Back to the on-screen lesson ·
What replacing a known standard deviation by one estimated from the same sample costs, why the critical value grows, and how fast the penalty disappears.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
You will count the degrees of freedom a one-sample interval has, take the critical value from the t distribution rather than the normal one, and build the interval it gives. You will also say why the estimated denominator widens the interval and how quickly the penalty falls away as the sample grows.
The last lesson assumed the population standard deviation was a known number. It almost never is. This lesson replaces it with an estimate from the same sample, and works out what that substitution costs — because it does cost something, and pretending otherwise makes the interval claim more than it delivers.
The t distribution with $k$ degrees of freedom is the distribution of a standard normal divided by the square root of an independent chi-square over its own degrees of freedom. It is symmetric, bell-shaped and has heavier tails than the normal, approaching it as $k$ grows. A t interval uses its critical value in place of the normal one.
With $\sigma$ known, the pivot $\dfrac{\bar X - \mu}{\sigma/\sqrt n}$ is standard normal. Replace $\sigma$ by the sample standard deviation $S$ and the pivot becomes
$$T = \frac{\bar X - \mu}{S/\sqrt n},$$
which for a normal population follows the t distribution on $n - 1$ degrees of freedom. One degree of freedom was spent estimating the mean, and the count is what is left.
The denominator is now random. On some samples $S$ comes out smaller than $\sigma$, and on those samples $T$ is larger than $Z$ would have been. Large values therefore happen more often than under the normal, which is what heavier tails means, and catching the middle $95\%$ needs a cut point further out. The interval is
$$\bar x \pm t_{n-1,\,\alpha/2}\,\frac{s}{\sqrt n},$$
and it is wider than the normal interval at every sample size.
How much wider depends on $n$. At $5$ observations the critical value is $2.776$ against $1.960$ — a $42\%$ penalty. At $30$ it is $2.045$, a penalty of four per cent. By a few hundred observations the difference is invisible, which is why large-sample intervals in print use $1.96$ and say nothing about degrees of freedom.
Another way: picture
Two bells drawn on the same axes. The t curve is a little lower in the middle and a little thicker in both tails; the normal is the sharper of the two. Sliding the cut points outward until the same $95\%$ is caught is exactly what a larger critical value does, and the two curves visibly merge as the degrees of freedom rise.
Another way: steps
Two-sided $95\%$ critical values, and how much wider the interval is than the one that would be built if $\sigma$ were known.
| Sample size | Degrees of freedom | Critical value | Wider by |
|---|---|---|---|
| 5 | 4 | 2.776 | 42% |
| 10 | 9 | 2.262 | 15% |
| 20 | 19 | 2.093 | 7% |
| 30 | 29 | 2.045 | 4% |
The penalty is large where samples are small, which is exactly where it matters and exactly where it is most often ignored. A five-observation study that quotes $1.96$ is claiming a precision it has not got by more than a third.
Using $n$ degrees of freedom instead of $n - 1$. The correction is small in the critical value and it is the whole point of the count.
Blaming the t distribution on the population's shape. The t distribution arises for a perfectly normal population. What makes it appear is the estimated denominator, not a non-normal numerator.
Reading the wider interval as a higher confidence level. The level is the same $95\%$. The interval is wider because the procedure knows less.
Using the normal critical value on a small sample for convenience. The interval then covers less often than it claims, and the shortfall is largest where the sample is smallest.
$n = 10$, $\bar x = 25$, $s = 6$, so the standard error is $6/\sqrt{10} \approx 1.897$.
The sample's own spread.
Nine degrees of freedom gives a critical value of $2.262$, and a half-width of about $4.29$.
One fewer than ten.
The interval is about $(20.71, 29.29)$; using $1.96$ would have given $(21.28, 28.72)$ and claimed too much.
The penalty, in numbers.
$n = 1000$, standard error $0.5$: the critical value on $999$ degrees of freedom is $1.962$.
Two thousandths above the normal value.
The half-width is $0.981$ against $0.980$ with the normal value.
A difference nothing would notice.
Reporting either is defensible; reporting neither and giving no standard error is not.
The approximation is fine; the omission is not.
The degrees of freedom are $15$.
One fewer than the sample size.
The critical value at $15$ degrees of freedom is $2.131$, so the half-width is $4.262$.
Critical value times standard error.
The interval is $(55.738, 64.262)$.
A sample of $13$ observations has mean $67$, and the standard error of that mean, computed from the sample's own standard deviation, is $3$. Fill in the four lines of the $95\%$ interval.
| Value | |
|---|---|
| Degrees of freedom | |
| Half-width of the interval | |
| Lower end | |
| Upper end |
A sample of $10$ observations has mean $32$ and a standard error of $2$, computed from its own standard deviation. Give the $95\%$ confidence interval for the population mean, with both endpoints included.
This task has no paper form; do it on a device.
A $95\%$ interval for a population mean is built from a sample of $33$, with the standard deviation estimated from the same data. How many degrees of freedom does the critical value use?
Answer:
The standard error of a mean is $4$, from a sample of $25$. Give the $95\%$ half-width if the population spread were known, and the half-width that is actually correct because it was estimated from these data.
Spread known: a. Spread estimated: b.
Four studies, each reporting a $95\%$ interval for a mean of about $48$. Match each to the critical value it should use.
| $1.960$, from the normal distribution | $2.306$, from the t distribution on $8$ degrees of freedom | $2.064$, from the t distribution on $24$ degrees of freedom | $1.960$ in practice, the t distribution being indistinguishable from the normal by now | |
|---|---|---|---|---|
| The population spread is known from long experience; $12$ observations | ||||
| The spread is estimated from the sample; $9$ observations | ||||
| The spread is estimated from the sample; $25$ observations | ||||
| The spread is estimated from the sample; $2000$ observations |
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
From a sample of $7$ with the spread estimated from the same data, the critical value used is larger than $1.960$. Why?
You can build an interval when the spread was estimated from the same data, count the degrees of freedom, and say how much wider the interval is than the known-spread one. Say in your own words why an estimated denominator gives the pivot heavier tails.
9. Your turn: $n = 16$, $\bar x = 60$, standard error $2$, at the $95\%$ level, step 3