Applied Statistics: Hypothesis testing WIP

Erasmus/Applied statistics/TeX_files/Estimating_parameters_2.tex

@@ -114,6 +114,24 @@ The \begriff{\person{Students} t distribution} is the most widely used distribut
 \end{center}
 
 The \begriff{chi-square distribution} is usually used for estimating the variance in a normal distribution.
+\begin{center}
+	\begin{tikzpicture}
+	\begin{axis}[
+	xmin=0, xmax=5, xlabel=$x$,
+	ymin=0, ymax=1, ylabel=$y$,
+	samples=400,
+	axis y line=middle,
+	axis x line=middle,
+	]
+	\addplot+[mark=none] {exp(-x/2)/(sqrt(2*pi) * sqrt(x))};
+	\addlegendentry{1 degree of freedom}
+	\addplot+[mark=none] {exp(-x/2) * sqrt(x)/sqrt(2*pi)};
+	\addlegendentry{3 degrees of freedom}
+	\addplot+[mark=none] {exp(-x/2) * x^(7/2)/(105 * sqrt(2*pi))};
+	\addlegendentry{9 degrees of freedom}
+	\end{axis}
+	\end{tikzpicture}
+\end{center}
 
 In a homogeneous \person{Poisson} process with a rate $\lambda$ events per unit time, the time until the first event happens has a distribution called an \begriff{exponential distribution}. All exponential distributions have their highest probability density at $x=0$ and steadily decrease as $x$ increases.
 \begin{center}

Erasmus/Applied statistics/TeX_files/Hypothesis_testing_1.tex

@@ -4,7 +4,7 @@ There are two types of questions in statistical interference:
     \item \textbf{Hypothesis testing:} Are the sample data consistent with some statement about the parameters?
 \end{itemize}
 
-The \begriff{Null Hypothesis} $H_0$ often specifies a single value for the unknown parameter such as "'$\alpha = \dots$"'. It is a default value that can be accepted as holding if there is no evidence against it. A researcher often collects data with the express hope of disapproving the null hypothesis.
+The \begriff{Null Hypothesis} $H_0$ often specifies a single value for the unknown parameter such as "$\alpha = \dots$". It is a default value that can be accepted as holding if there is no evidence against it. A researcher often collects data with the express hope of disapproving the null hypothesis.
 
 If the null hypothesis is not true, we say that the \begriff{alternative hypothesis} $H_A$ holds. If the data are not consistent with the null hypothesis, then we can conclude that the alternative hypothesis must be true. Either the null hypothesis or the alternative hypothesis must be true.
 
@@ -23,11 +23,11 @@ If the null hypothesis is not true, we say that the \begriff{alternative hypothe
 
 \subsection{The P-value (Probability value)}
 
-In an industrial process some measurement is normally distributed with standard deviation $\sigma = 10$ Its mean should be $\mu = 520$, but can differ a little bit. Samples of $n=10$ measurements are regularly collected as part of quality control. If a sample had $\bar{x}=529$, does the process need to be adjusted?
+In an industrial process some measurement is normally distributed with standard deviation $\sigma = 10$. Its mean should be $\mu = 520$, but can differ a little bit. Samples of $n=10$ measurements are regularly collected as part of quality control. If a sample had $\bar{x}=529$, does the process need to be adjusted?
 
 \input{./TeX_files/materials/samples_of_mean}
 
-From the 200 simulated samples above (Monte Carlo simulation), it seems very unlikely that a sample mean of 529 would have been recorded if $\mu = 529$. There is strong evidence that the industrial process no longer has a mean of $\mu = 520$ and needs to be adjusted.
+From the 200 simulated samples above (\person{Monte Carlo} simulation), it seems very unlikely that a sample mean of 529 would have been recorded if $\mu = 529$. There is strong evidence that the industrial process no longer has a mean of $\mu = 520$ and needs to be adjusted.
 
 \begin{definition}[p-value]
     A \begriff{p-value} describes the \textbf{evidence against} $H_0$. A p-value is evaluated from a random sample so it has a distribution in the same way that a sample mean has a distribution.

Erasmus/Applied statistics/TeX_files/Hypothesis_testing_2.tex

@@ -0,0 +1,117 @@
+\subsection{Likelihood ratio test}
+
+In some cases we need to perform a hypothesis test to compare two models: big "general" model ($M_B$) and small "simple" model ($M_S$) nested into the bigger model. \\
+$H_0$: $M_S$ fits the data \\
+$H_A$: $M_S$ does not fit the data and $M_B$ should be used instead. \\
+
+We need to verify if $M_B$ fits the data significantly better.
+\begin{itemize}
+	\item \textbf{Measure how well a model fits the data:} The fit of any model can be described by the maximum possible likelihood for that model:
+	\begin{align}
+		L(M) = \max\{P(data\vert model)\}\notag
+	\end{align}
+	Calculate the maximum likelihood estimates for all unknown parameters and insert them into the likelihood function.
+	\item \textbf{Work out the \begriff{likelihood ratio}:} 
+	\begin{align}
+		R = \frac{L(M_B)}{L(M_S)} \ge 1\notag
+	\end{align}
+	Big values of $R$ suggests that $M_S$ does not fit as well as $M_B$.
+	\item \textbf{Work out log of likelihood ratio:} 
+	\begin{align}
+		\log(R) = l(M_B) - l(M_S) \ge 0\notag
+	\end{align}
+	Big values of $R$ suggests that $M_S$ does not fit as well as $M_B$.
+\end{itemize}
+
+\begin{example}
+	There are a number of defective items on  a production line in 20 days that follow \person{Poisson}($\lambda$) distribution: 1, 2, 3, 4, 2, 3, 2, 5, 5, 2, 4, 3, 5, 1, 2, 4, 0, 2, 2, 6. \\
+	$M_S$: the sample comes from \person{Poission}(2) \\
+	$M_B$: the sample comes from \person{Poission}($\lambda$) \\
+\end{example}
+
+\begin{example}
+	Clinical records give the survival time for 30 people: 9.73,5.56, 4.28, 4.87, 1.55, 6.20, 1.08, 7.17, 28.65, 6.10, 16.16, 9.92, 2.40, 6.19. In a clinical trial of a new drug treatment 20 people had survival times of: 22.07, 12.47, 6.42, 8.15, 0.64, 20.04, 17.49, 2.22, 3.00. Is there any difference in survival times for those using the new drug? \\
+	$M_S$: Both examples come from the same exponential($\lambda$) distribution. \\
+	$M_B$: The first sample comes from exponential($\lambda_1$) and the second sample from exponential($\lambda_2$).
+\end{example}
+
+\begin{definition}
+	If the data come from $L(M_S)$, and $L(M_B)$ has $k$ more parameters than $L(M_S)$ then
+	\begin{align}
+		X^2 &= 2\log(R) \notag \\
+		&= 2\big(l(M_B) - l(M_S)\big) \notag \\
+		&\approx \chi^2(k \text{ degrees of freedom}) \notag
+	\end{align}
+\end{definition}
+
+The main steps for the likelihood ratio test are:
+\begin{enumerate}[label=\textbf{\arabic*.}]
+	\item Work out maximum likelihood estimates of all unknown parameters in $M_S$.
+	\item Work out maximum likelihood estimates of all unknown parameters in $M_B$.
+	\item Evaluate the test statistic: $\chi^2 = 2\big(l(M_B) - l(M_S)\big)$
+	\item The degrees of freedom for the test are the difference between the numbers of unknown parameters in two models. The p-value for the test is the upper tail probability of the $\chi^2(k \text{ degrees of freedom})$ distribution given the test statistic.
+	\item Interpret the p-value: small values give evidence that the null hypothesis ($M_S$ model) does not hold.
+\end{enumerate}
+
+\begin{example}
+	There are a number of defective items on  a production line in 20 days that follow \person{Poisson}($\lambda$) distribution: 1, 2, 3, 4, 2, 3, 2, 5, 5, 2, 4, 3, 5, 1, 2, 4, 0, 2, 2, 6.
+	\begin{center}
+		$\begin{array}{ccp{4cm}|p{7cm}}
+		&&null hypothesis & $H_0$: $\lambda = 2$ small model $M_S$ \\
+		\cline{3-4}
+		&&alternative hypothesis & $H_A$: $\lambda \neq 2$ big model $M_B$\\
+		\cline{3-4}
+		&&log-likelihood for the Poisson distribution & $l(\lambda) = \left(\sum_{i=1}^{20} x_i\right)\log(\lambda) - n\lambda$ \\
+		\cline{3-4}
+		\multirow{3.7}{3mm}{$M_B$}& \ldelim\{{3.5}{2mm}  & MLE for the unknown parameter& $\hat{\lambda} = \frac{\sum x_i}{n} = 2.9$ \\ \cline{3-4}
+		&  & Maximum possible value for the log-likelihood & $l(M_B) = 58\log(2.9) - 20\cdot 2.9 = 3.7532$ \\ \cline{3-4}
+		\multirow{3.7}{3mm}{$M_S$}& \ldelim\{{3.5}{2mm}  & MLE for the unknown parameter& no unknown parameter \\ \cline{3-4}
+		&  & Maximum possible value for the log-likelihood & $l(M_S) = 58\log(2) - 20\cdot 2 = 0.2025$ \\ \cline{3-4}
+		&&Likelihood ratio test & $\chi^2 = 2\big(l(M_B) - l(M_S)\big) = 7.101$ \\
+		\cline{3-4}
+		&&\multicolumn{2}{p{11cm}}{It should be compared to $\chi^2(1\text{ degree of freedom})$ since the difference in unknown parameters is equal to 1.} \\
+		\cline{3-4}
+		&&p-value & The p-value is 0.008 (the upper tail probability above 7.101) \\
+		\cline{3-4}
+		&&Interpreting p-value & The p-value is very small and we can conclude that there is  strong evidence that $M_B$ fits the data better than $M_S$: $\lambda\neq 2$.
+		\end{array}$
+	\end{center}
+	\begin{center}
+		\begin{tikzpicture}[scale=0.9]
+		\begin{axis}[
+		xmin=0, xmax=10, xlabel=$x$,
+		ymin=0, ymax=1, ylabel=$y$,
+		samples=50,
+		axis y line=middle,
+		axis x line=middle,
+		domain=0:10,
+		restrict y to domain=0:1,
+		]
+		\addplot[name path=f,blue] {exp(-x/2)/(sqrt(2*pi) * sqrt(x))};
+		\path[name path=axis] (axis cs:7.101,0) -- (axis cs:10,0);
+		\addplot [thick,color=blue,fill=blue,fill opacity=0.3] fill between[of=f and axis,soft clip={domain=7.101:10},];
+		\draw [dotted] (axis cs:7.101,0) -- (axis cs:7.101,0.6);
+		\node at (axis cs:8.5,0.4) (a) {p-value};
+		\draw (axis cs:8.5, 0.36) -- (axis cs: 7.5,0.0002);
+		\end{axis}
+		\end{tikzpicture}
+		\begin{tikzpicture}[scale=0.9]
+		\begin{axis}[
+		xmin=6, xmax=10, xlabel=$x$,
+		ymin=0, ymax=0.01, ylabel=$y$,
+		samples=50,
+		axis y line=middle,
+		axis x line=middle,
+		domain=0:10,
+		restrict y to domain=0:0.01,
+		]
+		\addplot[name path=f,blue] {exp(-x/2)/(sqrt(2*pi) * sqrt(x))};
+		\path[name path=axis] (axis cs:7.101,0) -- (axis cs:10,0);
+		\addplot [thick,color=blue,fill=blue,fill opacity=0.3] fill between[of=f and axis,soft clip={domain=7.101:10},];
+		\draw [dotted] (axis cs:7.101,0) -- (axis cs:7.101,0.006);
+		\node at (axis cs:8.5,0.006) (a) {p-value};
+		\draw (axis cs:8.5, 0.0056) -- (axis cs: 7.5,0.002);
+		\end{axis}
+		\end{tikzpicture}
+	\end{center}
+\end{example}

Erasmus/Applied statistics/TeX_files/Weibulls_Distribution.tex

@@ -101,6 +101,7 @@
 		    restrict y to domain=0:1,
 		    ]
 		    \addplot+[mark=none] {(1/exp(x))/(1-(1-1/exp(x)))};
+		   	\draw[blue] (axis cs: 0,1) -- (axis cs: 2,1);
 		    \end{axis}
         \end{tikzpicture} &
         \begin{tikzpicture}[scale=0.6]

Erasmus/Applied statistics/TeX_files/materials/samples_of_mean.tex

@@ -217,6 +217,6 @@
 		\draw[red,dotted] (529,1.2) -- (529,-0.2);
 		
 		\node at (530,-1.2) (axis) {Means of sample};
-		\node at (520,2) (top) {\large Means of samples of $n=10$ values from normal ($\mu=520$, $\sigma=10$)};
+		\node at (520,2) (top) {\textbf{Means of samples of $n=10$ values from normal ($\mu=520$, $\sigma=10$)}};
 	\end{tikzpicture}
 \end{center}

Erasmus/Applied statistics/mathscript.cls

@@ -70,6 +70,7 @@
 \RequirePackage{tabularx} 	%tabularx-environment (explicitly set width of columns)
 \RequirePackage{longtable} %Tabellen mit Seitenumbrüchen
 \RequirePackage{multirow}
+\RequirePackage{bigdelim}
 \RequirePackage{booktabs}	%improved rules
 \usepackage{colortbl} %einfärben von Spalten, Zeilen und Zellen