Exercises — Central limit theorem

The CLT says ${\bar{X}}_n\sim \mathcal{N}(\mu ,{\sigma }^2/n)$.

• Mean stays $\mu =10$.
• Variance shrinks: ${\sigma }^2/n=36/9=4$.
• So ${\bar{X}}_9\sim \mathcal{N}(10,4)$.
• Standard error $\mathrm{SE}=\sigma /\sqrt{n}=6/3=\boxed{2}$.

What it looks like: the bell of individual $X$ has spread $\sigma =6$; the bell of the average is narrower, spread $2$, centred on the same point.

The CLT requires finite variance ${\sigma }^2<\infty$.

• (a) Uniform: finite variance ✓ CLT applies.
• (b) Cauchy: infinite variance ✗ — CLT does not apply. (Its sample means stay Cauchy, never bell-shaped.)
• (c) Bernoulli: variance $p(1-p)$, finite ✓ CLT applies.

Answer: (b).

$\mathrm{SE}=10/\sqrt{25}=10/5=2$. Standardize the bound $52$: $Z=\frac{52-50}{2}=1.$ "How much of the bell is left of $z=1$?" That is $\Phi (1)$: $P({\bar{X}}_{25}\le 52)=\Phi (1)\approx \boxed{0.8413}.$

$\mathrm{SE}=1.708/\sqrt{49}=1.708/7=0.2440$. $Z=\frac{3.9-3.5}{0.2440}=\frac{0.4}{0.2440}=\mathrm{1.639.}$ We want the area to the right of $z$, which is $1-\Phi (z)$: $P({\bar{X}}_{49}\ge 3.9)=1-\Phi (1.639)\approx 1-0.9494=\boxed{0.0506}.$

Picture: $3.9$ sits $\approx 1.64$ SE above centre — right at the 95th-percentile line, so only about 5% of averages land beyond it.

$\mathrm{SE}=\sigma /\sqrt{n}$. To halve $\mathrm{SE}$ we need $\sqrt{n}$ to double, hence $n$ to quadruple. $n_{\text{new}}=4\times 100=\boxed{400}.$ Check: old $\mathrm{SE}=4/10=0.4$; new $\mathrm{SE}=4/20=0.2$. Exactly halved.

Why $\sqrt{n}$ and not $n$? Precision improves with the square root of samples — the classic law of diminishing returns behind sampling. See Law of Large Numbers.

${\mathrm{SE}}_A=8/\sqrt{64}=8/8=1.$ ${\mathrm{SE}}_B=20/\sqrt{400}=20/20=1.$ They are equally precise — ratio ${\mathrm{SE}}_A/{\mathrm{SE}}_B=1$.

Insight: B has more samples but noisier data; the two effects exactly cancel. Precision depends on the combination $\sigma /\sqrt{n}$, not either quantity alone.

95% two-sided confidence uses the critical value $z^{\star }=1.96$ (since $\Phi (1.96)=0.975$, leaving $2.5\%$ in each tail). We need the half-width $z^{\star }\cdot \mathrm{SE}$ to be $\le 0.02$: $1.96\cdot \frac{0.5}{\sqrt{n}}\le \mathrm{0.02.}$ Solve for $n$: $\sqrt{n}\ge \frac{1.96\times 0.5}{0.02}=49\Rightarrow n\ge 49^2=\boxed{2401}.$

What it means for training: to shrink the tolerance to a quarter (from $\pm 0.04$-type windows), the batch cost multiplies — a concrete instance of the compute-vs-precision tradeoff in Stochastic Gradient Descent and Sample Size Calculation.

${\sigma }^2=0.05\times 0.95=0.0475$. Required: $1.96\cdot \frac{\sqrt{0.0475}}{\sqrt{n}}\le 0.005$. $\sqrt{n}\ge \frac{1.96\sqrt{0.0475}}{0.005}.$ Square both sides (cleaner): $n\ge \frac{1.96^2\times 0.0475}{0.005^2}=\frac{3.8416\times 0.0475}{0.000025}=\frac{0.18248}{0.000025}=\mathrm{7299.3.}$ Round up (you can't fractionally sample a user): $n=\boxed{7300}$.

This is the standard formula behind A/B-test planning — see Confidence Intervals and Hypothesis Testing.

Not reliably valid. The CLT is a limit as $n\to \infty$; for small $n$ the sample mean of a skewed source is itself still skewed — the bell hasn't formed yet, so $\Phi$-based intervals are miscalibrated (true coverage $\ne$ 95%). Repairs:

• Increase $n$ so the approximation kicks in (skew slows convergence; you may need hundreds).
• Use a resampling approach that makes no normality assumption — Bootstrap Methods — to get the interval empirically.
• Or use exact/simulation methods like Monte Carlo Methods.

The key mastery point: CLT accuracy depends on both $n$ and the source's shape (skew/heavy tails).

(a) $\mathrm{SE}=\sigma /\sqrt{n}=3/\sqrt{144}=3/12=0.25$. 95% CI: $\bar{X}\pm z^{\star }\mathrm{SE}=7.4\pm 1.96(0.25)=7.4\pm 0.49$. $\boxed{[6.91,7.89]}.$

(b) Test statistic under $H_0$ (using ${\mu }_0=8$): $Z=\frac{\bar{X}-{\mu }_0}{\mathrm{SE}}=\frac{7.4-8}{0.25}=\frac{-0.6}{0.25}=-\mathrm{2.4.}$ $|Z|=2.4>1.96=z^{\star }$, so we reject $H_0$ at the 5% level. Consistency check: ${\mu }_0=8$ lies outside the CI $[6.91,7.89]$ — a two-sided test at level $\alpha$ rejects exactly when ${\mu }_0$ falls outside the $(1-\alpha )$ CI. The two methods agree, as they must. See Hypothesis Testing.

Two-sided p-value: $2(1-\Phi (2.4))=2(1-0.9918)=\boxed{0.0164}<0.05$. ✓

The standard error of the sample mean is ==$\sigma /\sqrt{n}$==. To quadruple precision (quarter the SE) you multiply $n$ by 16. A two-sided 95% test rejects $H_0$ when ${\mu }_0$ lies outside the 95% confidence interval. The CLT fails when the source variance is infinite (e.g. Cauchy).