Statistics Formula Sheet
Author
Elizabeth McDonald
Last Updated
3 年前
License
Creative Commons CC BY 4.0
Abstract
Formula sheet for a statistics course
Formula sheet for a statistics course
\documentclass{article}
\usepackage[landscape]{geometry}
\usepackage{url}
\usepackage{multicol}
\usepackage{amsmath}
\usepackage{esint}
\usepackage{amsfonts}
\usepackage{tikz}
\usetikzlibrary{decorations.pathmorphing}
\usepackage{amsmath,amssymb}
\usepackage{colortbl}
\usepackage{xcolor}
\usepackage{mathtools}
\usepackage{amsmath,amssymb}
\usepackage{enumitem}
\makeatletter
\newcommand*\bigcdot{\mathpalette\bigcdot@{.5}}
\newcommand*\bigcdot@[2]{\mathbin{\vcenter{\hbox{\scalebox{#2}{$\m@th#1\bullet$}}}}}
\makeatother
\title{STAT 251 Formula Sheet}
\usepackage[brazilian]{babel}
\usepackage[utf8]{inputenc}
\advance\topmargin-.8in
\advance\textheight3in
\advance\textwidth3in
\advance\oddsidemargin-1.45in
\advance\evensidemargin-1.45in
\parindent0pt
\parskip2pt
\newcommand{\hr}{\centerline{\rule{3.5in}{1pt}}}
%\colorbox[HTML]{e4e4e4}{\makebox[\textwidth-2\fboxsep][l]{texto}
\begin{document}
\begin{center}{\huge{\textbf{STAT 251 Formula Sheet}}}\\
\end{center}
\begin{multicols*}{2}
\tikzstyle{mybox} = [draw=black, fill=white, very thick,
rectangle, rounded corners, inner sep=10pt, inner ysep=10pt]
\tikzstyle{fancytitle} =[fill=black, text=white, font=\bfseries]
%------------ Measures of Center ---------------
\begin{tikzpicture}
\node [mybox] (box){%
\begin{minipage}{0.46\textwidth}
\begin{tabular}{lp{8cm} l}
Mean:
& $\overline{x} = \frac{\sum_{i=1}^{n}x_i}{n} = \frac{x_1 + x_2 + ... + x_n}{n} $ \\
Median: & If n is even then $ \tilde{x} = \frac{(\frac{n}{2})^\text{th} \text{ obs.} + (\frac{n+1}{2})^\text{th} \text{ obs.}}{2} $ \\
& If n is odd then $ \tilde{x} = \frac{n+1}{2}^\text{th}\text{ obs.}$ \\
\end{tabular}
\end{minipage}
};
%------------ Measures of Center Header ---------------------
\node[fancytitle, right=10pt] at (box.north west) {Measures of Center};
\end{tikzpicture}
%------------ Measures of Variability ---------------
\begin{tikzpicture}
\node [mybox] (box){%
\begin{minipage}{0.46\textwidth}
\begin{tabular}{lp{8cm} l}
Range: & $R = x_\text{largest} - x_\text{smallest}$\\
Variance: & $s^2 = \frac{\sum_{i=1}^{n}(x_i - \overline{x})^2}{n-1} = \frac{\sum_{i=1}^{n}x_i^2 - n\overline{x}^2}{n-1}$\\
Standard deviation: & $s = \sqrt{\frac{\sum_{i=1}^{n}(x_i - \overline{x})^2}{n-1}} = \sqrt{\frac{\sum_{i=1}^{n}x_i^2 - n\overline{x}^2}{n-1}}$\\
IQR: & $\text{IQR} = \text{Q3} - \text{Q1}$
\end{tabular}\\
\textbf{Method to compute $\text{Q}_{(p)}$:}
\begin{itemize}
\setlength\itemsep{0em}
\item Sort data from smallest to largest: $x_{(1)} \leq x_{(2)} \leq ... \leq x_{(n)}$
\item Compute the number $np + 0.5$
\item If $np + 0.5$ is an integer, $m$, then: $\text{Q}_{(p)} = x_{(m)}$
\item If $np + 0.5$ is not an integer, $m < np + 0.5 < m+1$ for some integer $m$, then:\\ $\text{Q}_{(p)} = \frac{x_{(m)}+x_{(m+1)}}{2}$
\end{itemize}
\textbf{Outliers:}
\begin{itemize}
\setlength\itemsep{0em}
\item Values smaller than $\text{Q1} - (1.5 \times \text{IQR})$ are outliers
\item Values greater than $\text{Q3} + (1.5 \times \text{IQR})$ are outliers
\end{itemize}
\end{minipage}
};
%------------ Measures of Variability Header ---------------------
\node[fancytitle, right=10pt] at (box.north west) {Measures of Variability};
\end{tikzpicture}
%------------ Discrete Random Variable and Distributions ---------------
\begin{tikzpicture}
\node [mybox] (box){%
\begin{minipage}{0.46\textwidth}
Consider a \textbf{discrete} random variable $X$ \\
\textbf{Probability Mass Function} (pmf): $f(x) = P(X = x)$
\begin{enumerate}
\setlength\itemsep{0em}
\item $f(x) \geq 0$ for all $x$ in $X$
\item $\sum_x f(x) = 1$
\end{enumerate}
\textbf{Cumulative Distributive Function} (cdf): $F(x) = P(X \leq x) = \sum_{k \leq x}f(k)$ \\
\begin{tabular}{lp{8cm} l}
Mean ($\mu$): & $E(X) = \sum_{x}xf(x)$ \\
Expected value: & $E(g(X)) = \sum_{x}g(x)f(x)$ \\
Variance ($\sigma^2$): & $Var(X) = \sum_x (x-\mu)^2f(x) = E(X^2)-[E(X)]^2$\\
SD ($\sigma$): & $SD(X) = \sqrt{Var(X)}$
\end{tabular} \\
\\
\end{minipage}
};
%------------ Discrete Random Variable and Distributions Header ---------------------
\node[fancytitle, right=10pt] at (box.north west) {Discrete Random Variables};
\end{tikzpicture}
%------------ Sets and Probability ---------------
\begin{tikzpicture}
\node [mybox] (box){%
\begin{minipage}{0.46\textwidth}
\textbf{Properties of Probability:}
\begin{itemize}
\setlength\itemsep{0em}
\item \textit{General Addition Rule:} $P(A \cup B) = P(A) + P(B) - P(A \cap B)$
\item \textit{Complement Rule:} $P(A^c) = 1 - P(A)$
\item If $A \subseteq B$ then $P(A \cap B) = P(A)$
\item If $A \subseteq B$ then $P(A) \leq P(B)$
\item $P(\emptyset) = 0$ and $P(S) = 1$
\item $0 \leq P(A) \leq 1$ for all $A$
\end{itemize}
\textbf{Conditional Probability:}
\begin{itemize}
\setlength\itemsep{0em}
\item $P(A|B) = \frac{P(A \cap B)}{P(B)}$ and $P(B|A) = \frac{P(A \cap B)}{P(A)}$
\item \textit{Multiplication Rule:} $P(A \cap B) = P(B) \times P(A|B)$ and \\ $P(A \cap B) = P(A) \times P(B|A)$
\item Events $A$ and $B$ are \textbf{independent} if and only if $P(A \cap B) = P(A)P(B)$ \\ and thus $P(A|B) = P(A)$ and $P(B|A) = P(B)$
\end{itemize}
\begin{tabular}{l l l}
\textbf{Bayes' Theorem:}
& $P(A_i|B)$ &$= \frac{P(B|A_i)P(A_i)}{\sum_{i=1}^{n}P(A_i)P(B|A_i)}$\\
&& $= \frac{P(B|A_i)P(A_i)}{P(B|A_1)P(A_1) + P(B|A_2)P(A_2)+ . . . + P(B|A_n)P(A_n)}$\\
\end{tabular}
\end{minipage}
};
%------------ Sets and Probability Header ---------------------
\node[fancytitle, right=10pt] at (box.north west) {Sets and Probability};
\end{tikzpicture}
%------------ Continuous Random Variable and Distributions ---------------
\begin{tikzpicture}
\node [mybox] (box){%
\begin{minipage}{0.46\textwidth}
Consider a \textbf{continuous} random variable $X$ \\
\textbf{Probability Density Function} (pdf): $P(a \leq X \leq b) = \int_{a}^{b} f(x) dx$
\begin{enumerate}
\setlength\itemsep{0em}
\item $f(x) \geq 0$ for all $x$
\item $\int^{\infty}_{-\infty} f(x) dx = 1$
\end{enumerate}
\textbf{Cumulative Distributive Function} (cdf): $F(x) = P(X \leq x) = \int_{-\infty}^{x} f(t) dt$ \\
\begin{tabular}{lp{8cm} l}
Median: & $x$ such that $F(x) = 0.5$ \\
$Q_1$ and $Q_3$: & $x$ such that $F(x) = 0.25$ and $x$ such that $F(x) = 0.75$\\
Mean ($\mu$): & $E(X) = \int^{\infty}_{-\infty} xf(x) dx$ \\
Expected value: & $E(g(X)) = \int^{\infty}_{-\infty} g(x)f(x) dx$ \\
Variance ($\sigma^2$): & $Var(X) = \int^{\infty}_{-\infty} (x-\mu)^2f(x) dx = E(X^2)-[E(X)]^2$\\
SD ($\sigma$): & $SD(X) = \sqrt{Var(X)}$
\end{tabular} \\
\end{minipage}
};
%------------ Continuous Random Variable and Distributions Header ---------------------
\node[fancytitle, right=10pt] at (box.north west) {Continuous Random Variables};
\end{tikzpicture}
%------------ Summarizing Main Features of f(x) ---------------
\begin{tikzpicture}
\node [mybox] (box){%
\begin{minipage}{0.46\textwidth}
Consider two random variables $X, Y$\\
\textbf{Properties of Probability:}
\begin{itemize}
\setlength\itemsep{0em}
\item $E(aX+b) = aE(X)+b$, for $a, b \in \mathbb{R}$
\item $E(X+Y)=E(X)+E(Y)$, for all pairs of $X$ and $Y$
\item $E(XY) = E(X)E(Y)$, for independent $X$ and $Y$
\item $Var(aX+b) = a^2Var(X)$, for $a, b \in \mathbb{R}$
\item $Var(X+Y) = Var(X) + Var(Y)$\\
$Var(X-Y) = Var(X) + Var(Y)$, for independent $X$ and $Y$
\end{itemize}
\textbf{Covariance:}
\begin{itemize}
\setlength\itemsep{0em}
\item $\text{Cov}(X,Y) = E{[X-E(X)][Y-E(Y)]} = E(XY)-E(X)E(Y)$\\
If $X$ and $Y$ are independent, $Cov(X,Y) = 0$
\item $Var(X+Y) = Var(X)+Var(Y)+2Cov(X,Y)$
\item $Var(aX+bY+c) = a^2Var(X)+b^2Var(Y)+2abCov(X,Y)$
\end{itemize}
\end{minipage}
};
%------------ Summarizing Main Features of f(x) ---------------------
\node[fancytitle, right=10pt] at (box.north west) {Summarizing Main Features of $f(x)$};
\end{tikzpicture}
%------------ Sum and Average of Independent Random Variables ---------------
\begin{tikzpicture}
\node [mybox] (box){%
\begin{minipage}{0.46\textwidth}
\textbf{Sum of Independent Random Variables:}\\
$Y = a_1X_1+a_2X_2+...+a_nX_n$, for $a_1, a_2,..., a_n \in \mathbb{R}$
\begin{itemize}
\setlength\itemsep{0em}
\item $E(Y) = a_1E(X_1)+a_2E(X_2)+...+a_nE(X_n)$
\item $Var(Y) = a_1^2Var(X_1) + a_2^2Var(X_2) + ... + a_n^2Var(X_n)$
\end{itemize}
If $n$ random variables $X_i$ have common mean $\mu$ and common variance $\sigma^2$ then,
\begin{itemize}
\setlength\itemsep{0em}
\item $E(Y) = (a_1 + a_2 + ... + a_n)\mu$
\item $Var(Y) = (a_1^2 + a_2^2 + ... + a_n^2)\sigma^2$
\end{itemize}
\textbf{Average of Independent Random Variables:}\\
$X_1,X_2,...,X_n$ are $n$ independent random variables
\begin{itemize}
\setlength\itemsep{0em}
\item $\overline{X} = \frac{X_1+X_2+...+X_n}{n}$
\item $E[\overline{X}] = \frac{1}{n}[E(X_1)+E(X_2)+...+E(X_n)]$
\item $Var[\overline{X}] = \frac{1}{n^2}[Var(X_1)+Var(X_2)+...+Var(X_n)]$
\end{itemize}
If $n$ random variables $X_i$ have common mean $\mu$ and common variance $\sigma^2$ then,
\begin{itemize}
\setlength\itemsep{0em}
\item $E[\overline{X}] = \mu$
\item $Var[\overline{X}] = \frac{\sigma^2}{n}$
\end{itemize}
\end{minipage}
};
%------------ Sum and Average of Independent Random Variables ---------------------
\node[fancytitle, right=10pt] at (box.north west) {Sum and Average of Independent Random Variables};
\end{tikzpicture}
%------------ Maximum and Minimum of Independent Variables ---------------
\begin{tikzpicture}
\node [mybox] (box){%
\begin{minipage}{0.46\textwidth}
Given $n$ independent random variables $X_1, X_2,...,X_n$.\\
For each $X_i$, cdf $F_X(x)$ and pdf is $f_X(x)$.\\
\textbf{Maximum of Independent Random Variables:}\\
Consider $V=max\{X_1, X_2, ... , X_n\}$\\
\underline{cdf of V}\\
\begin{tabular}{lp{12cm} l}
$F_V(v)$& $= P(V \leq v) = P(X_1 \leq v, X_2 \leq v,..., X_n \leq v)$\\
& $= P(X_1 \leq v)P(X_2 \leq v)...P(X_n \leq v) = F_{X_1}(v)F_{X_2}(v)...F_{X_n}(v)$\\
& $= [F_X(v)]^n$ ; if $X_i$'s are all identically distributed
\end{tabular}\\
\underline{pdf of V}\\
\begin{tabular}{lp{12cm} l}
$f_V(v)$& $= F_V'(v) = \frac{d}{dv}F_V(v) = \frac{d}{dv}[F_X(v)]^n = n[F_X(v)]^{n-1}\frac{d}{dv}F_X(v)$\\
& $= n[F_X(v)]^{n-1}f_X(v)$
\end{tabular}\\\\
\textbf{Minimum of Independent Random Variables:}\\
Consider $U=min\{X_1, X_2, ... , X_n\}$\\
\underline{cdf of U}\\
\begin{tabular}{lp{12cm} l}
$F_U(u)$& $= P(U \leq u) = 1-P(U>u) = 1-P(X_1>u, X_2>u,...,X_n>u)$\\
& $= 1-P(X_1>u)P(X_2>u)...P(X_n>u)$\\
& $= 1-[1-F_{X_1}(u)][1-F_{X_2}(u)]...[1-F_{X_n}(u)]$\\
& $= 1-[1-F_X(u)]^n$ ; if $X_i$'s are all identically distributed
\end{tabular}\\
\underline{pdf of U}\\
\begin{tabular}{lp{12cm} l}
$f_U(u)$& $= F_U'(u) = \frac{d}{du}\{1-[1-F_X(u)]^2\} = 0 - n[1-F_X(u)]^{n-1}\frac{d}{du}(-F_X(u))$\\
& $= n[1-F_X(u)]^{n-1}f_X(u)$
\end{tabular}
\end{minipage}
};
%------------ Maximum and Minimum of Independent Variables ---------------------
\node[fancytitle, right=10pt] at (box.north west) {Maximum and Minimum of Independent Variables};
\end{tikzpicture}
%------------ Some Continuous Distributions ---------------
\begin{tikzpicture}
\node [mybox] (box){%
\begin{minipage}{0.46\textwidth}
\textbf{Uniform Distribution:} $X \sim U(a, b)$\\
\begin{tabular}{lp{8cm} l}
Mean: &$\mu = E(X) = \frac{a+b}{2}$ \\
Variance: &$\sigma^2$ = $Var(X) = \frac{(b-a)^2}{12}$\\
\end{tabular}
\begin{multicols*}{2}
\underline{pdf of X}\\
$f(x)=$
$\begin{cases}
\frac{1}{b-a} & a \leq x \leq b\\
0 & \textrm{otherwise} \\
\end{cases}$\\
\columnbreak
\underline{cdf of X}\\
$F(x)=$
$\begin{cases}
0 & x < a\\
\frac{x-a}{b-a} & a \leq x \leq b\\
1 & x > b
\end{cases}$
\end{multicols*}
\textbf{Exponential Distribution:} $X \sim Exp(\lambda)$\\
\begin{tabular}{lp{8cm} l}
Mean: &$\mu = E(X) = \frac{1}{\lambda}$ \\
Variance: &$\sigma^2$ = $Var(X) = \frac{1}{\lambda^2}$\\
\end{tabular}
\begin{multicols*}{2}
\underline{pdf of X}\\
$f(x)=$
$\begin{cases}
\lambda e^{-\lambda x} & x \geq 0\\
0 & x < 0\\
\end{cases}$\\
\columnbreak
\underline{cdf of X}\\
$F(x)=$
$\begin{cases}
1 - e^{-\lambda x} & x \geq 0\\
0 & x < 0
\end{cases}$
\end{multicols*}
\end{minipage}
};
%------------ Some Continuous Distributions Header ---------------------
\node[fancytitle, right=10pt] at (box.north west) {Some Continuous Distributions};
\end{tikzpicture}
%------------ Normal Distribution ---------------
\begin{tikzpicture}
\node [mybox] (box){%
\begin{minipage}{0.46\textwidth}
\textbf{Normal Distribution:} $X \sim N(\mu, \sigma^2)$\\
Standardized Normal: $Z \sim N(0, 1)$ where $Z = \frac{X-\mu}{\sigma}$\\
\underline{68-95-99.7 Rule:}
\begin{itemize}
\setlength\itemsep{0em}
\item approximately 68\% of observations fall within $\sigma$ of $\mu$
\item approximately 95\% of observations fall within $2\sigma$ of $\mu$
\item approximately 99.7\% of observations fall within $3\sigma$ of $\mu$
\end{itemize}
\end{minipage}
};
%------------ Normal Distribution Header ---------------------
\node[fancytitle, right=10pt] at (box.north west) {Normal Distribution};
\end{tikzpicture}
%------------ Bernoulli and Binomial Random Variables ---------------
\begin{tikzpicture}
\node [mybox] (box){%
\begin{minipage}{0.46\textwidth}
\textbf{Bernoulli Random Variable:}\\
Bernoulli random variable $X$ has only two outcomes, success and failure.\\
$P(Success) = p$ and $P(Failure) = 1 - p$\\
\textbf{Bernoulli Distribution:}\\
$X \sim Bernoulli(p)$\\
\underline{pmf}: $P(X=x) = p^x(1-p)^{1-x}$ for $x=0,1$\\
\begin{tabular}{lp{8cm} l}
Mean: &$\mu = E(X) = p$ \\
Variance: &$\sigma^2 = Var(X) = p(1-p)$\\
\end{tabular}\\
\textbf{Binomial Random Variable:}\\
Binomial random variable $X$ is the number of successes for $n$ independent trials and each trial has the same probability of success $p$.\\
\textbf{Binomial Distribution:}\\
$X \sim Bin(n, p)$\\
\underline{pmf}: $P(X=x) = {n \choose x} p^x(1-p)^{n-x}$ for $x=0,1,2,...,n$\\
\underline{cdf}: $P(X\leq x) = \sum_{i=0}^{x} {n \choose i} p^i(1-p)^{n-i}$ for $x=0,1,2,...,n$\\
\textit{Note:} ${n \choose x} = \frac{n!}{x!(n-x)!}$\\
\begin{tabular}{lp{8cm} l}
Mean: &$\mu = E(X) = np$ \\
Variance: &$\sigma^2 = Var(X) = np(1-p)$\\
\end{tabular}
\end{minipage}
};
%------------ Bernoulli and Binomial Random Variables Header ---------------------
\node[fancytitle, right=10pt] at (box.north west) {Bernoulli and Binomial Random Variables};
\end{tikzpicture}
%------------ Geometric Distribution and Return Period ---------------
\begin{tikzpicture}
\node [mybox] (box){%
\begin{minipage}{0.46\textwidth}
\textbf{Geometric Random Variable:}\\
Geometric random variable X is the number of independent trials needed until the first success occurs.\\
\textbf{Geometric Distribution:}\\
$X \sim Geo(p)$ where $p$ is the probability of success\\
\underline{pmf}: $P(X=x)=p(1-p)^{x-1}$ for $x=1,2,3,...$\\
\underline{cdf}: $P(X \leq x)=1-(1-p)^x$ for $x=1,2,3,...$\\
\begin{tabular}{lp{8cm} l}
Mean: &$\mu = E(X) = \frac{1}{p}$ \\
Variance: &$\sigma^2 = Var(X) = \frac{1-p}{p^2}$\\
\end{tabular}\\
\end{minipage}
};
%------------ Geometric Distribution Header ---------------------
\node[fancytitle, right=10pt] at (box.north west) {Geometric Distribution};
\end{tikzpicture}
%------------ Poisson Distribution ---------------
\begin{tikzpicture}
\node [mybox] (box){%
\begin{minipage}{0.46\textwidth}
\textbf{Poisson Process:}\\
Random variable X is the number of occurrences in a given interval.\\
\textbf{Poisson Distribution:}\\
$X \sim Poisson(\lambda)$ where $\lambda$ is the rate of occurrences\\
\underline{pmf}: $P(X=x)=\frac{\lambda^x e^{-\lambda}}{x!}$ for $x=0,1,2,3,...$\\
\underline{cdf}: $P(X\leq x)=\sum_{i=0}^{x}\frac{\lambda^i e^{-\lambda}}{i!}$ for $x=0,1,2,3,...$\\
\begin{tabular}{lp{8cm} l}
Mean: &$\mu = E(X) = \lambda$ \\
Variance: &$\sigma^2 = Var(X) = \lambda$\\
\end{tabular}
\begin{itemize}
\setlength\itemsep{0em}
\item Let $T \sim Exp(\lambda)$ be the time between two consecutive occurrences of events. (Can also be the waiting time for first event.)
\end{itemize}
\end{minipage}
};
%------------ Poisson Distribution Header ---------------------
\node[fancytitle, right=10pt] at (box.north west) {Poisson Distribution};
\end{tikzpicture}
%------------ Poisson Approximation to the Binomial Distribution ---------------
\begin{tikzpicture}
\node [mybox] (box){%
\begin{minipage}{0.46\textwidth}
Let $X \sim Bin(n, p)$ be a binomial random variable. If $n$ is large $(n \geq 20)$ and $p$ or $1-p$ is small ($np<5$ or $n(1-p)<5$), then we can use a Poisson random variable with rate $\lambda = np$ to approximate the probabilistic behaviour of $X$.\\\\
$X \sim Poisson(np)$, approx. for $x=0,1,2,...n$
\end{minipage}
};
%------------ Poisson Approximation to the Binomial Distribution Header ---------------------
\node[fancytitle, right=10pt] at (box.north west) {Poisson Approximation to the Binomial Distribution};
\end{tikzpicture}
%------------ Central Limit Theorem ---------------
\begin{tikzpicture}
\node [mybox] (box){%
\begin{minipage}{0.46\textwidth}
Let $X_1$, $X_2$,..., $X_n$ be a random sample from an arbitrary population/distribution with mean $\mu$ and variance $\sigma^2$. When $n$ is large ($n \geq 20$) then\\\\
$\overline{X} = \frac{X_1 + X_2 + ... + X_n}{n} \sim N(\mu, \frac{\sigma^2}{n})$, approx.\\\\
When dealing with sum, the CLT can still be used. Then\\\\
$T = X_1 + X_2 + ... + X_n = n\overline{X}$ \\
$T \sim N(n\mu, n\sigma^2)$, approx.
\end{minipage}
};
%------------ Central Limit Theorem Header ---------------------
\node[fancytitle, right=10pt] at (box.north west) {Central Limit Theorem};
\end{tikzpicture}
%------------ Normal Approximation to the Binomial Distribution ---------------
\begin{tikzpicture}
\node [mybox] (box){%
\begin{minipage}{0.46\textwidth}
Let $X \sim Bin(n,p)$. When $n$ is large so that both $np \geq 5$ and $n(1-p) \geq 5$. We can use the normal distribution to get an approximate answer. Remember to use \textbf{continuity correction}.\\\\
$X \sim N(np, np(1-p))$, approx.
\end{minipage}
};
%------------ Normal Approximation to the Binomial Distribution Header ---------------------
\node[fancytitle, right=10pt] at (box.north west) {Normal Approximation to the Binomial Distribution};
\end{tikzpicture}
%------------ Normal Approximation to the Poisson Distribution ---------------
\begin{tikzpicture}
\node [mybox] (box){%
\begin{minipage}{0.46\textwidth}
Let $X \sim Poisson(\lambda)$. When $\lambda$ is large ($\lambda \geq 20$) then the Normal distribution can be used to approximate the Poisson distribution. Remember to use \textbf{continuity correction}.\\\\
$X \sim N(\lambda, \lambda)$, approx.
\end{minipage}
};
%------------ Normal Approximation to the Poisson Distribution Header ---------------------
\node[fancytitle, right=10pt] at (box.north west) {Normal Approximation to the Poisson Distribution};
\end{tikzpicture}
%------------ Continuity Correction ---------------
\begin{tikzpicture}
\node [mybox] (box){%
\begin{minipage}{0.46\textwidth}
Consider continuous random variable $Y$ and discrete random variable $X$.
\begin{itemize}
\setlength\itemsep{0em}
\item $P(X > 4) = P(X \geq 5) = P(Y \geq 4.5)$
\item $P(X \geq 4) = P(Y \geq 3.5)$
\item $P(X < 4) = P(X \leq 3) = P(Y \leq 3.5)$
\item $P(X \leq 4) = P(Y \leq 4.5)$
\item $P(X = 4) = P(3.5 \leq Y \leq 4.5)$
\end{itemize}
\end{minipage}
};
%------------ Continuity Correction Header ---------------------
\node[fancytitle, right=10pt] at (box.north west) {Continuity Correction};
\end{tikzpicture}
%------------ Point Estimators ---------------
\begin{tikzpicture}
\node [mybox] (box){%
\begin{minipage}{0.46\textwidth}
Suppose that $X_1$, $X_2$,..., $X_n$ are random samples from a population with mean $\mu$ and variance $\sigma^2$.
\begin{itemize}
\setlength\itemsep{0em}
\item $\overline{x}$ is an unbiased estimator of $\mu$\\
$\overline{x} = \frac{\sum_{i=1}^{n}x_i}{n}$
\item $s^2$ is an unbiased estimator of $\sigma^2$\\
$s^2 = \frac{\sum_{i=1}^{n}(x_i-\overline{x})^2}{n-1} = \frac{\sum_{i=1}^{n}x_i^2-n\overline{x}^2}{n-1}$
\item $\theta$ is the parameter, $\hat{\theta}$ is the point estimator. When $E(\hat{\theta}) = \theta$, $\hat{\theta}$ is an unbiased estimator. The bias of an estimator is $bias(\theta) = E(\hat{\theta}) - \theta$.
\end{itemize}
\end{minipage}
};
%------------ Point Estimators Header ---------------------
\node[fancytitle, right=10pt] at (box.north west) {Point Estimators};
\end{tikzpicture}
%------------ Confidence Interval ---------------
\begin{tikzpicture}
\node [mybox] (box){%
\begin{minipage}{0.46\textwidth}
\textbf{$(1-\alpha)100\%$ Confidence Interval for population mean $\mu$:}\\ (point estimator of $\mu$ is $\overline{x}$)\\
\textit{General Form:} point estimate $\pm$ margin of error\\
When $\sigma^2$ is \textbf{known}: $\overline{x} \pm z_{\frac{\alpha}{2}}\frac{\sigma}{\sqrt{n}}$\\
When $\sigma^2$ is \textbf{unknown}:
$\overline{x} \pm t_{\frac{\alpha}{2}, n-1}\frac{s}{\sqrt{n}}$\\
Typical $z$ values of $\alpha$:\\
\begin{tabular}{lp{1cm} ll}
$\alpha = 0.1$ & $90\%$ & $z_{\frac{\alpha}{2}} = z_{0.05} = 1.645$\\
$\alpha = 0.05$ & $95\%$ & $z_{\frac{\alpha}{2}} = z_{0.025} = 1.96$\\
$\alpha = 0.01$ & $99\%$ & $z_{\frac{\alpha}{2}} = z_{0.005} = 2.575$\\
\end{tabular}\\\\
\textbf{$(1-\alpha)100\%$ Confidence Interval for $\mu_1 - \mu_2$:}\\
$(\overline{x}_1 - \overline{x}_2) \pm t_{\frac{\alpha}{2}, n_1+n_2-2}s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}$
\end{minipage}
};
%------------ Confidence Interval Header ---------------------
\node[fancytitle, right=10pt] at (box.north west) {Confidence Interval};
\end{tikzpicture}
%------------ Pooled Standard Deviation ---------------
\begin{tikzpicture}
\node [mybox] (box){%
\begin{minipage}{0.46\textwidth}
Requires assumptions that population variances are equal: $\sigma_1^2 = \sigma_2^2 = \sigma^2$\\
The pooled standard deviation $s_p$ estimates the common standard deviation $\sigma$.\\
$s_p = \sqrt{\frac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}}$
\end{minipage}
};
%------------ Pooled Standard Deviation Header ---------------------
\node[fancytitle, right=10pt] at (box.north west) {Pooled Standard Deviation};
\end{tikzpicture}
%------------ Testing of Hypotheses about mu ---------------
\begin{tikzpicture}
\node [mybox] (box){%
\begin{minipage}{0.46\textwidth}
$H_o$: Null hypothesis is a tentative assumption about a population parameter.\\
$H_a$: Alternative hypothesis is what the test is attempting to establish.
\begin{itemize}
\setlength\itemsep{0em}
\item $H_o: \mu \geq \mu_o$ vs $H_a: \mu < \mu_o$ (one-tail test, lower-tail)
\item $H_o: \mu \leq \mu_o$ vs $H_a: \mu > \mu_o$ (one-tail test, upper-tail)
\item $H_o: \mu = \mu_o$ vs $H_a: \mu \neq \mu_o$ (two-tail test)
\end{itemize}
\textbf{Test Statistic:}\\
Case 1: $\sigma^2$ is \underline{known}\\
$z=\frac{\overline{x}-\mu_o}{\frac{\sigma}{\sqrt{n}}} \sim N(0,1)$\\
Case 2: $\sigma^2$ is \underline{unknown}\\
$t=\frac{\overline{x}-\mu_o}{\frac{s}{\sqrt{n}}} \sim t_{n-1}$\\
\textbf{Type I and Type II errors:}\\
\begin{tabular}{lp{8cm} l}
Type I error: & rejecting $H_o$ when $H_o$ is true\\
Type II error: & not rejecting $H_o$ with $H_o$ is false\\
\end{tabular}\\
\begin{tabular}{lp{8cm} l}
$P(\textrm{Type I error}) = \alpha$\\
$P(\textrm{Type II error}) = \beta$\\
\end{tabular}\\\\
\textbf{Power} is the probability of rejecting $H_o$, when $H_o$ is false.\\
$\textrm{Power} = 1-\beta$\\
\textbf{Comparison of two means:}\\
Two independent populations with means $\mu_1$ and $\mu_2$.\\
Assume random samples, normal distributions, and equal variances ($\sigma_1^2 = \sigma_2^2$).
\begin{itemize}
\setlength\itemsep{0em}
\item $H_o: \mu_1 - \mu_2 \geq \Delta_o$ vs $H_a: \mu_1 - \mu_2 < \Delta_o$ (lower-tail)
\item $H_o: \mu_1 - \mu_2 \leq \Delta_o$ vs $H_a: \mu_1 - \mu_2 > \Delta_o$ (upper-tail)
\item $H_o: \mu_1 - \mu_2 = \Delta_o$ vs $H_a: \mu_1 - \mu_2 \neq \Delta_o$ (two-tail)
\end{itemize}
\textbf{Test Statistic:}\\
$t=\frac{(\overline{\chi}_1-\overline{\chi}_2)-\Delta_o}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}} \sim t_{n_1+n_2-2}$\\
$s_p$ is the pooled standard deviation\\
\textbf{Rejection Rules:}\\
Consider test statistic $z$, and significance value $\alpha$.
\begin{itemize}
\setlength\itemsep{0em}
\item Lower-tail test: Reject $H_o$ if $z \leq z_{\alpha}$
\item Upper-tail test: Reject $H_o$ if $z \geq z_{\alpha}$
\item Two-tail test: Reject $H_o$ if $|z| \geq z_{\frac{\alpha}{2}}$
\end{itemize}
\end{minipage}
};
%------------ Testing of Hypotheses about mu Header ---------------------
\node[fancytitle, right=10pt] at (box.north west) {Testing of Hypotheses about $\mu$};
\end{tikzpicture}
%------------ ANOVA ---------------
\begin{tikzpicture}
\node [mybox] (box){%
\begin{minipage}{0.46\textwidth}
\textbf{One-way ANOVA:}\\
$k = $ number of populations or treatments being compared\\
$\mu_1 = $ mean of population 1 or true average response when treatment 1 is applied.\\
...\\
$\mu_k = $ mean of population $k$ or true average response when treatment $k$ is applied.\\
\textbf{Assumptions:}
\begin{itemize}
\setlength\itemsep{0em}
\item For each population, response variable is normally distributed
\item Variance of response variable, $\sigma^2$ is the same for all the populations
\item The observations must be independent
\end{itemize}
\textbf{Hypotheses:}\\
$H_o: \mu_1 = \mu_2 = ... = \mu_k$\\
$H_a: \mu_i \neq \mu_j$ for $i \neq j$\\
\textbf{Notation:}\\
$y_{ij}$ is the $j^{th}$ observed value from the $i^{th}$ population/treatment.\\
\begin{tabular}{lp{8cm} l}
Total mean: & $\overline{y}_{i\cdot} = \frac{y_{i\cdot}}{n_i} = \frac{\sum_{j=1}^{n_i}y_{ij}}{n_i}$\\
Total sample size: & $n = n_1 + n_2 + ... + n_k$ \\
Grand total: & $y_{\cdot\cdot} = \sum_{i=1}^{k}\sum_{j=1}^{n_i}y_{ij}$\\
Grand mean: & $\overline{y}_{\cdot\cdot} = \frac{y_{\cdot\cdot}}{n} = \frac{\sum_{i=1}^{k}\sum_{j=1}^{n_i}y_{ij}}{n}$\\
\end{tabular}\\
$s^2 = \frac{\sum_{j=1}^{k}(n_i-1)s_i^2}{n-k}$ = MSE, where $s_i^2 = \frac{\sum_{j=1}^{n_i}(y_{ij}-\overline{y}_{i\cdot})^2}{n_i-1}$\\
\textbf{The ANOVA Table:}
\begin{center}
\begin{tabular}{c|c|c|c|c}
Source of Variation & df & Sum of Squares & Mean Square & F-ratio \\
\hline
Treatment & $k-1$ & SSTr & MSTr = $\frac{\textrm{SSTr}}{k-1}$ & $\frac{\textrm{MSTr}}{\textrm{MSE}}$ \\
Error & $n-k$ & SSE & MSE = $\frac{\textrm{SSE}}{n-k}$ & \\
Total & $n-1$ & SST & & \\
\end{tabular}
\end{center}
\begin{tabular}{lp{12cm} l}
SST &= SSTr + SSE\\\\
SST &= $\sum_{i=1}^{k}\sum_{j=1}^{n_i}(y_{ij}-\overline{y}_{\cdot\cdot})^2 = \sum_{i=1}^{k}\sum_{j=1}^{n_i}y_{ij}^2-\frac{1}{n}y_{\cdot\cdot}^2$\\
SSTr &= $\sum_{i=1}^{k}\sum_{j=1}^{n_i}(\overline{y}_{i\cdot}-\overline{y}_{\cdot\cdot})^2 = \sum_{i=1}^{k}\frac{1}{n_i}y_{i\cdot}^2-\frac{1}{n}y_{\cdot\cdot}^2$\\
SSE &= $\sum_{i=1}^{k}\sum_{j=1}^{n_i}(y_{ij}-\overline{y}_{i\cdot})^2= \sum_{i=1}^{k}\sum_{j=1}^{n_i}y_{ij}^2-\sum_{i=1}^{k}\frac{y_{i\cdot}^2}{n_i} = \sum_{i=1}^{k}(n_i-1)s_i^2$\\
\end{tabular}\\\\
\textbf{Test Statistic:}\\
\begin{tabular}{lp{8cm} l}
$F_{obs} = \frac{\textrm{MSTr}}{\textrm{MSE}} \sim F_{v_1, v_2}$ & $v1 = df(\textrm{SSTr}) = k-1$\\
& $v2 = df(\textrm{SSE}) = n-k$\\
Reject $H_o$ if $F_{obs} \geq F_{\alpha, v_1, v_2}$
\end{tabular}\\
\end{minipage}
};
%------------ ANOVA Header ---------------------
\node[fancytitle, right=10pt] at (box.north west) {Analysis of Variance (ANOVA)};
\end{tikzpicture}
%------------ Covariance and Correlation Coefficient ---------------
\begin{tikzpicture}
\node [mybox] (box){%
\begin{minipage}{0.46\textwidth}
On a scatter plot, each observation is represented as a point with x-coord $x_i$ and y-coord $y_i$.\\
\textbf{Sample Covariance:} $Cov(x, y)$\\
\begin{tabular}{lp{12cm} l}
$Cov(x,y)$ & $=\frac{1}{n-1}\sum_{i=1}^{n}(x_i-\overline{x})(y_i-\overline{y})$\\
& $=\frac{1}{n-1}[\sum_{i=1}^{n}x_iy_i-\frac{\sum_{i=1}^{n}x_i\sum_{i=1}^{n}y_i}{n}]$\\
& $=\frac{1}{n-1}[\sum_{i=1}^{n}x_iy_i-n\overline{x}\overline{y}]$
\end{tabular}\\
\begin{itemize}
\setlength\itemsep{0em}
\item If $x$ and $y$ are positively associated, then $Cov(x, y)$ will be large and positive
\item If $x$ and $y$ are negatively associated, then $Cov(x, y)$ will be large and negative
\item If the variables are not positively nor negatively associated, then $Cov(x,y)$ will be small
\end{itemize}
\textbf{Sample Correlation Coefficient:} $r$\\
$r = \frac{1}{n-1}\sum_{i=1}^{n}(\frac{x_i-\overline{x}}{s_x})(\frac{y_i-\overline{y}}{s_y})$, where $s_x = \sqrt{\frac{\sum_{i=1}^{n}(x_i-\overline{x})^2}{n-1}}$ and $s_y = \sqrt{\frac{\sum_{i=1}^{n}(y_i-\overline{y})^2}{n-1}}$\\
$r = \frac{Cov(x,y)}{s_x s_y}$
\begin{itemize}
\setlength\itemsep{0em}
\item Always falls between -1 and +1
\item A positive $r$ value indicates a positive association
\item A negative $r$ value indicates a negative association
\item $r$ value close to +1 or -1 indicates a strong linear association
\item $r$ value close to 0 indicates a weak association
\end{itemize}
\end{minipage}
};
%------------ Covariance and Correlation Coefficient Header ---------------------
\node[fancytitle, right=10pt] at (box.north west) {Covariance and Correlation Coefficient};
\end{tikzpicture}
%------------ Simple Linear Regression ---------------
\begin{tikzpicture}
\node [mybox] (box){%
\begin{minipage}{0.46\textwidth}
\textbf{Regression Line:}\\
Simple linear regression model: $y = \beta_o + \beta_1x+\varepsilon$\\
$\beta_o$, $\beta_1$, and $\sigma^2$ are parameters, $y$ and $\varepsilon$ are random variables. $\varepsilon$ is the error term.\\
True regression line: $E(y) = \beta_o + \beta_1x$\\
Least squares regression line: $\hat{y} = \hat{\beta_o} + \hat{\beta_1}x$\\
$\hat{y}$, $\hat{\beta_o}$, and $\hat{\beta_1}$ are point estimates for $y$, $\beta_o$, and $\beta_1$.\\
Residual: $\varepsilon_i = y_i - \hat{y_i}$\\
$\hat{\beta_1} = \frac{\sum_{i=1}^{n}(x_i-\overline{x})(y_i-\overline{y})}{\sum_{i=1}^{n}(x_i-\overline{x})^2} = \frac{\sum_{i=1}^{n}x_iy_i-n\overline{x}\overline{y}}{\sum_{i=1}^{n}x_i^2-n\overline{x}^2} = r\frac{s_y}{s_x}$\\
$\hat{\beta_o} = \frac{\sum_{i=1}^{n}y_i-\hat{\beta_1}\sum_{i=1}^{n}x_i}{n} = \overline{y} - \hat{\beta_1}\overline{x}$\\
\textbf{Coefficient of Determination:} $r^2$\\
The proportion of observed $y$ variation that can be explained by the simple linear regression model.
\end{minipage}
};
%------------ Simple Linear Regression Header ---------------------
\node[fancytitle, right=10pt] at (box.north west) {Simple Linear Regression};
\end{tikzpicture}
%------------ Estimating Variance ---------------
\begin{tikzpicture}
\node [mybox] (box){%
\begin{minipage}{0.46\textwidth}
$\hat{\sigma}^2 = s^2 = \frac{\textrm{SSE}}{n-2} = \frac{\sum_{i=1}^{n}(y_i - \hat{y_i})^2}{n-2}$\\
\textbf{Error Sum of Squares} (SSE):\\
SSE = $\sum_{i=1}^{n}\varepsilon_i^2 = \sum_{i=1}^{n}(y_i - \hat{y_i})^2 = \sum_{i=1}^{n}[y_i-(\hat{\beta_o}+\hat{\beta_1}x_i)]^2$\\
SSE is a measure of variation in $y$ left unexplained by linear regression model.\\
\textbf{Total Sum of Squares} (SST):\\
SST = $\sum_{i=1}^{n}(y_i-\overline{y})^2$\\
SST is sum of squared deviations about sample mean of observed $y$ values.\\
\textbf{Regression Sum of Squares} (SSR):\\
SSR = $\sum_{i=1}^{n}(\hat{y_i}-\overline{y})^2$\\
SSR is total variation explained by the linear regression model. \\
SST = SSR + SSE\\
\textbf{Coefficient of Determination from SST, SSR, and SSE}:\\
$r^2 = 1 - \frac{SSE}{SST}$\\
or\\
$r^2 = \frac{SSR}{SST}$
\end{minipage}
};
%------------ Estimating Variance Header ---------------------
\node[fancytitle, right=10pt] at (box.north west) {Estimating $\sigma^2$ (SLR)};
\end{tikzpicture}
\vfill\null
\columnbreak
%------------ Slope Parameter ---------------
\begin{tikzpicture}
\node [mybox] (box){%
\begin{minipage}{0.46\textwidth}
When $\beta_1 = 0$ there is no linear relationship between the two variables.\\
\textbf{Hypotheses:}\\
$H_o: \beta_1 = 0$\\
$H_a: \beta_1 \neq 0$\\
\textbf{Test statistic:}\\
$t_{\textrm{obs}} = \frac{\hat{\beta_1} }{s_{\hat{\beta_1}}} \sim t_{n-2}$, where $s_{\hat{\beta_1}} = \frac{s}{s_x\sqrt{n-1}}$\\\\
Reject $H_o$ if $|t_{\textrm{obs}}| \geq t_{\frac{\alpha}{2},{n-2}}$\\
\textbf{$(1-\alpha)100\%$ Confidence Interval for $\beta_1$:}\\
$\hat{\beta_1} \pm t_{\frac{\alpha}{2}, n-2} s_{\hat\beta_1}$
\end{minipage}
};
%------------ Slope Parameter Header ---------------------
\node[fancytitle, right=10pt] at (box.north west) {Slope Parameter $\beta_1$ (SLR)};
\end{tikzpicture}
\end{multicols*}
\end{document}
Contact GitHub API Training Shop Blog About
© 2016 GitHub, Inc. Terms Privacy Security Status Help