Study guide: Biostatistics: Exponential and Inference

📋 Course Outline

1. Population et caractères statistiques
2. Tableaux et graphiques statistiques
3. Tendance centrale
4. Dispersion des données
5. Principes de dénombrement
6. 2.3 Permutations
7. 2.4 Combinations
8. 2.5 Combination Properties
9. 2.6 Choosing the Right Formula
10. Discrete Variables and Probability Laws
11. Poisson and Geometric Laws
12. Continuous Densities and Uniform Laws
13. Normal and Exponential Laws
14. Exponential Distribution
15. Chi-Square Distribution
16. Estimation and Confidence Intervals
17. Hypothesis Tests and Comparisons
18. Regression and Correlation

📖 1. Population et caractères statistiques

🔑 Key Concepts & Definitions

• Biostatistique: Méthodes de recueil, de description, d’analyse et d’interprétation des données de santé en tenant compte de leur variabilité.
• Population: Ensemble sur lequel porte l’étude, dont chacun des éléments est une unité statistique, ou individu.
• Échantillon: Un échantillon est une partie de la population réellement observée, de taille n ; un tirage au sort limite le biais lié au choix des unités, tandis qu’un choix par commodité peut biaiser l’échantillon.
• Caractère: Propriété étudiée sur chaque unité statistique, dont les valeurs possibles sont les modalités.
• Distribution statistique: La distribution statistique associe à chaque modalité distincte son effectif, et la somme des effectifs des modalités est égale à la taille n de la série.

★ Must-know

📌 La statistique descriptive résume les données observées sans généraliser au-delà des unités étudiées, tandis que l’inférence généralise à une population en s’appuyant sur les probabilités.

📌 Un caractère qualitatif nominal désigne des catégories sans ordre, un caractère qualitatif ordinal des catégories ordonnées, un caractère quantitatif discret des nombres comptés et un caractère quantitatif continu des mesures.

Further detail

• Avant tout calcul, il faut repérer les valeurs impossibles, les doublons et les données manquantes, puis vérifier et corriger les erreurs à la source.

📌 Un code numérique ne rend pas un caractère quantitatif, et un caractère continu reste continu même si sa mesure est arrondie.

💡 Memory Hook

Un caractère qualitatif classe les unités ; un caractère quantitatif les compte ou les mesure.

📖 2. Tableaux et graphiques statistiques

🔑 Key Concepts & Definitions

• Fréquence: La fréquence d’une modalité d’effectif $n_i$ dans une série de taille $n$ est $f_i = n_i/n$ ; les fréquences totalisent 1 et les effectifs totalisent n.
• Fréquence cumulée: Pour un caractère ordonné, la fréquence cumulée jusqu’à la modalité $x_i$ est $F_i = N_i/n$, où $N_i$ est l’effectif des unités dont la modalité est au plus $x_i$.
• Boîte à moustaches: Une boîte à moustaches représente une boîte allant de $Q_1$ à $Q_3$, un trait à la médiane et des moustaches allant jusqu’au minimum et au maximum.

★ Must-know

• 🔄 Le regroupement en classes se fait ainsi: repérer le minimum et le maximum, choisir de 5 à 10 classes contiguës fermées à gauche, choisir des amplitudes égales, sauf si un seuil médical le justifie, compter l’effectif de chaque classe

📐 Formula - Pour une classe d’amplitude $a_i = e_i - e_{i-1}$, le centre est $c_i = (e_{i-1}+e_i)/2$ et la densité d’effectif est $d_i = n_i/a_i$.

📌 Dans un histogramme de classes d’amplitudes inégales, la hauteur doit représenter la densité, car l’aire des rectangles doit être proportionnelle aux effectifs.

Further detail

📌 Les fréquences cumulées ne s’appliquent pas à un caractère nominal, car ses modalités ne sont pas ordonnées.

• Les classes d’IMC et leurs fréquences sont:
  • maigreur : 18 adultes, soit 9 %
  • normale : 112 adultes, soit 56 %
  • surpoids : 48 adultes, soit 24 %
  • obésité : 22 adultes, soit 11 %

📐 Formula - Dans un diagramme en secteurs, l’angle d’une modalité de fréquence $f_i$ est $\alpha_i = 360^\circ \times f_i$.

📌 Un diagramme en barres représente les modalités qualitatives par des barres séparées de même largeur, tandis qu’un diagramme en bâtons représente les valeurs d’un caractère discret par des traits verticaux séparés.

📌 Une courbe cumulative relie les points formés par la borne supérieure de chaque classe et sa fréquence cumulée, en partant du point correspondant à une fréquence nulle.

📖 3. Tendance centrale

🔑 Key Concepts & Definitions

• Mode: Le mode est la modalité ou valeur ayant le plus grand effectif ; pour des données regroupées en classes, la classe modale est celle de plus grande densité.
• Médiane: La médiane est la valeur qui partage une série ordonnée en deux moitiés, avec au moins la moitié des valeurs inférieures ou égales à elle et au moins la moitié supérieures ou égales à elle.

★ Must-know

📐 Formula - La moyenne d’une distribution est $\bar{x}=\frac{1}{n}\sum_i n_i x_i=\sum_i f_i x_i$ ; pour des données regroupées en classes, les centres des classes remplacent les valeurs observées.

📌 La moyenne est sensible aux valeurs extrêmes, tandis que la médiane y est robuste ; pour une distribution étalée à droite, la moyenne est généralement supérieure à la médiane.

Further detail

• Pour une série brute ordonnée, la médiane est la valeur de rang $(n+1)/2$ si n est impair et la moyenne des valeurs de rangs $n/2$ et $n/2+1$ si n est pair.

📐 Formula - Pour une série regroupée en classes, la médiane interpolée est $Me = e_{i-1}+a_i\frac{0{,}5-F_{i-1}}{F_i-F_{i-1}}$ dans la classe où la fréquence cumulée franchit 0,5.

📌 Le mode convient aux caractères nominaux, la médiane aux caractères ordinaux et aux distributions asymétriques ou comportant des valeurs extrêmes, et la moyenne aux caractères quantitatifs symétriques.

📐 Formula - La moyenne de deux groupes réunis est $\bar{x}=\frac{n_1\bar{x}_1+n_2\bar{x}_2}{n_1+n_2}$, et n’est généralement pas la moyenne simple des deux moyennes de groupe.

💡 Memory Hook

La moyenne suit la queue de la distribution ; la médiane résiste mieux aux valeurs extrêmes.

📖 4. Dispersion des données

🔑 Key Concepts & Definitions

• Écart interquartile: L’écart interquartile est $Q_3-Q_1$ et représente l’étendue de la moitié centrale des données ; il est robuste aux valeurs extrêmes.
• Coefficient de variation: Le coefficient de variation est $CV=\sigma/\bar{x}$, exprimé en pourcentage et sans unité pour un caractère positif ; il compare les dispersions de caractères d’unités ou de niveaux différents.

★ Must-know

📐 Formula - La variance est $V=\frac{1}{n}\sum_i n_i(x_i-\bar{x})^2=\frac{1}{n}\sum_i n_i x_i^2-\bar{x}^2$, et l’écart-type est $\sigma=\sqrt{V}$.

Further detail

• L’étendue d’une série est $x_{\max}-x_{\min}$ et ne dépend que des deux valeurs extrêmes.

📌 Dans la convention du cours pour une série brute ordonnée, $Q_1$ est la valeur de rang $\lceil n/4\rceil$ et $Q_3$ celle de rang $\lceil 3n/4\rceil$.

• La variance est nulle seulement si toutes les valeurs de la série sont égales.

📖 5. Principes de dénombrement

🔑 Key Concepts & Definitions

• Factorielle: Pour un entier $n\geq1$, la factorielle est $n!=n(n-1)\cdots2\times1$, avec $0!=1$.
• Arrangement avec répétition: An arrangement with repetition is an ordered sequence of p elements chosen from n possibilities, where an element may be chosen more than once.
• Arrangement sans répétition: An arrangement without repetition is an ordered sequence of p distinct elements chosen from n, where $p\le n$.

★ Must-know

📌 Pour un choix en étapes successives ayant respectivement $n_1,\ldots,n_k$ possibilités à chaque étape, le principe du produit donne $n_1\times\cdots\times n_k$ choix au total.

📌 Si des cas se répartissent en familles disjointes d’effectifs $m_1,\ldots,m_k$, le principe de la somme donne $m_1+\cdots+m_k$ cas au total.

• Le nombre d’ordres possibles pour ranger n objets distincts en file est $n!$.

• The number of arrangements with repetition is $n^p$ because each of the p positions has n possible choices.

📐 Formula - The number of arrangements without repetition is $A_n^p=\frac{n!}{(n-p)!}=n(n-1)\cdots(n-p+1)$.

Further detail

📌 Si tous les cas sont équiprobables, une probabilité est le rapport du nombre de cas favorables au nombre de cas possibles, ce qui nécessite de dénombrer ces cas.

• Un groupe ABO choisi parmi A, B, AB et O et un rhésus choisi parmi + et − donnent $4\times2=8$ groupes sanguins possibles.

📐 Formula - Pour deux ensembles qui peuvent se recouper, le nombre d’éléments de leur réunion est $\operatorname{card}(A\cup B)=\operatorname{card}(A)+\operatorname{card}(B)-\operatorname{card}(A\cap B)$.

• Ranger 3 patients distincts en file donne $3!=6$ ordres de passage.

• A codon is an ordered sequence of three RNA bases chosen from A, C, G, and U, giving $4^3=64$ possible codons.

• A four-digit code using the ten digits has $10^4=10\,000$ possible codes when digits may repeat.

• Assigning Monday, Tuesday, and Wednesday guard shifts to three different interns chosen from eight gives $A_8^3=8\times7\times6=336$ schedules.

📌 When p=n, an arrangement without repetition is a permutation, so $A_n^n=n!$.

💡 Memory Hook

« Et » multiplie les possibilités successives ; « ou » additionne les familles disjointes.

📖 6. 2.3 Permutations

🔑 Key Concepts & Definitions

• Permutation sans répétition: A permutation without repetition is an ordering of all n distinct objects, equivalently an arrangement of n objects taken all at once.
• Permutation avec répétition: A permutation with repetition orders n objects divided into k groups of identical objects, with group sizes $n_1,\ldots,n_k$ satisfying $n_1+\cdots+n_k=n$.

★ Must-know

📐 Formula - The number of permutations of n distinct objects is $P_n=A_n^n=n!$.

📐 Formula - The number of permutations of n objects with group sizes $n_1,\ldots,n_k$ is $\frac{n!}{n_1!n_2!\cdots n_k!}$.

Further detail

• Five distinct patients can be ordered in $5!=120$ ways; if the most urgent patient goes first, the remaining four can be ordered in $4!=24$ ways.

• If a mother and her child must pass consecutively among five patients, treating them as a block that can be ordered in two ways gives $2\times4!=48$ orders.

• Six DNA bases containing two A, two T, one C, and one G can be arranged in $\frac{6!}{2!2!1!1!}=180$ distinct sequences.

• Nine boxes comprising four identical paracetamol boxes, three identical amoxicillin boxes, and two identical metronidazole boxes can be shelved in $\frac{9!}{4!3!2!}=1\,260$ ways.

• If the four identical paracetamol boxes stay together, treating them as one block gives $\frac{6!}{1!3!2!}=60$ arrangements.

• If a paracetamol box must be first in the nine-box arrangement, the remaining boxes can be arranged in $\frac{8!}{3!3!2!}=560$ ways, which is $\frac49$ of all arrangements.

📖 7. 2.4 Combinations

🔑 Key Concepts & Definitions

• Combinaison sans remise: A combination without replacement is an unordered subset of p distinct elements chosen from n, where $0\le p\le n$.
• Combinaison avec remise: A combination with repetition is an unordered choice of p objects from n types in which a type may be selected multiple times; only the count of each type matters.

★ Must-know

📐 Formula - The number of combinations without replacement is $C_n^p=\binom{n}{p}=\frac{n!}{p!(n-p)!}=\frac{A_n^p}{p!}$.

📐 Formula - The number of combinations with repetition is $\binom{n+p-1}{p}=C_{n+p-1}^p$.

Further detail

• The notations for choosing p elements from n without regard to order are $C_n^p$ and $\binom{n}{p}$, read as “p among n.”

• The boundary values are $C_n^0=C_n^n=1$ and $C_n^1=n$.

• Choosing three people from ten without regard to order gives $C_{10}^3=\frac{A_{10}^3}{3!}=120$ groups.

• The combinations-with-repetition formula can be represented by p object-symbols and n−1 separators, with the p object positions chosen among n+p−1 positions.

• Choosing two boxes from types X and Y with repetition and without regard to order gives XX, XY, and YY, for $\binom{3}{2}=3$ choices.

📌 With repetition, p may exceed n, and the unordered outcomes need not be equally probable even when each ordered draw is equally likely.

• A team of three chosen from ten people has $C_{10}^3=120$ possible teams; exactly two women among six women and four men gives $C_6^2C_4^1=60$ teams, and at least one man gives $120-C_6^3=100$ teams.

• Distributing six identical boxes among three generic types gives $\binom{8}{6}=28$ distributions, and requiring every type to appear gives $\binom{5}{3}=10$ distributions.

💡 Memory Hook

Arrangements count order; combinations ignore it.

📖 8. 2.5 Combination Properties

★ Must-know

📐 Formula - Combination symmetry gives $\binom{n}{p}=\binom{n}{n-p}$.

📐 Formula - For $1\le p\le n-1$, Pascal’s relation is $\binom{n}{p}=\binom{n-1}{p-1}+\binom{n-1}{p}$.

📐 Formula - For real a and b and integer $n\ge1$, the binomial theorem is $(a+b)^n=\sum_{k=0}^{n}\binom{n}{k}a^k b^{n-k}$.

Further detail

• To prove Pascal’s relation, divide the p-element subsets into those containing a fixed element and those not containing it, then count the two disjoint cases and add them.

• In Pascal’s triangle, row n and position p counted from zero give $\binom{n}{p}$; each interior entry is the sum of the two entries above it, each row is symmetric, and row n sums to $2^n$.

• The binomial expansion for n=3 is $(a+b)^3=a^3+3a^2b+3ab^2+b^3$.

📐 Formula - Setting a=b=1 in the binomial theorem gives $\sum_{k=0}^{n}\binom{n}{k}=2^n$.

• A set with n elements has $2^n$ subsets because each element is either included or excluded.

• For four symptoms that can each be present or absent, there are $2^4=16$ profiles: one with none, four with one, six with two, four with three, and one with all four.

💡 Memory Hook

In Pascal’s triangle, each interior number is supported by the two numbers above it.

📖 9. 2.6 Choosing the Right Formula

★ Must-know

• 🔄 To choose a counting formula:
  1. Split the problem into stages or disjoint cases.
  2. Identify n and p.
  3. Check whether exchanging two elements changes the outcome.
  4. Check whether an element can recur.

📌 A simultaneous selection uses combinations; successive ordered draws without replacement use arrangements without repetition, and successive ordered draws with replacement use arrangements with repetition.

Further detail

• When all distinct objects are arranged, use $n!$; when objects are identical within groups of sizes $n_1,\ldots,n_k$, use $\frac{n!}{n_1!\cdots n_k!}$.

📌 For a problem asking for at least one occurrence, count the complement with none and subtract that count from the total.

• For a compound counting problem, use the product principle for successive choices and the sum principle for disjoint cases.

📖 10. Discrete Variables and Probability Laws

🔑 Key Concepts & Definitions

• Discrete random variable: A variable that assigns a real number to each outcome of a random experiment and has finitely many possible values or values that can be numbered.
• Probability law: A list of each possible value xi with its probability pi = P(X = xi), where 0 ≤ pi ≤ 1 and the probabilities sum to 1.
• Bernoulli law: A Bernoulli trial has a success with probability p and a failure with probability 1 − p; its indicator X is 1 for success and 0 for failure, with $E(X)=p$ and $V(X)=p(1-p)$.
• Binomial law: If n independent Bernoulli trials have the same success probability p, their success count X follows a binomial law with values 0 through n and $P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}$.

★ Must-know

📐 Formula - For a discrete variable, $E(X)=\sum_i x_i p_i$, $V(X)=\sum_i p_i(x_i-E(X))^2=E(X^2)-E(X)^2$, and $\sigma(X)=\sqrt{V(X)}$.

📐 Formula - For $X\sim B(n;p)$, $E(X)=np$ and $V(X)=np(1-p)$.

Further detail

• For two independent children of parents AS × AS, each with probability 1/4 of being SS, the number X of SS children has probabilities P(X = 0) = 9/16, P(X = 1) = 6/16, and P(X = 2) = 1/16.

📐 Formula - For constants a and b, $E(aX+b)=aE(X)+b$, $V(aX+b)=a^2V(X)$, and $\sigma(aX+b)=|a|\sigma(X)$.

📌 A binomial model requires a fixed number of trials, two outcomes per trial, the same success probability on every trial, and independent trials.

📖 11. Poisson and Geometric Laws

🔑 Key Concepts & Definitions

• Poisson law: A Poisson variable with parameter λ > 0 counts events in an interval and takes values 0, 1, 2, … with $P(X=k)=e^{-\lambda}\frac{\lambda^k}{k!}$ and $E(X)=V(X)=\lambda$.
• Geometric law: In independent Bernoulli trials with success probability p, the rank X of the first success follows a geometric law on 1, 2, 3, … with $P(X=k)=(1-p)^{k-1}p$, $E(X)=1/p$, and $V(X)=(1-p)/p^2$.
• Hypergeometric law: From a population of N individuals containing K = Np carriers, a sample of n drawn without replacement has carrier count X with $P(X=k)=\frac{\binom{K}{k}\binom{N-K}{n-k}}{\binom{N}{n}}$ for $\max(0,n-N+K)\leq k\leq\min(n,K)$.

★ Must-know

📐 Formula - For a hypergeometric variable with p = K/N, $E(X)=np$ and $V(X)=np(1-p)\frac{N-n}{N-1}$.

Further detail

📌 The Poisson law models events occurring independently, one at a time, at a constant average rate; λ is the mean count for the interval studied and scales with its duration or size.

📌 When n is large and p is small, with the practical conditions n ≥ 30 and p ≤ 0.1, the binomial law B(n;p) can be approximated by a Poisson law with parameter np.

📐 Formula - For a geometric variable, $P(X>k)=(1-p)^k$ and $P(X\leq k)=1-(1-p)^k$.

📌 Because $\frac{N-n}{N-1}<1$ when n > 1, hypergeometric sampling without replacement has less variance than sampling with replacement at the same n and p.

📌 When the sampling fraction n/N is at most 0.1, the hypergeometric law H(N;n;p) can be approximated by the binomial law B(n;p).

💡 Memory Hook

Poisson counts events in an interval; geometric counts trials until the first success.

📖 12. Continuous Densities and Uniform Laws

🔑 Key Concepts & Definitions

• Continuous random variable: A variable whose probabilities are represented by areas under a density, as for measurements such as glucose, height, or duration rather than counts.
• Probability density: A density f is nonnegative and has total area $\int_{-\infty}^{+\infty}f(x)\,dx=1$; the probability of an interval is $P(a<X\leq b)=\int_a^b f(x)\,dx$.
• Uniform law: A variable uniform on [a;b] has constant density $f(x)=\frac{1}{b-a}$ on that interval, cumulative distribution $F(x)=\frac{x-a}{b-a}$ for a ≤ x ≤ b, and interval probability $P(c<X\leq d)=\frac{d-c}{b-a}$.

★ Must-know

• For a continuous variable, the probability of any exact value is zero, so probabilities are calculated over intervals.

📐 Formula - The continuous cumulative distribution function is $F(x)=P(X\leq x)=\int_{-\infty}^{x}f(t)\,dt$, so $P(a<X\leq b)=F(b)-F(a)$ and $P(X>a)=1-F(a)$.

Further detail

📐 Formula - For a continuous variable, $E(X)=\int x f(x)\,dx$, $V(X)=\int (x-E(X))^2f(x)\,dx=E(X^2)-E(X)^2$, and $\sigma(X)=\sqrt{V(X)}$.

📐 Formula - For a uniform variable on [a;b], $E(X)=\frac{a+b}{2}$, $V(X)=\frac{(b-a)^2}{12}$, and $\sigma(X)=\frac{b-a}{\sqrt{12}}$.

💡 Memory Hook

Picture probability as area beneath a density curve.

📖 13. Normal and Exponential Laws

🔑 Key Concepts & Definitions

• Normal law: A normal variable with mean μ and standard deviation σ has density $f(x)=\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}$, expectation μ, and standard deviation σ.
• Exponential law: An exponential variable with rate λ > 0 has density $f(x)=\lambda e^{-\lambda x}$ for x ≥ 0 and zero for x < 0, cumulative distribution $F(x)=1-e^{-\lambda x}$ for x ≥ 0, and survival probability $P(X>x)=e^{-\lambda x}$.

★ Must-know

📐 Formula - If $X\sim N(\mu;\sigma)$, then $Z=\frac{X-\mu}{\sigma}\sim N(0;1)$ and $P(X\leq x)=\Phi\left(\frac{x-\mu}{\sigma}\right)$.

Further detail

• The normal density is a bell-shaped curve symmetric about μ, where its mean, median, and mode coincide; its inflection points are μ − σ and μ + σ.

• For a normal law, the probabilities within one, two, and three standard deviations of the mean are 0.6827, 0.9545, and 0.9973, respectively.

📌 A binomial variable B(n;p) can be approximated by a normal variable with mean np and standard deviation $\sqrt{np(1-p)}$ when np ≥ 5 and n(1 − p) ≥ 5; continuity correction represents integer k by [k − 0.5;k + 0.5].

📖 14. Exponential Distribution

★ Must-know

📐 Formula - For an exponential distribution $E(\lambda)$ with $\lambda>0$, the mean and standard deviation are $E(X)=\sigma(X)=\frac{1}{\lambda}$, the variance is $V(X)=\frac{1}{\lambda^2}$, and the median is $\frac{\ln 2}{\lambda}$.

📌 The exponential distribution is memoryless: for $s,t>0$, $P(X>s+t\mid X>s)=P(X>t)$, so having already waited for time $s$ does not change the distribution of the remaining wait.

• If arrivals follow a Poisson distribution with mean $\lambda t$ during a period of length $t$, the waiting time $T$ to the first arrival satisfies $P(T>t)=P(\text{no arrivals})=e^{-\lambda t}$ and therefore follows $E(\lambda)$.

Further detail

• Exponential waiting times can model intervals between emergency-department arrivals and radioactive disintegrations, but not human lifetimes when the risk of death increases with age.

💡 Memory Hook

Past waiting does not change the remaining wait; aging does change the human-death risk.

📖 15. Chi-Square Distribution

🔑 Key Concepts & Definitions

• Chi-square distribution: The chi-square distribution with $\nu$ degrees of freedom is the distribution of $K=Z_1^2+\cdots+Z_\nu^2$ when the $Z_i$ are independent standard normal variables.
• Chi-square critical value: The value $\chi^2_{1-\alpha}(\nu)$ is exceeded with probability $\alpha$ by a chi-square variable with $\nu$ degrees of freedom, so $P(K>\chi^2_{1-\alpha}(\nu))=\alpha$.

📝 Essential Points

📐 Formula - For $K\sim\chi^2(\nu)$, $E(K)=\nu$ and $V(K)=2\nu$.

• A chi-square curve is right-skewed and becomes more symmetric as its degrees of freedom increase.

• For one degree of freedom, $\chi^2_{1-\alpha}(1)=u_\alpha^2$, where $u_\alpha$ is the two-sided standard-normal quantile.

• The tabulated upper-tail chi-square critical values for degrees of freedom 1 through 5 at $\alpha=0.10{,}0.05{,}0.01$ are respectively: 1: 2.706, 3.841, 6.635; 2: 4.605, 5.991, 9.210; 3: 6.251, 7.815, 11.345; 4: 7.779, 9.488, 13.277; 5: 9.236, 11.070, 15.086.

• The chi-square test compares observed and theoretical counts; larger discrepancies produce larger observed chi-square statistics.

💡 Memory Hook

Imagine independent standard-normal values being squared and stacked into a right-skewed curve.

📖 16. Estimation and Confidence Intervals

🔑 Key Concepts & Definitions

• Population parameter: A population parameter is a fixed but unknown value describing a target population, such as its mean, proportion, or standard deviation.
• Confidence interval: A confidence interval at level $1-\alpha$ is calculated from a sample using a method that produces intervals containing the unknown parameter in proportion $1-\alpha$ of possible samples.

★ Must-know

• For a random independent sample of size $n$, the point estimates of the population mean, proportion, variance, and standard deviation are respectively $\bar{x}=\sum x_i/n$, $\hat p=k/n$, $s^2=\sum(x_i-\bar{x})^2/(n-1)$, and $s=\sqrt{s^2}$.

📐 Formula - For independent identically distributed observations with mean $\mu$ and standard deviation $\sigma$, the sample mean satisfies $E(\bar X)=\mu$ and $\sigma(\bar X)=\frac{\sigma}{\sqrt n}$; $\sigma/\sqrt n$ is its standard error.

📐 Formula - For a large sample, a 95% confidence interval for a population mean is $\bar{x}\pm1.96\frac{s}{\sqrt n}$, replacing the unknown population standard deviation with the sample standard deviation.

📐 Formula - A 95% confidence interval for a proportion is $\hat p\pm1.96\sqrt{\frac{\hat p(1-\hat p)}{n}}$, provided $n\ge30$, $n\hat p\ge5$, and $n(1-\hat p)\ge5$.

📐 Formula - For a desired half-width $e$ at 95% confidence, the required sample size is rounded up from $n\ge(1.96\sigma/e)^2$ for estimating a mean or $n\ge1.96^2p(1-p)/e^2$ for estimating a proportion.

Further detail

📐 Formula - The sample variance can also be calculated as $s^2=\frac{\sum x_i^2-n\bar{x}^2}{n-1}$.

• The sample variance divides the squared deviations from the sample mean by $n-1$ because those deviations tend to underestimate population variance; equivalently, $s^2=nV/(n-1)$ when $V=\sum(x_i-\bar{x})^2/n$.

• By the central limit theorem, for sufficiently large samples (approximately $n\ge 30$), the sample mean is approximately normal with mean $\mu$ and standard deviation $\sigma/\sqrt n$, whatever the distribution of the observations.

📐 Formula - For an observed sample proportion, $E(\hat P)=p$ and $\sigma(\hat P)=\sqrt{p(1-p)/n}$; its distribution is approximately normal when $np\ge5$ and $n(1-p)\ge5$.

• For a 95% confidence interval, the standard-normal two-sided quantiles are $u_{0.10}=1.645$, $u_{0.05}=1.96$, and $u_{0.01}=2.576$; use the quantile matching the chosen confidence level.

📌 For a mean interval with $n<30$, the data are assumed normal and the Student distribution is used.

• Before sampling, a 95% confidence interval has a 95% chance of containing the fixed population parameter; after the interval is calculated, it either contains the parameter or it does not.

• A confidence interval becomes wider as confidence level or sample dispersion increases and becomes narrower in proportion to $1/\sqrt n$; halving its width requires four times as many subjects.

📖 17. Hypothesis Tests and Comparisons

🔑 Key Concepts & Definitions

• Significance level probability: The significance probability is the probability, assuming $H_0$ is true, of observing an outcome at least as extreme as the one obtained; for a standard-normal statistic it is $2[1-\Phi(|z|)]$ in a two-sided test and $1-\Phi(|z|)$ in a one-sided test in the direction of $H_1$.

★ Must-know

• The null hypothesis $H_0$ states no effect or difference, whereas the alternative $H_1$ specifies the effect or difference being sought; alternatives may be two-sided or one-sided, with a one-sided direction chosen before examining data.

• A type I error rejects a true $H_0$ with risk $\alpha$, while a type II error fails to reject a false $H_0$ with risk $\beta$; test power is $1-\beta$.

• Failing to reject $H_0$ does not demonstrate that $H_0$ is true.

• A hypothesis test proceeds by stating $H_0$ and $H_1$ and choosing $\alpha$, checking sampling and count conditions, calculating a test statistic, identifying the critical region, and rejecting $H_0$ if the statistic falls in that region before interpreting the result in context.

• For a chi-square independence test, calculate expected counts under independence as $T_{ij}=L_iC_j/n$, compute $\chi^2_{obs}=\sum_{i,j}(O_{ij}-T_{ij})^2/T_{ij}$, and reject independence at level $\alpha$ when $\chi^2_{obs}>\chi^2_{1-\alpha}(\nu)$, where $\nu=(r-1)(c-1)$.

Further detail

• The significance level $\alpha$ is fixed in advance, often at 5%; $\beta$ is not fixed and decreases when sample size increases or the true effect becomes larger.

📌 For a standard-normal test statistic, a two-sided test rejects at level 5% if $|z|>1.96$ and at level 1% if $|z|>2.576$; a one-sided test rejects in the prespecified direction if $|z|>1.645$ at 5% or $|z|>2.326$ at 1%.

📌 A result is significant at level $\alpha$ when its significance probability is below $\alpha$.

📐 Formula - For testing one proportion against $p_0$, $z=\frac{\hat p-p_0}{\sqrt{p_0(1-p_0)/n}}$, provided $np_0\ge5$ and $n(1-p_0)\ge5$.

📐 Formula - For independent large samples, the test statistic for two means is $z=\frac{\bar{x}_1-\bar{x}_2}{\sqrt{s_1^2/n_1+s_2^2/n_2}}$; the stated normal approximation requires $n_1,n_2\ge30$.

📐 Formula - For independent samples, the two-proportion test uses pooled proportion $\hat p=(k_1+k_2)/(n_1+n_2)$ and statistic $z=\frac{\hat p_1-\hat p_2}{\sqrt{\hat p(1-\hat p)(1/n_1+1/n_2)}}$, with expected pooled successes and failures of at least 5 in each group.

📐 Formula - For a chi-square test of qualitative characteristics, $\chi^2_{obs}=\sum(O-T)^2/T$, with degrees of freedom $\nu=(r-1)(c-1)$ for a table with $r$ rows and $c$ columns; the approximate chi-square distribution requires every expected count $T\ge5$.

📌 For comparing two means with either sample size below 30, use Student's test assuming normally distributed data and similar variances, with $n_1+n_2-2$ degrees of freedom.

📐 Formula - A 95% confidence interval for the difference between two means is $(\bar{x}_1-\bar{x}_2)\pm1.96\sqrt{s_1^2/n_1+s_2^2/n_2}$.

📌 The chi-square homogeneity test uses the same calculation as the independence test to compare the distribution of a characteristic across groups.

💡 Memory Hook

Rejecting H₀ supports evidence against it; failing to reject H₀ does not prove it.

📖 18. Regression and Correlation

🔑 Key Concepts & Definitions

• Covariance: The covariance of paired quantitative variables is $\operatorname{Cov}(x,y)=\frac1n\sum_{i=1}^n(x_i-\bar{x})(y_i-\bar{y})=\frac1n\sum_{i=1}^n x_iy_i-\bar{x}\bar{y}$; it is positive when the variables tend to vary in the same direction and negative when they tend to vary in opposite directions.

★ Must-know

📐 Formula - The least-squares regression line of $y$ on $x$ minimizes the sum of squared vertical deviations $\sum(y_i-ax_i-b)^2$ and has coefficients $a=\operatorname{Cov}(x,y)/V(x)$ and $b=\bar{y}-a\bar{x}$.

📐 Formula - The linear correlation coefficient is $r=\operatorname{Cov}(x,y)/(\sigma_x\sigma_y)$, where $\sigma_x=\sqrt{V(x)}$ and $\sigma_y=\sqrt{V(y)}$, and $-1\le r\le1$.

📌 A statistically significant correlation does not prove that $x$ causes $y$, because a third factor may influence both; a randomized trial is the reference method for establishing a causal effect.

Further detail

• The least-squares regression line passes through the point of means $(\bar{x},\bar{y})$ and summarizes the trend for predicting $y$ from $x$ within the observed range.

• Values of $|r|$ near 1 indicate that points are nearly aligned, values of $r$ near 0 indicate little linear association, and $r^2$ represents the proportion of variance in $y$ explained by the linear relationship.

📌 To test for population correlation, use $\nu=n-2$ degrees of freedom and reject the null hypothesis of no correlation when $|r|>r_\alpha$ from the critical-value table; the pairs must be independent and approximately normally distributed.

• A correlation near zero does not rule out a nonlinear relationship between the variables.

💡 Memory Hook

Correlation measures association; randomization, not correlation alone, supports causal inference.