A Confidence Interval for a Mean
| English | Chinese | Pinyin |
|---|---|---|
| degrees of freedom | 自由度 | zì yóu dù |
Check the conditions
- Before a $t$-interval for $\mu$, verify three conditions:
- Random: the data come from a random sample or randomized experiment.
- 10%: $n \le 0.10N$ if sampling without replacement.
- Normal/large sample: the population is normal, or $n \ge 30$ (CLT), or the data show no strong skew/outliers.
Degrees of freedom
- The $t$-distribution's exact shape depends on the degrees of freedom 自由度.
- For a one-sample $t$-interval, $df = n - 1$.
- More degrees of freedom → the $t$ looks more like the normal.
- The $df$ picks which $t$-curve (and thus which $t^{*}$) to use.
Build the interval
- The interval is again estimate $\pm$ margin of error:
-
$$\bar{x} \pm t^{*}\frac{s_x}{\sqrt{n}}$$
- $\dfrac{s_x}{\sqrt{n}}$ is the standard error of the mean; $t^{*}$ comes from the $t$-table at $df = n-1$.
- The result is a range of plausible values for $\mu$.
Margin of error and level
- The margin of error is $t^{*} \times \dfrac{s_x}{\sqrt{n}}$.
- A higher confidence level → larger $t^{*}$ → wider interval.
- A larger sample → smaller $SE$ (and larger $df$) → narrower interval.
- Same trade-off as the proportion interval, now with $t$.
Use $t^{*}$ with $df = n-1$, not $z^{*}$. The critical value comes from the $t$-distribution at the right degrees of freedom — using $z^{*}=1.96$ for a mean makes the interval too narrow. And the standard error is $s_x/\sqrt{n}$ (sample SD over $\sqrt{n}$), not $s_x/n$.
$n = 25$ apples, $\bar{x} = 150$ g, $s_x = 10$ g; build a $95\%$ interval. ($df = 24$, so $t^{*} \approx 2.06$.)
- SE: $\dfrac{10}{\sqrt{25}} = \dfrac{10}{5} = 2$.
- Margin of error: $2.06 \times 2 = 4.12$.
- Interval: $150 \pm 4.12 = (145.9,\ 154.1)$ g.
After checking random, 10%, and normal/large sample, a one-sample $t$-interval for $\mu$ is $\bar{x} \pm t^{*}\frac{s_x}{\sqrt{n}}$, using the $t$-distribution with $df = n-1$. The margin of error is $t^{*} \times SE$; higher confidence widens it, and a larger sample narrows it.
The t-interval area
The interval uses t* at df = n − 1, not z*.
For a one-sample t-interval with n = 25, what are the degrees of freedom (n − 1)?
df = n − 1 = 25 − 1 = 24.
s = 10, n = 25. Compute the standard error of the mean s/√n.
10/√25 = 10/5 = 2.
With SE = 2 and t* = 2.06, find the margin of error t*·SE. Two decimals.
2.06 × 2 = 4.12.
For a confidence interval for a mean (σ unknown), the critical value comes from...
Use t* with n − 1 degrees of freedom.
The standard error of the mean is s divided by n.
It's s/√n, not s/n.