diff --git a/Wahrscheinlichkeitstheorie_und_Statistik/Wahrscheinlichkeitstheorie_und_Statistik.tex b/Wahrscheinlichkeitstheorie_und_Statistik/Wahrscheinlichkeitstheorie_und_Statistik.tex
index bcf1cbc..7a9d099 100644
--- a/Wahrscheinlichkeitstheorie_und_Statistik/Wahrscheinlichkeitstheorie_und_Statistik.tex
+++ b/Wahrscheinlichkeitstheorie_und_Statistik/Wahrscheinlichkeitstheorie_und_Statistik.tex
@@ -1224,4 +1224,78 @@
 	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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new file mode 100644
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diff --git a/Wahrscheinlichkeitstheorie_und_Statistik/bilder/1-6_guetefunktion.xml b/Wahrscheinlichkeitstheorie_und_Statistik/bilder/1-6_guetefunktion.xml
new file mode 100644
index 0000000..15a2d61
--- /dev/null
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