Exercises — Confidence intervals — derivation for mean, proportion

(i) $\sigma$ is known → use the z-mean formula $\bar{X}\pm z_{\alpha /2}\sigma /\sqrt{n}$. (ii) $\sigma$ unknown and $n$ small → use the t-mean formula with $n-1=11$ degrees of freedom. (iii) Count of successes out of trials → this is a proportion, use $\hat{p}\pm z_{\alpha /2}\sqrt{\hat{p}(1-\hat{p})/n}$.

Confidence level is $1-\alpha =0.90$, so $\alpha =0.10$. We split $\alpha$ into two tails, so each tail has $\alpha /2=0.05$. The critical value is $z_{0.05}=1.645$. (Note: $1.645$ is the two-tailed 90% value — it is also the one-sided 95% value, which is the source of a famous trap; see below.)

Step — standard error. $\text{SE}=\frac{\sigma }{\sqrt{n}}=\frac{0.8}{\sqrt{64}}=\frac{0.8}{8}=0.1$. Why $\sqrt{n}$? We bound the mean, whose spread is $\sigma /\sqrt{n}$, not one reading's spread. Step — critical value. 99% → $\alpha =0.01$ → $\alpha /2=0.005$ → $z_{0.005}=2.576$. Step — margin. $E=2.576\times 0.1=0.2576$. Answer. $21.5\pm 0.2576=[21.2424,21.7576]{}^{\circ }$C.

Step — why t. $\sigma$ unknown and $n=25$ is smallish, so use t with $n-1=24$ df. Step — standard error. $\text{SE}=s/\sqrt{n}=5/\sqrt{25}=5/5=1$. Step — margin. $E=t_{0.025,24}\times \text{SE}=2.064\times 1=2.064$. Answer. $40\pm 2.064=[37.936,42.064]$ hours.

Step — point estimate. $\hat{p}=48/200=0.24$. Check normality: $n\hat{p}=48\ge 5$ ✓ and $n(1-\hat{p})=152\ge 5$ ✓. Step — standard error. $\text{SE}=\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}=\sqrt{\frac{0.24\times 0.76}{200}}=\sqrt{\frac{0.1824}{200}}=\sqrt{0.000912}=0.030199\dots$ Step — margin. $E=1.96\times 0.030199=0.059191$. Answer. $0.24\pm 0.0592=[0.1808,0.2992]$.

The margin is $E=z_{\alpha /2}\sigma /\sqrt{n}$. Only $\sqrt{n}$ changes. $\frac{E_B}{E_A}=\frac{1/\sqrt{400}}{1/\sqrt{100}}=\frac{\sqrt{100}}{\sqrt{400}}=\frac{10}{20}=\frac{1}{2}.$ Lab B's margin is half of Lab A's. Lesson: to halve the margin you must quadruple the sample size — precision costs $n$ quadratically. (This is exactly the lever used in Sample Size Determination.)

Not correct. Once the numbers $[65.74,74.26]$ are fixed and the true $\mu$ is a fixed (if unknown) constant, $\mu$ is either inside or outside — the probability is $0$ or $1$, there is no "95%" left to speak of for this interval. Correct interpretation: "If we repeated the whole sampling procedure many times, about 95% of the intervals produced this way would contain the true mean $\mu$." The 95% describes the long-run reliability of the method, not one realized interval.

Standard error is $s/\sqrt{n}=8/\sqrt{16}=2$ (unchanged by which multiplier you pick).

• Honest ($t$): $E_t=2.131\times 2=4.262$.
• Wrong ($z$): $E_z=1.96\times 2=3.92$. Understatement $=4.262-3.92=0.342$ score points on the half-width. The $t$-interval is honest. Because $s$ is only an estimate of $\sigma$ and can run low for small $n$, the true uncertainty is larger; the fatter t-tails widen the interval to compensate. Using $z$ pretends you know $\sigma$ exactly.

Start from $E=z_{\alpha /2}\frac{\sigma }{\sqrt{n}}$. Solve for $n$: $\sqrt{n}=\frac{z_{\alpha /2}\sigma }{E}⟹n={\left(\frac{z_{\alpha /2}\sigma }{E}\right)}^2.$ Plug in $z_{0.025}=1.96$, $\sigma =3$, $E=0.5$: $n={\left(\frac{1.96\times 3}{0.5}\right)}^2={\left(\frac{5.88}{0.5}\right)}^2=(11.76{)}^2=\mathrm{138.2976.}$ Sample size must be a whole number and the margin must be at most $0.5$, so round up: $n=139$. (Why round up, never down? $n=138$ gives a slightly larger margin than $0.5$, violating the requirement.)

Margin: $E=z_{\alpha /2}\sqrt{\hat{p}(1-\hat{p})/n}$. The product $\hat{p}(1-\hat{p})$ is largest at $\hat{p}=0.5$, where it equals $0.25$. Using this worst case guarantees the margin no matter what $p$ turns out to be. $n=\frac{z_{\alpha /2}^2\hat{p}(1-\hat{p})}{E^2}=\frac{(1.96{)}^2(0.25)}{(0.03{)}^2}=\frac{3.8416\times 0.25}{0.0009}=\frac{0.9604}{0.0009}=\mathrm{1067.11.}$ Round up: $n=1068$. (This is why "±3% margin" national polls sample roughly a thousand people.)

Rule: a two-sided level-$\alpha$ test rejects ${\mu }_0$ exactly when ${\mu }_0$ lies outside the corresponding $(1-\alpha )$ CI.

• ${\mu }_0=50$: is $50\in [48.2,51.8]$? Yes. So we fail to reject — 50 is a plausible mean.
• ${\mu }_0=47$: is $47\in [48.2,51.8]$? No ($47<48.2$). So we reject ${\mu }_0=47$ at $\alpha =0.01$. The interval is the set of null values you would not reject — that is the CI↔test duality.

(a) Conditions. $\hat{p}=25/400=0.0625$. Check $n\hat{p}=400\times 0.0625=25\ge 5$ ✓ and $n(1-\hat{p})=400\times 0.9375=375\ge 5$ ✓. Normal approximation is valid. (b) 90% CI. $\alpha =0.10$, $z_{0.05}=1.645$. $\text{SE}=\sqrt{\frac{0.0625\times 0.9375}{400}}=\sqrt{\frac{0.05859375}{400}}=\sqrt{0.0001464844}=\mathrm{0.0121030.}$ $E=1.645\times 0.0121030=0.019909$. CI $=0.0625\pm 0.019909=[0.042591,0.082409]$. (c) Is 8% ruled out? The interval reaches up to $0.0824>0.08$. Since $0.08$ lies inside the interval, we cannot rule out a true defect rate of 8% at this confidence — it remains plausible. The guarantee is not statistically contradicted (nor confirmed) by this sample.

Standard error is common to both: $\text{SE}=s/\sqrt{n}=12/\sqrt{9}=12/3=4$. (a) Correct (t): $E_t=2.306\times 4=9.224$. CI $=100\pm 9.224=[90.776,109.224]$. (b) Wrong (z): $E_z=1.96\times 4=7.84$. CI $=100\pm 7.84=[92.16,107.84]$. Understatement percentage: $\frac{E_t-E_z}{E_t}=\frac{9.224-7.84}{9.224}=\frac{1.384}{9.224}=0.15004\dots \approx 15.0\%$. The $z$-interval is about 15% too narrow — a serious overconfidence for $n=9$. Small samples punish the $z$-shortcut hardest, since $t_{n-1}$ has much fatter tails when $n-1$ is tiny.

• Halve $\alpha$ before looking up $z_{\alpha /2}$ (L1).
• The width lives in the standard error $\sigma /\sqrt{n}$, not $\sigma$ (L2).
• Confidence is about the procedure, and $t$ is the honest choice when $\sigma$ is estimated (L3).
• Invert the margin formula and round $n$ up to hit a target precision; worst-case $p=0.5$ for proportions (L4).
• Always check normality conditions and judge claims against the whole interval (L5).