Statistics.

1 · What probability actually is

Start with the picture, not the formula. An experiment is any situation whose result you can't predict for sure, but where you do know the list of possible results. Tossing a coin is an experiment; so is "which operating system will my next laptop run?" or "what will Apple's share price be on Monday?"

Each single result is an outcome. The full list of possible outcomes is the sample space, written \(S\).

• Coin toss: \(S=\{\text{Heads},\text{Tails}\}\).
• Roll a die: \(S=\{1,2,3,4,5,6\}\).
• Next laptop OS: \(S=\{\text{Windows},\text{MacOS},\text{Linux},\dots\}\).

An event is just a statement about the result — "the die shows an even number", "I get Heads". Formally it's a subset of \(S\): the even-number event is the set \(\{2,4,6\}\). We say the event occurs when the actual outcome is one of the outcomes inside it.

So what is a probability? The most useful picture is long-run frequency: if you repeated the experiment over and over, the probability of an event is the proportion of times it would happen. Flip a fair coin thousands of times and the fraction landing Tails settles near \(0.5\) — that limiting fraction is the probability. (Real example: across the world about 105 boys are born for every 100 girls, year after year — so \(P(\text{newborn is male})\approx0.51\).) A probability is always a number between 0 and 1: 0 = never, 1 = certain.

2 · Combining events: AND, OR, NOT

Events are sets, so we combine them like sets. Picture a rectangle for \(S\) and circles inside it for events (a Venn diagram).

• OR — union \(A\cup B\): the outcomes in \(A\), or in \(B\), or in both. It occurs if at least one of them happens.
• AND — intersection \(A\cap B\): the outcomes in both. It occurs only if they happen together.
• NOT — complement \(A^c\): everything in \(S\) that is not in \(A\). It occurs exactly when \(A\) doesn't.

Mutually exclusive (disjoint) events can't happen at the same time — their intersection is empty, \(A\cap B=\varnothing\). "The die shows 2" and "the die shows 5" are mutually exclusive. The empty set \(\varnothing\) is the impossible event.

One identity worth seeing now, because it drives a lot of exam problems: "in \(A\) but not in \(B\)" is \(A\cap B^c\), and it equals what's in \(A\) minus the overlap. We'll turn that into numbers next.

3 · The rules of probability

Three basic rules (they just codify the frequency picture):

1. \(0\le P(A)\le1\) — proportions live between 0 and 1.
2. \(P(S)=1\) — something in the list always happens.
3. If \(A,B\) are mutually exclusive, \(P(A\cup B)=P(A)+P(B)\) — non-overlapping chances just add.

From these you derive the two you'll actually use:

Complement rule. Since \(A\) and \(A^c\) split \(S\): \(P(A^c)=1-P(A)\). (Heads has probability 0.4 ⇒ Tails has 0.6.) This is the engine behind "at least one" problems — it's usually easier to compute the opposite and subtract.

Addition rule (when events can overlap): \[ P(A\cup B)=P(A)+P(B)-P(A\cap B). \] Why subtract? If you just add \(P(A)+P(B)\), the overlap \(A\cap B\) gets counted twice, so you remove it once.

Concrete example (Ross 4.3). A shop takes Amex or VISA. 22% of customers carry Amex, 58% carry VISA, 14% carry both. Probability a customer has at least one card: \(0.22+0.58-0.14=0.66\). And "VISA but not Amex" \(=0.58-0.14=0.44\) — exactly the \(A\cap B^c\) idea from section 2.

4 · Equally likely outcomes — just count

When every outcome in \(S\) is equally likely (a fair die, a well-shuffled deck, "a person chosen at random"), probability becomes pure counting:

The phrase "chosen at random" is your signal that outcomes are equally likely.

• Fair die: \(P(\text{even})=3/6=1/2\).
• European roulette, bet on odd: numbers \(\{0,1,\dots,36\}\), 18 are odd ⇒ \(18/37\).
• Retirement centre (Ross 4.4): 420 members, 144 smokers ⇒ \(P(\text{smoker})=144/420=12/35\).

This is why counting techniques (section 9) matter: to get a probability you often just need to count the favourable outcomes and divide by the total.

5 · Conditional probability — updating on information

Often you learn something partway through. Conditional probability \(P(B\mid A)\) is "the probability of \(B\) given that \(A\) has happened."

The intuition (no formula yet). Roll two dice; you're told the first die is a 4. That knowledge shrinks the world: only six outcomes are still possible — \((4,1),\dots,(4,6)\). Among those, only \((4,6)\) makes the sum 10, so the chance is \(1/6\). You re-computed the probability inside a reduced sample space: \(A\) became your new \(S\).

Turning that into a formula — measure the overlap relative to the thing you now know:

The famous trap (Ross 4.10 / the two-children problem). A couple has two children; you learn at least one is a girl. Probability both are girls? Equally likely families are \(\{(g,g),(g,b),(b,g),(b,b)\}\). "At least one girl" rules out only \((b,b)\), leaving three equally likely cases; just one is \((g,g)\). So the answer is \(1/3\), not \(1/2\) — the extra information reshapes the sample space. (This exact reasoning powers exam exercise 5 below.)

Rearranging the formula gives the multiplication rule: \(P(A\cap B)=P(A)\,P(B\mid A)\) — useful for "draw two without replacement" problems, where the second draw's odds depend on the first.

6 · Independence — when information doesn't help

Sometimes knowing \(A\) tells you nothing about \(B\). Then \(P(B\mid A)=P(B)\), and the multiplication rule simplifies to the test you'll use:

Signals of independence: "with replacement", "i.i.d.", "each toss is fair" — separate trials that don't influence each other.

Why it's not automatic (Ross 4.13). Two fair dice. Let \(A=\)"first die is 3". Compare two events: \(B=\)"sum is 8" and \(C=\)"sum is 7". Knowing the first die is 3 changes the chance of an 8 (now you just need a 5 next) — so \(A,B\) are dependent. But the chance of a 7 stays \(1/6\) whatever the first die shows (there's always exactly one matching second die) — so \(A,C\) are independent. Same first event, opposite verdicts: independence is something you check, not assume.

For "at least one" across independent trials, lean on the complement: three children, \(P(\text{at least one girl})=1-P(\text{all boys})=1-(1/2)^3=7/8\).

7 · Total probability — averaging over cases

Often the thing you want depends on a hidden "case" or "cause". Split the problem by the case, compute each piece, and combine. If \(B\) either happens or not, then for any \(A\):

Read it as a weighted average: the chance of \(A\) in each case, weighted by how likely that case is. (It generalises to any set of mutually-exclusive cases \(B_1,\dots,B_k\): \(P(A)=\sum_i P(A\mid B_i)P(B_i)\).)

Concrete example (insurance). 30% of drivers are high-risk, 70% low-risk. A high-risk driver has an accident this year with probability 0.4, a low-risk one with probability 0.2. Overall chance a random driver has an accident: \[ P(\text{accident})=0.4(0.30)+0.2(0.70)=0.12+0.14=0.26. \] You couldn't answer without splitting by risk type — that's the whole move.

8 · Bayes' theorem — reasoning backwards

Total probability runs cause → effect. Bayes runs it backwards: you observed the effect, and you want the probability of the cause. "It rained — how likely is it that the morning had been sunny?" "The test is positive — how likely is the disease?"

Start from the definition \(P(\text{cause}\mid\text{effect})=\dfrac{P(\text{cause}\cap\text{effect})}{P(\text{effect})}\), write the top as \(P(\text{effect}\mid\text{cause})P(\text{cause})\), and expand the bottom with total probability:

The procedure: (1) name the causes \(H\) and the observed effect \(E\); (2) write the priors \(P(H)\) and the likelihoods \(P(E\mid H)\) straight from the text; (3) total probability gives the denominator; (4) divide.

The showcase example — why a positive test can still be reassuring (Ross 4.17). A blood test is 99% accurate when the disease is present (\(P(E\mid H)=0.99\)) and gives a false positive 2% of the time (\(P(E\mid H^c)=0.02\)). Only 0.5% of people have the disease (\(P(H)=0.005\)). You test positive — what's the chance you're actually sick? \[ P(H\mid E)=\frac{0.99(0.005)}{0.99(0.005)+0.02(0.995)}\approx0.199. \] About 20% — surprisingly low, because the huge healthy population produces many false positives that swamp the few true cases. This is exactly the engine behind exam exercises 3 and 4.

9 · Counting — tools for equally-likely problems

Section 4 said probability is often just counting favourable ÷ total. Here are the tools, introduced only because we need them.

Basic principle. If step 1 has \(n\) options and step 2 has \(m\), together there are \(n\cdot m\). (One man from 8, one woman from 12 → \(96\) pairs.)

Permutations / factorial. The number of orderings of \(n\) distinct objects is \(n!=n(n-1)\cdots2\cdot1\) (with \(0!=1\)). Order matters here.

Combinations. When order does not matter — choosing a group of \(k\) from \(n\):

Two reflexes for the exam: "all different" → count the complement (the few repeats) and subtract; "these specific items are included" → fix them, then count the choices for the remaining slots. Watch for repeated elements — e.g. the two identical P's in APPLE: \(\binom{5}{2}=10\) selections, only one is the pair PP.

10 · Which tool? — recognition guide

Once the concepts are in place, exam problems are about spotting which one applies.

pmf, expectation, variance

A discrete random variable \(X\) is described by its pmf \(p(x)=P(X=x)\), with \(\sum_x p(x)=1\). Two summaries do most of the work:

The variance shortcut \(E[X^2]-(E[X])^2\) is almost always faster than \(\sum (x-\mu)^2 p(x)\) on the exam — compute \(E[X]\) and \(E[X^2]\) in one pass over the table, then subtract.

Normalising trick. If a pmf is given up to a constant \(c\) (e.g. \(p(i)=c\,\lambda^i/i!\)), find \(c\) by forcing \(\sum_x p(x)=1\). Recognising the series (here \(\sum \lambda^i/i! = e^\lambda\)) names the distribution.

Expectation & variance of linear combinations

These rules are the engine of the estimator-theory chapter (bias/MSE) — learn them cold.

If \(X,Y\) are independent then \(\mathrm{Cov}(X,Y)=0\), so \(\mathrm{Var}(aX+bY)=a^2\mathrm{Var}(X)+b^2\mathrm{Var}(Y)\). Note expectation is linear unconditionally, but a constant added to \(X\) never changes the variance: \(\mathrm{Var}(Y+4)=\mathrm{Var}(Y)\).

The four key discrete models

Geometric on the exam: its pmf is usually given inside the problem (e.g. "draw with replacement until a black ball"); you just identify \(p\) and plug in. The MLE \(\hat\theta=1/\bar X\) is derived later (estimator chapter).

Poisson tell: "average number per week/page/area" and a count question → Poisson with \(\lambda\)=that average. \(P(X\ge1)=1-e^{-\lambda}\) is the most common ask.

Which model? — recognition guide

Standardize and read the table

If \(X\sim N(\mu,\sigma^2)\), convert to the standard normal \(Z\sim N(0,1)\) by

Everything is one of: a tail probability, an inverse (recover \(\sigma\) or a percentile), or a comparison. Two identities you use every time:

• Symmetry: \( \Phi(-z)=1-\Phi(z) \), so \( P(X\ge a)=1-\Phi\!\big(\tfrac{a-\mu}{\sigma}\big)=\Phi\!\big(\tfrac{\mu-a}{\sigma}\big) \).
• Central band: \( P(\mu-d\le X\le \mu+d)=2\Phi\!\big(\tfrac{d}{\sigma}\big)-1 \).

Table values worth memorising: \(\Phi(0.5)=0.6915\), \(\Phi(1)=0.8413\), \(\Phi(1/3)=0.6293\); inverses \(\Phi^{-1}(0.975)=1.96\), \(\Phi^{-1}(0.75)=0.675\).

The four exam flavours

1. Tail / comparison ("probability the return is at least 50", "which is more likely"): standardize, read \(\Phi\); for a comparison the bigger \(\Phi\big(\tfrac{\mu-a}{\sigma}\big)\) wins.
2. Inverse-σ ("95% are between 140 and 160, find σ"): write the band as \(2\Phi(d/\sigma)-1=\text{coverage}\), solve \(d/\sigma=z\), then \(\sigma=d/z\).
3. Percentiles ("25th pct = 3, 75th pct = 7"): \(\mu\) is the midpoint of symmetric percentiles; get \(\sigma\) from \(\Phi\big(\tfrac{x_p-\mu}{\sigma}\big)=p\).
4. Moments→probability ("E[X]=2, E[X(X−1)]=6, find Var and P(X≤4)"): recover \(\sigma^2=E[X^2]-(E[X])^2\) using \(E[X(X-1)]=E[X^2]-E[X]\), then standardize.

Which flavour? — recognition guide

Sample mean, variance, and the two sums you reconstruct

For a sample \(x_1,\dots,x_n\):

Exam problems rarely give you the raw data — they give summaries and ask you to recover the two totals:

Once you hold those two sums you can add a point, merge groups, or back-solve a missing piece.

Weighted / grouped mean & back-solving

An overall mean is the weighted average of subgroup means, weights = subgroup proportions (or sizes):

Forward: proportions and group means given → plug in. Back-solve: the overall mean is given and a group mean / group size is unknown → set up one linear equation (or a 2×2 system if a difference between group means is also stated) and solve.

Typical back-solve: total grade \(=\) (class-A mean)·\(n_A+\)(class-B mean)·\((N-n_A)\), divided by \(N\), equals the stated overall mean → solve for \(n_A\).

Augmenting a sample with a new observation

Adding one point \(x_{n+1}\): recover \(\sum x_i=n\bar x\) and \(\sum(x_i-\bar x)^2=(n-1)s^2\), add the new value, recompute.

• New mean: \( \bar x_{new}=\dfrac{n\bar x + x_{n+1}}{n+1} \).
• If the new point equals the old mean, the mean is unchanged and the sum of squared deviations is unchanged; the variance still shifts because the divisor grows from \(n-1\) to \(n\). Example: \(n=12\), \(\bar x=10\), \(s^2=12\), add \(x_{13}=10\) → mean stays 10, SS stays \(132\), new \(s^2=132/12=11\).

Reading dispersion from a histogram (no computation)

When several histograms share the same mean and you must rank their standard deviations without computing: SD measures spread about the mean, i.e. the typical size of \((x_i-\bar x)^2\).

• Mass concentrated near the centre (mound/bell shape) → smallest SD.
• Mass spread evenly (flat/uniform) → middle SD.
• Mass pushed to the extremes (U-shape / bimodal) → largest SD.

So for shapes {mound B, uniform A, U-shape C} with SDs {1.38, 1.71, 1.98}: B = 1.38 (smallest), A = 1.71, C = 1.98 (largest). You cannot actually compute an SD from a histogram — the argument is purely about where the mass sits.

Sample correlation — spot the line

The sample correlation coefficient is

Exam shortcut: if the points lie exactly on a straight line \(y=a+bx\), then \(r=\operatorname{sign}(b)\) — \(+1\) for positive slope, \(-1\) for negative — with no arithmetic. The hint "try to plot the points" is the giveaway. Only compute the full formula if the points are not perfectly collinear.

Which tool? — recognition guide

The Central Limit Theorem

Take \(n\) independent, identically distributed units \(X_1,\dots,X_n\), each with mean \(\mu\) and variance \(\sigma^2\). For large \(n\) (rule of thumb \(n\ge30\)) their sum and mean are approximately Normal:

The shape of the individual \(X_i\) does not matter — only \(\mu\), \(\sigma^2\), and \(n\). This is what lets you answer "probability that the total exceeds a threshold" without knowing the per-unit distribution.

The standardization template

Almost every CLT exam problem is: many i.i.d. units, per-unit \(\mu\) and \(\sigma\) given, asked the probability a total crosses a threshold \(s\). The recipe:

1. Total parameters: mean \(n\mu\), variance \(n\sigma^2\), sd \(\sigma\sqrt n\).
2. Standardize: \( Z=\dfrac{s-n\mu}{\sigma\sqrt n} \).
3. Read \(\Phi\): \( P(S_n\le s)=\Phi(Z) \), \( P(S_n\ge s)=1-\Phi(Z) \).

Watch the direction ("sufficient to cover" / "exceed" → upper tail). A standardized value \(|Z|\gtrsim4\) means the probability is effectively 0 or 1.

Binomial → Normal approximation

A Binomial count is a sum of \(n\) Bernoulli's, so for large \(n\) the CLT gives

Standardize with \(\mu=np\), \(\sigma=\sqrt{np(1-p)}\): \( P(X\le k)\approx\Phi\!\big(\tfrac{k-np}{\sqrt{np(1-p)}}\big) \).

Continuity correction. Because \(X\) is discrete, the more accurate version replaces \(k\) with \(k+\tfrac12\) (for \(\le\)) or \(k-\tfrac12\) (for \(\ge\)). Exams often say "without (half) continuity correction" — then use \(k\) as-is. Always check the wording.

Recognition guide

The estimator vocabulary

An estimator \(T\) for a parameter \(\theta\) is a function of the sample. Three numbers grade it:

Unbiased means \(E[T]=\theta\) (Bias \(=0\)), and then \(\mathrm{MSE}=\mathrm{Var}\). More efficient = smaller MSE (smaller variance, among unbiased estimators).

The algebra engine (from the discrete-RV chapter): \(E\) is linear always; for independent pieces \( \mathrm{Var}(aU+bW)=a^2\mathrm{Var}(U)+b^2\mathrm{Var}(W) \). Population moments you reuse: Bernoulli \(E=p,\mathrm{Var}=p(1-p)\); Poisson \(E=\mathrm{Var}=\lambda\); sample mean of \(m\) obs \(\mathrm{Var}(\bar X_m)=\sigma^2/m\); and \(\sigma^2=E[X^2]-\mu^2\).

Unbiasedness & finding the constant

Weights that sum to 1 → automatically unbiased for the mean. If \(T=c\bar X_1+(1-c)\bar X_2\) then \(E[T]=\mu\) for every \(c\).

Find a constant. When asked "for what \(a\) is \(T\) unbiased for \(\sigma^2\)", set \(E[T]=\theta\) and solve. The recurring move is \(E[X_i^2]=\sigma^2+\mu^2\), and for independent \(X_i,X_j\), \(E[X_iX_j]=\mu^2\). Example: \(T=\frac1n\sum X_i^2+a\) with \(\mu\) known → \(E[T]=\sigma^2+\mu^2+a=\sigma^2\Rightarrow a=-\mu^2\).

Efficiency & minimum-MSE weighting

Compare efficiency: if all candidates are unbiased, compute each variance and pick the smallest.

Optimal weight. For \(M_c=c\bar X_1+(1-c)\bar X_2\) (independent), \(\mathrm{MSE}=c^2\mathrm{Var}_1+(1-c)^2\mathrm{Var}_2\); differentiate and set to 0. The minimiser is inverse-variance weighting

With sample means of sizes \(n_1,n_2\) (\(\mathrm{Var}_i=\sigma^2/n_i\)) this becomes \(c^*=\dfrac{n_1}{n_1+n_2}\) — weight each sample by its size. Put more weight on the more precise (larger / lower-variance) sample.

Nonlinear transforms break unbiasedness (Jensen)

If \(T\) is unbiased for \(\sigma^2\), is \(\sqrt T\) unbiased for \(\sigma\)? No. For a nonlinear \(g\), \(E[g(T)]\neq g(E[T])\) in general (Jensen's inequality). Since \(\sqrt{\cdot}\) is concave, \(E[\sqrt T]\le\sqrt{E[T]}=\sigma\), with strict inequality unless \(T\) is constant → \(\sqrt T\) is biased low for \(\sigma\). Whenever an exam takes a square root, log, or reciprocal of an unbiased estimator, the answer to "still unbiased?" is no.

Maximum likelihood (MLE)

Recipe: write the likelihood \(L(\theta)=\prod_i f(x_i\mid\theta)\), take \(\log\), differentiate, set to 0, solve.

Geometric example \(f(x\mid\theta)=\theta(1-\theta)^{x-1}\): \(L=\theta^n(1-\theta)^{\sum(x_i-1)}\), \(\ell=n\log\theta+\big(\sum(x_i-1)\big)\log(1-\theta)\), giving \(\hat\theta=\dfrac{n}{\sum x_i}=\dfrac1{\bar X}\). On the exam the pmf/pdf is given in the problem — you just run the recipe.

Method-of-moments (MoM)

Equate the population moment to the sample moment and solve for the parameter. For one parameter, set \(E[X]=\bar X\) (using the theoretical mean as a function of \(\theta\)) and invert.

Example density \(f(x)=2x/\theta^2\) on \([0,\theta]\): \(E[X]=\int_0^\theta x\frac{2x}{\theta^2}dx=\frac{2\theta}{3}\). Set \(\bar X=\frac{2\theta}{3}\Rightarrow \hat\theta_M=\frac{3}{2}\bar X\). It is unbiased here (\(E[\hat\theta_M]=\theta\)) with \(\mathrm{MSE}=\theta^2/(8n)\to0\) — consistent. MoM and MLE can differ; MoM is usually the quicker algebra.

Recognition guide

The CI machinery

Every confidence interval is estimate ± margin, the margin being a critical value times a standard error.

Critical values: \(z_{0.025}=1.96\) (95%), \(z_{0.005}=2.576\) (99%), \(z_{0.05}=1.645\) (90%). The total length of a symmetric CI is twice the margin, \(2z_{\alpha/2}\cdot\text{SE}\).

Sample size: back-solve from a target precision

Given a target on the interval (total length \(L\), or half-width / accuracy \(d=L/2\)), set the length condition and solve for \(n\), then round up.

When \(\hat p\) is unknown (sample not yet taken), use the worst case \(p(1-p)\le\frac14\) — it guarantees the precision whatever the true \(p\).

Recover n from a published interval

If a poll reports "with \(C\%\) confidence, support is between \(a\) and \(b\)", read off \(\hat p=\tfrac{a+b}{2}\) and half-width \(d=\tfrac{b-a}{2}\), then invert the margin:

E.g. \((48\%,52\%)\) at 90%: \(\hat p=0.5\), \(d=0.02\), \(n=1.645^2(0.25)/0.02^2\approx1691\).

Asymmetric tails (non-standard split)

A "95% CI" normally splits \(\alpha=0.05\) as \(2.5\%\) per tail. If the problem asks for an unequal split — say \(3\%\) in the lower tail, \(2\%\) in the upper — use a different z on each side:

The total tail probability still sums to \(\alpha\); only the per-side allocation changed, so the interval is no longer symmetric about \(\bar x\).

z or t?

Default to z: known σ, or large \(n\) (≥30) so the CLT applies and \(s\) is a good plug-in. Use t\(_{n-1}\) only for a small sample from a normal population with unknown σ. For \(n=100\) the two barely differ (e.g. \(z_{0.025}=1.96\) vs \(t_{99,0.025}\approx1.98\)) and either is acceptable — the exams lean on z.

Recognition guide

The test skeleton

Every test is the same four moves; only the standard error changes.

1. State \(H_0,H_1\) — the direction comes from the wording (below).
2. Pick the statistic and its null distribution: \( \text{TS}=\dfrac{\text{estimate}-\text{null}}{\text{SE}} \), \(\sim N(0,1)\) (large n / known σ) or \(t_{df}\) (small n, normal).
3. Compute the observed value \(ts_{obs}\).
4. Decide: reject if \(ts_{obs}\) falls in the rejection region — two-sided \(|ts|\ge z_{\alpha/2}\), one-sided \(ts\ge z_\alpha\) (or \(\le-z_\alpha\)). Or report the p-value and reject when \(\text{p-value}\le\alpha\).

Default to z (large samples / known σ); §8.6–8.7 proportion and Poisson tests are asymptotic-normal too. Use t only for small-n normal data (and the paired test).

The standard-error table

Pooled proportion: \( \hat p_p=\dfrac{X_1+X_2}{n_1+n_2} \). Critical values: \(z_{0.025}=1.96\), \(z_{0.05}=1.645\); \(t_{9,0.05}=1.833\).

One-sided or two-sided? Read the wording

• One-sided: "more than", "over 30%", "more effective", "improved", "larger mean" → \(H_1\) points one way; reject only in that tail with \(z_\alpha\) (1.645 at 5%).
• Two-sided: "changed", "differ significantly", "need to recalibrate", "is there a difference" → \(H_1\neq\); reject in both tails with \(z_{\alpha/2}\) (1.96 at 5%).

For a two-group test, set \(H_0:\) difference \(=0\). The direction of \(H_1\) decides which tail; e.g. "B more effective" with statistic \((\bar x_A-\bar x_B)/SE\) rejects for small (negative) values.

p-value & the threshold significance level

The p-value is the null-probability of a statistic at least as extreme as observed, in the direction of \(H_1\): one-sided \(P(Z\ge ts_{obs})\); two-sided \(2P(Z\ge|ts_{obs}|)\). Reject whenever \(\alpha\ge\) p-value.

"Find the significance levels at which \(H_0\) is rejected" is just the p-value: e.g. a one-sided \(ts=0.93\) gives \(P(Z\ge0.93)=1-\Phi(0.93)=1-0.8238=0.1762\), so reject for \(\alpha\ge17.62\%\). A \(ts=2.4\) gives p \(=1-\Phi(2.4)=0.0082\) → reject for \(\alpha\ge0.82\%\).

Recognition guide

Exponential distribution (memoryless)

In the program but its inference (§7.6 exp-CI) is skipped, and it never appeared on an exam. Know only the basics: \(f(x)=\lambda e^{-\lambda x}\) for \(x\ge0\), with \(E[X]=1/\lambda\), \(\mathrm{Var}(X)=1/\lambda^2\), and tail \(P(X>t)=e^{-\lambda t}\).

Memoryless property: \(P(X>s+t\mid X>s)=P(X>t)\) — a used component is as good as new. This is the one fact most likely to be asked.

Chebyshev's inequality & the Weak Law

Distribution-free bound on how far a variable strays from its mean:

It is loose (works for any distribution) but needs no normality. The Weak Law of Large Numbers follows: \(\bar X_n\to\mu\) in probability as \(n\to\infty\), since \(\mathrm{Var}(\bar X_n)=\sigma^2/n\to0\).

The t and χ² distributions

t\(_{df}\): bell-shaped, symmetric, heavier tails than the normal; \(df=n-1\). Used for the mean when \(\sigma\) is unknown and \(n\) is small; as \(df\to\infty\) it converges to \(N(0,1)\). On these exams it appears only once (the paired test).

χ²\(_{df}\): the distribution of a sum of squared standard normals; right-skewed, positive. It is the sampling distribution behind variance inference: \(\dfrac{(n-1)s^2}{\sigma^2}\sim\chi^2_{n-1}\).

Confidence interval & test for a variance

Author-light — plausibly examinable, never seen. Using \(\dfrac{(n-1)s^2}{\sigma^2}\sim\chi^2_{n-1}\):

Test \(H_0:\sigma^2=\sigma_0^2\): statistic \(\chi^2=\dfrac{(n-1)s^2}{\sigma_0^2}\), compared to \(\chi^2_{n-1}\) critical values. (Note the chi-square table is asymmetric — you read two different critical values for the two tails.)

Out of scope — recognise and move on

Chi-squared test of homogeneity / independence (contingency table). Appeared once (EBf1-P6, no solution provided) but it is Ross chapter 9, outside this program (ch4–8). If you ever see a contingency table: build expected counts \(E_{ij}=\dfrac{\text{row}_i\cdot\text{col}_j}{N}\), statistic \(\chi^2=\sum\dfrac{(O-E)^2}{E}\), \(df=(r-1)(c-1)\). Just a margin note — do not study deeply.

F-test (equality of two variances, §8.5.1). The F-distribution (§5.8.3) is explicitly skipped, so treat the F-test as out of scope — recognition only, never required on these papers.

Which procedure? — top-level routing

Inference: which CI / which test?

Walk the questions in order:

1. CI or test? "compute an interval / how large a sample" → CI. "is there reason / more than / changed" → test.
2. Mean or proportion? averages/measurements → mean. percentages/counts of successes → proportion.
3. One sample or two? one group vs a target → one-sample. two groups compared → two-sample. same units before/after → paired.
4. z or t? known σ or large n (≥30) → z. small n, normal, unknown σ → t. (Proportions always z.)
5. One- or two-sided? "more / over / improved" → one-sided \(z_\alpha\). "changed / differ" → two-sided \(z_{\alpha/2}\).

CI & test formula bank (T1 + T2 spine)

\(\hat p_p=\dfrac{X_1+X_2}{n_1+n_2}\). Reject: two-sided \(|ts|\ge z_{\alpha/2}\); one-sided \(ts\ge z_\alpha\) (or \(\le-z_\alpha\)). p-value: one-sided \(1-\Phi(ts)\), two-sided \(2(1-\Phi(|ts|))\); reject when \(\alpha\ge\) p-value.

Sample size (round UP)

Critical values & Φ shortcuts

Φ values: \(\Phi(0.5)=0.6915\), \(\Phi(1)=0.8413\), \(\Phi(2)=0.9772\), \(\Phi(2.4)=0.9918\), \(\Phi(1/3)=0.6293\). Symmetry \(\Phi(-z)=1-\Phi(z)\). Central band \(2\Phi(d/\sigma)-1\). Inverse: \(\Phi^{-1}(0.975)=1.96\), \(\Phi^{-1}(0.75)=0.675\). Standardize \(Z=(X-\mu)/\sigma\).

Estimator-theory recipes (T3 / T10)

• MSE: \(\mathrm{MSE}=\mathrm{Var}+\mathrm{Bias}^2\); Bias \(=E[T]-\theta\). Unbiased ⇒ MSE=Var.
• Var of combo: \(\mathrm{Var}(aU+bW)=a^2\mathrm{Var}(U)+b^2\mathrm{Var}(W)\) (independent).
• Unbiased weights: any \(c\bar X_1+(1-c)\bar X_2\) is unbiased for μ (weights sum to 1).
• Min-MSE weight: inverse-variance, \(c^*=\dfrac{n_1}{n_1+n_2}\) for sample means.
• Find constant: set \(E[T]=\theta\); use \(E[X^2]=\sigma^2+\mu^2\), \(E[X_iX_j]=\mu^2\) (indep).
• Jensen: \(\sqrt T\) (or log, 1/T) of an unbiased T is biased.
• MLE: \(L=\prod f\) → \(\log\) → \(d/d\theta=0\). Geometric → \(\hat\theta=1/\bar X\).
• Method-of-moments: set \(E[X]=\bar X\) (as a function of θ), invert.
• Moments: Bernoulli \(E{=}p,V{=}p(1{-}p)\); Poisson \(E{=}V{=}\lambda\); \(\mathrm{Var}(\bar X_m)=\sigma^2/m\).

Probability templates (T4 / counting / CLT)

• Total probability: \(P(E)=\sum_i P(E\mid H_i)P(H_i)\).
• Bayes: \(P(H_j\mid E)=\dfrac{P(E\mid H_j)P(H_j)}{\sum_i P(E\mid H_i)P(H_i)}\).
• Coin mixture: \(P(X=0\mid N=n)=(\tfrac12)^n\).
• Conditional in Bernoulli: disjoint arrangements ÷ binomial; \(p\) cancels.
• Inclusion–exclusion: \(P(B\cup C)=P(B)+P(C)-P(B\cap C)\); \(P(B\cap C^c)=P(B)-P(B\cap C)\).
• Counting: \(P=\#\text{fav}/\binom{n}{k}\); complement for "different", fix items for "included".
• CLT (sum): \(S_n\approx N(n\mu,n\sigma^2)\), \(Z=\dfrac{s-n\mu}{\sigma\sqrt n}\) (SUM uses \(\sigma\sqrt n\), not \(\sigma/\sqrt n\)).
• Binomial→Normal: \(N(np,np(1-p))\); skip \(\pm\tfrac12\) if "no continuity correction".
• Normal models→prob: \(\mathrm{Var}=E[X^2]-(E[X])^2\), \(E[X(X-1)]=E[X^2]-E[X]\).