🎯Standard error of a proportion

$$SE(\hat{p}) = \sqrt{\frac{\hat{p},(1-\hat{p})}{n}}$$

• p̂: the proportion observed in the sample
• n: the number of independent observations
• SE: how much p̂ would wobble from sample to sample if I could draw the sample again

It is the spread of the estimate, not of the data. See 🔤 Standard error, where I have written that distinction out for the third time.

Worked number. 214 dispatchers, of whom 62 percent did the thing being counted:

• 0.62 × 0.38 = 0.2356
• 0.2356 / 214 = 0.001101
• square root: 0.033, a little over three percentage points

To halve that I would need four times as many dispatchers, 856, which Halden does not have lying about.

The term p(1 − p) is largest at p = 0.5, so a proportion near the middle is the hardest to pin down. From 🎓 Lecture 5 - Sampling distributions, second attempt. First used in anger for 📌 Present the triage test readout.