Exercises — Confidence intervals

What decides it: two questions — is $\sigma$ known, and is $n$ large ($\ge 30$)? Here $\sigma$ is unknown and $n=12<30$, so we must use the ==$t$-distribution==. Its heavier tails pay for the extra uncertainty in using $s$ instead of $\sigma$. Degrees of freedom $=n-1=12-1=11$.

(a) Point estimate is the centre of the interval (the coral line in Figure s01): $\bar{x}=\frac{0.804+0.936}{2}=\mathrm{0.870.}$ (b) Margin of error $E$ is the half-width — the length of one mint arrow, the distance from centre to either end: $E=\frac{0.936-0.804}{2}=\mathrm{0.066.}$

99% confidence means $\alpha =1-0.99=0.01$. We split that leftover probability equally into two tails, so each tail holds $\alpha /2=0.005$. The value cutting off the upper 0.5% tail is $z_{0.005}=2.576.$ If we used $\alpha$ not $\alpha /2$ we'd only guard one side and the net would leak on the other.

Step 1 — standard error $SE$ (spread of the sample mean): $SE=\frac{\sigma }{\sqrt{n}}=\frac{4.0}{\sqrt{64}}=\frac{4.0}{8}=\mathrm{0.5.}$ Step 2 — margin of error $E$ ($z_{0.025}=1.96$): $E=1.96\times 0.5=\mathrm{0.98.}$ Step 3 — the interval ($\bar{x}\pm E$, exactly the s01 picture): $50.0\pm 0.98=[49.02,50.98].$

Step 0 — check the approximation is valid. The normal approximation to a proportion needs enough "successes" and "failures". The usual rule of thumb is $n\hat{p}\ge 10$ and $n(1-\hat{p})\ge 10$ (a looser version uses $\ge 5$): $n\hat{p}=200\times 0.72=144\ge 10,n(1-\hat{p})=200\times 0.28=56\ge 10.$ Both comfortably clear the bar, so the normal approximation is safe. (If either were below 5–10 — e.g. a rare event — we'd switch to an exact binomial or Wilson interval instead.) Step 1 — point estimate: $\hat{p}=\frac{144}{200}=\mathrm{0.72.}$ Step 2 — standard error $SE$ for a proportion (binomial variance $p(1-p)/n$): $SE=\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}=\sqrt{\frac{0.72\times 0.28}{200}}=\sqrt{0.001008}=\mathrm{0.03175.}$ Step 3 — margin $E$: $E=1.96\times 0.03175=0.06222$. ${\text{CI}}_{95\%}=0.72\pm 0.0622=[0.6578,0.7822].$

$n=7$, so $df=6$; $\sigma$ unknown and $n<30\Rightarrow$ use $t$. $SE=\frac{s}{\sqrt{n}}=\frac{0.07}{\sqrt{7}}=\frac{0.07}{2.6458}=\mathrm{0.02646.}$ $E=2.447\times 0.02646=\mathrm{0.06474.}$ ${\text{CI}}_{95\%}=0.50\pm 0.0647=[0.4353,0.5647].$

Start from the definition of the margin and isolate $n$ step by step. Step 1 — write what we know: $E=z_{\alpha /2}\cdot \frac{\sigma }{\sqrt{n}}.$ Step 2 — move $\sqrt{n}$ up to meet $E$. $n$ is buried inside $\sqrt{n}$ in the denominator, so we first multiply both sides by $\sqrt{n}$ and divide both sides by $E$ to get $\sqrt{n}$ alone on one side: $\sqrt{n}=\frac{z_{\alpha /2}\sigma }{E}.$ Step 3 — undo the square root. The unknown is still trapped under a  ​, and the inverse of "square root" is "square", so we square both sides (squaring is legal here because every quantity — $z$, $\sigma$, $E$, $n$ — is positive): $n={\left(\frac{z_{\alpha /2}\sigma }{E}\right)}^2.$ Step 4 — plug numbers: $n={\left(\frac{1.96\times 4.0}{0.5}\right)}^2=(15.68{)}^2=\mathrm{245.86.}$ Always round up (a fractional person doesn't exist and rounding down misses the target): $n=246.$

Width $\propto 1/\sqrt{n}$. To halve width, we need $\sqrt{n_{\text{new}}}=2\sqrt{n_{\text{old}}}$, i.e. $n_{\text{new}}=4n_{\text{old}}=4\times 100=400.$ The lesson: precision is expensive — cutting error in half costs 4× the data. Figure s02 makes this concrete: it plots interval width against $n$. The curve falls like $1/\sqrt{n}$, and the two marked dots ($n=100$ in coral, $n=400$ in mint) sit at exactly half the height of each other — you slide four times to the right just to drop halfway down.

Difference of independent estimates → variances add: $S{E}_{\text{diff}}=\sqrt{S{E}_A^2+S{E}_B^2}=\sqrt{0.020^2+0.018^2}=\sqrt{0.000724}=\mathrm{0.02691.}$ ${\text{CI}}_{95\%}=(0.87-0.82)\pm 1.96\times 0.02691=0.05\pm 0.05274=[-0.0027,0.1027].$ The interval includes 0, so we cannot conclude B is truly better — see Hypothesis Testing and A/B Testing for the matched view.

Mean: $\bar{x}=(10+12+11+13+14)/5=60/5=12$. Deviations & squares: $(-2{)}^2,(0{)}^2,(-1{)}^2,(1{)}^2,(2{)}^2=4,0,1,1,4$; sum $=10$. Sample variance (Bessel $n-1=4$): $s^2=10/4=2.5$, so $s=1.5811$. Standard error: $SE=s/\sqrt{5}=1.5811/2.2361=0.7071$. Margin: $E=t_{0.05,4}\times SE=2.132\times 0.7071=1.5075$. Interval — centre $\bar{x}=12$, cushion $E=1.5075$ on each side: ${\text{CI}}_{90\%}=12\pm 1.5075=[12-1.5075,12+1.5075]=[10.492,13.508].$

The $t$-interval assumes approximate normality (or large $n$ so the CLT smooths things out); with $n=8$, skew and an outlier violate both, so the interval will be mis-centred and mis-sized. The safer choice is a percentile bootstrap interval — resample the 8 scores with replacement thousands of times and read off the 5th/95th percentiles (see Bootstrap Methods and Cross-validation).

Once computed, the interval $[10.49,13.51]$ is fixed and $\mu$ is a fixed (unknown) constant — the true mean is either in it or not, no probability left. The 90% describes the procedure: if we repeated this sampling-and-interval recipe many times, about 90% of the resulting intervals would contain $\mu$.

Planning (use the expected 0.20). Proportion SE gives $n=\frac{z_{\alpha /2}^{2}p(1-p)}{E^2}=\frac{1.96^2\times 0.20\times 0.80}{0.01^2}=\frac{3.8416\times 0.16}{0.0001}=6146.56\Rightarrow 6147.$ Observed CI with $\hat{p}=0.22$, $n=6147$ (quick validity check: $n\hat{p}=1352\ge 10$, $n(1-\hat{p})=4795\ge 10$, so the normal approximation is fine): $SE=\sqrt{\frac{0.22\times 0.78}{6147}}=\sqrt{2.7916\times 10^{-5}}=0.005284,$ ${\text{CI}}_{95\%}=0.22\pm 1.96\times 0.005284=0.22\pm 0.01036=[0.2096,0.2304].$ The realised margin ($\pm 1.04\%$) slightly exceeds the target because observed $\hat{p}=0.22$ has larger variance than the planned $0.20$ — a lesson in planning with the most conservative $p$ ($=0.5$) when unsure.

Frequentist net: the interval is random, $\mu$ is a fixed fish; 95% of the nets we throw catch it, but any one net either did or didn't. Bayesian pond: $\mu$ is treated as random with a probability distribution; the 95% credible interval genuinely holds 95% of that posterior probability, so "95% probability $\mu$ is inside" is legitimate there. Coincidence: with a flat (uninformative) prior and large $n$, the posterior is dominated by the likelihood and the two intervals become numerically almost identical.

Red flags:

1. $n=4$ is tiny. The critical value is $t_{0.025,3}=3.182$ (far above $z=1.96$), which alone makes the net very wide — the study is badly under-powered.
2. $\sigma$ unknown and data non-normal. The $t$-interval leans on approximate normality (or large $n$ via the CLT); with $n=4$ and skew, neither holds, so the interval is both mis-centred and mis-sized.
3. The CI contains 0. Since $[-2,8]$ straddles zero, the effect is not statistically distinguishable from "no effect" — you cannot claim a real effect exists.
4. Reporting only the point estimate "3.0". Quoting the crisp centre hides that the honest interval spans $[-2,8]$, a huge range; that is false precision (the L3-style "wide interval is honest" lesson).

What to do instead: collect more data — recall from L3·Q2 that halving the width needs 4× the sample; use a bootstrap percentile interval given the non-normality; back the interval up with a formal hypothesis test on whether the effect differs from 0; and always report the full interval, never the point estimate alone.