Use the power sandbox as a confidence-interval calculator, and predict — before you read it off the screen — the 95% multiplier z*, the margin of error z*·SE at one sample size, how that margin shrinks when you quadruple n, and how it widens when σ is unknown and you must use t*.
A confidence interval is just x̄ ± margin, and the margin is a multiplier times a standard error: m = z*·SE, with SE = σ/√n. This sandbox already prints both pieces — its critical value is the multiplier (z* when σ is known, t* when it is estimated), and its SE = σ/√n line is the standard error. So the width of an interval is not a mystery: it falls like 1/√n — to halve it you must quadruple n — and it grows when you pay the t penalty for not knowing σ. This lab is about that width; point estimation and the bootstrap are the other two M5 lessons.
Open the sandbox. It opens on n = 25 with the population SD at σ = 1. Each step names the one knob to change; leave everything else at these values. The critical-value readout and the SE = σ/√n line print the two pieces you multiply.
Test statistic -> One-sample z Tails -> Two-tailed
Population SD -> 1 n -> 25 alpha -> 0.05
Watch the critical-value readout — with two-tailed and σ known it is ±z* — and the SE = σ/√n line in the derivation panel. Their product is the margin of error.
Keep the one-sample z test (σ known), two-tailed, α = 0.05. The 95% multiplier is the 0.975 quantile of the standard normal, z* = Φ⁻¹(0.975). Predict it, then read the critical-value readout.
The readout shows ±1.960. That 1.96 is the number behind every 95% interval you have ever seen — it does not depend on n or σ, only on the 95% confidence level.
With σ = 1 and n = 25, the standard error is SE = σ/√n = 1/5 = 0.20. The margin of error is m = z*·SE = 1.960 × 0.20. Predict it, then read the SE line and multiply by the z* you just found.
The margin is 0.392, so the 95% interval is x̄ ± 0.392. Everything about its width is in those two readouts: a fixed multiplier and a standard error that carries all the n.
Drag n from 25 up to 100, leaving σ, the test and α alone. Now SE = 1/√100 = 0.10. Predict the new margin z*·SE, then read the SE line again.
The margin drops to 0.196 — exactly half of 0.392. Quadrupling n (25 → 100) halves the interval's width, because SE ∝ 1/√n. Twice the precision costs four times the data.
Put n back to 25 and switch the test to one-sample t (σ estimated from the sample). The multiplier becomes t*, the 0.975 quantile of t on n − 1 = 24 df. Predict it, then read the critical-value readout — the panel also prints what z would be.
The multiplier is ±2.064 — wider than 1.960. Not knowing σ and estimating it from 24 df costs you a fatter multiplier, so the honest interval is wider. As n grows, t* shrinks back toward 1.960.
Still on the t test at n = 25 with the same SE = σ/√n = 0.20. The t-interval margin is m = t*·SE = 2.064 × 0.20. Predict it, then compare with the z-interval margin 0.392 from step two.
The t-interval margin is 0.413, against 0.392 for the z-interval — about 5% wider at the same data, purely the price of estimating σ. Same standard error, a wider multiplier, a wider interval.
Everything above is waiting in the sandbox. Drag n and watch the SE = σ/√n line — and therefore the margin — fall like 1/√n; drag α from 0.05 toward 0.01 and watch the multiplier grow as a 99% interval buys more confidence with more width; switch between z and t to see the small-sample penalty appear and then fade as n climbs.
- Read the 95% multiplier off the critical-value readout: z* = 1.960
- Multiplied it by SE = σ/√n = 0.20 to get the margin z*·SE = 0.392 at n = 25
- Saw quadrupling n (25 → 100) halve the margin to 0.196 — the 1/√n law
- Switched to unknown σ and read the wider multiplier t* = 2.064 at 24 df
- Found the t-interval margin 0.413 — wider than the z-interval's 0.392, the price of estimating σ