Wahrscheinlichkeit

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Thomas Ba. 14 years ago
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%TEST ENDE
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Zugehörige Verteilungsfunktion \(F(x) = \phi(x) = \phi_{0,1}(x) \rightarrow\) hierfür: \underline{Tabellenwerte}
\end{mydef}
\includegraphics{bilder/1-6_dichtefunktion.eps}
Aus der Tabelle:\\
\bsp \(\phi(1.27) = 0.89796 \)\\
\( \phi(0) = \phi(0.00) = 0.5 \)\\
\( \phi(-1.27) = 1- \phi(1.27) = 1-0.89796 \)
Berechnung von \(\phi(-x)\):\\
\( \phi(-x) = 1 -\phi(x) \)
Wenn $X$ $N(0,1)$ verteilt ist, dann ist \( P(X\le b) = \phi(b), P(x\ge a) = 1-\phi(a), P(a\le x\le b) = \phi(b)-\phi(a), P(X=a)=0\)
Wenn $X$ $N(\mu,0)$ verteilt ist, dann ist…\\
\includegraphics{bilder/1-6_dichtefunktion_2.eps}\\
… \( \frac{X-\mu}{\rho} \) \( N(0,1)\) verteilt. Standardverteilung \(\to\) Tabelle Anwendbar.
\bsp Bauteile aus der Produktion $\to$ \underline{Länge des Bauteils} ist Normalverteilt. Mit \( \mu = 10cm\) (Normalverteilt), \( \sigma = 0.2cm \) (Streuung $\to$ Maß für die mittlere Abweichung von $\mu$)\\
$\to 10cm\pm 0.2cm$
Ausschuss liegt vor, wenn die Länge unter $9.5cm$ liegt.\\
Gesucht: Auschussanteil\\
\begin{align*}
\to P(X\le 9.5) &= P(\frac{x-\mu}{\sigma} \le \frac{0.9-\mu}{\sigma})\\
&= P(\underbrace{\frac{x-\mu}{\sigma}}_{\text{ist }N(0,1)\text{ verteilt}} \le \frac{9.5-10}{0.2})\\
&= P(\frac{x-\mu}{\sigma} \le -2.5) = \phi(-2.5)\\
&= 1-\phi(2.5) = 1-0.99379\\
&= 0.00621
\end{align*}
Wie viel \% der Bauteile nimmt uns ein Kunde ab, wenn er eine Länge zwischen $9.7$ und $10.25$ benötigt?
\begin{align*}
P(9.7\le X<10.25) &= P\left( \frac{9.7-\mu}{\sigma} \le \frac{x-\mu}{\sigma} < \frac{10.25-\mu}{\sigma}\right)\\
&= P\left(\frac{-0.3}{0.2}\le \frac{x-\mu}{\sigma}<\frac{0.25}{0.2}\right)\\
&= \phi(\frac{0.25}{0.2}) - \phi(\frac{-0.3}{0.2})\\
&= \phi(1.25) - \phi(-1.5) = \phi(1.25) - (1-\phi(1.5)) \\
&= 0.89435 - (1-0.93319)\\
&= 0.89435 - 0.06681\\
&= 0.82754 = \underline{\underline{82.754\%}}
\end{align*}
Berechne die Wahrscheinlichkeit, dass ein Bauteil länger als $10.7cm$ ist.
\begin{align*}
P(X>10.7) &= 1 - P(X<10.7)\\
&= 1-P(\frac{x-\mu}{\sigma} < \frac{10.7-\mu}{\sigma})\\
&= 1-P(\frac{x-\mu}{\sigma} < \frac{0.7}{0.2})\\
&= 1-P(\frac{x-\mu}{\sigma} < 3.5)\\
&= 1 - \phi(3.5) = 1-0.99977\\
&= 0.00023 = \underline{\underline{0.023\%}}
\end{align*}
\subsection{Schätzungen und Tests}
Test z.\,B. Liefert die Maschine im Mittel die geforderten $10cm$ Länge?\\
\(\to\) mit einer Stichprobe.
\subsubsection{Stichproben}
Stichprobenumfang $n$, Stichprobenwerte \(x_1,x_2,…,x_n\).\\
$\to$ arithmetisches Mittel (Durchschnitt)
\( \overline{x} = \frac1n * (x_1+x_2+…+x_n) = \frac1n \sum\limits_{i=1}^n x_i \)\\
\( \to\) Schätzung für $\mu$
Stichprobenvarianz
\begin{align*}
s_{\color{Orange}n}^2 &= \frac{1}{\color{Orange}n} * \left( (x_1-\overline{x})^2 + (x_2-\overline{x})^2+ … + (x_n-\overline{x})^2 \right) \\
&= \frac1n * \left( x_1^2 + x_2^2 + … + x_n^2 \right) - \overline{x}^2\\
s^2 = s_{\color{Orange}n-1}^2 &= \frac{1}{\color{Orange}n-1} + \left( (x_1 - \overline{x})^2 + (x_2 - \overline{x})^2 + … + (x_n - \overline{x})^2 \right)\\
&= \frac{1}{n-1} * (x_1^2 + … + x_n^2) - \frac{n}{n-1} * \overline{x}^2
\end{align*}
% Datum: 2011-12-21
\begin{align*}
\overline{x} &= \frac1n * \left( x_1+x_2+…+x_n \right) & \textcolor{Orange}{\to \text{ Schätzwert für }\mu}\\
s^2 = s_{n-1}^2 &= \frac{1}{n-1} \left( x_1^2 + x_2^2 + … + x_n^2 \right) - \frac{n}{n-1}*\overline{x}^2\\
s = s_{n-1} &= \sqrt{s_{n-1}^2} & \textcolor{Orange}{\to \text{ Schätzwert für }\sigma}
\end{align*}
\bsp Stichprobe mit 100 Bauteilen aus der Produktion\\
$\to$ gemessene Lösungen: 50 mal 10cm, 20 mal 10.1cm, 10 mal 10.2cm, 20 mal 9.9cm.
\begin{align*}
\overline{x} &= \frac{1}{100} * (50*10 + 20*10.1 + 10*10.2 + 20*9.9)\\
&= 10.02 \\
s^2 &= \frac{1}{100} * (50*10^2 + 20*10.1^2 + 10*10.2^2 + 20*9.9^2) - \frac{100}{99} * 10.02^2\\
&= \frac{19}{2475} \approx 0.007677\\
s &= \sqrt{\frac{19}{2475}} = \frac{\sqrt{209}}{165} \approx 0.08762
\end{align*}
\textcolor{Green}{MaĂź fĂĽr die Streuung: $0.0876cm$\\($\to\ 10.02\pm 0.08762cm$)}
\subsubsection{Zentraler Grenzwertsatz}\label{sec:zentraler_grenzwertsatz}
Wenn die Zufallsvariablen $X_i\ , i=1,2,3,…$ unabhängig ($X_1,X_2$ sind unabhängig, wenn $P(X_1=a \land X_2=b) = P(X_1=a)*P(X_2=b)$) und identisch Verteilt (alle $X_i$ besitzen die gleiche Verteilungsfunktion) sind mit Erwartungswert \(\mu=E(X_i)\) und Varianz \(\sigma^2 = Var(X_i) \), dann gilt für die Zufallsvariable \(\overline{X}=\frac1n*\left(X_1+X_2+…+X_n\right)\):\\
\( \frac{\overline{X}-\mu}{\sigma} * \sqrt{n} \) ist für große $n$ näherungsweise $N(0,1)$ verteilt.\\
(und: Konvergenz fĂĽr \(n\to\infty\))
\includegraphics{bilder/1-6_dichtefunktion_3.eps}
\subsubsection{Tests (Erwartungswerttests)}
\underline{Nullhypothese} $\to$ soll getestet werden beim Erwartungswerttest:\\
\(H_0: \mu = \mu_0 \) z.\,B. Sollwert der Bauteillängen\\
\(\to\) Zweiseitiger Test\\
oder \(H_0: \mu\le\mu_0\) oder \(H_0: \mu\ge\mu_0 \)\\
\(\to\) Einseitige Tests
Testgröße \(T=\frac{\overline{X}-\mu_0}{\sigma} * \sqrt{n} \)\\
i\(\to\) ist für große $n$ näherungsweise $N(0,1)$ verteilt (siehe \ref{sec:zentraler_grenzwertsatz}) wenn $H_0$ stimmt.
\includegraphics{bilder/1-6_dichtefunktion_4.eps}
Man muss sich eine \underline{Irrtumswahrscheinlichkeit} $\alpha$ vorgeben (z.\,B. \(\alpha = 1\%, 2\% \text{ oder } 5\%\))\\
Testentscheidung: $H_0$ wird \( \begin{cases} \text{angenommen}, &\text{wenn } -c\le T\le c\\ \text{abgelehnt}, &\text{wenn } T> c \text{ oder } T< -c \end{cases} \)
Bestimmung von $c$:
\begin{align*}
\phi(c) &= 1-\frac{\alpha}{2}\\
c &= \phi^{-1}(1-\frac{\alpha}{2}) & \text{in der Tabelle: rückwärts}
\end{align*}
\underline{Vorgehensweise beim Test}\\
\begin{itemize}
\item aus den vorgegebenen $\alpha$ wird $c$ bestimmt: \(\phi(c) = 1-\frac{\alpha}{2} \to \)\underline{$c$ aus der Tabelle}
\item in der Stichprobe wird $\overline{x}$ (und evtl. $s$) bestimmt\\
\(\to\) Testgröße \(T=\frac{\overline{x}-\mu_0}{\sigma} * \sqrt{n} \) ($\sigma$ kann durch $s$ angenähert werden)
\item Testentscheidung: $H_0$ wird \( \begin{cases} \text{angenommen}, &\text{wenn } -c\le T\le c\\ \text{abgelehnt}, &\text{wenn } T> c \text{ oder } T< -c \end{cases} \)
\end{itemize}
\bsp aus \ref{sec:zentraler_grenzwertsatz}:\\
zusätzliche Angaben: Sollwert: $10cm$ ($\mu_0$); Irrtumswahrscheinlichkeit: $2\%$ ($\alpha$)
\begin{multicols}{2}
\begin{align*}
c &= \phi^{-1}\left(1-\frac{\alpha}{2}\right)\\
&= \phi^{-1}\left( 1- \frac{0.02}{2}\right)\\
&= \phi^{-1}( 1- 0.01 )\\
&= \phi^{-1}( 0.99) \\
&= 2.33
\end{align*}
\begin{align*}
T &= \frac{10.02-10}{0.08762} * \sqrt{100}\\
&= \frac{0.02}{0.08762} * 10 = \frac{0.2}{0.08762}\\
&= 2.28258
\end{align*}
\end{multicols}
Testentscheidung: \(\underbrace{-c}_{-2.33}\le \underbrace{T}_{2.283}\le \underbrace{c}_{2.33}\)\\
\(\Rightarrow H_0\) annehmen, d.\,h. es spricht nichts dagegen, dass die Maschine im Mittel den Sollwert produziert.\\
\textcolor{Orange}{Nullhypothese: $\mu=\mu_0=10$}
Bisher: zweiseitiger Test: \(H_0: \mu=\mu0\)\\
Jetzt: \underline{einseitige} Tests:\\
\(H_0: \mu\le\mu_0\) (vorgegebener Maximalwert)
\includegraphics{bilder/1-6_dichtefunktion_5.eps}
Testentscheidung:\\
$H_0$ wird \(\begin{cases}\text{angenommen},&\text{wenn } T\le c\\\text{abgelehnt},&\text{wenn }T>c\end{cases}\)
im \bsp wenn die Bauteile \underline{maximal $10cm$} lang sein dĂĽrfen.\\
\(\to H_0: \mu \le \underbrace{\mu_0}_{=10} \)
\(T=2.283\) (wie beim zweiseitigen Test)\\
fĂĽr $\alpha = 2\%$:
\( \phi(c) = 1-\alpha = 1-0.02 = 0.98 \Rightarrow c=2.06 \)
Testentscheidung:\\
\( \underbrace{T}_{2.283} > \underbrace{c}_{2.06} \to H_0\) ablehnen $\to$ die Maschine liefert zu lange Teile (wahrscheinlich)
\end{document}

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