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https://github.com/vale981/TUD_MATH_BA
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478 lines
15 KiB
TeX
478 lines
15 KiB
TeX
\documentclass[british,a4paper,order=firstname]{mathscript}
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\usepackage{mathoperators}
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\title{\textbf{Applied statistics: Coursework 1}}
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\author{\person{Henry Haustein}}
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\begin{document}
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\pagenumbering{roman}
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\pagestyle{plain}
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\maketitle
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\hypertarget{tocpage}{}
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\tableofcontents
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\bookmark[dest=tocpage,level=1]{Table of contents}
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\pagebreak
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\pagenumbering{arabic}
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\pagestyle{fancy}
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\section{Task 1}
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\subsection{Part (1)}
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In the given data were two out of 26 data points with an Al/Be ratio of more than 4.5. That means
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\begin{align}
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\hat{p} = \frac{2}{26} = \frac{1}{13}\notag
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\end{align}
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\subsection{Part (2)}
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Using the following formula from the lecture we get the 95\% confidence interval:
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\begin{align}
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\hat{p}&\pm 2\cdot\sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \notag \\
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\frac{1}{13} &\pm \underbrace{2\cdot\sqrt{\frac{\frac{1}{13}\cdot\frac{12}{13}}{26}}}_{0.1045} \notag
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\end{align}
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Our 95\% confidence interval is [-0.0276,0.1814] which means that we are 95\% sure that the true proportion lies between -0.0276 and 0.1814.
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\subsection{Part (3)}
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To get the 95\% confidence interval via bootstrap I want to use the \texttt{bootci} function in MATLAB.
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\begin{lstlisting}
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data = [3.75, 4.05, 3.81, 3.23, 3.13, 3.3, 3.21, 3.32, ...
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4.09, 3.9, 5.06, 3.85, 3.88, 4.06, 4.56, 3.6, 3.27, ...
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4.09, 3.38, 3.37, 2.73, 2.95, 2.25, 2.73, 2.55, 3.06];
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parameter = @(y) length(find(y > 4.5))/length(y);
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bootci(10000,{parameter, data},'alpha',0.05,'type',...
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'percentile')
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\end{lstlisting}
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That gives the 95\% confidence interval: [0,0.1923]
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\subsection{Part (4)}
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Yes, the confidence interval from the bootstrap procedure is more appropriate because it's not containing Al/Be ratios that are not possible like -0.0276. A negative ratio would suggest that there is a negative amount of data points in the sample which exceed 4.5. That is not possible.
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\pagebreak
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\section{Task 2}
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\subsection{Part (1)}
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\begin{lstlisting}
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x = [-4.5, -1, -0.5, -0.15, 0, 0.01, 0.02, 0.05, ...
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0.15, 0.2, 0.5, 0.5, 1, 2, 3];
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m = mean(x);
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s = std(x);
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\end{lstlisting}
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\begin{center}
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\begin{tabular}{p{4cm}|p{7cm}}
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null hypothesis & $H_0$: $\mu = 0$ \\
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\hline
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alternative hypothesis & $H_A$: $\mu\neq 0$ \\
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\hline
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t-test for $\mu$ & $t=\frac{m-0}{\frac{s}{\sqrt{15}}} =\frac{0.0853}{\frac{1.6031}{\sqrt{15}}} = 0.2062$ \\
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\hline
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rejection region & \texttt{tinv(0.05,15)} = -1.7531 \\
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\hline
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conclusion & $t$ is not in the rejection region so $H_0$ is accepted at the 10\% significance level.
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\end{tabular}
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\end{center}
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\begin{center}
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\begin{tikzpicture}
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\begin{axis}[
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xmin=-3, xmax=3, xlabel=$x$,
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ymin=0, ymax=1, ylabel=$y$,
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samples=400,
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axis y line=middle,
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axis x line=middle,
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domain=-3:3,
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restrict y to domain=0:1,
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width = 16cm,
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height = 8cm,
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]
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\addplot[name path=f,blue] {116640000000*sqrt(15)/(143*pi*(x^2+15)^(8))};
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\path[name path=axis] (axis cs:1.7531,0) -- (axis cs:3,0);
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\path[name path=axis2] (axis cs:-3,0) -- (axis cs:-1.7531,0);
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\draw (axis cs:1.7531,0) -- (axis cs:1.7531,1);
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\draw (axis cs:-1.7531,0) -- (axis cs:-1.7531,1);
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\draw [dotted] (axis cs:0.2062,0) -- (axis cs:0.2062,0.6);
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\node at (axis cs:1,0.5) (a) {0.05};
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\node at (axis cs:-1,0.5) (a2) {0.05};
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\draw (axis cs:1, 0.46) -- (axis cs: 2.2,0.02);
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\draw (axis cs:-1, 0.46) -- (axis cs: -2.2,0.02);
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\node at (axis cs: 2.0,0.95) (b) {1.7531};
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\node at (axis cs: -2.0,0.95) (b2) {-1.7531};
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\node at (axis cs: 0.45,0.55) (c) {0.2062};
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\node[red] at (axis cs: 2.4,0.78) (d) {rejection region};
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\node[red] at (axis cs: -2.4,0.78) (d2) {rejection region};
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\begin{scope}[transparency group]
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\begin{scope}[blend mode=multiply]
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\addplot [thick,color=blue,fill=blue,fill opacity=0.3] fill between[of=f and axis,soft clip={domain=1.7531:3},];
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\addplot [thick,color=blue,fill=blue,fill opacity=0.3] fill between[of=f and axis2,soft clip={domain=-3:-1.7531},];
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\draw[red,fill=red,opacity=0.2] (axis cs: 1.7531,0) -- (axis cs: 1.7531,1) -- (axis cs: 3,1) -- (axis cs: 3,0) -- (axis cs: 1.7531,0);
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\draw[red,fill=red,opacity=0.2] (axis cs: -1.7531,0) -- (axis cs: -1.7531,1) -- (axis cs: -3,1) -- (axis cs: -3,0) -- (axis cs: -1.7531,0);
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\end{scope}
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\end{scope}
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\end{axis}
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\end{tikzpicture}
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\end{center}
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\subsection{Part (2)}
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If we reduce the significance level our rejection region gets smaller. With $\alpha = 0.05$ the rejection region will start at \texttt{tinv(0.025,15)} = -2.1314. The $t$ calculated in part (1) won't change $\Rightarrow$ our decision won't change too.
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To get the type 2 error we use the MATLAB function \texttt{sampsizepwr} and $type\, 2\, error = 1-power$.
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\begin{lstlisting}
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testtype = 't';
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p0 = [0 1.6031];
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p1 = 0.0853;
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n = 15;
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power = sampsizepwr(testtype,p0,p1,[],n)
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\end{lstlisting}
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This gives $power = 0.0542\Rightarrow type\, 2\, error = 0.9458$. This is the probability of wrongly accepting $H_0$ when it is false.
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\subsection{Part (3)}
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$H_0$: $\mu=0$, normal distribution, small model $M_S$ \\
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$H_A$: $\mu\neq 0$, normal distribution, big model $M_B$ \\
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The log-likelihood function for normal distribution is
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\begin{align}
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\label{log-likelihood}
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-\frac{n}{2}\log(2\pi)-\frac{n}{2}\log(\sigma^2)-\frac{1}{2\sigma^2}\sum_{j=1}^{n} (x_j-\mu)^2
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\end{align}
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Let's start with the MLEs for $\mu$ and $\sigma$ in $M_B$:
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\begin{align}
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\hat{\mu} &= \frac{1}{n}\sum_{j=1}^n x_j \notag \\
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&= 0.0853 \notag \\
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\widehat{\sigma^2} &= \frac{1}{n}\sum_{j=1}^n (x_j-\hat{\mu})^2 \notag \\
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&= 2.3986 \notag
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\end{align}
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Maximum possible value for the log-likelihood $\xRightarrow{\cref{log-likelihood}}$ -27.8457. \\
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Now we'll calculate the MLE for $\sigma$ in $M_S$:
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\begin{align}
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\widehat{\sigma^2} &= \frac{1}{n}\sum_{j=1}^n (x_j-\hat{\mu})^2 \notag \\
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&= 2.3986 \notag
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\end{align}
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Maximum possible value for the log-likelihood $\xRightarrow{\cref{log-likelihood}}$ -27.8684. \\
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Likelihood ratio test:
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\begin{align}
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\chi^2 &= 2\Big(l(M_B) - l(M_S)\Big) \notag \\
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&= 0.0454 \notag
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\end{align}
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It should be compared to $\chi^2$(1 degree of freedom) since the difference in unknown parameters is equal to 1. The following piece of MATLAB code will calculate the p-value.
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\begin{lstlisting}
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p = chi2cdf(0.0454,1,'upper')
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\end{lstlisting}
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The p-value is 0.8313 which means that we accept $H_0$: The small model $M_S$ fits the data good enough. This is the same result as in part (1) and (2).
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\pagebreak
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\section{Task 3}
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\subsection{Part (1)}
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First of all we need to prepare the data:
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\begin{lstlisting}
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raw = load('input_data.txt');
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data = reshape(raw,[1 500]); %produce a single vector
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\end{lstlisting}
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After that we do for every distribution (normal, exponential, uniform, lognormal, \person{Rayleigh}, gamma) the same procedure:
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\begin{enumerate}[label=\textbf{\arabic*.}]
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\item Estimate the parameter. This is often done with the function \texttt{<distribution>fit} but for estimating the parameters in the gamma distribution I used \texttt{fitdist(data', 'Gamma')} because \texttt{gamfit} doesn't work.
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\item Creating the CDF with \texttt{makedist}.
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\item Run the \person{Kolmogorov-Smirnov} test with \texttt{kstest}.
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\end{enumerate}
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\begin{lstlisting}
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%normal distribution
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[mu,sigma] = normfit(data)
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norm_cdf = makedist('Normal','mu',mu,'sigma',sigma);
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[h,p] = kstest(data,'CDF',norm_cdf)
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%exponential distribution
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mu = expfit(data)
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exp_cdf = makedist('Exponential','mu',mu);
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[h,p] = kstest(data,'CDF',exp_cdf)
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%uniform distribution
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[low,up] = unifit(data)
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uni_cdf = makedist('Uniform','lower',low,'upper',up);
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[h,p] = kstest(data,'CDF',uni_cdf)
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%lognormal distribution
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logmu = mean(log(data))
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logsigma = std(log(data))
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logn_cdf = makedist('Lognormal','mu',logmu,'sigma',logsigma);
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[h,p] = kstest(data,'CDF',logn_cdf)
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%rayleigh distribution
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b = raylfit(data)
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rayl_cdf = makedist('Rayleigh','b',b);
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[h,p] = kstest(data,'CDF',rayl_cdf)
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%gamma distribution
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distribution = fitdist(data','Gamma');
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a = distribution.a
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b = distribution.b
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gamma_cdf = makedist('Gamma','a',a,'b',b);
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[h,p] = kstest(data,'CDF',gamma_cdf)
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\end{lstlisting}
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Running this gives the following output. The best fitting distribution is marked green, the worst red.
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\begin{center}
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\begin{tabular}{c|l|p{3cm}|p{3cm}}
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\textbf{distribution}& \textbf{estimated parameters} & \multicolumn{2}{c}{\person{Kolmogorov-Smirnov} \textbf{test}} \\
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& & $h$ & $p$ \\
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\hline
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normal & $\mu=2.3804$, $\sigma=1.2486$ & $h=1$ & $p=0.0158$ \\
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\hline
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exponential & $\mu=2.3804$ & $h=1$ & $p=2.2618\cdot 10^{-23}$ \\
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\hline
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\rowcolor{red}uniform & $lower=0.1478$, $upper=7.8807$ & $h=1$ & $p=1.5096\cdot 10^{-72}$ \\
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\hline
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lognormal & $\log(\mu)=0.7050$, $\log(\sigma)=0.6243$ & $h=1$ & $p=0.0017$ \\
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\hline
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\rowcolor{green}\person{Rayleigh} & $b=1.9003$ & $h=0$ & $p=0.8939$ \\
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\hline
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gamma & $a=3.2378$, $b=0.7352$ & $h=0$ & $p=0.2771$ \\
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\end{tabular}
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\end{center}
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\subsection{Part (2)}
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\input{./TeX_files/materials/CW1_kstest}
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\pagebreak
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\section{Task 4}
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\subsection{Part (1)}
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The probability density function $f(t)$ is
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\begin{align}
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f(t) = \frac{2t\cdot\frac{\exp(-t^2)}{100}}{100} = \frac{t\cdot\exp(-t^2)}{5000}\notag
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\end{align}
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\textcolor{red}{After the PDF was changed we have}
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{\color{red}\begin{align}
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f(t) = \frac{2t\cdot\exp\left(\frac{-t^2}{100} \right) }{100} \notag
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\end{align}}
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\begin{center}
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\begin{tikzpicture}[scale=0.9]
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\begin{axis}[
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xmin=0, xmax=5, xlabel=$x$,
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ymin=0, ymax=0.0001, ylabel=$y$,
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samples=400,
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axis y line=middle,
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axis x line=middle,
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restrict y to domain=0:1,
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]
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\addplot+[mark=none] {(x*exp(-x^2))/5000};
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\end{axis}
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\end{tikzpicture}
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\begin{tikzpicture}[scale=0.9]
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\begin{axis}[
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xmin=0, xmax=20, xlabel=$x$,
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ymin=0, ymax=0.2, ylabel=$y$,
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samples=400,
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axis y line=middle,
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axis x line=middle,
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restrict y to domain=0:1,
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domain=0:20
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]
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\addplot+[mark=none, color=red] {(2*x*exp(-x^2/100))/100};
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\end{axis}
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\end{tikzpicture}
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\end{center}
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The cumulative distribution function $F(t)$ is then
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\begin{align}
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F(t) &= \int_0^t f(\xi)\,\diff\xi \notag \\
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&= \int_0^t \frac{\xi\cdot\exp(-\xi^2)}{5000}\,\diff\xi\notag \\
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&= \frac{\exp(-t^2)\Big(\exp(t^2)-1\Big)}{10000} \notag
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\end{align}
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{\color{red}\begin{align}
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F(t) &= \int_0^t f(\xi)\,\diff\xi \notag \\
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&= \int_0^t \frac{2\xi\cdot\exp\left(\frac{-\xi^2}{100} \right) }{100}\,\diff\xi \notag \\
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&= 1-\exp\left(\frac{-t^2}{100} \right) \notag
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\end{align}}
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\begin{center}
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\begin{tikzpicture}[scale=0.9]
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\begin{axis}[
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xmin=0, xmax=5, xlabel=$x$,
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ymin=0, ymax=0.0001, ylabel=$y$,
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samples=400,
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axis y line=middle,
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axis x line=middle,
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restrict y to domain=0:1,
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]
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\addplot+[mark=none] {(exp(-x^2)*(exp(x^2)-1))/10000};
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\end{axis}
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\end{tikzpicture}
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\begin{tikzpicture}[scale=0.9]
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\begin{axis}[
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xmin=0, xmax=20, xlabel=$x$,
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ymin=0, ymax=1, ylabel=$y$,
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samples=400,
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axis y line=middle,
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axis x line=middle,
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restrict y to domain=0:1,
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domain=0:20
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]
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\addplot+[mark=none,color=red] {1-exp(-x^2/100)};
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\end{axis}
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\end{tikzpicture}
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\end{center}
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For the survival function we get
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\begin{align}
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R(t) &= 1 - F(t) \notag \\
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&= \frac{\exp(-t^2)+9999}{10000} \notag
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\end{align}
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{\color{red}\begin{align}
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R(t) &= 1-F(t) \notag \\
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&= \exp\left(\frac{-t^2}{100} \right) \notag
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\end{align}}
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\begin{center}
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\begin{tikzpicture}[scale=0.9]
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\begin{axis}[
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xmin=0, xmax=5, xlabel=$x$,
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ymin=0, ymax=1, ylabel=$y$,
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samples=400,
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axis y line=middle,
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axis x line=middle,
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restrict y to domain=0:1,
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]
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\addplot+[mark=none] {(exp(-x^2)+9999)/10000};
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\end{axis}
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\end{tikzpicture}
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\begin{tikzpicture}[scale=0.9]
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\begin{axis}[
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xmin=0, xmax=5, xlabel=$x$,
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ymin=0.9999, ymax=1, ylabel=$y$,
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samples=400,
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axis y line=middle,
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axis x line=middle,
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restrict y to domain=0:1,
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y tick label style={
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/pgf/number format/.cd,
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precision=5,
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/tikz/.cd
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},
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]
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\addplot+[mark=none] {(exp(-x^2)+9999)/10000};
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\end{axis}
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\end{tikzpicture}
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\end{center}
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\begin{center}
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\begin{tikzpicture}
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\begin{axis}[
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xmin=0, xmax=20, xlabel=$x$,
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ymin=0, ymax=1, ylabel=$y$,
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samples=400,
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axis y line=middle,
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axis x line=middle,
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restrict y to domain=0:1,
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domain=0:20
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]
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\addplot+[mark=none,color=red] {exp(-x^2/100)};
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\end{axis}
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\end{tikzpicture}
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\end{center}
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To get the reliability of the component at $t=7$ we simply evaluate $R(7)$ which is 0.9999 \textcolor{red}{(0.6126)}.
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The hazard function is defined as
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\begin{align}
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h(t) &= \frac{f(t)}{1-F(t)} \notag \\
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&= \frac{2t}{9999\cdot \exp(t^2)+1} \notag
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\end{align}
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{\color{red}\begin{align}
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h(t) &= \frac{f(t)}{1-F(t)} \notag \\
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&= \frac{t}{50} \notag
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\end{align}}
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\begin{center}
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\begin{tikzpicture}[scale=0.9]
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\begin{axis}[
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xmin=0, xmax=5, xlabel=$x$,
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ymin=0, ymax=0.0001, ylabel=$y$,
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samples=400,
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axis y line=middle,
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axis x line=middle,
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restrict y to domain=0:1,
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]
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\addplot+[mark=none] {(2*x)/(9999*exp(x^2)+1)};
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\end{axis}
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\end{tikzpicture}
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\begin{tikzpicture}[scale=0.9]
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|
\begin{axis}[
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xmin=0, xmax=20, xlabel=$x$,
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ymin=0, ymax=1, ylabel=$y$,
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samples=400,
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axis y line=middle,
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axis x line=middle,
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restrict y to domain=0:1,
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domain=0:20
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]
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\addplot+[mark=none,color=red] {x/50};
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\end{axis}
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\end{tikzpicture}
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\end{center}
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The hazard function describes how an item ages where $t$ affects the risk of failure. It is the frequency with which the item fails, expressed in failures per unit of time.
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\subsection{Part (2)}
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Given $h(x)\sim(\sqrt{x})^{-1}$ we will try to find out the $shape$-parameter of the \person{Weibull} distribution first.
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\begin{center}
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\begin{tikzpicture}
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\begin{axis}[
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xmin=0, xmax=5, xlabel=$x$,
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ymin=0, ymax=1, ylabel=$y$,
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samples=400,
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axis y line=middle,
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axis x line=middle,
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restrict y to domain=0:1,
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yticklabels={,,},
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xticklabels={,,}
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]
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\addplot+[mark=none] {1/sqrt(x)};
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\end{axis}
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\end{tikzpicture}
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\end{center}
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Comparing this graph to graphs of the hazard function with different $shape$-parameters we see that $shape=0.5$ fits best.
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\begin{center}
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\begin{tabular}{p{5cm}|p{5cm}|p{5cm}}
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$shape = 0.5$ & $shape = 1$ & $shape = 2$ \\
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\hline
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\multicolumn{3}{c}{\cellcolor{gray!50}\textbf{Hazard function} $\left(h = \frac{\text{PDF}}{1-\text{CDF}}\right)$} \\
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\hline
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\begin{tikzpicture}[scale=0.6]
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\begin{axis}[
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xmin=0, xmax=2, xlabel=$x$,
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ymin=0, ymax=1, ylabel=$y$,
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samples=400,
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axis y line=middle,
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axis x line=middle,
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]
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\addplot+[mark=none] {(0.5*x^(-0.5)*exp(-x^0.5))/(1-(1-exp(-x^0.5)))};
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\end{axis}
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\end{tikzpicture} &
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\begin{tikzpicture}[scale=0.6]
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\begin{axis}[
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xmin=0, xmax=2, xlabel=$x$,
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ymin=0, ymax=1, ylabel=$y$,
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samples=400,
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axis y line=middle,
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axis x line=middle,
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restrict y to domain=0:1,
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]
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\addplot+[mark=none] {(1/exp(x))/(1-(1-1/exp(x)))};
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\draw[blue] (axis cs: 0,1) -- (axis cs: 2,1);
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\end{axis}
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\end{tikzpicture} &
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\begin{tikzpicture}[scale=0.6]
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\begin{axis}[
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xmin=0, xmax=2, xlabel=$x$,
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ymin=0, ymax=1, ylabel=$y$,
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samples=400,
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axis y line=middle,
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axis x line=middle,
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]
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\addplot+[mark=none] {(2*x^(1)*exp(-x^2))/(1-(1-exp(-x^2)))};
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\end{axis}
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\end{tikzpicture} \\
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\end{tabular}
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\end{center}
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To get the $scale$-parameter of the distribution we use the other provided information:
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\begin{align}
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5 &= \mu \notag \\
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&= scale\cdot\Gamma\left(1+\frac{1}{shape}\right) \notag \\
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&= scale\cdot\Gamma(3) \notag \\
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\Rightarrow scale &= \frac{5}{2} \notag
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\end{align}
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Let's build the survival function:
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\begin{align}
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R(t) &= 1-\Bigg(1-\exp\left(-\sqrt{\frac{x}{\nicefrac{5}{2}}}\right)\Bigg) \notag \\
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&= \exp(-\sqrt{x}\cdot\sqrt{2.5})\notag
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\end{align}
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That mean that the probability of surviving 6 years (30 years) is $R(6) = 0.0208$ ($R(30) = 0.0002$).
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\end{document}
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