Estimation: the band around the number
Introducido| English | Significado |
|---|---|
| central limit theorem/ˈsentrəl ˈlɪmɪt ˈθɪərəm/ | central limit theorem |
| standard error/ˈstændəd ˈerə/ | The standard deviation of a statistic's sampling distribution, often estimated from a sample. |
| confidence interval/ˈkɒnfɪdəns ˈɪntəvl/ | An interval calculated from sample data using a method with a stated long-run probability of containing the population parameter. |
The mean is itself a random variable
- Measure a different Saturday and the mean moves. Estimation is the discipline of saying what ONE sample says about the truth you cannot see — with an honest band.
- The one number everyone quotes has a distribution of its own, and that distribution is the whole machinery.
The central limit theorem
- Draw 16 orders, take the mean; draw again; again. The means pile into their own bell — narrower than the data, centred on the truth. That is the central limit theorem 中心极限定理.
- The width shrinks with $\sqrt{n}$: the data's sd 4.91 becomes the mean's sd 2.45 at n = 4 and 1.23 at n = 16 ✓.
The central limit theorem says the raw delivery times become normal as the sample grows.
Dificultad 1/5 · 100 monedas + un pequeño bono en el primer intento de hoy
The SAMPLE MEANS become normal as n grows - the data keep whatever shape they have. That is why means, not data, earn the bell.
The standard error
- The standard error 标准误 is $SE = s/\sqrt{n}$ — the sd OF THE MEAN, not of the data. The seed-11 sample: $s = 4.70$, $n = 16$ → $SE = \mathbf{1.175}$.
- Quadruple the data, halve the SE. One more decimal of precision costs up to a hundred times the data — the most expensive exchange rate in statistics.
Sample: n = 16, s = 4.70. The standard error of the mean, to three decimals?
Dificultad 2/5 · 140 monedas + un pequeño bono en el primer intento de hoy
$SE = 4.70/\sqrt{16} = 4.70/4 = 1.175$ - the sd of the sample mean, not of orders.
Sample mean 28.66, SE 1.175, t(15) = 2.131. The 95% CI lower bound, to one decimal?
Dificultad 2/5 · 140 monedas + un pequeño bono en el primer intento de hoy
$28.66 - 2.131 \times 1.175 = 28.66 - 2.50 = 26.16 \approx 26.2$. The upper bound is 31.2.
The sample quadruples to n = 64 with the same s = 4.70. The new SE, to three decimals?
Dificultad 2/5 · 140 monedas + un pequeño bono en el primer intento de hoy
$4.70/\sqrt{64} = 4.70/8 = 0.588$ - four times the data, half the uncertainty.
The standard error 1.175 means most orders take within about 1.2 minutes of 28.66.
Dificultad 1/5 · 100 monedas + un pequeño bono en el primer intento de hoy
That is the sd's job (4.91). The SE describes how much SAMPLE MEANS wobble across hypothetical repeats - a different random variable entirely.
The confidence interval
- The 95% confidence interval 置信区间: $\bar x \pm t \times SE = 28.66 \pm 2.131 \times 1.175 = \mathbf{[26.2,\ 31.2]}$ ✓.
- What 95% means: the METHOD captures the truth in 95% of its uses — not "the truth is 95% likely in this band", and not "95% of orders".
Read the interval correctly
Distinguish correct and incorrect readings of a confidence interval, and SE from sd.
The 95% CI for mean delivery is [26.2, 31.2]. Which sentence is correct?
Dificultad 1/5 · 100 monedas + un pequeño bono en el primer intento de hoy
Confidence attaches to the PROCEDURE, not the band. Individual orders vary with sd 4.91 - far wider than the mean's interval.
Which change makes the interval NARROWER, all else equal?
Dificultad 1/5 · 100 monedas + un pequeño bono en el primer intento de hoy
Width is $2t \cdot s/\sqrt{n}$: up with confidence and spread, down only with sample size - and only at its square root.
One interval, three readings
- Wrong: "the true mean has a 95% chance of being in [26.2, 31.2]". Wrong: "95% of orders take 26–31 minutes".
- Right: "bands built this way capture the true mean 95% of the time — this is one such band". The interval is about the MEAN; the confidence is about the PROCEDURE.
To know the mean within +/-0.5 minutes (t about 2, s about 4.7), roughly what n is needed?
Dificultad 2/5 · 140 monedas + un pequeño bono en el primer intento de hoy
$n \ge (2 \times 4.7/0.5)^2 = 18.8^2 \approx 354$ - one decimal of precision costs nine times the 40-order dataset.
Three estimation slips
- Confusing sd with SE. sd 4.91 describes orders; SE 1.175 describes the mean's wobble.
- Reading confidence as probability about this interval. The method has the rate; the band does not.
- Forgetting the design. The arithmetic assumes the sample was random; it cannot repair a biased one.
Recap
- $SE = s/\sqrt{n}$; the CLT makes means roughly normal.
- $\bar x \pm t \cdot SE$, with confidence read as a property of the method.
- Precision costs $\sqrt{n}$; unbiasedness costs good design.