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\input{tcilatex}

\begin{document}


\section{Exam}

\section{Text}

\textbf{\Large Stockholms Universitet, Statistisk -- Finansiell statistik}

\medskip

\textbf{Examinator}: P\"{a}r Stockhammar/Mikael M\"{o}ller

\textbf{Godk\"{a}nt}: F\"{o}r godk\"{a}nd hemtentamen kr\"{a}vs godk\"{a}nt p%
\aa\ alla tal. Ett tal \"{a}r godk\"{a}nt om det erh\aa llit 8 po\"{a}ng av
10 m\"{o}jliga.

\textbf{Rest}: Om mer \"{a}n 25 och mindre \"{a}n 40 po\"{a}ng erh\aa lls f%
\aa r f\"{o}rnyad inl\"{a}mning g\"{o}ras fram till dagen innan ordinarie
tentamensdag.

L\"{o}sningarna skall vara v\"{a}l motiverade och l\"{a}tta att f\"{o}lja.
Saknas motivering g\"{o}rs avdrag. Om antaganden och analys ej klart framg%
\aa r kan det bli fullt avdrag. Endast ett tal:s l\"{o}sning per pappersark.

Uppgiften skall redovisas skriftligt och l\"{a}mnas till \"{o}vningsl\"{a}%
raren. Inl\"{a}mning sker i samband med undervisningen eller i brevl\aa dan
utanf\"{o}r hissarna p\aa\ plan 7 i B-huset. Senaste inl\"{a}mningsdag
meddelas av \"{o}vningsl\"{a}raren och uppgiften skall vara godk\"{a}nd
innan ordinarie tentamen. F\"{o}r att bli godk\"{a}nd p\aa\ den
obligatoriska delen -- v\"{a}rd 2 po\"{a}ng -- m\aa ste alla inl\"{a}%
mningsuppgifter vara godk\"{a}nda innan ordinarie tentamenstillf\"{a}lle; om
detta ej \"{a}r uppfyllt m\aa ste den obligatoriska delen g\"{o}ras om n\"{a}%
sta termin. Det \"{a}r laborantens ansvar att ha kopia p\aa\ inl\"{a}mnad
uppgift om denna skulle f\"{o}rkomma.

\textbf{Observera}: Inl\"{a}mningsuppgiften l\"{a}mnas ej tillbaks. Inget
annat besked \"{a}n underk\"{a}nd/rest/godk\"{a}nd ges.

\bigskip

\subsection{Setup}

Title: Hemtentamen

Submit:Click to Grade

$\limfunc{nplaces}(x,n)=1.0\left\lfloor 10^{n}x+0.5\right\rfloor /10^{n}$

$\limfunc{uniform}=\func{rand}(1000000)/1000000.0$

$\limfunc{svar}:=(\limfunc{accept},\limfunc{reject})$

$\limfunc{rejectarea}:=(\limfunc{outside},\limfunc{inside})$

Points: 10

CSTFile:Help

Choices: Break, Permute

Print Choices: a,b,c,d,e,f,g,h

\section{Question}

\section{Comment}

Grundl\"{a}ggande sannolikhetsl\"{a}ra: Klassisk sannolikhetsdefinition,
Kombinatorik, Betingad sannolikhet.

\section{Variant}

\subsection{Setup}

$n:=\func{rand}(6000,10000)$

$k:=\func{rand}(3,10)$

$\lambda :=\func{rand}(1,100)/10.0$

$p:=\limfunc{nplaces}(\frac{\lambda }{n},5)$

$\limfunc{svar}:=\limfunc{nplaces}(1-\func{PoissonDist}(k-1;\lambda ),3)$

\subsection{Statement}

Ett f\"{o}rs\"{a}kringsbolag s\"{a}ljer ansvarsf\"{o}rs\"{a}kringar. I f\"{o}%
rs\"{a}kringsportf\"{o}ljen finns $%
%TCIMACRO{\FORMULA{n}{n}{evaluate}}%
%BeginExpansion
n%
%EndExpansion
$ f\"{o}rs\"{a}kringar och erfarenheten har visat att sannolikheten f\"{o}r
att en s\aa dan f\"{o}rs\"{a}kring skall drabbas av skada \"{a}r under ett 
\aa r $%
%TCIMACRO{\FORMULA{p}{p}{evaluate}}%
%BeginExpansion
p%
%EndExpansion
$. Best\"{a}m sannolikheten att portf\"{o}ljen ett \aa r drabbas av minst $%
%TCIMACRO{\FORMULA{k}{k}{evaluate}}%
%BeginExpansion
k%
%EndExpansion
$ skador.

\subsection{Solution}

S\"{a}tt%
\begin{eqnarray*}
X_{i} &=&\left\{ 
\begin{array}{cl}
1 & \text{om f\"{o}rs\"{a}kring }i\text{ drabbas av skada} \\ 
0 & \text{om f\"{o}rs\"{a}kring }i\text{ ej drabbas av skada}%
\end{array}%
\right. \quad i=1,2,\ldots ,%
%TCIMACRO{\FORMULA{n}{n}{evaluate} }%
%BeginExpansion
n
%EndExpansion
\\
X &=&\sum_{i=1}^{%
%TCIMACRO{\FORMULA{n}{n}{evaluate}}%
%BeginExpansion
n%
%EndExpansion
}X_{i}=\text{antalet skador}
\end{eqnarray*}%
d\aa\ g\"{a}ller att $X\in Bin(%
%TCIMACRO{\FORMULA{n}{n}{evaluate}}%
%BeginExpansion
n%
%EndExpansion
,%
%TCIMACRO{\FORMULA{p}{p}{evaluate}}%
%BeginExpansion
p%
%EndExpansion
)$. S\"{o}kt sannolikhet kan nu skrivas%
\[
P(X\geq 
%TCIMACRO{\FORMULA{k}{k}{evaluate}}%
%BeginExpansion
k%
%EndExpansion
)=\sum_{i=%
%TCIMACRO{\FORMULA{k}{k}{evaluate}}%
%BeginExpansion
k%
%EndExpansion
}^{%
%TCIMACRO{\FORMULA{n}{n}{evaluate}}%
%BeginExpansion
n%
%EndExpansion
}\binom{%
%TCIMACRO{\FORMULA{n}{n}{evaluate}}%
%BeginExpansion
n%
%EndExpansion
}{i}%
%TCIMACRO{\FORMULA{p}{p}{evaluate}}%
%BeginExpansion
p%
%EndExpansion
^{i}(1-%
%TCIMACRO{\FORMULA{p}{p}{evaluate}}%
%BeginExpansion
p%
%EndExpansion
)^{%
%TCIMACRO{\FORMULA{n}{n}{evaluate}}%
%BeginExpansion
n%
%EndExpansion
-i} 
\]%
men vi hamnar helt utanf\"{o}r binomialf\"{o}rdelningstabellen. Vi
approximerar d\"{a}rf\"{o}r med en Poissonf\"{o}rdelning och erh\aa ller d%
\aa\ att $Bin(%
%TCIMACRO{\FORMULA{n}{n}{evaluate}}%
%BeginExpansion
n%
%EndExpansion
,%
%TCIMACRO{\FORMULA{p}{p}{evaluate}}%
%BeginExpansion
p%
%EndExpansion
)\approx Po(%
%TCIMACRO{\FORMULA{n}{n}{evaluate}}%
%BeginExpansion
n%
%EndExpansion
\times 
%TCIMACRO{\FORMULA{p}{p}{evaluate}}%
%BeginExpansion
p%
%EndExpansion
)=Po(%
%TCIMACRO{\FORMULA{\lambda }{\lambda }{evaluate}}%
%BeginExpansion
\lambda %
%EndExpansion
)$. Tabell ger nu att%
\[
P(X\geq 
%TCIMACRO{\FORMULA{k}{k}{evaluate}}%
%BeginExpansion
k%
%EndExpansion
)=1-%
%TCIMACRO{%
%\FORMULA{\func{PoissonDist}(k-1;\lambda )}{\left( \func{stats}\func{poissonCDF}\left( \lambda \right) \right) \left( k-1\right) }{evaluate}}%
%BeginExpansion
\left( \func{stats}\func{poissonCDF}\left( \lambda \right) \right) \left( k-1\right) %
%EndExpansion
=%
%TCIMACRO{\FORMULA{\limfunc{svar}}{\limfunc{svar}}{evaluate}}%
%BeginExpansion
\limfunc{svar}%
%EndExpansion
\text{.} 
\]

\section{Variant}

\subsection{Setup}

$q_{1}:=\func{rand}(1,5)/10.0$

$q_{2}:=\func{rand}(10q_{1}+2,9)/10.0$

$p:=\func{rand}(35,45)/100.0$

$n:=\func{rand}(1000000,2000000)$

$\limfunc{svar}_{1}:=q_{1}p+q_{2}(1-p)$

$\limfunc{svar}_{2}:=q_{1}pn+q_{2}(1-p)n$

\subsection{Statement}

Varje h\"{o}st \"{a}r det stora kampanjer f\"{o}r att vaccinera sig mot \aa %
rets influensa. Erfarenheten visar att sannolikheten att f\aa\ influensan
efter vaccination \"{a}r $%
%TCIMACRO{\FORMULA{q_{1}}{q_{1}}{evaluate}}%
%BeginExpansion
q_{1}%
%EndExpansion
$ och utan vaccination \"{a}r den $%
%TCIMACRO{\FORMULA{q_{2}}{q_{2}}{evaluate}}%
%BeginExpansion
q_{2}%
%EndExpansion
$. Ett \aa r vaccineras $%
%TCIMACRO{%
%\FORMULA{\left\lfloor 100p\right\rfloor }{\left\lfloor 100p\right\rfloor }{evaluate}}%
%BeginExpansion
\left\lfloor 100p\right\rfloor %
%EndExpansion
$ procent av befolkningen.

\begin{enumerate}
\item Om en person dras helt slumpm\"{a}ssigt vad \"{a}r d\aa\ dennes
sannolikhet att f\aa\ influensan?

\item Vad \"{a}r det f\"{o}rv\"{a}ntade antalet smittade personer om
populationen best\aa r av $%
%TCIMACRO{\FORMULA{n}{n}{evaluate}}%
%BeginExpansion
n%
%EndExpansion
$ individer?

\item \"{A}r h\"{a}ndelserna att 'f\aa\ influensan' och 'bli vaccinerad'
oberoende h\"{a}ndelser?

\item Vad \"{a}r sannolikheten att en person som f\aa r influensan \"{a}r
vaccinerad?
\end{enumerate}

\subsection{Solution}

Nedanst\aa ende samband \"{a}r givna:%
\begin{eqnarray*}
P(\,I &\mid &V\,)=P(\text{Influensa givet Vaccinerad)}=%
%TCIMACRO{\FORMULA{q_{1}}{q_{1}}{evaluate} }%
%BeginExpansion
q_{1}
%EndExpansion
\\
P(\,I &\mid &\overline{V}\,)=P(\text{Influensa givet Ej Vaccinerad})=%
%TCIMACRO{\FORMULA{q_{2}}{q_{2}}{evaluate} }%
%BeginExpansion
q_{2}
%EndExpansion
\\
P(\,\bar{I} &\mid &V\,)=1-P(\,I\mid V\,)=%
%TCIMACRO{\FORMULA{1-q_{1}}{1-q_{1}}{evaluate} }%
%BeginExpansion
1-q_{1}
%EndExpansion
\\
P(\,\bar{I} &\mid &\overline{V}\,)=1-P(\,I\mid \overline{V}\,)=%
%TCIMACRO{\FORMULA{1-q_{2}}{1-q_{2}}{evaluate} }%
%BeginExpansion
1-q_{2}
%EndExpansion
\\
P(V) &=&%
%TCIMACRO{\FORMULA{p}{p}{evaluate}}%
%BeginExpansion
p%
%EndExpansion
\quad P(\overline{V})=%
%TCIMACRO{\FORMULA{1-p}{1-p}{evaluate}}%
%BeginExpansion
1-p%
%EndExpansion
\end{eqnarray*}%
Med hj\"{a}lp av dessa erh\aa lls f\"{o}ljande:

\begin{enumerate}
\item sannolikheten f\"{o}r att en slumpm\"{a}ssigt tagen person f\aa r
influensan \"{a}r%
\begin{eqnarray*}
P(I) &=&P(\,I\mid V\,)P(V)+P(\,I\mid \overline{V}\,)P(\overline{V})= \\
&=&%
%TCIMACRO{\FORMULA{q_{1}}{q_{1}}{evaluate}}%
%BeginExpansion
q_{1}%
%EndExpansion
\times 
%TCIMACRO{\FORMULA{p}{p}{evaluate}}%
%BeginExpansion
p%
%EndExpansion
+%
%TCIMACRO{\FORMULA{q_{2}}{q_{2}}{evaluate}}%
%BeginExpansion
q_{2}%
%EndExpansion
\times (%
%TCIMACRO{\FORMULA{1-p}{1-p}{evaluate}}%
%BeginExpansion
1-p%
%EndExpansion
)=%
%TCIMACRO{\FORMULA{\limfunc{svar}_{1}}{\limfunc{svar}_{1}}{evaluate}}%
%BeginExpansion
\limfunc{svar}_{1}%
%EndExpansion
\text{.}
\end{eqnarray*}

\item det f\"{o}rv\"{a}ntade antalet smittade personer blir $=%
%TCIMACRO{\FORMULA{\limfunc{svar}_{2}}{\limfunc{svar}_{2}}{evaluate}}%
%BeginExpansion
\limfunc{svar}_{2}%
%EndExpansion
$

\item vi m\aa ste testa likheten $P(\,A\mid B\,)=P(A)$ f\"{o}r alla m\"{o}%
jliga kombinationer av\ $V$ och $I$. F\"{o}rst erh\aa lls%
\begin{eqnarray*}
P(\,I &\mid &V\,)=%
%TCIMACRO{\FORMULA{q_{1}}{q_{1}}{evaluate} }%
%BeginExpansion
q_{1}
%EndExpansion
\\
P(I) &=&%
%TCIMACRO{%
%\FORMULA{\limfunc{svar}_{1}}{\left( \limfunc{svar}\right) 1}{evaluate}}%
%BeginExpansion
\left( \limfunc{svar}\right) 1%
%EndExpansion
\end{eqnarray*}%
och eftersom det g\"{a}ller 
\[
P(\,I\mid V\,)\neq P(I) 
\]%
\"{a}r de tv\aa\ h\"{a}ndelserna ej oberoende.

\item s\"{o}kt sannolikhet blir 
\[
P(\,V\mid I\,)=\frac{P(\,I\mid V\,)P(V)}{P(I)}=\frac{%
%TCIMACRO{\FORMULA{q_{1}p}{pq_{1}}{evaluate}}%
%BeginExpansion
pq_{1}%
%EndExpansion
}{%
%TCIMACRO{%
%\FORMULA{q_{1}p+q_{2}(1-p)}{pq_{1}+q_{2}\left( 1-p\right) }{evaluate}}%
%BeginExpansion
pq_{1}+q_{2}\left( 1-p\right) %
%EndExpansion
}=%
%TCIMACRO{%
%\FORMULA{\frac{q_{1}p}{q_{1}p+q_{2}(1-p)}}{p\frac{q_{1}}{pq_{1}+q_{2}\left( 1.0-1.0p\right) }}{evaluatenum}}%
%BeginExpansion
p\frac{q_{1}}{pq_{1}+q_{2}\left( 1.0-1.0p\right) }%
%EndExpansion
\]
\end{enumerate}

\section{Variant}

\subsection{Setup}

$\mu :=\func{rand}(130,140)$

$\sigma :=\func{rand}(10,15)$

$v_{1}:=\func{rand}(110,120)$

$v_{2}:=\func{rand}(150,170)$

$m:=\func{rand}(45,55)$

$n_{1}:=\func{NormalDist}(1.0\times v_{1};\mu ,\sigma )$

$n_{2}:=\func{NormalDist}(1.0\times v_{2};\mu ,\sigma )$

$N:=\frac{m}{n_{1}}$

$x:=N\times n_{2}$

\subsection{Statement}

En kontrollant testar tegelpannors h\aa llfasthet. Dessa skall t\aa l en
vindstyrka som \"{a}r normalf\"{o}rdelad med v\"{a}ntev\"{a}rdet $%
%TCIMACRO{\FORMULA{\mu }{\mu }{evaluate}}%
%BeginExpansion
\mu %
%EndExpansion
$ km/timme och standardavvikelsen $%
%TCIMACRO{\FORMULA{\sigma }{\sigma }{evaluate}}%
%BeginExpansion
\sigma %
%EndExpansion
$ km/timme. Vid ett vindtunnelexperiment simulerades vindhastigheter upp
till $%
%TCIMACRO{\FORMULA{v_{1}}{v_{1}}{evaluate}}%
%BeginExpansion
v_{1}%
%EndExpansion
$ km/timme och d\aa\ gick $%
%TCIMACRO{\FORMULA{m}{m}{evaluate}}%
%BeginExpansion
m%
%EndExpansion
$ tegelpannor s\"{o}nder. Hur m\aa nga tegelpannor ber\"{a}knas klara
vindhastigheter upp till $%
%TCIMACRO{\FORMULA{v_{2}}{v_{2}}{evaluate}}%
%BeginExpansion
v_{2}%
%EndExpansion
$ km/timme?

\subsection{Solution}

S\"{a}tt 
\begin{eqnarray*}
X &=&\text{vindhastighet i km/timme} \\
\mu &=&%
%TCIMACRO{\FORMULA{\mu }{\mu }{evaluate} }%
%BeginExpansion
\mu 
%EndExpansion
\\
\sigma &=&%
%TCIMACRO{\FORMULA{\sigma }{\sigma }{evaluate}}%
%BeginExpansion
\sigma %
%EndExpansion
\end{eqnarray*}%
d\"{a}r $X\in N\left( 
%TCIMACRO{\FORMULA{\mu }{\mu }{evaluatenum}}%
%BeginExpansion
\mu %
%EndExpansion
,%
%TCIMACRO{\FORMULA{\sigma }{\sigma }{evaluatenum}}%
%BeginExpansion
\sigma %
%EndExpansion
\right) $.

Enligt texten \"{a}r h\"{a}ndelserna $\left\{ X\leq 
%TCIMACRO{\FORMULA{v_{1}}{v_{1}}{evaluate}}%
%BeginExpansion
v_{1}%
%EndExpansion
\right\} $ och $\left\{ 
%TCIMACRO{\FORMULA{m}{m}{evaluate}}%
%BeginExpansion
m%
%EndExpansion
\text{ tegelpannor gick s\"{o}nder}\right\} $ ekvivalenta och d\"{a}rmed 
\"{a}r deras sannolikheter lika 
\[
P\left( X\leq 
%TCIMACRO{\FORMULA{v_{1}}{v_{1}}{evaluate}}%
%BeginExpansion
v_{1}%
%EndExpansion
\right) =P\left( 
%TCIMACRO{\FORMULA{m}{m}{evaluate}}%
%BeginExpansion
m%
%EndExpansion
\text{ tegelpannor gick s\"{o}nder}\right) \text{.} 
\]%
Om vi antar att tegelpannorna g\aa r s\"{o}nder oberoende av varandra samt
att de alla har samma sannolikhet att g\aa\ s\"{o}nder (vi antar likformig f%
\"{o}rdelning) s\aa\ g\"{a}ller att 
\[
P\left( 
%TCIMACRO{\FORMULA{m}{m}{evaluate}}%
%BeginExpansion
m%
%EndExpansion
\text{ tegelpannor gick s\"{o}nder}\right) =\frac{g}{m}=\frac{%
%TCIMACRO{\FORMULA{m}{m}{evaluate}}%
%BeginExpansion
m%
%EndExpansion
}{N} 
\]%
d\"{a}r $N$ \"{a}r det totala antalet tegelpannor. Detta ger oss ekvationen 
\[
%TCIMACRO{\FORMULA{n_{1}}{n_{1}}{evaluate}}%
%BeginExpansion
n_{1}%
%EndExpansion
=\frac{%
%TCIMACRO{\FORMULA{m}{m}{evaluate}}%
%BeginExpansion
m%
%EndExpansion
}{N} 
\]%
ty $P\left( X\leq 
%TCIMACRO{\FORMULA{v_{1}}{v_{1}}{evaluate}}%
%BeginExpansion
v_{1}%
%EndExpansion
\right) =%
%TCIMACRO{\FORMULA{n_{1}}{n_{1}}{evaluate}}%
%BeginExpansion
n_{1}%
%EndExpansion
$. Vi finner d\"{a}rf\"{o}r att $N=%
%TCIMACRO{\FORMULA{N}{N}{evaluate}}%
%BeginExpansion
N%
%EndExpansion
$ (sj\"{a}lvklart finns det ej en br\aa kdels-tegelpanna men eventuell
avrundning g\"{o}rs senare).

Om vi nu g\aa r tillbaks till texten s\aa\ finner vi att h\"{a}ndelserna $%
\left\{ X\leq 
%TCIMACRO{\FORMULA{v_{2}}{v_{2}}{evaluate}}%
%BeginExpansion
v_{2}%
%EndExpansion
\right\} $ och $\left\{ x\text{ tegelpannor g\aa r s\"{o}nder}\right\} $ ocks%
\aa\ \"{a}r ekvivalenta och detta ger ekvationen 
\[
P\left( X\leq 
%TCIMACRO{\FORMULA{v_{2}}{v_{2}}{evaluate}}%
%BeginExpansion
v_{2}%
%EndExpansion
\right) =P\left( x\text{ tegelpannor gick s\"{o}nder}\right) 
\]%
d\"{a}r $x$ \"{a}r antalet trasiga tegelpannor. Denna g\aa ng \"{a}r $N$ k%
\"{a}nd och vi erh\aa ller ekvationen

\[
%TCIMACRO{\FORMULA{n_{2}}{n_{2}}{evaluatenum}}%
%BeginExpansion
n_{2}%
%EndExpansion
=\frac{x}{%
%TCIMACRO{\FORMULA{N}{N}{evaluate}}%
%BeginExpansion
N%
%EndExpansion
} 
\]%
varur vi finner 
\[
x=%
%TCIMACRO{\FORMULA{N}{N}{evaluate}}%
%BeginExpansion
N%
%EndExpansion
\times 
%TCIMACRO{\FORMULA{n_{2}}{n_{2}}{evaluatenum}}%
%BeginExpansion
n_{2}%
%EndExpansion
=%
%TCIMACRO{\FORMULA{x}{x}{evaluate}}%
%BeginExpansion
x%
%EndExpansion
\]%
Antalet tegelpannor som rider ut en storm med vindhastigheter p\aa\ upp till 
$%
%TCIMACRO{\FORMULA{v_{2}}{v_{2}}{evaluate}}%
%BeginExpansion
v_{2}%
%EndExpansion
$ km/timme blir d\"{a}rf\"{o}r $N-x=%
%TCIMACRO{\FORMULA{N}{N}{evaluate}}%
%BeginExpansion
N%
%EndExpansion
-%
%TCIMACRO{\FORMULA{x}{x}{evaluate}}%
%BeginExpansion
x%
%EndExpansion
$ vilket korrekt avrundat till $2$ decimalers noggranhet blir $%
%TCIMACRO{%
%\FORMULA{\limfunc{nplaces}(N-x,2)}{%
%\begin{array}{cc}
%\left( \limfunc{nplaces}\right) \left( N-x\right) & 2\limfunc{nplaces}%
%\end{array}}{evaluate}}%
%BeginExpansion
%
\begin{array}{cc}
\left( \limfunc{nplaces}\right) \left( N-x\right) & 2\limfunc{nplaces}%
\end{array}%
%EndExpansion
$.

Detta betyder att $%
%TCIMACRO{%
%\FORMULA{\left\lfloor N-x\right\rfloor }{\left\lfloor N-x\right\rfloor }{evaluate}}%
%BeginExpansion
\left\lfloor N-x\right\rfloor %
%EndExpansion
$ av $%
%TCIMACRO{%
%\FORMULA{\left\lceil N\right\rceil }{\left\lceil N\right\rceil }{evaluate}}%
%BeginExpansion
\left\lceil N\right\rceil %
%EndExpansion
$ tegelpannor kan anses klara en vindstyrka om $%
%TCIMACRO{\FORMULA{v_{2}}{v_{2}}{evaluate}}%
%BeginExpansion
v_{2}%
%EndExpansion
$ km/timme.

\section{Question}

\section{Comment}

Diskreta och kontinuerliga f\"{o}rdelningar

\section{Variant}

\subsection{Setup}

$\lambda :=\func{rand}(450,550)$

$\mu :=\frac{1.0}{\lambda }$

$n:=\func{rand}(5,15)$

$t_{1}:=\func{rand}(10,15)$

$t_{2}:=\func{rand}(1,t_{1}-1)$

\subsection{Statement}

Huddinge sjukhus kirurgiska enhet kan aldrig till\aa tas vara utan
elektricitet. Sjukhusets ledning planerar d\"{a}rf\"{o}r att ink\"{o}pa en
eller fler elgeneratorer som automatiskt kopplas in vid str\"{o}mavbrott.
Ledningen best\"{a}mmer sig f\"{o}r att tolka ordet 'aldrig' som att chansen
f\"{o}r att den kirurgiska avdelningen skall bli utan str\"{o}m f\aa r vara h%
\"{o}gst $1$ procent (av $100$ str\"{o}mavbrott f\aa r h\"{o}gst $1$ bli
totalt).

De modeller p\aa\ elgeneratorer som finns p\aa\ marknaden har alla en
medeltid mellan fel p\aa\ $%
%TCIMACRO{\FORMULA{\lambda }{\lambda }{evaluate}}%
%BeginExpansion
\lambda %
%EndExpansion
$ timmar. Best\"{a}m sannolikheten f\"{o}r ett totalt (\"{a}ven
generatorerna har slutat fungera) str\"{o}mavbrott p\aa\ kirurgiska
avdelningen i f\"{o}ljande fall:

\begin{enumerate}
\item en generator ink\"{o}ps och str\"{o}mmen f\"{o}rsvinner under $%
%TCIMACRO{\FORMULA{t_{1}}{t_{1}}{evaluate}}%
%BeginExpansion
t_{1}%
%EndExpansion
$ timmar en g\aa ng under \aa ret.

\item tv\aa\ generator ink\"{o}ps och str\"{o}mmen f\"{o}rsvinner under $%
%TCIMACRO{\FORMULA{t_{1}}{t_{1}}{evaluate}}%
%BeginExpansion
t_{1}%
%EndExpansion
$ timmar en g\aa ng under \aa ret. B\aa da generatorerna s\"{a}tts ig\aa ng
samtidigt.

\item en generator ink\"{o}ps och str\"{o}mmen f\"{o}rsvinner under $%
%TCIMACRO{\FORMULA{t_{2}}{t_{2}}{evaluate}}%
%BeginExpansion
t_{2}%
%EndExpansion
$ timmar $%
%TCIMACRO{\FORMULA{n}{n}{evaluate}}%
%BeginExpansion
n%
%EndExpansion
$ g\aa nger under \aa ret.

\item tv\aa\ generator ink\"{o}ps och str\"{o}mmen f\"{o}rsvinner under $%
%TCIMACRO{\FORMULA{t_{2}}{t_{2}}{evaluate}}%
%BeginExpansion
t_{2}%
%EndExpansion
$ timmar $%
%TCIMACRO{\FORMULA{n}{n}{evaluate}}%
%BeginExpansion
n%
%EndExpansion
$ g\aa nger under \aa ret. B\aa da generatorerna s\"{a}tts ig\aa ng
samtidigt.
\end{enumerate}

Ledning: Tid mellan fel kan approximeras med en exponentialf\"{o}rdelningen.

\subsection{Solution}

S\"{a}tt%
\begin{eqnarray*}
X_{i} &=&\text{tid mellan fel p\aa\ elgenerator }i\quad i=1,2 \\
\mu &=&\text{felintensiteten}=\text{antal fel per tidsenhet.}
\end{eqnarray*}%
d\"{a}r $X_{i}\in Exp(\mu )$, $i=1,2$ och $\mu =%
%TCIMACRO{\FORMULA{\mu }{\mu }{evaluatenum}}%
%BeginExpansion
\mu %
%EndExpansion
$ \"{a}r generatorernas angivna felintensitet (antal fel per tidsenhet).

\begin{enumerate}
\item Den s\"{o}kta sannolikheten kan skrivas%
\[
P(X_{1}\leq 
%TCIMACRO{\FORMULA{t_{1}}{t_{1}}{evaluate}}%
%BeginExpansion
t_{1}%
%EndExpansion
)=1-e^{-%
%TCIMACRO{\FORMULA{\mu }{\mu }{evaluatenum}}%
%BeginExpansion
\mu %
%EndExpansion
\times 
%TCIMACRO{\FORMULA{t_{1}}{t_{1}}{evaluate}}%
%BeginExpansion
t_{1}%
%EndExpansion
}=%
%TCIMACRO{\FORMULA{1-e^{-\mu t_{1}}}{1-e^{-\mu t_{1}}}{evaluate}}%
%BeginExpansion
1-e^{-\mu t_{1}}%
%EndExpansion
\]

\item Om str\"{o}mmen g\aa r startar vi b\aa da generatorerna p\aa\ en g\aa %
ng. Sannolikheten f\"{o}r att avdelningen skall bli utan str\"{o}m kan d\aa\ %
skrivas%
\[
P(\left\{ X_{1}\leq 
%TCIMACRO{\FORMULA{t_{1}}{t_{1}}{evaluate}}%
%BeginExpansion
t_{1}%
%EndExpansion
\right\} \cap \left\{ X_{2}\leq 
%TCIMACRO{\FORMULA{t_{1}}{t_{1}}{evaluate}}%
%BeginExpansion
t_{1}%
%EndExpansion
\right\} )=P(X_{1}\leq 
%TCIMACRO{\FORMULA{t_{1}}{t_{1}}{evaluate}}%
%BeginExpansion
t_{1}%
%EndExpansion
)P(X_{2}\leq 
%TCIMACRO{\FORMULA{t_{1}}{t_{1}}{evaluate}}%
%BeginExpansion
t_{1}%
%EndExpansion
) 
\]%
ty avdelningen blir utan str\"{o}m f\"{o}rst n\"{a}r b\aa da har fallerat
och generatorerna kan anses fungera oberoende av varandra. S\"{o}kt
sannolikhet kan skrivas%
\[
P(\left\{ X_{1}\leq 
%TCIMACRO{\FORMULA{t_{1}}{t_{1}}{evaluate}}%
%BeginExpansion
t_{1}%
%EndExpansion
\right\} \cap \left\{ X_{2}\leq 
%TCIMACRO{\FORMULA{t_{1}}{t_{1}}{evaluate}}%
%BeginExpansion
t_{1}%
%EndExpansion
\right\} )=%
%TCIMACRO{\FORMULA{1-e^{-\mu t_{1}}}{1-e^{-\mu t_{1}}}{evaluate}}%
%BeginExpansion
1-e^{-\mu t_{1}}%
%EndExpansion
\times 
%TCIMACRO{\FORMULA{1-e^{-\mu t_{1}}}{1-e^{-\mu t_{1}}}{evaluate}}%
%BeginExpansion
1-e^{-\mu t_{1}}%
%EndExpansion
=%
%TCIMACRO{%
%\FORMULA{\left( 1-e^{-\mu t_{1}}\right) ^{2}}{\left( 1-e^{-\mu t_{1}}\right) ^{2}}{evaluate}}%
%BeginExpansion
\left( 1-e^{-\mu t_{1}}\right) ^{2}%
%EndExpansion
\]

\item Sannolikheten f\"{o}r fel vid \textbf{ett} str\"{o}mavbrott om $%
%TCIMACRO{\FORMULA{t_{2}}{t_{2}}{evaluate}}%
%BeginExpansion
t_{2}%
%EndExpansion
$ timmar blir (j\"{a}mf\"{o}r 1. ovan)%
\[
p=P(X_{1}\leq 
%TCIMACRO{\FORMULA{t_{2}}{t_{2}}{evaluate}}%
%BeginExpansion
t_{2}%
%EndExpansion
)=1-e^{-%
%TCIMACRO{\FORMULA{\mu }{\mu }{evaluate}}%
%BeginExpansion
\mu %
%EndExpansion
\times 
%TCIMACRO{\FORMULA{t_{2}}{t_{2}}{evaluate}}%
%BeginExpansion
t_{2}%
%EndExpansion
}=%
%TCIMACRO{\FORMULA{1-e^{-\mu t_{2}}}{1-e^{-\mu t_{2}}}{evaluate}}%
%BeginExpansion
1-e^{-\mu t_{2}}%
%EndExpansion
\text{.} 
\]%
S\"{a}tt%
\[
Y_{j}=\left\{ 
\begin{array}{cl}
1 & \text{utan str\"{o}m tillf\"{a}lle }j \\ 
0 & \text{annars}%
\end{array}%
\right. 
\]%
och $Y=\sum Y_{j}=$ totala antalet fel under de $%
%TCIMACRO{\FORMULA{n}{n}{evaluate}}%
%BeginExpansion
n%
%EndExpansion
$ str\"{o}mavbrotten. Det g\"{a}ller att $Y\in Bin(%
%TCIMACRO{\FORMULA{n}{n}{evaluate}}%
%BeginExpansion
n%
%EndExpansion
,p)$. Den s\"{o}kta sannolikheten kan nu skrivas%
\[
P(Y=0)=\binom{%
%TCIMACRO{\FORMULA{n}{n}{evaluate}}%
%BeginExpansion
n%
%EndExpansion
}{0}\left( 
%TCIMACRO{\FORMULA{1-e^{-\mu t_{2}}}{1-e^{-\mu t_{2}}}{evaluate}}%
%BeginExpansion
1-e^{-\mu t_{2}}%
%EndExpansion
\right) ^{0}(1-%
%TCIMACRO{\FORMULA{1-e^{-\mu t_{2}}}{1-e^{-\mu t_{2}}}{evaluate}}%
%BeginExpansion
1-e^{-\mu t_{2}}%
%EndExpansion
)^{%
%TCIMACRO{\FORMULA{n}{n}{evaluate}}%
%BeginExpansion
n%
%EndExpansion
}=%
%TCIMACRO{\FORMULA{e^{-n\mu t_{2}}}{e^{-n\mu t_{2}}}{evaluate}}%
%BeginExpansion
e^{-n\mu t_{2}}%
%EndExpansion
\]

\item I detta fall erh\aa ller vi%
\[
p=P(\left\{ X_{1}\leq 
%TCIMACRO{\FORMULA{t_{2}}{t_{2}}{evaluate}}%
%BeginExpansion
t_{2}%
%EndExpansion
\right\} \cap \left\{ X_{2}\leq 
%TCIMACRO{\FORMULA{t_{2}}{t_{2}}{evaluate}}%
%BeginExpansion
t_{2}%
%EndExpansion
\right\} )=%
%TCIMACRO{%
%\FORMULA{\left( 1-e^{-\mu t_{2}}\right) ^{2}}{\left( 1-e^{-\mu t_{2}}\right) ^{2}}{evaluate}}%
%BeginExpansion
\left( 1-e^{-\mu t_{2}}\right) ^{2}%
%EndExpansion
\text{.} 
\]%
I \"{o}vrigt som 3. varf\"{o}r%
\[
P(Y=0)=\binom{%
%TCIMACRO{\FORMULA{n}{n}{evaluate}}%
%BeginExpansion
n%
%EndExpansion
}{0}\left( 
%TCIMACRO{%
%\FORMULA{\left( 1-e^{-\mu t_{2}}\right) ^{2}}{\left( 1-e^{-\mu t_{2}}\right) ^{2}}{evaluate}}%
%BeginExpansion
\left( 1-e^{-\mu t_{2}}\right) ^{2}%
%EndExpansion
\right) ^{0}\left( 1-%
%TCIMACRO{%
%\FORMULA{\left( 1-e^{-\mu t_{2}}\right) ^{2}}{\left( 1-e^{-\mu t_{2}}\right) ^{2}}{evaluate}}%
%BeginExpansion
\left( 1-e^{-\mu t_{2}}\right) ^{2}%
%EndExpansion
\right) ^{%
%TCIMACRO{\FORMULA{n}{n}{evaluate}}%
%BeginExpansion
n%
%EndExpansion
}=%
%TCIMACRO{%
%\FORMULA{(1-\left( 1-e^{-\mu t_{2}}\right) ^{2})^{n}}{\left( 1-\left( 1-e^{-\mu t_{2}}\right) ^{2}\right) ^{n}}{evaluate}}%
%BeginExpansion
\left( 1-\left( 1-e^{-\mu t_{2}}\right) ^{2}\right) ^{n}%
%EndExpansion
\text{.} 
\]
\end{enumerate}

\section{Question}

\section{Comment}

Konfidensintervall

\section{Variant}

\subsection{Setup}

$x:=\func{rand}(6400,6700)$

$s:=\func{rand}(3000,3200)$

$\alpha :=\func{rand}(\{0.01,0.02,0.05,0.1,0.2\})$

$n:=\func{rand}(15,29)$

$u:=x-\func{TInv}(0.5\alpha ;n-1)\frac{s}{\sqrt{n}}$

$l:=x+\func{TInv}(0.5\alpha ;n-1)\frac{s}{\sqrt{n}}$

$q:=\limfunc{nplaces}\left( \func{TInv}(1-0.5\alpha ;n-1)\frac{s}{\sqrt{n}}%
,1\right) $

\subsection{Statement}

Ett f\"{o}rs\"{a}kringsbolag beh\"{o}ver veta medelskadan p\aa\ sina
vagnskadef\"{o}rs\"{a}kringar f\"{o}r att kunna s\"{a}tta en korrekt premie.
Tidigare erfarenheter har visat att dessa skadors f\"{o}rdelning kan
approximeras med en normalf\"{o}rdelning. Bolaget tar nu slumpm\"{a}ssigt ut 
$%
%TCIMACRO{\FORMULA{n}{n}{evaluate}}%
%BeginExpansion
n%
%EndExpansion
$ f\"{o}rs\"{a}kringar som drabbats av skador och finner att medelv\"{a}rdet
av deras skador bel\"{o}per sig p\aa\ $%
%TCIMACRO{\FORMULA{x}{x}{evaluate}}%
%BeginExpansion
x%
%EndExpansion
$ kronor med en skattad standardavvikelse om $%
%TCIMACRO{\FORMULA{s}{s}{evaluate}}%
%BeginExpansion
s%
%EndExpansion
$ kronor. Konstruera ett $%
%TCIMACRO{%
%\FORMULA{\left\lfloor 100(1-\alpha )\right\rfloor }{100+\left\lfloor -100\alpha \right\rfloor }{evaluate}}%
%BeginExpansion
100+\left\lfloor -100\alpha \right\rfloor %
%EndExpansion
$ procentigt symmetriskt konfidensintervall f\"{o}r vagnskadeportf\"{o}ljens
medelskada. Redovisa gjorda antaganden.

\subsection{Solution}

S\"{a}tt%
\[
X_{i}=\text{skadestorlek f\"{o}rs\"{a}kring }i\quad i=1,2,\ldots ,%
%TCIMACRO{\FORMULA{n}{n}{evaluate}}%
%BeginExpansion
n%
%EndExpansion
\text{.} 
\]%
d\"{a}r $X_{i}\in N(\mu ,\sigma )$. Vi antar att de olika f\"{o}rs\"{a}%
kringarna drabbas av skada oberoende av varandra. Eftersom vi har ett litet
stickprov ($n<30$) och standardavvikelsen $\sigma $ \"{a}r ok\"{a}nd skall
vi anv\"{a}nda intervallet%
\begin{eqnarray*}
\bar{x}\pm t_{\alpha /2}(n-1)\frac{s}{\sqrt{n}} &=&%
%TCIMACRO{\FORMULA{x}{x}{evaluate}}%
%BeginExpansion
x%
%EndExpansion
\pm t_{%
%TCIMACRO{\FORMULA{1-0.5\alpha }{1.0-0.5\alpha }{evaluatenum}}%
%BeginExpansion
1.0-0.5\alpha %
%EndExpansion
}(%
%TCIMACRO{\FORMULA{n}{n}{evaluate}}%
%BeginExpansion
n%
%EndExpansion
-1)\frac{%
%TCIMACRO{\FORMULA{s}{s}{evaluate}}%
%BeginExpansion
s%
%EndExpansion
}{\sqrt{%
%TCIMACRO{\FORMULA{n}{n}{evaluate}}%
%BeginExpansion
n%
%EndExpansion
}} \\
&=&%
%TCIMACRO{\FORMULA{x}{x}{evaluate}}%
%BeginExpansion
x%
%EndExpansion
\pm 
%TCIMACRO{\FORMULA{q}{q}{evaluate} }%
%BeginExpansion
q
%EndExpansion
\\
&=&\left( 
%TCIMACRO{\FORMULA{l}{l}{evaluatenum}}%
%BeginExpansion
l%
%EndExpansion
,%
%TCIMACRO{\FORMULA{u}{u}{evaluatenum}}%
%BeginExpansion
u%
%EndExpansion
\right) \text{.}
\end{eqnarray*}

\section{Variant}

\subsection{Setup}

$M:=\func{rand}(9000,11000)$

$n_{1}:=\func{rand}(350,450)$

$n_{2}:=n_{1}$

$p_{1}:=\func{rand}(50,70)/1000.0$

$p_{2}:=\func{rand}(15,40)/1000.0$

$\alpha :=\func{rand}(10,50)/1000.0$

$q:=\func{NormalInv}\left( 1-\frac{\alpha }{2.0};0,1\right) $

$l:=\limfunc{nplaces}(p_{1}-p_{2}-q\sqrt{\frac{p_{1}(1-p_{1})}{n_{1}}+\frac{%
p_{2}(1-p_{2})}{n_{2}}},3)$

$u:=\limfunc{nplaces}(p_{1}-p_{2}+q\sqrt{\frac{p_{1}(1-p_{1})}{n_{1}}+\frac{%
p_{2}(1-p_{2})}{n_{2}}},3)$

\subsection{Statement}

En legotillverkare av batterier till mobiltelefoner vill, f\"{o}re ink\"{o}%
p, testa tv\aa\ olika system f\"{o}r kvalitetskontroll. Man anv\"{a}nder de
tv\aa\ systemen p\aa\ tv\aa\ olika men i \"{o}vrigt tekniskt likv\"{a}rdiga
'linor'. En slumpm\"{a}ssigt utvald dag tar man slumpm\"{a}ssigt $%
%TCIMACRO{\FORMULA{n_{1}}{n_{1}}{evaluate}}%
%BeginExpansion
n_{1}%
%EndExpansion
$ batterier fr\aa n lina 1:s produktion och lika m\aa nga fr\aa n lina 2,
total produktion per lina \"{a}r $%
%TCIMACRO{\FORMULA{M}{M}{evaluate}}%
%BeginExpansion
M%
%EndExpansion
$ batterier. Man fann d\aa\ att proportionen defekta batterier blev $%
%TCIMACRO{\FORMULA{p_{1}}{p_{1}}{evaluate}}%
%BeginExpansion
p_{1}%
%EndExpansion
$ p\aa\ lina 1 och $%
%TCIMACRO{\FORMULA{p_{2}}{p_{2}}{evaluate}}%
%BeginExpansion
p_{2}%
%EndExpansion
$ p\aa\ lina 2. Bilda ett symmetriskt $%
%TCIMACRO{%
%\FORMULA{\left\lfloor 100(1-\alpha )\right\rfloor }{\left\lfloor 100-100\alpha \right\rfloor }{evaluate}}%
%BeginExpansion
\left\lfloor 100-100\alpha \right\rfloor %
%EndExpansion
$ procentigt konfidensintervall f\"{o}r skillnaden av proportionen defekta
mellan de tv\aa\ kvalitetskontrollsystemen. Svara med tre decimalers
noggranhet. Vilken rekommendation ger du?

\subsection{Solution}

S\"{a}tt%
\begin{eqnarray*}
X_{1i} &=&\left\{ 
\begin{array}{cl}
1 & \text{batteri }i\text{ lina 1 defekt} \\ 
0 & \text{annars}%
\end{array}%
\right. \\
X_{2j} &=&\left\{ 
\begin{array}{cl}
1 & \text{batteri }j\text{ lina 2 defekt} \\ 
0 & \text{annars}%
\end{array}%
\right.
\end{eqnarray*}%
Det g\"{a}ller nu att $X_{1}=\sum_{i=1}^{%
%TCIMACRO{\FORMULA{n_{1}}{n_{1}}{evaluate}}%
%BeginExpansion
n_{1}%
%EndExpansion
}X_{1i}\in Bin(%
%TCIMACRO{\FORMULA{n_{1}}{n_{1}}{evaluate}}%
%BeginExpansion
n_{1}%
%EndExpansion
,p_{1})$ och $X_{2}=\sum_{i=1}^{%
%TCIMACRO{\FORMULA{n_{2}}{n_{2}}{evaluate}}%
%BeginExpansion
n_{2}%
%EndExpansion
}X_{2i}\in Bin(%
%TCIMACRO{\FORMULA{n_{2}}{n_{2}}{evaluate}}%
%BeginExpansion
n_{2}%
%EndExpansion
,p_{2})$. Centrala gr\"{a}nsv\"{a}rdessatsen ger nu att%
\[
\bar{X}_{1}\approx N\left( p_{1},\frac{p_{1}(1-p_{1})}{n_{1}}\right) \text{
och }\bar{X}_{2}\approx N\left( p_{2},\frac{p_{2}(1-p_{2})}{n_{2}}\right) 
\]%
ty $n_{2}p_{2}=%
%TCIMACRO{\FORMULA{n_{2}}{n_{2}}{evaluate}}%
%BeginExpansion
n_{2}%
%EndExpansion
\times 
%TCIMACRO{\FORMULA{p_{2}}{p_{2}}{evaluate}}%
%BeginExpansion
p_{2}%
%EndExpansion
\approx 
%TCIMACRO{\FORMULA{n_{2}p_{2}}{n_{2}p_{2}}{evaluate}}%
%BeginExpansion
n_{2}p_{2}%
%EndExpansion
\geq 5$ och $n_{2}(1-p_{2})=%
%TCIMACRO{\FORMULA{n_{2}}{n_{2}}{evaluate}}%
%BeginExpansion
n_{2}%
%EndExpansion
\times 
%TCIMACRO{\FORMULA{1-p_{2}}{1-p_{2}}{evaluate}}%
%BeginExpansion
1-p_{2}%
%EndExpansion
\approx 
%TCIMACRO{\FORMULA{n_{2}(1-p_{2})}{n_{2}\left( 1-p_{2}\right) }{evaluate}}%
%BeginExpansion
n_{2}\left( 1-p_{2}\right) %
%EndExpansion
\geq 5$. Motsvarande g\"{a}ller f\"{o}r lina 1.

Ett $%
%TCIMACRO{%
%\FORMULA{\left\lfloor 100(1-\alpha )\right\rfloor }{\left\lfloor 100-100\alpha \right\rfloor }{evaluate}}%
%BeginExpansion
\left\lfloor 100-100\alpha \right\rfloor %
%EndExpansion
$ procentigt symmetriskt konfidensintervall f\"{o}r skillnaden $p_{1}-p_{2}$
kan nu allm\"{a}nt skrivas%
\[
I=\bar{x}_{1}-\bar{x}_{2}\pm \lambda _{\alpha /2}\sqrt{\frac{p_{1}(1-p_{1})}{%
n_{1}}+\frac{p_{2}(1-p_{2})}{n_{2}}}\text{.} 
\]%
H\"{a}rav f\"{o}ljer nu att%
\begin{eqnarray*}
I &=&%
%TCIMACRO{\FORMULA{p_{1}}{p_{1}}{evaluate}}%
%BeginExpansion
p_{1}%
%EndExpansion
-%
%TCIMACRO{\FORMULA{p_{2}}{p_{2}}{evaluate}}%
%BeginExpansion
p_{2}%
%EndExpansion
\pm 
%TCIMACRO{\FORMULA{q}{q}{evaluatenum}}%
%BeginExpansion
q%
%EndExpansion
\sqrt{\frac{%
%TCIMACRO{\FORMULA{p_{1}}{p_{1}}{evaluate}}%
%BeginExpansion
p_{1}%
%EndExpansion
(1-%
%TCIMACRO{\FORMULA{p_{1}}{p_{1}}{evaluate}}%
%BeginExpansion
p_{1}%
%EndExpansion
)}{%
%TCIMACRO{\FORMULA{n_{1}}{n_{1}}{evaluate}}%
%BeginExpansion
n_{1}%
%EndExpansion
}+\frac{%
%TCIMACRO{\FORMULA{p_{2}}{p_{2}}{evaluate}}%
%BeginExpansion
p_{2}%
%EndExpansion
(1-%
%TCIMACRO{\FORMULA{p_{2}}{p_{2}}{evaluate}}%
%BeginExpansion
p_{2}%
%EndExpansion
)}{%
%TCIMACRO{\FORMULA{n_{2}}{n_{2}}{evaluate}}%
%BeginExpansion
n_{2}%
%EndExpansion
}} \\
&=&%
%TCIMACRO{\FORMULA{p_{1}}{p_{1}}{evaluate}}%
%BeginExpansion
p_{1}%
%EndExpansion
-%
%TCIMACRO{\FORMULA{p_{2}}{p_{2}}{evaluate}}%
%BeginExpansion
p_{2}%
%EndExpansion
\pm 
%TCIMACRO{\FORMULA{\frac{u-l}{2}}{0.5u-0.5l}{evaluatenum} }%
%BeginExpansion
0.5u-0.5l
%EndExpansion
\\
&=&(%
%TCIMACRO{\FORMULA{l}{l}{evaluatenum}}%
%BeginExpansion
l%
%EndExpansion
,%
%TCIMACRO{\FORMULA{u}{u}{evaluatenum}}%
%BeginExpansion
u%
%EndExpansion
)\text{.}
\end{eqnarray*}

\section{Question}

\section{Comment}

Klassisk hypotespr\"{o}vning

\subsection{Setup}

$n:=6$

$\mu :=\func{rand}(10,15)$

$\sigma :=\func{rand}(3,10)/10.0$

$\epsilon :=\func{rand}(\{0,1\})$

$\limfunc{normal}=\func{NormalInv}\left( \func{rand}(1000000)/1000000.0;\mu
,\sigma \right) $

$x=\limfunc{nplaces}(\limfunc{normal},1)$

$A:=\left( 
\begin{array}{cccccc}
x+\epsilon & x+\epsilon & x+\epsilon & x+\epsilon & x+\epsilon & x+\epsilon
\\ 
x & x & x & x & x & x%
\end{array}%
\right) $

$D_{1}:=A_{1,1}-A_{2,1}$

$D_{2}:=A_{1,2}-A_{2,2}$

$D_{3}:=A_{1,3}-A_{2,3}$

$D_{4}:=A_{1,4}-A_{2,4}$

$D_{5}:=A_{1,5}-A_{2,5}$

$D_{6}:=A_{1,6}-A_{2,6}$

$m:=\frac{1}{6}(D_{1}+D_{2}+D_{3}+D_{4}+D_{5}+D_{6})$

$s_{2}:=\frac{1}{5}\left(
D_{1}^{2}+D_{2}^{2}+D_{3}^{2}+D_{4}^{2}+D_{5}^{2}+D_{6}^{2}-6m^{2}\right) $

$\alpha :=\func{rand}(\{0.01,0.02,0.05,0.1,0.2\})$

$\tau :=\func{TInv}(1-\frac{\alpha }{2};5)$

$T:=\limfunc{nplaces}\left( \frac{m}{\sqrt{\frac{s_{2}}{6}}},3\right) $

$f:=\left\{ 
\begin{array}{ccc}
2 & if & \limfunc{abs}(T)\geq \tau \\ 
1 & if & \limfunc{abs}(T)<\tau%
\end{array}%
\right. $

\subsection{Statement}

En produktionsingenj\"{o}r vill testa om det f\"{o}religger n\aa gon
skillnad i tids\aa tg\aa ng mellan tv\aa\ metoder, $A$ och $B$, f\"{o}r att
utf\"{o}ra ett visst arbetsmoment. Med hj\"{a}lp av slumpens hj\"{a}lp v\"{a}%
ljs $%
%TCIMACRO{\FORMULA{n}{n}{evaluate}}%
%BeginExpansion
n%
%EndExpansion
$ arbetare vilka f\"{o}rst utf\"{o}r arbetsmomentet enligt metod $A$ och d%
\"{a}refter med metod $B$. Resultatet m\"{a}tt i minuter ges i nedanst\aa %
ende tabell 
\[
\begin{tabular}{l|l|l}
& \multicolumn{2}{|l}{\textbf{Arbetstid}} \\ \cline{2-3}
\textbf{Arbetare} & $A$ & $B$ \\ \hline
\multicolumn{1}{c|}{1} & \multicolumn{1}{|r|}{$%
%TCIMACRO{\FORMULA{A_{1,1}}{A_{1,1}}{evaluate}}%
%BeginExpansion
A_{1,1}%
%EndExpansion
$} & \multicolumn{1}{|r}{$%
%TCIMACRO{\FORMULA{A_{2,1}}{A_{2,1}}{evaluate}}%
%BeginExpansion
A_{2,1}%
%EndExpansion
$} \\ 
\multicolumn{1}{c|}{2} & \multicolumn{1}{|r|}{$%
%TCIMACRO{\FORMULA{A_{1,2}}{A_{1,2}}{evaluate}}%
%BeginExpansion
A_{1,2}%
%EndExpansion
$} & \multicolumn{1}{|r}{$%
%TCIMACRO{\FORMULA{A_{2,2}}{A_{2,2}}{evaluate}}%
%BeginExpansion
A_{2,2}%
%EndExpansion
$} \\ 
\multicolumn{1}{c|}{3} & \multicolumn{1}{|r|}{$%
%TCIMACRO{\FORMULA{A_{1,3}}{A_{1,3}}{evaluate}}%
%BeginExpansion
A_{1,3}%
%EndExpansion
$} & \multicolumn{1}{|r}{$%
%TCIMACRO{\FORMULA{A_{2,3}}{A_{2,3}}{evaluate}}%
%BeginExpansion
A_{2,3}%
%EndExpansion
$} \\ 
\multicolumn{1}{c|}{4} & \multicolumn{1}{|r|}{$%
%TCIMACRO{\FORMULA{A_{1,4}}{A_{1,4}}{evaluate}}%
%BeginExpansion
A_{1,4}%
%EndExpansion
$} & \multicolumn{1}{|r}{$%
%TCIMACRO{\FORMULA{A_{2,4}}{A_{2,4}}{evaluate}}%
%BeginExpansion
A_{2,4}%
%EndExpansion
$} \\ 
\multicolumn{1}{c|}{5} & \multicolumn{1}{|r|}{$%
%TCIMACRO{\FORMULA{A_{1,5}}{A_{1,5}}{evaluate}}%
%BeginExpansion
A_{1,5}%
%EndExpansion
$} & \multicolumn{1}{|r}{$%
%TCIMACRO{\FORMULA{A_{2,5}}{A_{2,5}}{evaluate}}%
%BeginExpansion
A_{2,5}%
%EndExpansion
$} \\ 
\multicolumn{1}{c|}{6} & \multicolumn{1}{|r|}{$%
%TCIMACRO{\FORMULA{A_{1,6}}{A_{1,6}}{evaluate}}%
%BeginExpansion
A_{1,6}%
%EndExpansion
$} & \multicolumn{1}{|r}{$%
%TCIMACRO{\FORMULA{A_{2,6}}{A_{2,6}}{evaluate}}%
%BeginExpansion
A_{2,6}%
%EndExpansion
$}%
\end{tabular}%
\]%
Under antagande om normalf\"{o}rdelning testa p\aa\ signifikansniv\aa n $%
%TCIMACRO{%
%\FORMULA{\left\lfloor 100\alpha \right\rfloor }{\left\lfloor 100\alpha \right\rfloor }{evaluate}}%
%BeginExpansion
\left\lfloor 100\alpha \right\rfloor %
%EndExpansion
$ procent om det f\"{o}religger n\aa gon skillnad mellan\ de tv\aa\ %
metoderna $A$ och $B$.

\subsection{Solution}

S\"{a}tt%
\begin{eqnarray*}
X_{Ai} &=&\text{arbetstid vid metod }A\text{ arbetare }i \\
X_{Bi} &=&\text{arbetstid vid metod }B\text{ arbetare }i
\end{eqnarray*}%
d\"{a}r $X_{Ai}\in N(\mu _{A},\sigma _{A})$ och $X_{Bi}\in N(\mu _{B},\sigma
_{B})$. Bilda den nya variabeln $D_{i}=X_{Ai}-X_{Bi}$ f\"{o}r vilken det g%
\"{a}ller att $D_{i}\in N(\mu _{A}-\mu _{B},\sigma )$ d\"{a}r $\sigma =\sqrt{%
\sigma _{A}^{2}+\sigma _{B}^{2}}$. F\"{o}r denna nya variabel g\"{a}ller f%
\"{o}ljande observerade v\"{a}rden%
\[
\begin{tabular}{l|llllll}
& 1 & 2 & 3 & 4 & 5 & 6 \\ \hline
$d$ & $%
%TCIMACRO{\FORMULA{D_{1}}{D_{1}}{evaluate}}%
%BeginExpansion
D_{1}%
%EndExpansion
$ & $%
%TCIMACRO{\FORMULA{D_{2}}{D_{2}}{evaluate}}%
%BeginExpansion
D_{2}%
%EndExpansion
$ & $%
%TCIMACRO{\FORMULA{D_{3}}{D_{3}}{evaluate}}%
%BeginExpansion
D_{3}%
%EndExpansion
$ & $%
%TCIMACRO{\FORMULA{D_{4}}{D_{4}}{evaluate}}%
%BeginExpansion
D_{4}%
%EndExpansion
$ & $%
%TCIMACRO{\FORMULA{D_{5}}{D_{5}}{evaluate}}%
%BeginExpansion
D_{5}%
%EndExpansion
$ & $%
%TCIMACRO{\FORMULA{D_{6}}{D_{6}}{evaluate}}%
%BeginExpansion
D_{6}%
%EndExpansion
$%
\end{tabular}%
\]%
och f\"{o}r dessa finner vi%
\begin{eqnarray*}
\bar{d} &=&\frac{\left( 
%TCIMACRO{\FORMULA{D_{1}}{D_{1}}{evaluate}}%
%BeginExpansion
D_{1}%
%EndExpansion
\right) +\left( 
%TCIMACRO{\FORMULA{D_{2}}{D_{2}}{evaluate}}%
%BeginExpansion
D_{2}%
%EndExpansion
\right) +\left( 
%TCIMACRO{\FORMULA{D_{3}}{D_{3}}{evaluate}}%
%BeginExpansion
D_{3}%
%EndExpansion
\right) +\left( 
%TCIMACRO{\FORMULA{D_{4}}{D_{4}}{evaluate}}%
%BeginExpansion
D_{4}%
%EndExpansion
\right) +\left( 
%TCIMACRO{\FORMULA{D_{5}}{D_{5}}{evaluate}}%
%BeginExpansion
D_{5}%
%EndExpansion
\right) +\left( 
%TCIMACRO{\FORMULA{D_{6}}{D_{6}}{evaluate}}%
%BeginExpansion
D_{6}%
%EndExpansion
\right) }{6}=%
%TCIMACRO{\FORMULA{m}{m}{evaluate} }%
%BeginExpansion
m
%EndExpansion
\\
s_{d}^{2} &=&\frac{1}{5}\left( \left( 
%TCIMACRO{\FORMULA{D_{1}}{D_{1}}{evaluate}}%
%BeginExpansion
D_{1}%
%EndExpansion
\right) ^{2}+\left( 
%TCIMACRO{\FORMULA{D_{2}}{D_{2}}{evaluate}}%
%BeginExpansion
D_{2}%
%EndExpansion
\right) ^{2}+\left( 
%TCIMACRO{\FORMULA{D_{3}}{D_{3}}{evaluate}}%
%BeginExpansion
D_{3}%
%EndExpansion
\right) ^{2}+\left( 
%TCIMACRO{\FORMULA{D_{4}}{D_{4}}{evaluate}}%
%BeginExpansion
D_{4}%
%EndExpansion
\right) ^{2}+\left( 
%TCIMACRO{\FORMULA{D_{5}}{D_{5}}{evaluate}}%
%BeginExpansion
D_{5}%
%EndExpansion
\right) ^{2}+\left( 
%TCIMACRO{\FORMULA{D_{6}}{D_{6}}{evaluate}}%
%BeginExpansion
D_{6}%
%EndExpansion
\right) ^{2}-6\left( 
%TCIMACRO{\FORMULA{m}{m}{evaluate}}%
%BeginExpansion
m%
%EndExpansion
\right) ^{2}\right) =%
%TCIMACRO{\FORMULA{s_{2}}{s_{2}}{evaluate}}%
%BeginExpansion
s_{2}%
%EndExpansion
\text{.}
\end{eqnarray*}

\begin{description}
\item[\textbf{Steg 1}] Vi skall testa om n\aa gon skillnad mellan metoderna f%
\"{o}religger dvs 
\[
H_{0}:\mu _{A}-\mu _{B}=0\quad H_{1}:\mu _{A}-\mu _{B}\neq 0 
\]

\item[\textbf{Steg 2}] Eftersom vi har ett litet stickprov och normalf\"{o}%
rdelning blir v\aa r testvariabel%
\[
T=\frac{\bar{D}}{s_{D}/\sqrt{n}} 
\]%
och denna \"{a}r $t$-f\"{o}rdelad med $n-1=5$ frihetsgrader.

\item[\textbf{Steg 3}] F\"{o}rkastelseomr\aa det erh\aa lls ur ekvationen%
\begin{eqnarray*}
\alpha &=&P(\text{f\"{o}rkasta }H_{0}\text{ givet }H_{0}\text{ sann}) \\
&=&P\left( T<t_{\alpha /2}(n-1)\text{ och }T>t_{1-\alpha /2}(n-1)\right)
\end{eqnarray*}%
H\"{a}r erh\aa lls $t_{\alpha /2}(n-1)$ till $t_{%
%TCIMACRO{\FORMULA{0.5\times \alpha }{0.5\alpha }{evaluatenum}}%
%BeginExpansion
0.5\alpha %
%EndExpansion
}(5)=%
%TCIMACRO{\FORMULA{\tau }{\tau }{evaluate}}%
%BeginExpansion
\tau %
%EndExpansion
$.

\item[\textbf{Steg 4}] Det observerade v\"{a}rdet \"{a}r%
\[
T_{\text{obs}}=\frac{%
%TCIMACRO{\FORMULA{m}{m}{evaluatenum}}%
%BeginExpansion
m%
%EndExpansion
}{\sqrt{%
%TCIMACRO{\FORMULA{s_{2}}{s_{2}}{evaluatenum}}%
%BeginExpansion
s_{2}%
%EndExpansion
/6}}=%
%TCIMACRO{%
%\FORMULA{\frac{m}{\sqrt{\frac{s_{2}}{6}}}}{\frac{m}{\sqrt{0.166\,67s_{2}}}}{evaluatenum}}%
%BeginExpansion
\frac{m}{\sqrt{0.166\,67s_{2}}}%
%EndExpansion
\]%
och eftersom detta v\"{a}rde ligger $%
%TCIMACRO{%
%\FORMULA{\limfunc{rejectarea}_{f}}{\limfunc{rejectarea}\left( f\right) }{evaluate}}%
%BeginExpansion
\limfunc{rejectarea}\left( f\right) %
%EndExpansion
$ f\"{o}rkastelseomr\aa det $\left\vert T_{\text{obs}}\right\vert >%
%TCIMACRO{\FORMULA{\tau }{\tau }{evaluate}}%
%BeginExpansion
\tau %
%EndExpansion
$ g\"{o}r vi $%
%TCIMACRO{%
%\FORMULA{\limfunc{svar}_{f}}{\limfunc{svar}\left( f\right) }{evaluate}}%
%BeginExpansion
\limfunc{svar}\left( f\right) %
%EndExpansion
$ av nollhypotesen p\aa\ signifikansniv\aa n $%
%TCIMACRO{%
%\FORMULA{\left\lceil 100\alpha \right\rceil }{\left\lceil 100\alpha \right\rceil }{evaluate}}%
%BeginExpansion
\left\lceil 100\alpha \right\rceil %
%EndExpansion
$ procent.
\end{description}

\section{Variant}

\subsection{Setup}

$s:=\func{rand}(150,170)/10.0$

$n:=\func{rand}(10,30)$

$\sigma _{0}:=\func{rand}(12,17)$

$\alpha :=\func{rand}(\{0.10,0.05,0.025,0.01\})$

$\tau :=\func{ChiSquareInv}(1-\alpha ;n-1)$

$T:=\limfunc{nplaces}\left( \frac{(n-1)s^{2}}{\sigma _{0}^{2}},3\right) $

$f:=\left\{ 
\begin{array}{ccc}
2 & if & T<\tau \\ 
1 & if & T\geq \tau%
\end{array}%
\right. $

\subsection{Statement}

Malm\"{o} aviation s\"{a}ger i sin reklam att deras plan mellan Visby och
Bromma har en ankomsttid vars standardavvikelse \"{a}r h\"{o}gst $%
%TCIMACRO{\FORMULA{\sigma _{0}}{\sigma _{0}}{evaluate}}%
%BeginExpansion
\sigma _{0}%
%EndExpansion
$ minuter. Konkurrenten SAS h\"{a}vdar i marknadsdomstolen att detta \"{a}r
fel och att Malm\"{o} aviation skall \aa d\"{o}mas skadest\aa nd f\"{o}r
falsk marknadsf\"{o}ring. SAS baserar sin st\"{a}mning p\aa\ att de slumpm%
\"{a}ssigt valt ut $%
%TCIMACRO{\FORMULA{n}{n}{evaluate}}%
%BeginExpansion
n%
%EndExpansion
$ flygningar samt att man d\aa\ erh\aa llit den observerade
standardavvikelsen $%
%TCIMACRO{\FORMULA{s}{s}{evaluate}}%
%BeginExpansion
s%
%EndExpansion
$ minuter. Hj\"{a}lp domstolen genom att g\"{o}ra ett l\"{a}mpligt test, p%
\aa\ signifikansniv\aa n\ $%
%TCIMACRO{%
%\FORMULA{\left\lfloor 100\alpha \right\rfloor }{\left\lfloor 100\alpha \right\rfloor }{evaluate}}%
%BeginExpansion
\left\lfloor 100\alpha \right\rfloor %
%EndExpansion
$ procent, som avg\"{o}r vilket av de tv\aa\ flygbolagen som har r\"{a}tt.
Dina antaganden skall klart framg\aa .

\subsection{Solution}

S\"{a}tt%
\[
X_{i}=\text{tidsskillnad flygning nummer }i\quad i=1,2,\ldots ,%
%TCIMACRO{\FORMULA{n}{n}{evaluate}}%
%BeginExpansion
n%
%EndExpansion
\]%
och antag att $X_{i}\in N(\mu ,\sigma )$ (ett inte orimligt antagande med
tanke p\aa\ att sena/tidiga ankomster borde vara lika vanliga).

\begin{description}
\item[\textbf{Steg 1}] Vi g\"{o}r ett ensidigt test med $\sigma _{0}=%
%TCIMACRO{\FORMULA{\sigma _{0}}{\sigma _{0}}{evaluate}}%
%BeginExpansion
\sigma _{0}%
%EndExpansion
$ ty Malm\"{o} aviation h\"{a}vdar just denna standardavvikelse. Eftersom
det \"{a}r ett m\aa l som g\"{a}ller falsk marknadsf\"{o}ring och ett falskt
p\aa st\aa ende drabbar konsumenterna v\"{a}ljer domstolen testet%
\[
H_{0}:\sigma >\sigma _{0}=%
%TCIMACRO{\FORMULA{\sigma _{0}}{\sigma _{0}}{evaluate}}%
%BeginExpansion
\sigma _{0}%
%EndExpansion
\quad H_{1}:\sigma \leq 
%TCIMACRO{\FORMULA{\sigma _{0}}{\sigma _{0}}{evaluate}}%
%BeginExpansion
\sigma _{0}%
%EndExpansion
\]%
ty sannolikheten att f\"{o}rkasta denna hypotes, om den \"{a}r sann, kan v%
\"{a}ljas oberoende av antalet m\"{a}tningar. Sannolikheten att p\aa st\aa\ n%
\aa got som missgynnar konsumenten kan d\"{a}rmed v\"{a}ljas och h\aa llas l%
\aa g.

\item[\textbf{Steg 2}] Vi v\"{a}ljer som testvariabel%
\[
T=\frac{(n-1)S^{2}}{\sigma _{0}^{2}} 
\]%
f\"{o}r vilken det g\"{a}ller att $T\in \chi ^{2}(n-1)$.

\item[\textbf{Steg 3}] F\"{o}rkastelseomr\aa det erh\aa lls ur ekvationen%
\begin{eqnarray*}
\alpha &=&P(\text{f\"{o}rkasta }H_{0}\text{ givet }H_{0}\text{ sann}) \\
&=&P\left( T<\chi _{1-\alpha }^{2}(n-1)\right)
\end{eqnarray*}%
d\"{a}r $\chi _{1-\alpha }^{2}(n-1)=\chi _{%
%TCIMACRO{\FORMULA{1-\alpha }{1.0-1.0\alpha }{evaluatenum}}%
%BeginExpansion
1.0-1.0\alpha %
%EndExpansion
}^{2}(%
%TCIMACRO{\FORMULA{n-1}{n-1}{evaluate}}%
%BeginExpansion
n-1%
%EndExpansion
)=%
%TCIMACRO{\FORMULA{\tau }{\tau }{evaluatenum}}%
%BeginExpansion
\tau %
%EndExpansion
$.

\item[\textbf{Steg 4}] Det observerade v\"{a}rdet \"{a}r%
\[
T_{\text{obs}}=\frac{(%
%TCIMACRO{\FORMULA{n}{n}{evaluate}}%
%BeginExpansion
n%
%EndExpansion
-1)%
%TCIMACRO{\FORMULA{s^{2}}{s^{2}}{evaluatenum}}%
%BeginExpansion
s^{2}%
%EndExpansion
}{%
%TCIMACRO{\FORMULA{\sigma _{0}^{2}}{\sigma _{0}^{2}}{evaluatenum}}%
%BeginExpansion
\sigma _{0}^{2}%
%EndExpansion
}=%
%TCIMACRO{\FORMULA{T}{T}{evaluate}}%
%BeginExpansion
T%
%EndExpansion
\]%
och eftersom detta v\"{a}rde ligger $%
%TCIMACRO{%
%\FORMULA{\limfunc{rejectarea}_{f}}{\limfunc{rejectarea}\left( f\right) }{evaluate}}%
%BeginExpansion
\limfunc{rejectarea}\left( f\right) %
%EndExpansion
$ f\"{o}rkastelseomr\aa det $T_{\text{obs}}<%
%TCIMACRO{\FORMULA{\tau }{\tau }{evaluate}}%
%BeginExpansion
\tau %
%EndExpansion
$ g\"{o}r vi $%
%TCIMACRO{%
%\FORMULA{\limfunc{svar}_{f}}{\limfunc{svar}\left( f\right) }{evaluate}}%
%BeginExpansion
\limfunc{svar}\left( f\right) %
%EndExpansion
$ av nollhypotesen p\aa\ signifikansniv\aa n $%
%TCIMACRO{%
%\FORMULA{\left\lfloor 100\alpha \right\rfloor }{\left\lfloor 100\alpha \right\rfloor }{evaluate}}%
%BeginExpansion
\left\lfloor 100\alpha \right\rfloor %
%EndExpansion
$ procent.
\end{description}

\section{Question}

\section{Comment}

Enkel linj\"{a}r regression och logistisk regression.

\section{Variant}

\subsection{Setup}

$n:=10$

$p:=2$

$\sigma :=45$

$\alpha :=\func{rand}(\{0.01,0.02,0.05,0.1\})$

$\beta :=\left( 
\begin{array}{c}
90 \\ 
16%
\end{array}%
\right) $

$X:=\left( 
\begin{array}{cccccccccc}
1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ 
1.1 & 0.9 & 2.0 & 0.7 & 1.2 & 0.8 & 3.1 & 3.2 & 7.0 & 1.3%
\end{array}%
\right) ^{T}$

$\limfunc{normal}=\func{NormalInv}\left( \func{rand}(1000000)/1000000.0;0,%
\sigma \right) $

$\epsilon :=\left( 
\begin{array}{cccccccccc}
\limfunc{normal} & \limfunc{normal} & \limfunc{normal} & \limfunc{normal} & 
\limfunc{normal} & \limfunc{normal} & \limfunc{normal} & \limfunc{normal} & 
\limfunc{normal} & \limfunc{normal}%
\end{array}%
\right) ^{T}$

$y:=X\beta +\epsilon $

$z:=\left( 
\begin{array}{cc}
1 & \func{rand}(2,4)%
\end{array}%
\right) $

$b:=\left( X^{T}X\right) ^{-1}X^{T}y$

$M:=X^{T}X$

$s:=\frac{1}{n-p}\left( y-Xb\right) ^{T}\left( y-Xb\right) $

$v:=s_{1,1}\times M^{-1}$

$\tau :=\limfunc{nplaces}\left( \func{TInv}(1-\frac{\alpha }{2}%
;n-p),3\right) $

$l_{0}:=b_{1,1}-\tau \sqrt{v_{1,1}}$

$r_{0}:=b_{1,1}+\tau \sqrt{v_{1,1}}$

$l_{1}:=b_{2,1}-\tau \sqrt{v_{2,2}}$

$r_{1}:=b_{2,1}+\tau \sqrt{v_{2,2}}$

$q:=s_{1,1}+\left( z\times v\times z^{T}\right) _{1,1}$

$l_{p}:=b_{1,1}+z_{1,2}b_{2,1}-\tau \sqrt{q}$

$r_{p}:=b_{1,1}+z_{1,2}b_{2,1}+\tau \sqrt{q}$

\subsection{Statement}

Ledningen f\"{o}r f\"{o}rs\"{a}kringsbolaget \textit{If\ldots } misst\"{a}%
nker att avst\aa ndet mellan en brandh\"{a}rd och n\"{a}rmaste brandstation 
\"{a}r av betydelse f\"{o}r en brandskada:s storlek ($y)$. F\"{o}r att f\aa\ %
bel\"{a}gg f\"{o}r denna tes tar man ut de senasted $%
%TCIMACRO{\FORMULA{n}{n}{evaluate}}%
%BeginExpansion
n%
%EndExpansion
$ brandskadorna och f\"{o}r dessa tar man reda p\aa\ avst\aa ndet till n\"{a}%
rmaste brandstation m\"{a}tt i mil ($x$). Man erh\"{o}ll d\"{a}rvid f\"{o}%
ljande tabell 
\[
\begin{tabular}{l|rrrrrrrrrr}
$y$ & $%
%TCIMACRO{\FORMULA{y_{1,1}}{y_{1,1}}{evaluate}}%
%BeginExpansion
y_{1,1}%
%EndExpansion
$ & $%
%TCIMACRO{\FORMULA{y_{2,1}}{y_{2,1}}{evaluate}}%
%BeginExpansion
y_{2,1}%
%EndExpansion
$ & $%
%TCIMACRO{\FORMULA{y_{3,1}}{y_{3,1}}{evaluate}}%
%BeginExpansion
y_{3,1}%
%EndExpansion
$ & $%
%TCIMACRO{\FORMULA{y_{4,1}}{y_{4,1}}{evaluate}}%
%BeginExpansion
y_{4,1}%
%EndExpansion
$ & $%
%TCIMACRO{\FORMULA{y_{5,1}}{y_{5,1}}{evaluate}}%
%BeginExpansion
y_{5,1}%
%EndExpansion
$ & $%
%TCIMACRO{\FORMULA{y_{6,1}}{y_{6,1}}{evaluate}}%
%BeginExpansion
y_{6,1}%
%EndExpansion
$ & $%
%TCIMACRO{\FORMULA{y_{7,1}}{y_{7,1}}{evaluate}}%
%BeginExpansion
y_{7,1}%
%EndExpansion
$ & $%
%TCIMACRO{\FORMULA{y_{8,1}}{y_{8,1}}{evaluate}}%
%BeginExpansion
y_{8,1}%
%EndExpansion
$ & $%
%TCIMACRO{\FORMULA{y_{9,1}}{y_{9,1}}{evaluate}}%
%BeginExpansion
y_{9,1}%
%EndExpansion
$ & $%
%TCIMACRO{\FORMULA{y_{10,1}}{y_{10,1}}{evaluate}}%
%BeginExpansion
y_{10,1}%
%EndExpansion
$ \\ 
$x$ & $%
%TCIMACRO{\FORMULA{X_{1,2}}{X_{1,2}}{evaluate}}%
%BeginExpansion
X_{1,2}%
%EndExpansion
$ & $%
%TCIMACRO{\FORMULA{X_{2,2}}{X_{2,2}}{evaluate}}%
%BeginExpansion
X_{2,2}%
%EndExpansion
$ & $%
%TCIMACRO{\FORMULA{X_{3,2}}{X_{3,2}}{evaluate}}%
%BeginExpansion
X_{3,2}%
%EndExpansion
$ & $%
%TCIMACRO{\FORMULA{X_{4,2}}{X_{4,2}}{evaluate}}%
%BeginExpansion
X_{4,2}%
%EndExpansion
$ & $%
%TCIMACRO{\FORMULA{X_{5,2}}{X_{5,2}}{evaluate}}%
%BeginExpansion
X_{5,2}%
%EndExpansion
$ & $%
%TCIMACRO{\FORMULA{X_{6,2}}{X_{6,2}}{evaluate}}%
%BeginExpansion
X_{6,2}%
%EndExpansion
$ & $%
%TCIMACRO{\FORMULA{X_{7,2}}{X_{7,2}}{evaluate}}%
%BeginExpansion
X_{7,2}%
%EndExpansion
$ & $%
%TCIMACRO{\FORMULA{X_{8,2}}{X_{8,2}}{evaluate}}%
%BeginExpansion
X_{8,2}%
%EndExpansion
$ & $%
%TCIMACRO{\FORMULA{X_{9,2}}{X_{9,2}}{evaluate}}%
%BeginExpansion
X_{9,2}%
%EndExpansion
$ & $%
%TCIMACRO{\FORMULA{X_{10,2}}{X_{10,2}}{evaluate}}%
%BeginExpansion
X_{10,2}%
%EndExpansion
$%
\end{tabular}%
\]%
P\aa\ grundval av ekonomiska \"{o}verv\"{a}ganden best\"{a}mde man sig f\"{o}%
r en beslutsos\"{a}kerhet om $%
%TCIMACRO{%
%\FORMULA{\left\lceil 100\times \alpha \right\rceil }{\left\lceil 100\alpha \right\rceil }{evaluate}}%
%BeginExpansion
\left\lceil 100\alpha \right\rceil %
%EndExpansion
$ procent. Best\"{a}m

\begin{enumerate}
\item en l\"{a}mplig modell f\"{o}r kostnaden som funktion av avst\aa ndet.

\item konfidensintervall f\"{o}r de i modellen ing\aa ende parametrarna.

\item ett konfidensintervall f\"{o}r den f\"{o}rv\"{a}ntade kostnaden som
funktion av avst\aa ndet.

\item en prognos f\"{o}r kostnaden om avst\aa ndet \"{a}r $%
%TCIMACRO{\FORMULA{z_{1,2}}{z_{1,2}}{evaluate}}%
%BeginExpansion
z_{1,2}%
%EndExpansion
$ mil.

\item ett prognosintervall f\"{o}r kostnaden om avst\aa ndet \"{a}r $%
%TCIMACRO{\FORMULA{z_{1,2}}{z_{1,2}}{evaluate}}%
%BeginExpansion
z_{1,2}%
%EndExpansion
$ mil.
\end{enumerate}

\subsection{Solution}

En plot av data ger att en enkel line\"{a}r regressionsmodell \"{a}r l\"{a}%
mplig. S\"{a}tt d\"{a}rf\"{o}r%
\[
y_{i}=\beta _{0}+\beta _{1}x_{i}+\epsilon _{i}\quad i=1,2,\ldots ,%
%TCIMACRO{\FORMULA{n}{n}{evaluate}}%
%BeginExpansion
n%
%EndExpansion
\]%
d\"{a}r $\epsilon _{i}\in N(0,\sigma )$ \"{a}r oberoende.

\begin{enumerate}
\item Den s\"{o}kta regressionsekvationen \"{a}r%
\[
y=%
%TCIMACRO{\FORMULA{b_{1,1}}{b_{1,1}}{evaluate}}%
%BeginExpansion
b_{1,1}%
%EndExpansion
+%
%TCIMACRO{\FORMULA{b_{2,1}}{b_{2,1}}{evaluate}}%
%BeginExpansion
b_{2,1}%
%EndExpansion
\times x 
\]

\item Konfidensintervall f\"{o}r interceptet blir%
\begin{eqnarray*}
\beta _{0}\pm t_{1-\alpha /2}(n-p)\times s\sqrt{\frac{1}{n}+\frac{\bar{x}^{2}%
}{\sum (x_{i}-\bar{x})^{2}}} &=&%
%TCIMACRO{\FORMULA{b_{1,1}}{b_{1,1}}{evaluate}}%
%BeginExpansion
b_{1,1}%
%EndExpansion
\pm t_{%
%TCIMACRO{\FORMULA{1-\alpha /2}{1-\frac{1}{2}\alpha }{evaluate}}%
%BeginExpansion
1-\frac{1}{2}\alpha %
%EndExpansion
}(%
%TCIMACRO{\FORMULA{n}{n}{evaluate}}%
%BeginExpansion
n%
%EndExpansion
-%
%TCIMACRO{\FORMULA{p}{p}{evaluate}}%
%BeginExpansion
p%
%EndExpansion
)\times 
%TCIMACRO{\FORMULA{\sqrt{s_{1,1}}}{\sqrt{s_{1,1}}}{evaluatenum}}%
%BeginExpansion
\sqrt{s_{1,1}}%
%EndExpansion
\times 
%TCIMACRO{%
%\FORMULA{\sqrt{M_{1,1}^{-1}}}{\sqrt{\frac{1}{M_{1,1}}}}{evaluatenum} }%
%BeginExpansion
\sqrt{\frac{1}{M_{1,1}}}
%EndExpansion
\\
&=&(%
%TCIMACRO{\FORMULA{l_{0}}{l_{0}}{evaluate}}%
%BeginExpansion
l_{0}%
%EndExpansion
,%
%TCIMACRO{\FORMULA{r_{0}}{r_{0}}{evaluate}}%
%BeginExpansion
r_{0}%
%EndExpansion
)
\end{eqnarray*}

Konfidensintervall f\"{o}r lutningen blir%
\begin{eqnarray*}
\hat{\beta}_{1}\pm t_{1-\alpha /2}(n-p)\times \frac{s}{\sum (x_{i}-\bar{x}%
)^{2}} &=&%
%TCIMACRO{\FORMULA{b_{2,1}}{b_{2,1}}{evaluate}}%
%BeginExpansion
b_{2,1}%
%EndExpansion
\pm t_{%
%TCIMACRO{\FORMULA{1-\alpha /2}{1-\frac{1}{2}\alpha }{evaluate}}%
%BeginExpansion
1-\frac{1}{2}\alpha %
%EndExpansion
}(%
%TCIMACRO{\FORMULA{n}{n}{evaluate}}%
%BeginExpansion
n%
%EndExpansion
-%
%TCIMACRO{\FORMULA{p}{p}{evaluate}}%
%BeginExpansion
p%
%EndExpansion
)\times 
%TCIMACRO{\FORMULA{\sqrt{s_{1,1}}}{\sqrt{s_{1,1}}}{evaluatenum}}%
%BeginExpansion
\sqrt{s_{1,1}}%
%EndExpansion
\times 
%TCIMACRO{%
%\FORMULA{\sqrt{M_{2,2}^{-1}}}{\sqrt{\frac{1}{M_{2,2}}}}{evaluatenum} }%
%BeginExpansion
\sqrt{\frac{1}{M_{2,2}}}
%EndExpansion
\\
&=&\left( 
%TCIMACRO{\FORMULA{l_{1}}{l_{1}}{evaluate}}%
%BeginExpansion
l_{1}%
%EndExpansion
,%
%TCIMACRO{\FORMULA{r_{1}}{r_{1}}{evaluate}}%
%BeginExpansion
r_{1}%
%EndExpansion
\right)
\end{eqnarray*}

\item Konfidensintervall f\"{o}r den f\"{o}rv\"{a}ntade kostnaden som
funktion av $x$ kan skrivas%
\begin{eqnarray*}
I_{\mu (x)} &=&\beta _{0}+\hat{\beta}_{1}x\pm t_{1-\alpha /2}(n-p)\times s%
\sqrt{\left( 
\begin{array}{cc}
1 & x%
\end{array}%
\right) \left( X^{T}X\right) ^{-1}\left( 
\begin{array}{c}
1 \\ 
x%
\end{array}%
\right) } \\
&=&%
%TCIMACRO{\FORMULA{b_{1,1}}{b_{1,1}}{evaluate}}%
%BeginExpansion
b_{1,1}%
%EndExpansion
+%
%TCIMACRO{\FORMULA{b_{2,1}}{b_{2,1}}{evaluate}}%
%BeginExpansion
b_{2,1}%
%EndExpansion
\times x\pm t_{%
%TCIMACRO{\FORMULA{1-\alpha /2}{1-\frac{1}{2}\alpha }{evaluate}}%
%BeginExpansion
1-\frac{1}{2}\alpha %
%EndExpansion
}(%
%TCIMACRO{\FORMULA{n}{n}{evaluate}}%
%BeginExpansion
n%
%EndExpansion
-%
%TCIMACRO{\FORMULA{p}{p}{evaluate}}%
%BeginExpansion
p%
%EndExpansion
)\times \sqrt{%
%TCIMACRO{%
%\FORMULA{\left( %
%\begin{array}{cc}
%1 & x%
%\end{array}\right) v\left( %
%\begin{array}{c}
%1 \\ 
%x%
%\end{array}\right) }{v+vx^{2}}{expand}}%
%BeginExpansion
v+vx^{2}%
%EndExpansion
}
\end{eqnarray*}

\item En prognos av kostnaden n\"{a}r $x=%
%TCIMACRO{\FORMULA{z_{1,2}}{z_{1,2}}{evaluate}}%
%BeginExpansion
z_{1,2}%
%EndExpansion
$ blir%
\[
y(%
%TCIMACRO{\FORMULA{z_{1,2}}{z_{1,2}}{evaluate}}%
%BeginExpansion
z_{1,2}%
%EndExpansion
)=\beta _{0}+\hat{\beta}_{1}x=%
%TCIMACRO{\FORMULA{b_{1,1}}{b_{1,1}}{evaluate}}%
%BeginExpansion
b_{1,1}%
%EndExpansion
+%
%TCIMACRO{\FORMULA{z_{1,2}}{z_{1,2}}{evaluate}}%
%BeginExpansion
z_{1,2}%
%EndExpansion
\times (%
%TCIMACRO{\FORMULA{b_{2,1}}{b_{2,1}}{evaluate}}%
%BeginExpansion
b_{2,1}%
%EndExpansion
)=%
%TCIMACRO{%
%\FORMULA{b_{1,1}+z_{1,2}b_{2,1}}{b_{1,1}+b_{2,1}z_{1,2}}{evaluate}}%
%BeginExpansion
b_{1,1}+b_{2,1}z_{1,2}%
%EndExpansion
\]

\item Ett prognosintervall f\"{o}r kostnaden n\"{a}r $x=%
%TCIMACRO{\FORMULA{z_{1,2}}{z_{1,2}}{evaluate}}%
%BeginExpansion
z_{1,2}%
%EndExpansion
$ blir%
\begin{eqnarray*}
I_{\hat{y}(x)} &=&\beta _{0}+\hat{\beta}_{1}x\pm t_{1-\alpha /2}(n-p)\times s%
\sqrt{1+\left( 
\begin{array}{cc}
1 & x%
\end{array}%
\right) \left( X^{T}X\right) ^{-1}\left( 
\begin{array}{c}
1 \\ 
x%
\end{array}%
\right) } \\
&=&%
%TCIMACRO{\FORMULA{b_{1,1}}{b_{1,1}}{evaluate}}%
%BeginExpansion
b_{1,1}%
%EndExpansion
+(%
%TCIMACRO{\FORMULA{b_{2,1}}{b_{2,1}}{evaluate}}%
%BeginExpansion
b_{2,1}%
%EndExpansion
)%
%TCIMACRO{\FORMULA{z_{1,2}}{z_{1,2}}{evaluate}}%
%BeginExpansion
z_{1,2}%
%EndExpansion
\pm t_{%
%TCIMACRO{\FORMULA{1-\alpha /2}{1-\frac{1}{2}\alpha }{evaluate}}%
%BeginExpansion
1-\frac{1}{2}\alpha %
%EndExpansion
}(%
%TCIMACRO{\FORMULA{n}{n}{evaluate}}%
%BeginExpansion
n%
%EndExpansion
-%
%TCIMACRO{\FORMULA{p}{p}{evaluate}}%
%BeginExpansion
p%
%EndExpansion
)\times \sqrt{%
%TCIMACRO{\FORMULA{q}{q}{evaluate}}%
%BeginExpansion
q%
%EndExpansion
} \\
&=&\left( 
%TCIMACRO{\FORMULA{l_{p}}{l_{p}}{evaluate}}%
%BeginExpansion
l_{p}%
%EndExpansion
,%
%TCIMACRO{\FORMULA{r_{p}}{r_{p}}{evaluate}}%
%BeginExpansion
r_{p}%
%EndExpansion
\right)
\end{eqnarray*}
\end{enumerate}

\section{Variant}

\subsection{Setup}

$n:=14$

$p:=2$

$\sigma :=2.5$

$\beta :=\left( 
\begin{array}{c}
1 \\ 
0.1%
\end{array}%
\right) $

$X:=\left( 
\begin{array}{cccccccccccccc}
1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ 
20 & 80 & 100 & 99 & 125 & 140 & 110 & 99 & 99 & 125 & 130 & 180 & 120 & 200%
\end{array}%
\right) ^{T}$

$\alpha :=\func{rand}(\{0.01,0.02,0.05,0.1\})$

$\limfunc{normal}=\func{NormalInv}\left( \func{rand}(1000000)/1000000.0;0,%
\sigma \right) $

$\epsilon :=\left( 
\begin{array}{cccccccccccccc}
\limfunc{normal} & \limfunc{normal} & \limfunc{normal} & \limfunc{normal} & 
\limfunc{normal} & \limfunc{normal} & \limfunc{normal} & \limfunc{normal} & 
\limfunc{normal} & \limfunc{normal} & \limfunc{normal} & \limfunc{normal} & 
\limfunc{normal} & \limfunc{normal}%
\end{array}%
\right) ^{T}$

$y:=X\beta +\epsilon $

$z:=\left( 
\begin{array}{cc}
1 & \func{rand}(50,250)%
\end{array}%
\right) $

$b:=\left( X^{T}X\right) ^{-1}X^{T}y$

$M:=X^{T}X$

$s:=\frac{1}{n-p}\left( y-Xb\right) ^{T}\left( y-Xb\right) $

$v:=s_{1,1}\times M^{-1}$

$\tau :=\limfunc{nplaces}\left( \func{TInv}(1-\frac{\alpha }{2}%
;n-p),3\right) $

$q:=s_{1,1}+\left( z\times v\times z^{T}\right) _{1,1}$

$l_{p}:=b_{1,1}+z_{1,2}b_{2,1}-\tau \sqrt{q}$

$r_{p}:=b_{1,1}+z_{1,2}b_{2,1}+\tau \sqrt{q}$

$\tau :=\limfunc{nplaces}\left( \func{TInv}(1-\frac{\alpha }{2}%
;n-p),3\right) $

$t:=\limfunc{nplaces}\left( \frac{b_{2,1}}{\sqrt{v_{2,2}}},3\right) $

$f:=\left\{ 
\begin{array}{ccc}
2 & if & \limfunc{abs}(t)\geq \tau \\ 
1 & if & \limfunc{abs}(t)<\tau%
\end{array}%
\right. $

\subsection{Statement}

\textit{Svensk fastighetsbyr\aa } vill veta hur f\"{o}rs\"{a}ljningstiden
(veckor) f\"{o}rh\aa ller sig i f\"{o}rh\aa llande till det satta priset p%
\aa\ fastigheten (i $10000$-tals kronor). Detta f\"{o}r att b\aa de optimera
sin egen men ocks\aa\ kundens vinst. Ledningen misst\"{a}nker att det f\"{o}%
religger ett linj\"{a}rt f\"{o}rh\aa llande och vill testa denna tes. Man
tar d\"{a}rf\"{o}r ut $%
%TCIMACRO{\FORMULA{n}{n}{evaluate}}%
%BeginExpansion
n%
%EndExpansion
$ stycken slutf\"{o}rda f\"{o}rs\"{a}ljningar och noterar det pris som beg%
\"{a}rts samt tiden fr\aa n uppdragets b\"{o}rjan till det aff\"{a}ren var
avslutad. Man erh\"{o}ll d\"{a}rvid f\"{o}ljande tabell%
\[
\begin{tabular}{l|rrrrrrr}
Tid & $%
%TCIMACRO{\FORMULA{y_{1,1}}{y_{1,1}}{evaluate}}%
%BeginExpansion
y_{1,1}%
%EndExpansion
$ & $%
%TCIMACRO{\FORMULA{y_{2,1}}{y_{2,1}}{evaluate}}%
%BeginExpansion
y_{2,1}%
%EndExpansion
$ & $%
%TCIMACRO{\FORMULA{y_{3,1}}{y_{3,1}}{evaluate}}%
%BeginExpansion
y_{3,1}%
%EndExpansion
$ & $%
%TCIMACRO{\FORMULA{y_{4,1}}{y_{4,1}}{evaluate}}%
%BeginExpansion
y_{4,1}%
%EndExpansion
$ & $%
%TCIMACRO{\FORMULA{y_{5,1}}{y_{5,1}}{evaluate}}%
%BeginExpansion
y_{5,1}%
%EndExpansion
$ & $%
%TCIMACRO{\FORMULA{y_{6,1}}{y_{6,1}}{evaluate}}%
%BeginExpansion
y_{6,1}%
%EndExpansion
$ & $%
%TCIMACRO{\FORMULA{y_{7,1}}{y_{7,1}}{evaluate}}%
%BeginExpansion
y_{7,1}%
%EndExpansion
$ \\ 
Pris & $%
%TCIMACRO{\FORMULA{X_{1,2}}{X_{1,2}}{evaluate}}%
%BeginExpansion
X_{1,2}%
%EndExpansion
$ & $%
%TCIMACRO{\FORMULA{X_{2,2}}{X_{2,2}}{evaluate}}%
%BeginExpansion
X_{2,2}%
%EndExpansion
$ & $%
%TCIMACRO{\FORMULA{X_{3,2}}{X_{3,2}}{evaluate}}%
%BeginExpansion
X_{3,2}%
%EndExpansion
$ & $%
%TCIMACRO{\FORMULA{X_{4,2}}{X_{4,2}}{evaluate}}%
%BeginExpansion
X_{4,2}%
%EndExpansion
$ & $%
%TCIMACRO{\FORMULA{X_{5,2}}{X_{5,2}}{evaluate}}%
%BeginExpansion
X_{5,2}%
%EndExpansion
$ & $%
%TCIMACRO{\FORMULA{X_{6,2}}{X_{6,2}}{evaluate}}%
%BeginExpansion
X_{6,2}%
%EndExpansion
$ & $%
%TCIMACRO{\FORMULA{X_{7,2}}{X_{7,2}}{evaluate}}%
%BeginExpansion
X_{7,2}%
%EndExpansion
$ \\ \hline
Tid & $%
%TCIMACRO{\FORMULA{y_{8,1}}{y_{8,1}}{evaluate}}%
%BeginExpansion
y_{8,1}%
%EndExpansion
$ & $%
%TCIMACRO{\FORMULA{y_{9,1}}{y_{9,1}}{evaluate}}%
%BeginExpansion
y_{9,1}%
%EndExpansion
$ & $%
%TCIMACRO{\FORMULA{y_{10,1}}{y_{10,1}}{evaluate}}%
%BeginExpansion
y_{10,1}%
%EndExpansion
$ & $%
%TCIMACRO{\FORMULA{y_{11,1}}{y_{11,1}}{evaluatenum}}%
%BeginExpansion
y_{11,1}%
%EndExpansion
$ & $%
%TCIMACRO{\FORMULA{y_{12,1}}{y_{12,1}}{evaluatenum}}%
%BeginExpansion
y_{12,1}%
%EndExpansion
$ & $%
%TCIMACRO{\FORMULA{y_{13,1}}{y_{13,1}}{evaluatenum}}%
%BeginExpansion
y_{13,1}%
%EndExpansion
$ & $%
%TCIMACRO{\FORMULA{y_{14,1}}{y_{14,1}}{evaluatenum}}%
%BeginExpansion
y_{14,1}%
%EndExpansion
$ \\ 
Pris & $%
%TCIMACRO{\FORMULA{X_{8,2}}{X_{8,2}}{evaluate}}%
%BeginExpansion
X_{8,2}%
%EndExpansion
$ & $%
%TCIMACRO{\FORMULA{X_{9,2}}{X_{9,2}}{evaluate}}%
%BeginExpansion
X_{9,2}%
%EndExpansion
$ & $%
%TCIMACRO{\FORMULA{X_{10,2}}{X_{10,2}}{evaluate}}%
%BeginExpansion
X_{10,2}%
%EndExpansion
$ & $%
%TCIMACRO{\FORMULA{X_{11,2}}{X_{11,2}}{evaluate}}%
%BeginExpansion
X_{11,2}%
%EndExpansion
$ & $%
%TCIMACRO{\FORMULA{X_{12,2}}{X_{12,2}}{evaluate}}%
%BeginExpansion
X_{12,2}%
%EndExpansion
$ & $%
%TCIMACRO{\FORMULA{X_{13,2}}{X_{13,2}}{evaluate}}%
%BeginExpansion
X_{13,2}%
%EndExpansion
$ & $%
%TCIMACRO{\FORMULA{X_{14,2}}{X_{14,2}}{evaluate}}%
%BeginExpansion
X_{14,2}%
%EndExpansion
$%
\end{tabular}%
\]

G\"{o}r en analys av data f\"{o}r att avg\"{o}ra om ledningens misstanke 
\"{a}r befogad samt g\"{o}r en prognos och ett prognosintervall, med
konfidensgraden $%
%TCIMACRO{%
%\FORMULA{\left\lceil 100\times \alpha \right\rceil }{\left\lceil 100\alpha \right\rceil }{evaluate}}%
%BeginExpansion
\left\lceil 100\alpha \right\rceil %
%EndExpansion
$ procent, f\"{o}r f\"{o}rs\"{a}ljningstiden n\"{a}r beg\"{a}rt pris \"{a}r $%
%TCIMACRO{\FORMULA{z_{1,2}}{z_{1,2}}{evaluate}}%
%BeginExpansion
z_{1,2}%
%EndExpansion
$-tusen kronor.

\subsection{Solution}

En plot av data ger att en enkel line\"{a}r regressionsmodell \"{a}r l\"{a}%
mplig. S\"{a}tt d\"{a}rf\"{o}r%
\[
y_{i}=\beta _{0}+\beta _{1}x_{i}+\epsilon _{i}\quad i=1,2,\ldots ,%
%TCIMACRO{\FORMULA{n}{n}{evaluate}}%
%BeginExpansion
n%
%EndExpansion
\]%
d\"{a}r $\epsilon _{i}\in N(0,\sigma )$ \"{a}r oberoende.

\begin{enumerate}
\item Den s\"{o}kta regressionsekvationen \"{a}r%
\[
y=%
%TCIMACRO{\FORMULA{b_{1,1}}{b_{1,1}}{evaluate}}%
%BeginExpansion
b_{1,1}%
%EndExpansion
+%
%TCIMACRO{\FORMULA{b_{2,1}}{b_{2,1}}{evaluate}}%
%BeginExpansion
b_{2,1}%
%EndExpansion
\times x\text{.} 
\]%
F\"{o}r att avg\"{o}ra om denna linj\"{a}ra modell \"{a}r l\"{a}mplig g\"{o}%
rs ett $t$-test

\begin{enumerate}
\item $H_{0}:\beta _{1}=0\quad H_{1}:\beta _{1}\neq 0$

\item Som testvariabel tar vi%
\[
T=\frac{\beta _{1}}{V(\beta _{1})} 
\]%
d\"{a}r $T\in t(%
%TCIMACRO{\FORMULA{n-p}{n-p}{evaluate}}%
%BeginExpansion
n-p%
%EndExpansion
)$.

\item Beslutsregeln erh\aa lls ur ekvationen%
\begin{eqnarray*}
\alpha &=&%
%TCIMACRO{\FORMULA{\alpha }{\alpha }{evaluate}}%
%BeginExpansion
\alpha %
%EndExpansion
=P(\text{f\"{o}rkasta }H_{0}\text{ givet }H_{0}\text{ sann}) \\
&=&P(T>a\text{ eller }T<-a)
\end{eqnarray*}%
vilket ger $a=t_{1-\frac{\alpha }{2}}(%
%TCIMACRO{\FORMULA{n-p}{n-p}{evaluate}}%
%BeginExpansion
n-p%
%EndExpansion
)=%
%TCIMACRO{\FORMULA{\tau }{\tau }{evaluatenum}}%
%BeginExpansion
\tau %
%EndExpansion
$.

\item Observerat \"{a}r $T_{\text{obs}}=%
%TCIMACRO{\FORMULA{t}{t}{evaluatenum}}%
%BeginExpansion
t%
%EndExpansion
$ och eftersom detta v\"{a}rde ligger $%
%TCIMACRO{%
%\FORMULA{\limfunc{rejectarea}_{f}}{\limfunc{rejectarea}\left( f\right) }{evaluate}}%
%BeginExpansion
\limfunc{rejectarea}\left( f\right) %
%EndExpansion
$ f\"{o}rkastelseomr\aa det $\left\vert T_{\text{obs}}\right\vert >%
%TCIMACRO{\FORMULA{\tau }{\tau }{evaluatenum}}%
%BeginExpansion
\tau %
%EndExpansion
$ g\"{o}r vi $%
%TCIMACRO{%
%\FORMULA{\limfunc{svar}_{f}}{\limfunc{svar}\left( f\right) }{evaluate}}%
%BeginExpansion
\limfunc{svar}\left( f\right) %
%EndExpansion
$ av nollhypotesen p\aa\ signifikansniv\aa n $%
%TCIMACRO{%
%\FORMULA{\left\lceil 100\times \alpha \right\rceil }{\left\lceil 100\alpha \right\rceil }{evaluate}}%
%BeginExpansion
\left\lceil 100\alpha \right\rceil %
%EndExpansion
$ procent.
\end{enumerate}

\item En prognos av f\"{o}rs\"{a}ljningstiden n\"{a}r $x=%
%TCIMACRO{\FORMULA{z_{1,2}}{z_{1,2}}{evaluate}}%
%BeginExpansion
z_{1,2}%
%EndExpansion
$ blir%
\[
y(%
%TCIMACRO{\FORMULA{z_{1,2}}{z_{1,2}}{evaluate}}%
%BeginExpansion
z_{1,2}%
%EndExpansion
)=\hat{\beta}_{0}+\hat{\beta}_{1}x=%
%TCIMACRO{\FORMULA{b_{1,1}}{b_{1,1}}{evaluate}}%
%BeginExpansion
b_{1,1}%
%EndExpansion
+%
%TCIMACRO{\FORMULA{z_{1,2}}{z_{1,2}}{evaluate}}%
%BeginExpansion
z_{1,2}%
%EndExpansion
\times 
%TCIMACRO{\FORMULA{b_{2,1}}{b_{2,1}}{evaluate}}%
%BeginExpansion
b_{2,1}%
%EndExpansion
=%
%TCIMACRO{%
%\FORMULA{b_{1,1}+z_{1,2}b_{2,1}}{b_{1,1}+b_{2,1}z_{1,2}}{evaluate}}%
%BeginExpansion
b_{1,1}+b_{2,1}z_{1,2}%
%EndExpansion
\]

\item Ett prognosintervall f\"{o}r f\"{o}rs\"{a}ljningstiden n\"{a}r $x=%
%TCIMACRO{\FORMULA{z_{1,2}}{z_{1,2}}{evaluate}}%
%BeginExpansion
z_{1,2}%
%EndExpansion
$ blir%
\begin{eqnarray*}
I_{\hat{y}(x)} &=&\hat{\beta}_{0}+\hat{\beta}_{1}x\pm t_{1-\alpha
/2}(n-p)\times s\sqrt{1+\left( 
\begin{array}{cc}
1 & x%
\end{array}%
\right) \left( X^{T}X\right) ^{-1}\left( 
\begin{array}{c}
1 \\ 
x%
\end{array}%
\right) } \\
&=&%
%TCIMACRO{\FORMULA{b_{1,1}}{b_{1,1}}{evaluate}}%
%BeginExpansion
b_{1,1}%
%EndExpansion
+(%
%TCIMACRO{\FORMULA{b_{2,1}}{b_{2,1}}{evaluate}}%
%BeginExpansion
b_{2,1}%
%EndExpansion
)%
%TCIMACRO{\FORMULA{z_{1,2}}{z_{1,2}}{evaluate}}%
%BeginExpansion
z_{1,2}%
%EndExpansion
\pm t_{%
%TCIMACRO{\FORMULA{1-\alpha /2}{1-\frac{1}{2}\alpha }{evaluate}}%
%BeginExpansion
1-\frac{1}{2}\alpha %
%EndExpansion
}(%
%TCIMACRO{\FORMULA{n}{n}{evaluate}}%
%BeginExpansion
n%
%EndExpansion
-%
%TCIMACRO{\FORMULA{p}{p}{evaluate}}%
%BeginExpansion
p%
%EndExpansion
)\times \sqrt{%
%TCIMACRO{\FORMULA{q}{q}{evaluate}}%
%BeginExpansion
q%
%EndExpansion
} \\
&=&\left( 
%TCIMACRO{\FORMULA{l_{p}}{l_{p}}{evaluate}}%
%BeginExpansion
l_{p}%
%EndExpansion
,%
%TCIMACRO{\FORMULA{r_{p}}{r_{p}}{evaluate}}%
%BeginExpansion
r_{p}%
%EndExpansion
\right)
\end{eqnarray*}
\end{enumerate}

\section{Variant}

\subsection{Setup}

$\beta _{0}:=-\func{rand}(3,7)/10.0$

$\beta _{1}:=\func{rand}(8,16)/1000.0$

$\beta _{2}:=\func{rand}(35,45)/10.0$

$\beta _{3}:=\func{rand}(10,17)/10.0$

$v:=\func{rand}(1000,2000)/100.0$

$k:=\func{rand}(300,450)$

$x_{1}:=\func{rand}(5,15)$

$x_{2}:=\frac{v}{k}$

$x_{3}:=\func{rand}(\{0,1\})$

\subsection{Statement}

En b\"{o}rsanalytiker vill studera sannolikheten f\"{o}r att en aktie g\aa r
upp under veckan, efter en bokslutskommunik\'{e}, som en funktion av vissa
bakgrundsv\"{a}rden ur kommunik\'{e}n. Utifr\aa n gamla data skattar han%
\[
P(\text{h\"{o}gre kurs})=\frac{1}{1+e^{%
%TCIMACRO{\FORMULA{\beta _{0}}{\beta _{0}}{evaluate}}%
%BeginExpansion
\beta _{0}%
%EndExpansion
-%
%TCIMACRO{\FORMULA{\beta _{1}}{\beta _{1}}{evaluate}}%
%BeginExpansion
\beta _{1}%
%EndExpansion
x_{1}-%
%TCIMACRO{\FORMULA{\beta _{2}}{\beta _{2}}{evaluate}}%
%BeginExpansion
\beta _{2}%
%EndExpansion
x_{2}-%
%TCIMACRO{\FORMULA{\beta _{3}}{\beta _{3}}{evaluate}}%
%BeginExpansion
\beta _{3}%
%EndExpansion
x_{3}}} 
\]%
d\"{a}r $x_{1}=$ soliditeten i procent, $x_{2}=$ vinsten/b\"{o}rsv\"{a}rdet
och $x_{3}=1$ om bolaget aviserat n\aa gon form av emission eller split och $%
x_{3}=0$ annars.

\begin{enumerate}
\item Vad \"{a}r sannolikheten att kursen g\aa r upp f\"{o}r ett bolag med
soliditeten $%
%TCIMACRO{\FORMULA{x_{1}}{x_{1}}{evaluate}}%
%BeginExpansion
x_{1}%
%EndExpansion
$ procent, vinsten per aktie $%
%TCIMACRO{\FORMULA{v}{v}{evaluate}}%
%BeginExpansion
v%
%EndExpansion
$, b\"{o}rskursen $%
%TCIMACRO{\FORMULA{k}{k}{evaluate}}%
%BeginExpansion
k%
%EndExpansion
$ SEK och $x_{3}=%
%TCIMACRO{\FORMULA{x_{3}}{x_{3}}{evaluate}}%
%BeginExpansion
x_{3}%
%EndExpansion
$?

\item Avg\"{o}r, genom att derivera uttrycket ovan, hur mycket st\"{o}rre
sannolikheten hade varit om soliditeten varit $1$ procentenhet st\"{o}rre.
\end{enumerate}

\subsection{Solution}

Den bakomliggande modellen \"{a}r en logistisk regressionsmodell%
\[
Y=E\!\left( \,Y\mid x_{1},x_{2},x_{3}\,\right) +\epsilon =p+\epsilon 
\]%
d\"{a}r $p=\frac{1}{1+e^{%
%TCIMACRO{\FORMULA{\beta _{0}}{\beta _{0}}{evaluate}}%
%BeginExpansion
\beta _{0}%
%EndExpansion
-%
%TCIMACRO{\FORMULA{\beta _{1}}{\beta _{1}}{evaluate}}%
%BeginExpansion
\beta _{1}%
%EndExpansion
x_{1}-%
%TCIMACRO{\FORMULA{\beta _{2}}{\beta _{2}}{evaluate}}%
%BeginExpansion
\beta _{2}%
%EndExpansion
x_{2}-%
%TCIMACRO{\FORMULA{\beta _{3}}{\beta _{3}}{evaluate}}%
%BeginExpansion
\beta _{3}%
%EndExpansion
x_{3}}}$. Den s\"{o}kta sannolikheten f\"{o}r $x_{1}=%
%TCIMACRO{\FORMULA{x_{1}}{x_{1}}{evaluate}}%
%BeginExpansion
x_{1}%
%EndExpansion
$, $x_{2}=%
%TCIMACRO{\FORMULA{x_{2}}{x_{2}}{evaluate}}%
%BeginExpansion
x_{2}%
%EndExpansion
$ och $x_{3}=%
%TCIMACRO{\FORMULA{x_{3}}{x_{3}}{evaluate}}%
%BeginExpansion
x_{3}%
%EndExpansion
$ blir%
\begin{eqnarray*}
p &=&\frac{1}{1+e^{-\beta _{0}-\beta _{1}x_{1}-\beta _{2}x_{2}-\beta
_{3}x_{3}}} \\
&=&\frac{1}{1+e^{%
%TCIMACRO{\FORMULA{\beta _{0}}{\beta _{0}}{evaluate}}%
%BeginExpansion
\beta _{0}%
%EndExpansion
-%
%TCIMACRO{\FORMULA{\beta _{1}}{\beta _{1}}{evaluate}}%
%BeginExpansion
\beta _{1}%
%EndExpansion
\times 
%TCIMACRO{\FORMULA{x_{1}}{x_{1}}{evaluate}}%
%BeginExpansion
x_{1}%
%EndExpansion
-%
%TCIMACRO{\FORMULA{\beta _{2}}{\beta _{2}}{evaluate}}%
%BeginExpansion
\beta _{2}%
%EndExpansion
\times 
%TCIMACRO{\FORMULA{x_{2}}{x_{2}}{evaluate}}%
%BeginExpansion
x_{2}%
%EndExpansion
-%
%TCIMACRO{\FORMULA{\beta _{3}}{\beta _{3}}{evaluate}}%
%BeginExpansion
\beta _{3}%
%EndExpansion
\times 
%TCIMACRO{\FORMULA{x_{3}}{x_{3}}{evaluate}}%
%BeginExpansion
x_{3}%
%EndExpansion
}} \\
&=&\frac{1}{1+e^{%
%TCIMACRO{%
%\FORMULA{-\beta _{0}-\beta _{1}x_{1}-\beta _{2}x_{2}-\beta _{3}x_{3}}{-\beta _{0}-\beta _{1}x_{1}-\beta _{2}x_{2}-\beta _{3}x_{3}}{evaluate}}%
%BeginExpansion
-\beta _{0}-\beta _{1}x_{1}-\beta _{2}x_{2}-\beta _{3}x_{3}%
%EndExpansion
}} \\
&=&%
%TCIMACRO{%
%\FORMULA{\limfunc{nplaces}\left( \frac{1}{1+e^{-\beta _{0}-\beta _{1}x_{1}-\beta _{2}x_{2}-\beta _{3}x_{3}}},3\right) }{\left( \frac{\limfunc{nplaces}}{1+\exp \left( -\beta _{0}-\beta _{1}x_{1}-\beta _{2}x_{2}-\beta _{3}x_{3}\right) },3\limfunc{nplaces}\right) \allowbreak }{evaluate}}%
%BeginExpansion
\left( \frac{\limfunc{nplaces}}{1+\exp \left( -\beta _{0}-\beta _{1}x_{1}-\beta _{2}x_{2}-\beta _{3}x_{3}\right) },3\limfunc{nplaces}\right) \allowbreak %
%EndExpansion
\end{eqnarray*}

Derivatan av sannolikheten med avseende p\aa\ soliditeten blir%
\begin{eqnarray*}
\frac{\partial p}{\partial x_{1}} &=&\frac{\partial }{\partial x_{1}}\frac{1%
}{1+e^{-\alpha -\beta _{1}x_{1}-\beta _{2}x_{2}-\beta _{3}x_{3}}} \\
&=&\beta _{1}\frac{e^{-\alpha -\beta _{1}x_{1}-\beta _{2}x_{2}-\beta
_{3}x_{3}}}{\left( 1+e^{-\alpha -\beta _{1}x_{1}-\beta _{2}x_{2}-\beta
_{3}x_{3}}\right) ^{2}} \\
&=&%
%TCIMACRO{\FORMULA{\beta _{1}}{\beta _{1}}{evaluate}}%
%BeginExpansion
\beta _{1}%
%EndExpansion
\frac{e^{%
%TCIMACRO{%
%\FORMULA{-\beta _{0}-\beta _{1}x_{1}-\beta _{2}x_{2}-\beta _{3}x_{3}}{-\beta _{0}-x_{1}\beta _{1}-x_{2}\beta _{2}-x_{3}\beta _{3}}{evaluate}}%
%BeginExpansion
-\beta _{0}-x_{1}\beta _{1}-x_{2}\beta _{2}-x_{3}\beta _{3}%
%EndExpansion
}}{\left( 1+e^{%
%TCIMACRO{%
%\FORMULA{-\beta _{0}-\beta _{1}x_{1}-\beta _{2}x_{2}-\beta _{3}x_{3}}{-\beta _{0}-x_{1}\beta _{1}-x_{2}\beta _{2}-x_{3}\beta _{3}}{evaluate}}%
%BeginExpansion
-\beta _{0}-x_{1}\beta _{1}-x_{2}\beta _{2}-x_{3}\beta _{3}%
%EndExpansion
}\right) ^{2}} \\
&=&%
%TCIMACRO{%
%\FORMULA{\limfunc{nplaces}\left( \beta _{1}\frac{e^{-\beta _{0}-\beta _{1}x_{1}-\beta _{2}x_{2}-\beta _{3}x_{3}}}{\left( 1+e^{-\beta _{0}-\beta _{1}x_{1}-\beta _{2}x_{2}-\beta _{3}x_{3}}\right) ^{2}},3\right) }{\left( %
%\begin{array}{cc}
%\left( \limfunc{nplaces}\right) \beta _{1}\frac{\exp \left( -1.0\beta _{0}-1.0x_{1}\beta _{1}-1.0x_{2}\beta _{2}-1.0x_{3}\beta _{3}\right) }{\left( \exp \left( -1.0\beta _{0}-1.0x_{1}\beta _{1}-1.0x_{2}\beta _{2}-1.0x_{3}\beta _{3}\right) +1.0\right) ^{2}} & 3.0\limfunc{nplaces}%
%\end{array}\right) \allowbreak }{evaluatenum}}%
%BeginExpansion
\left( %
\begin{array}{cc}
\left( \limfunc{nplaces}\right) \beta _{1}\frac{\exp \left( -1.0\beta _{0}-1.0x_{1}\beta _{1}-1.0x_{2}\beta _{2}-1.0x_{3}\beta _{3}\right) }{\left( \exp \left( -1.0\beta _{0}-1.0x_{1}\beta _{1}-1.0x_{2}\beta _{2}-1.0x_{3}\beta _{3}\right) +1.0\right) ^{2}} & 3.0\limfunc{nplaces}%
\end{array}\right) \allowbreak %
%EndExpansion
\end{eqnarray*}%
Om soliditeten var en procent st\"{o}rre hade sannolikheten \"{o}kat med $%
%TCIMACRO{%
%\FORMULA{\limfunc{nplaces}\left( \beta _{1}\frac{e^{-\beta _{0}-\beta _{1}x_{1}-\beta _{2}x_{2}-\beta _{3}x_{3}}}{\left( 1+e^{-\beta _{0}-\beta _{1}x_{1}-\beta _{2}x_{2}-\beta _{3}x_{3}}\right) ^{2}},3\right) }{\left( %
%\begin{array}{cc}
%\left( \limfunc{nplaces}\right) \beta _{1}\frac{\exp \left( -1.0\beta _{0}-1.0x_{1}\beta _{1}-1.0x_{2}\beta _{2}-1.0x_{3}\beta _{3}\right) }{\left( \exp \left( -1.0\beta _{0}-1.0x_{1}\beta _{1}-1.0x_{2}\beta _{2}-1.0x_{3}\beta _{3}\right) +1.0\right) ^{2}} & 3.0\limfunc{nplaces}%
%\end{array}\right) \allowbreak }{evaluatenum}}%
%BeginExpansion
\left( %
\begin{array}{cc}
\left( \limfunc{nplaces}\right) \beta _{1}\frac{\exp \left( -1.0\beta _{0}-1.0x_{1}\beta _{1}-1.0x_{2}\beta _{2}-1.0x_{3}\beta _{3}\right) }{\left( \exp \left( -1.0\beta _{0}-1.0x_{1}\beta _{1}-1.0x_{2}\beta _{2}-1.0x_{3}\beta _{3}\right) +1.0\right) ^{2}} & 3.0\limfunc{nplaces}%
\end{array}\right) \allowbreak %
%EndExpansion
$ enheter.

\end{document}