean $\mu$ and standard deviation $\dfrac{\sigma}{\sqrt{n}}$ . When that sampling distribution is approximately normal, the standardized statistic

z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}}

follows the standard normal distribution. The catch is $\sigma$ . If you don't know the population mean $\mu$ , you almost certainly don't know the population standard deviation either.

The natural fix is to replace $\sigma$ with the sample standard deviation $s$ . The result is called the standard error of the sample mean.

Definition

Standard error of the mean

When $\sigma$ is unknown, the standard error of $\bar{x}$ is

SE_{\bar{x}} = \frac{s}{\sqrt{n}},

where $s$ is the sample standard deviation and $n$ is the sample size. It estimates how far $\bar{x}$ typically falls from $\mu$ .

Now the standardized statistic is

t = \frac{\bar{x} - \mu}{s / \sqrt{n}}.

This is not a $z$ -score anymore. The numerator varies from sample to sample, and so does the denominator, because $s$ changes from sample to sample too. Sometimes $s$ underestimates $\sigma$ , which makes $t$ larger in size than $z$ would have been. The result is a distribution that is more spread out than the standard normal, especially for small samples.

Meet the t-distributions

William Gosset worked out the distribution of this statistic in 1908 while studying small samples at a brewery. There isn't just one $t$ -distribution. There is a whole family, and each member is identified by its degrees of freedom.

Properties of the t-distributions

When you draw an SRS of size $n$ from a normal population, the statistic $t = \dfrac{\bar{x} - \mu}{s/\sqrt{n}}$ has a $t$ -distribution with $df = n - 1$ degrees of freedom. Every $t$ -distribution:

is symmetric, single-peaked and centered at $0$ , like the standard normal;
has heavier tails and a lower peak than the standard normal, so more of its area lies far from $0$ ;
gets closer and closer to the standard normal as $df$ increases.

Why $n - 1$ ? The sample standard deviation is built from the deviations $x_i - \bar{x}$ , and those deviations always add to $0$ . Once you know $n - 1$ of them, the last one is forced. Only $n - 1$ deviations are free to vary, so that's how much independent information $s$ contains.

The heavier tails are the whole point. Because $s$ is only an estimate, you need to go farther from the center to capture the same middle area. For example, the middle 95% of the standard normal lies between $-1.96$ and $1.96$ . The middle 95% of the $t$ -distribution with $4$ degrees of freedom lies between $-2.776$ and $2.776$ .

Reading a t-table

A $t$ -table lists critical values $t^*$ . Each row is a number of degrees of freedom, and each column is an upper-tail area (or, equivalently, a confidence level for the middle area). Here is an excerpt.

$df$	tail 0.10	tail 0.05	tail 0.025	tail 0.01	tail 0.005
confidence	80%	90%	95%	98%	99%
4	1.533	2.132	2.776	3.747	4.604
7	1.415	1.895	2.365	2.998	3.499
9	1.383	1.833	2.262	2.821	3.250
11	1.363	1.796	2.201	2.718	3.106
14	1.345	1.761	2.145	2.624	2.977
15	1.341	1.753	2.131	2.602	2.947
19	1.328	1.729	2.093	2.539	2.861
20	1.325	1.725	2.086	2.528	2.845
24	1.318	1.711	2.064	2.492	2.797
29	1.311	1.699	2.045	2.462	2.756
30	1.310	1.697	2.042	2.457	2.750
40	1.303	1.684	2.021	2.423	2.704
60	1.296	1.671	2.000	2.390	2.660
$z^*$ ( $\infty$ )	1.282	1.645	1.960	2.326	2.576

Read down any column and the critical values shrink toward the $z^*$ value in the last row. That's the "approaches the normal" property in numbers.

For a confidence interval with confidence level $C$ , the area $C$ sits in the middle and the leftover area $1 - C$ is split between the two tails. So a 95% interval uses the column with upper-tail area $0.025$ .

Worked example: Finding a critical value

You plan a 95% confidence interval for a mean from a random sample of $n = 20$ observations. What critical value $t^*$ should you use?

Solution. The degrees of freedom are $df = 20 - 1 = 19$ . A 95% confidence level leaves $0.05$ in the two tails together, so $0.025$ in the upper tail. In the row for $df = 19$ and the $0.025$ column, $t^* = 2.093$ .

On a TI-83/84, invT(0.975, 19) gives the same value: $2.093$ . (You enter the area to the left of $t^*$ , which is $0.95 + 0.025 = 0.975$ .)

Common mistake

Don't use $df = n$ . The degrees of freedom for one-sample $t$ procedures are $n - 1$ . Also watch the column: for a 95% interval you need upper-tail area $0.025$ , not $0.05$ . Using the $0.05$ column gives a 90% interval.

Tail areas and p-values

In a significance test you go the other direction: you have a $t$ statistic and need the area beyond it. A table can only bracket that area. Technology gives the exact value.

Worked example: Finding a tail area

Find $P(t > 2.1)$ for a $t$ -distribution with $9$ degrees of freedom.

Solution with the table. In the $df = 9$ row, $2.1$ falls between $1.833$ (tail $0.05$ ) and $2.262$ (tail $0.025$ ). So the upper-tail area is between $0.025$ and $0.05$ .

Solution with technology. tcdf(2.1, 1E99, 9) gives about $0.0326$ . That fits inside the bracket from the table.

Worked example: A lower tail

Find $P(t < -1.5)$ when $df = 15$ .

Solution. By symmetry, $P(t < -1.5) = P(t > 1.5)$ . In the $df = 15$ row, $1.5$ lies between $1.341$ (tail $0.10$ ) and $1.753$ (tail $0.05$ ), so the area is between $0.05$ and $0.10$ . Technology, tcdf(-1E99, -1.5, 15), gives about $0.0772$ .

Compare with the standard normal: $P(z < -1.5) \approx 0.0668$ . The $t$ area is larger because of the heavier tails.

Tip

A quick sanity check: for the same confidence level, $t^*$ is always a bit bigger than $z^*$ . If your $t^*$ comes out smaller than $1.96$ for a 95% interval, you've used the wrong column or the wrong row.

Why this matters for the rest of the unit

Every procedure in this unit (one-sample intervals, one-sample tests, paired data and two-sample comparisons) uses the same recipe:

\text{statistic} \pm t^* \cdot SE \qquad \text{or} \qquad t = \frac{\text{statistic} - \text{null value}}{SE}.

What changes from lesson to lesson is the statistic, the standard error and the degrees of freedom. Once you're comfortable with $t^*$ and tail areas, the rest is careful bookkeeping.

Practice

Practice 1

A researcher takes a random sample of $18$ bags of trail mix and plans to use a one-sample $t$ procedure. How many degrees of freedom does the procedure use?