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\author{Dylan (dylablo)}
\pdfinfo{
  /Title (foundation-of-statistics-with-michael-cronin-ch-1.pdf)
  /Creator (Cheatography)
  /Author (Dylan (dylablo))
  /Subject (Foundation of Statistics with Michael Cronin Ch 1 Cheat Sheet)
}

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    {\parbox{\dimexpr\textwidth-2\fboxsep\relax}{\noindent
        \hspace*{-6pt}\includegraphics[width=5.8cm]{/web/www.cheatography.com/public/images/cheatography_logo.pdf}}
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\begin{tabulary}{11cm}{L}
    \vspace{-2pt}\large{\bf{\textcolor{DarkBackground}{\textrm{Foundation of Statistics with Michael Cronin Ch 1 Cheat Sheet}}}} \\
    \normalsize{by \textcolor{DarkBackground}{Dylan (dylablo)} via \textcolor{DarkBackground}{\uline{cheatography.com/68322/cs/17324/}}}
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\noindent
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  \mymulticolumn{2}{p{5.377cm}}{\bf\textcolor{white}{Cheatographer}}  \\
  \vspace{-2pt}Dylan (dylablo) \\
  \uline{cheatography.com/dylablo} \\
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  \mymulticolumn{1}{p{5.377cm}}{\bf\textcolor{white}{Cheat Sheet}}  \\
   \vspace{-2pt}Not Yet Published.\\
   Updated 4th October, 2018.\\
   Page {\thepage} of \pageref{LastPage}.
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  \mymulticolumn{1}{p{5.377cm}}{\bf\textcolor{white}{Sponsor}}  \\
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  \vspace{-5pt}
  %\includegraphics[width=48px,height=48px]{dave.jpeg}
  Measure your website readability!\\
  www.readability-score.com
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\begin{multicols*}{2}

\begin{tabularx}{8.4cm}{x{3.76 cm} x{4.24 cm} }
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\mymulticolumn{2}{x{8.4cm}}{\bf\textcolor{white}{Simple Linear Regression}}  \tn
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Regression & Studies the relationship between quantitative variables. \tn 
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Simple Linear Regression & Only considers 2 variables \tn 
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Response Variable & Usually denoted Y. We attempt to predict this. \tn 
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Predictor Variable & Usually denoted X. We use this to predict Y. \tn 
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(x`i`,y`i`) & The values for X and Y at case i. We usually denote n to be the number of cases. \tn 
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\mymulticolumn{2}{x{8.4cm}}{} \tn 
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Outline of Simple Linear Regression & Assume a linear relationship between X and Y: Y = β`0`+β`1` \tn 
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β`0` & The intercept ie. the value of Y when X=0, ie where the line crosses the Y axis. \tn 
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β`1` & The slope. The change in Y for a single unit change in X. \tn 
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 & We estimate β`0`and β`1`from the data and use the model to predict Y for any given X. \tn 
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\begin{tabularx}{8.4cm}{x{3.76 cm} x{4.24 cm} }
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\mymulticolumn{2}{x{8.4cm}}{\bf\textcolor{white}{Simple Linear Regression (cont)}}  \tn
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\mymulticolumn{2}{x{8.4cm}}{Methods of Linear Regression} \tn 
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Scatter Plot & Put all points on a scatter plot and gauge visually whether or not the relationship looks linear. \tn 
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Line of Closest Fit & If the relationship looks linear then we find the line fo closest fit and use it to estimate β`0` and β`1` \tn 
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\mymulticolumn{2}{x{8.4cm}}{\bf\textcolor{white}{Co-Variance and Independent Variables}}  \tn
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Independent Events & P(A|B)=P(A) \tn 
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Independent Discrete Variables & P(X = x and Y=y) = P(X=x)P(Y=y) \tn 
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Independent Continuous Variables & The joint pdf of X and Y = h(x,y) = fx(x)gy(y) - the product of individual pdfs. \tn 
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Covariance & "the mean value of the product of the deviations of two variates from their respective means"\{\{nl\}\}Covariance of X and Y = cov(X,Y) = E(X - μ`1`)(Y - μ`2`) where μ`1`=E(X) and μ`2`=E(Y) \tn 
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Covariance of independent variables & cov(X,Y)=0 \tn 
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Covariance as defined by the book & Measures the association between X and Y, the extent to which they vary together.\{\{nl\}\}If large X occurs with large Y and small x with small y, there is a positive association ie. cov(X,Y) \textgreater{} 0.\{\{nl\}\}If large X occurs with small y and Large Y occurs with small x, there is a negative associationie. cov(X,Y) \textless{} 0. \tn 
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\mymulticolumn{2}{x{8.4cm}}{\bf\textcolor{white}{Co-Variance and Independent Variables (cont)}}  \tn
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Direction of association & + indicates postive direction, - indicates negative direction. \tn 
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\begin{tabularx}{8.4cm}{x{3.36 cm} x{4.64 cm} }
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\mymulticolumn{2}{x{8.4cm}}{\bf\textcolor{white}{Least Squares Criterion}}  \tn
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Intro & In a scatter plot there could be many potential lines that could fit the data. We use the Least SquaresCriterion to select the best line. \tn 
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e`i`(error) & THe difference between what the line says the value should be and what it actually is. \tn 
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e`i`(residual) & Difference between the fitted line and actual reality \tn 
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Residual Sum of Squares (RSS) & We chose β`0` and β`1` so as to minimize RSS. \tn 
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\mymulticolumn{2}{x{8.4cm}}{\textasciicircum{} above a letter indicates we are using an estimator} \tn 
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\begin{tabularx}{8.4cm}{X}
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\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Least Sum of Squares Important Formula}}  \tn
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\mymulticolumn{1}{p{8.4cm}}{\vspace{1px}\centerline{\includegraphics[width=5.1cm]{/web/www.cheatography.com/public/uploads/dylablo_1538659112_leastsquares.PNG}}} \tn 
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\mymulticolumn{1}{x{8.4cm}}{RSS=min(\textbackslash{}sum\_\{i=1\}\textasciicircum{}n\textbackslash{}hat\{e\}\_i\textasciicircum{}2)\textbackslash{}\textbackslash{}Through\textbackslash{} Partial\textbackslash{} differentiation\textbackslash{} we\textbackslash{} derive\textbackslash{} the\textbackslash{} estimators...\textbackslash{}\textbackslash{}\textbackslash{}hat\{β\}\_\{1\}=\textbackslash{}frac\{SXY\}\{SXX\}\textbackslash{}\textbackslash{}\textbackslash{}hat\{β\}\_\{0\}=\textbackslash{}bar\{y\}-\textbackslash{}hat\{β\}\_\{1\}\textbackslash{}bar\{x\}}  \tn 
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\begin{tabularx}{8.4cm}{X}
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\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Errors}}  \tn
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\mymulticolumn{1}{x{8.4cm}}{Real data almost never falls in a perfectly straight line. ie. Real data rarely has a perfectly linear relationship. As such real data has errors which could be... \newline % Row Count 4 (+ 4)
- Measurement Errors: Continuous Variables cannot be measured with 100\% accuracy. \newline % Row Count 6 (+ 2)
- An effect of variables not included in  the model \newline % Row Count 8 (+ 2)
- Natural variability. \newline % Row Count 9 (+ 1)
We should incorporate them into our simple linear regression models. eg. \newline % Row Count 11 (+ 2)
y`i` = β`0` + β`1`x`i` + e`i` where e`i` is the error on the ith case \newline % Row Count 13 (+ 2)
and \newline % Row Count 14 (+ 1)
y`i` = β`0` + β`1`x`i` is the true regression line \newline % Row Count 16 (+ 2)
{\emph{*}} \newline % Row Count 17 (+ 1)
Assumptions about errors: \newline % Row Count 18 (+ 1)
We make these assumptions as we need them to... \newline % Row Count 19 (+ 1)
- prove the optimaity ofthe estimates for β`0` and β`1` \newline % Row Count 21 (+ 2)
- prove the confidence intervals for β`0` and β`1` \newline % Row Count 23 (+ 2)
e`i` \textasciitilde{} NID(0,σ\textasciicircum{}2\textasciicircum{}) \newline % Row Count 24 (+ 1)
- N: Normally distributed with mean 0 \newline % Row Count 25 (+ 1)
- I: Independent variables \newline % Row Count 26 (+ 1)
- D: Distributed. \newline % Row Count 27 (+ 1)
- σ\textasciicircum{}2\textasciicircum{}: Common Variance. \newline % Row Count 28 (+ 1)
- "e`i` is normally distributed with mean 0 and common variance of σ\textasciicircum{}2\textasciicircum{}" \newline % Row Count 30 (+ 2)
} \tn 
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\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Errors (cont)}}  \tn
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\mymulticolumn{1}{x{8.4cm}}{These assumption can also be expressed in terms of "Co-Variance" \newline % Row Count 2 (+ 2)
E(e`i`) = 0, var(e`i`) = σ\textasciicircum{}2\textasciicircum{}, cov(e`i`,e`j`) = 0, for i ≠ j \newline % Row Count 4 (+ 2)
- "Expected value e`i` is 0, variance is σ\textasciicircum{}2\textasciicircum{}, covariance of e`i` and e`j` is 0 where i is not j" \newline % Row Count 6 (+ 2)
Combined with the normality assumptions, this implies e`i`s are independent. \newline % Row Count 8 (+ 2)
Assumptions must be verified when applying to a regression model.% Row Count 10 (+ 2)
} \tn 
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\begin{tabularx}{8.4cm}{x{2.96 cm} x{5.04 cm} }
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\mymulticolumn{2}{x{8.4cm}}{\bf\textcolor{white}{Sample Correlation Coefficient r`xy`}}  \tn
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r`xy` = & SXY/sqrt((SXX)(SYY)) = {[}SXY/(n-1{]}/{[}sqrt((SXX/n-1)(SYY/n-1)){]} \tn 
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Correlation Coefficient & r`xy` is the sample covariance scaled to lie in{[}-1,1{]}. ie. -1\textless{}=r`xy`\textless{}=1 \tn 
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r`xy`\textgreater{}0 & Positive association \tn 
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r`xy`\textless{}0 & Negative association \tn 
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r`xy`=1 & All points lie on positive slope. The closer r`xy` is to 1, the closer all points are to lying on the positive line. \tn 
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r`xy`=-1 & All points lie on negative slope. The closer r`xy` is to -1, the closer all points are to lying on the negative line. \tn 
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Bivariate Regression & rr and/or its square r\textasciicircum{}2\textasciicircum{} is used to measure howwell the linear model fits the data. \tn 
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Multiple Regression & The multiple correlation coefficient (R\textasciicircum{}2\textasciicircum{}) is used to measure how well the linear model fits the data. \tn 
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x-bar/x̅ & Indicates the sample mean of x \tn 
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SXY & The standard deviation of X on Y \tn 
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\begin{tabularx}{8.4cm}{x{2.96 cm} x{5.04 cm} }
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\mymulticolumn{2}{x{8.4cm}}{\bf\textcolor{white}{Sample Correlation Coefficient r`xy` (cont)}}  \tn
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Linearity & Linearity cannot be deduuced from correlation coefficient. It should be paired with the scatter plot and never be considered in isolation. \tn 
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\begin{tabularx}{8.4cm}{x{3.12 cm} x{4.88 cm} }
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\mymulticolumn{2}{x{8.4cm}}{\bf\textcolor{white}{The X\textasciicircum{}2\textasciicircum{} Distribution (Chi-Squared)}}  \tn
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Degrees of Freedom (df) & The number of different values/quantities which a distribution can be assigned. \tn 
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X\textasciicircum{}2\textasciicircum{}(v) & A chi-squared distribution with v df. \tn 
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E(X\textasciicircum{}2\textasciicircum{}(v)) = v & ie. The expeced value of a X\textasciicircum{}2\textasciicircum{} distribution with v df, is v. \tn 
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RSS/σ\textasciicircum{}2\textasciicircum{} \textasciitilde{} X\textasciicircum{}2\textasciicircum{}(n-2) & So...\{\{nl\}\}E(RSS/σ\textasciicircum{}2\textasciicircum{}) =\textgreater{} E(X\textasciicircum{}2\textasciicircum{}(n-2)) = n-2 and so E(RSS/n-2)=σ\textasciicircum{}2\textasciicircum{} \tn 
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RSS/n-2 & An unbiased estimate of σ\textasciicircum{}2\textasciicircum{}. \tn 
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sqrt(σ\textasciicircum{}2\textasciicircum{}) = σ & Estimate of Standard Error of Regression/Residual Standard Error(in R) \tn 
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sqrt(estimated variance) = & standard error \tn 
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% That's all folks
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