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\begin{document}

  \section{Statistics}

  \subsection*{Continuous random variables}

  A continuous random variable \(X\) has a pdf \(f\) such that:

  \begin{enumerate}
    \item \(f(x) \ge 0 \forall x \)
    \item \(\int^\infty_{-\infty} f(x) \> dx = 1\)
  \end{enumerate}

  \begin{align*}
    E(X) &= \int_\textbf{X} (x \cdot f(x)) \> dx \\
    \operatorname{Var}(X) &= E\left[(X-\mu)^2\right]
  \end{align*}

  \[ \Pr(X \le c) = \int^c_{-\infty} f(x) \> dx \]
  

  \subsection*{Two random variables \(X, Y\)}

  If \(X\) and \(Y\) are independent:
  \begin{align*}
    \operatorname{E}(aX+bY) & = a\operatorname{E}(X)+b\operatorname{E}(Y) \\
    \operatorname{Var}(aX \pm bY \pm c) &= a^2 \operatorname{Var}(X) + b^2 \operatorname{Var}(Y)
  \end{align*}

  \subsection*{Linear functions \(X \rightarrow aX+b\)}

  \begin{align*}
    \Pr(Y \le y) &= \Pr(aX+b \le y) \\
    &= \Pr\left(X \le \dfrac{y-b}{a}\right) \\
    &= \int^{\frac{y-b}{a}}_{-\infty} f(x) \> dx
  \end{align*}

  \begin{align*}
    \textbf{Mean:} && \operatorname{E}(aX+b) & = a\operatorname{E}(X)+b \\
    \textbf{Variance:} && \operatorname{Var}(aX+b) &= a^2 \operatorname{Var}(X) \\
  \end{align*}

  \subsection*{Expectation theorems}

  For some non-linear function \(g\), the expected value \(E(g(X))\) is not equal to \(g(E(X))\).

  \begin{align*}
    E(X^2) &= \operatorname{Var}(X) - \left[E(X)\right]^2 \\
    E(X^n) &= \Sigma x^n \cdot p(x) \tag{non-linear} \\
    &\ne [E(X)]^n \\
    E(aX \pm b) &= aE(X) \pm b \tag{linear} \\
    E(b) &= b \tag{\(\forall b \in \mathbb{R}\)}\\
    E(X+Y) &= E(X) + E(Y) \tag{two variables}
  \end{align*}

  \subsection*{Sample mean}

  Approximation of the \textbf{population mean} determined experimentally.

  \[ \overline{x} = \dfrac{\Sigma x}{n} \]

  where
  \begin{description}[nosep, labelindent=0.5cm]
    \item \(n\) is the size of the sample (number of sample points)
    \item \(x\) is the value of a sample point
  \end{description}

\begin{cas}
  \begin{enumerate}[leftmargin=3mm]
    \item Spreadsheet
    \item In cell A1:\\ \path{mean(randNorm(sd, mean, sample size))}
    \item Edit \(\rightarrow\) Fill \(\rightarrow\) Fill Range
    \item Input range as A1:An where \(n\) is the number of samples
    \item Graph \(\rightarrow\) Histogram
  \end{enumerate}
  \end{cas}

  \subsubsection*{Sample size of \(n\)}

  \[ \overline{X} = \sum_{i=1}^n \frac{x_i}{n} = \dfrac{\sum x}{n} \]

  Sample mean is distributed with mean \(\mu\) and sd \(\frac{\sigma}{\sqrt{n}}\) (approaches these values for increasing sample size \(n\)).

  For a new distribution with mean of \(n\) trials, \(\operatorname{E}(X^\prime) = \operatorname{E}(X), \quad \operatorname{sd}(X^\prime) = \dfrac{\operatorname{sd}(X)}{\sqrt{n}}\)

  \begin{cas}
  
    \begin{itemize}
      \item Spreadsheet \(\rightarrow\) Catalog \(\rightarrow\) \verb;randNorm(sd, mean, n); where \verb;n; is the number of samples. Show histogram with Histogram key in top left
      \item To calculate parameters of a dataset: Calc \(\rightarrow\) One-variable
    \end{itemize}

  \end{cas}
  
  \subsection*{Normal distributions}


  \[ Z = \frac{X - \mu}{\sigma} \]

  Normal distributions must have area (total prob.) of 1 \(\implies \int^\infty_{-\infty} f(x) \> dx = 1\) \\
  \(\text{mean} = \text{mode} = \text{median}\)

  \begin{warning}
    Always express \(z\) as +ve. Express confidence \textit{interval} as ordered pair.
  \end{warning}

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                  \end{figure*}

  \subsection*{Central limit theorem}

  If \(X\) is randomly distributed with mean \(\mu\) and sd \(\sigma\), then with an adequate sample size \(n\) the distribution of the sample mean \(\overline{X}\) is approximately normal with mean \(E(\overline{X})\) and \(\operatorname{sd}(\overline{X}) = \frac{\sigma}{\sqrt{n}}\).

  \subsection*{Confidence intervals}

  \begin{itemize}
    \item \textbf{Point estimate:} single-valued estimate of the population mean from the value of the sample mean \(\overline{x}\)
    \item \textbf{Interval estimate:} confidence interval for population mean \(\mu\)
    \item \(C\)\% confidence interval \(\implies\) \(C\)\% of samples will contain population mean \(\mu\)
  \end{itemize}

  \subsubsection*{95\% confidence interval}

  For 95\% c.i. of population mean \(\mu\):

  \[ x \in \left(\overline{x} \pm 1.96 \dfrac{\sigma}{\sqrt{n}} \right)\]

  where:
  \begin{description}[nosep, labelindent=0.5cm]
    \item \(\overline{x}\) is the sample mean
    \item \(\sigma\) is the population sd
    \item \(n\) is the sample size from which \(\overline{x}\) was calculated
  \end{description}

  \begin{cas}
    Menu \(\rightarrow\) Stats \(\rightarrow\) Calc \(\rightarrow\) Interval \\
    Set \textit{Type = One-Sample Z Int} \\ \-\hspace{1em} and select \textit{Variable}
  \end{cas}

  \subsection*{Margin of error}

  For 95\% confidence interval of \(\mu\):
  \begin{align*}
    M &= 1.96 \times \dfrac{\sigma}{\sqrt{n}} \\
    \implies n &= \left( \dfrac{1.96 \sigma}{M} \right)^2
  \end{align*}

  Always round \(n\) up to a whole number of samples.

  \subsection*{General case}

  For \(C\)\% c.i. of population mean \(\mu\):

  \[ x \in \left( \overline{x} \pm k \dfrac{\sigma}{\sqrt{n}} \right) \]
  \hfill where \(k\) is such that \(\Pr(-k < Z < k) = \frac{C}{100}\)

  \subsection*{Confidence interval for multiple trials}

  For a set of \(n\) confidence intervals (samples), there is \(0.95^n\) chance that all \(n\) intervals contain the population mean \(\mu\).

  \section{Hypothesis testing}

  \begin{warning}
    Note hypotheses are always expressed in terms of population parameters
  \end{warning}

  \subsection*{Null hypothesis \(H_0\)}

  Sample drawn from population has same mean as control population, and any difference can be explained by sample variations.

  \subsection*{Alternative hypothesis \(H_1\)}

  Amount of variation from control is significant, despite standard sample variations.

  \subsection*{\(p\)-value}


  \begin{align*}
    p &= \Pr(\overline{X} \lessgtr \mu(H_1)) \\
    &= 2 \cdot \Pr(\overline{X} <> \mu(H_1) | \mu = 8)
  \end{align*}

  Probability of observing a value of the sample statistic as significant as the one observed, assuming null hypothesis is true.

  \vspace{0.5em}
  \begin{tabularx}{23em}{|l|X|}
    \hline
    \rowcolor{cas}
    \(\boldsymbol{p}\) & \textbf{Conclusion} \\
    \hline
    \(> 0.05\) & insufficient evidence against \(H_0\) \\
    \(< 0.05\) (5\%) & good evidence against \(H_0\) \\
    \(< 0.01\) (1\%) & strong evidence against \(H_0\) \\
    \(< 0.001\) (0.1\%) & very strong evidence against \(H_0\) \\
    \hline
  \end{tabularx}

  \subsection*{Statistical significance}

  Significance level is denoted by \(\alpha\).

  \-\hspace{1em} If \(p<\alpha\), null hypothesis is \textbf{rejected} \\
  \-\hspace{1em} If \(p>\alpha\), null hypothesis is \textbf{accepted}

  \subsection*{\(z\)-test}

  Hypothesis test for a mean of a sample drawn from a normally distributed population with a known standard deviation.

  \begin{cas}
  Menu \(\rightarrow\) Statistics \(\rightarrow\) Calc \(\rightarrow\) Test. \\
  Select \textit{One-Sample Z-Test} and \textit{Variable}, then input:
    \begin{description}[nosep, style=multiline, labelindent=0.5cm, leftmargin=2cm, font=\normalfont]
    \item[\(\mu\) cond:] same operator as \(H_1\)
    \item[\(\mu_0\):] expected sample mean (null hypothesis)
    \item[\(\sigma\):] standard deviation (null hypothesis)
    \item[\(\overline{x}\):] sample mean
    \item[\(n\):] sample size
  \end{description}
  \end{cas}

  \subsection*{One-tail and two-tail tests}

  \subsubsection*{One tail}

  \begin{itemize}
    \item \(\mu\) has changed in one direction
    \item State ``\(H_1: \mu \lessgtr \) known population mean''
  \end{itemize}

  \subsubsection*{Two tail}

  \begin{itemize}
    \item Direction of \(\Delta \mu\) is ambiguous
    \item State ``\(H_1: \mu \ne\) known population mean''
  \end{itemize}

  For two tail tests:
  \begin{align*}
    p\text{-value} &= \Pr(|\overline{X} - \mu| \ge |\overline{x}_0 - \mu|) \\
    &= \left( |Z| \ge \left|\dfrac{\overline{x}_0 - \mu}{\sigma \div \sqrt{n}} \right| \right)
  \end{align*}

  \subsection*{Modulus notation for two tail}

  \(\Pr(|\overline{X} - \mu| \ge a) \implies\) ``the probability that the distance between \(\overline{\mu}\) and \(\mu\) is \(\ge a\)''

  \subsection*{Inverse normal}

  \begin{cas}
    \verb;invNormCdf("L", ;\(\alpha\)\verb;, ;\(\dfrac{\sigma}{n^\alpha}\)\verb;, ;\(\mu\)\verb;);
  \end{cas}

  \subsection*{Errors}

  \begin{description}[labelwidth=2.5cm, labelindent=0.5cm]
    \item [Type I error] \(H_0\) is rejected when it is \textbf{true}
    \item [Type II error] \(H_0\) is \textbf{not} rejected when it is \textbf{false}
  \end{description}

% \subsection*{Using c.i. to find \(p\)}
% need more here

\end{document}