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Wahrscheinlichkeitstheorie_und_Statistik/Wahrscheinlichkeitstheorie_und_Statistik.tex
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Testentscheidung:\\
\( \underbrace{T}_{2.283} > \underbrace{c}_{2.06} \to H_0\) ablehnen $\to$ die Maschine liefert zu lange Teile (wahrscheinlich)
einseitig umgekehrt: $H_0 \mu \ge \mu_0$\\
\includegraphics{bilder/1-6_dichtefunktion_6.eps}\\
$\hookrightarrow$ mit der Tabelle: $c = -\phi^{-1} (1-\alpha)$\\
Testentscheidung: $ T \ge c \to H_0 $ annehmen\\
$T<c \to H_0 ablehnen$
(Unterschied zum einseitigen Test $H_0: \mu \le \mu_0$)
\bsp Haltbarkeit von Birnen\\
gefordert: mindestens 10 Tage (durchschnittlich) $\to H_0: \mu\ge 10$\\
Stichprobe: 1000 Birnen $\to$ Haltbarkeit: bei 800 Birnen 10 Tage, bei 100 Birnen 11 Tage, bei 100 Birnen 8 Tage\\
Irrtumswahrscheinlichkeit: $\alpha = 2\%$\\
Spricht die Stichprobe gegen die Aussage, das die Haltbarkeit mindestens 10 Tage beträgt?
Bestimmung von $T$:\\
$n=1000$\\
$\overline{x} = \frac1n * (x_1+x_2+…+x_n)$\\
$\sigma = s = \sqrt{s^2}$\\
($\sigma$ bisher unbekannt $\to$ aus der Stichprobe)\\
mit $s^2 = s_{n-1}^2 = \frac{1}{n-1} * (x_1^2,x_2^2+…+x_n^2) - \frac{n}{n-1} * \overline{x}^2$
$\overline{x} = \frac{1}{1000} * (800 * 10 + 100 * 11 + 100 * 8) = 9.9$\\
$s^2 = \frac{1}{999} * (800 * 10^2 + 100 * 11^2 + 100 * 8^2 ) - \frac{1000}{999} * 9.9^2 = \frac{490}{999} = 0.\overline{490}$\\
$s = \sqrt{\frac{490}{999}} = \frac{7*\sqrt{1110}}{333} \approx 0.7$
$T = \frac{\overline{x}-\mu_0}{\sigma} * \sqrt{n} = \frac{9.9-10}{0.7} * \sqrt{1000} = \frac{-10*\sqrt{10}}{7} \approx -4.5175$
Bestimmung von $c$:\\
$ c = -\phi^{-1} (1-\alpha) = -\phi^{-1}(0.98) = -2.05 $ (Rückwärts in Tabelle)
Testentscheidung:\\
\( \underbrace{T}_{-4.5175} < \underbrace{c}_{-2.05} \to H_0\) ablehnen $\to$ Man davon ausgehen, dass die Haltbarkeit (im Schnitt) \underline{unter} 10 Tagen liegt.
Im Überblick:\\
\begin{tabular}{c|c|c|c|c}
& Testgröße & Grenze $c$ & $H_0$ annehmen, wenn & $\mu_0$\\\hline
$H_0: \mu = \mu_0$ &
immer gleich &
$c = \phi^{-1}(1-\frac \alpha 2)$ &
$-c \le T \le c$ & Sollwert\\
$H_0: \mu = \mu_0$ &
$T = \frac{\overline{x}-\mu_0}{\color{Orange}\sigma} * \sqrt{n}$ &
$c = \phi^{-1}(1-\alpha)$ &
$T\le c$ & Höchstens\\
$H_0: \mu = \mu_0$ &
\textcolor{Orange}{evtl. $s$ aus der Stichprobe} &
$c = -\phi^{-1}(1-\alpha)$ &
$T\ge c$ & Mindestens
\end{tabular}
\subsubsection{Fehler 1. Art, Fehler 2. Art}
Vorgeben muss man $\alpha = $ Fehler 1. Art $ = P( H_0 \text{ ablehnen } \mid H_0 \text{ stimmt} )$\\
es ergibt sich $\beta = $ Fehler 2. Art $ = P( H_0 \text{ annehmen } \mid H_0 \text{ stimmt nicht})$
Problem: je kleiner man $\alpha$ wählt, umso größer ist $\beta$.
\includegraphics{bilder/1-6_dichtefunktion_7.eps}
\textbf{Gütefunktion eines Tests:}
beim zweiseitigen Test: $H_0: \mu = \mu_0$\\
wenn $H_0$ nicht stimmt, dann ist das wahre $\mu = \mu_1$
\includegraphics{bilder/1-6_guetefunktion.eps}
Berechnung von $\beta$ im \bsp von vorhin:\\
$H_0: \mu \ge 10 ; \alpha ? 2\% \to c = -2.05 $, $H_0$ annehmen, wenn $T\ge c$\\
wenn das wahre $\mu$ z.\,B. gleich $0.8$ ist \textcolor{Orange}{$\to H_0$ stimmt nicht}
$\beta = P(H_0 \text{ annehmen} \mid \mu = 9.8 ) = P(T \ge -2.05 \mid \mu = 9.8)$\\
$= P(\frac{\overline{x}-\mu_0}{\sigma} * \sqrt{n} \ge -2.05 \mid \mu = 9.8) $
\end{document}

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192 708 m
176.003 708.077 l
176.111 704 l
192.168 704 l
192 708 l
192 708 l
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<text matrix="1 0 0 1 -16 -4" transformations="translations" pos="80 696" stroke="yellow" type="label" width="7.57" height="6.931" depth="3.43" halign="center" valign="center">$\frac{\alpha}{2}$</text>
<text matrix="1 0 0 1 16 -4" transformations="translations" pos="176 696" stroke="yellow" type="label" width="7.57" height="6.931" depth="3.43" halign="center" valign="center">$\frac{\alpha}{2}$</text>
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64 708 m
96 708
112 712
120 720
124 752
128 756
128 756 s
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64 708 m
96 708
112 712
120 720
124 752
128 756
128 756 s
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80 712 m
80 696 l
80 696 l
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80 700 m
176 700 l
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<text transformations="translations" pos="80 692" stroke="darkgreen" type="label" width="12.06" height="5.812" depth="0.83" halign="center" valign="center">$-c$</text>
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<text transformations="translations" pos="168 748" stroke="turquoise" type="label" width="48.917" height="6.926" depth="1.93" valign="baseline">Fl\&quot;ache = $\beta$</text>
<text transformations="translations" pos="168 736" stroke="turquoise" type="label" width="68.549" height="6.918" depth="0" valign="baseline">= Fehler 2. Art</text>
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80 708 m
120 708
136 712
144 720
148 752
152 756
152 756
156 752
160 720
168 712
176 710.341
176 710.341
176.111 704
176.111 704
80.3468 704
80.3468 704
80.2511 708.082 s
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<text transformations="translations" pos="168 724" stroke="turquoise" type="label" width="76.076" height="6.926" depth="1.93" valign="baseline">$\to$ hier sehr gross</text>
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</ipe>

280
Wahrscheinlichkeitstheorie_und_Statistik/bilder/1-6_guetefunktion.xml
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