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\newcommand{\ch}{Chapitre 13}
\newcommand{\Ch}{Des stats inf\'erentielles pour vous porter bonheur !}	
\newcommand{\num}{13}	
\newcommand{\Annee}{2012-2013}

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%\vspace{3cm}

\begin{act}
 En Syldavie, Norbert fait des études statistiques sur le daltonisme des éléphants syldaves pratiquant le saut en parachute.\\
 Il pense que $46\%$ d'entre eux sont des mâles et $18\%$ des éléphants ont plus de $60$ ans.\\
 Il a prélevé un échantillon de $400$ éléphants parmi lesquels $195$ sont des mâles et $313$ ont moins de $60$ ans.\\
 \textit{L'échantillon a été réalisé par un tirage au hasard et il peut être assimilé à un tirage avec remise.}
 \begin{enumerate}
  \item L'échantillon de Norbert est-il représentatif ?
  \item L'étude de Norbert montre que dans cet échantillon, $29\%$ des éléphants sont daltoniens.\\
  Estimer la proportion d'éléphants daltoniens dans la population totale.
 \end{enumerate}
\end{act}

\vspace{3cm}


\begin{act}
On reprend le contexte de l'activité précédente.\\
Norbert veut effectuer une étude plus précise avec un échantillon plus grand.\\
 Il a un échantillon de $1200$ éléphants parmi lesquels $562$ sont des mâles et $951$ ont moins de $60$ ans.
 \begin{enumerate}
  \item \begin{enumerate}
         \item Pourquoi aimeront-on trouver une autre méthode que celles employées précédemment pour savoir si 
  l'échantillon de Norbert est représentatif ?
 \item Quel nouvel outil de Terminale pourrait-on utiliser ?
 \item Proposer un nouvel intervalle de fluctuation au seuil de $95\%$. \\
 Quelle différence a-t-il avec les précédents ? Quels sont ses avantages ? ses inconvénients ?
        \end{enumerate}
  \item L'étude de Norbert montre que dans cet échantillon, $32\%$ des éléphants sont daltoniens.\\
  Estimer la proportion d'éléphants daltoniens dans la population totale.
 \end{enumerate}
\end{act}

\vspace{3cm}
 
 \Cadre[Conclusion]{Soit $p$ la probabilité d'apparition d'un caractère dans une population totale.
\begin{itemize}
 \item \textbf{Lorsque l'on connait $p$}, ou l'on fait une hypothèse sur la valeur de $p$, 
  on utilise un \textbf{intervalle de \ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots
 à $95\%$} pour estimer la
 fréquence d'apparition $f$ du caractère dans un échantillon de taille $n$.
 $$\text{Si } \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad I=\Big[ \ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots , \ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots \Big]$$
 $$\text{ ou } \qquad\qquad\qquad\qquad
 I'=\Big[ \ldots\ldots\ldots , \ldots\ldots\ldots \Big]$$ 
  où pour $X\hookrightarrow \ldots\ldots\ldots\ldots\ldots\ldots\ldots$ et :
  \begin{itemize}
   \item $a$ désigne le plus petit entier $k$ à partir duquel $P(X\leq k)$ dépasse $\ldots\ldots\ldots$
   \item $b$ désigne le plus petit entier $k$ à partir duquel $P(X\leq k)$ dépasse $\ldots\ldots\ldots$
  \end{itemize}
  \item \textbf{Lorsque l'on ne connait pas $p$}, 
    on utilise un \textbf{intervalle de \ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots
  à $95\%$} pour estimer la valeur de $p$ à partir de la fréquence $f$ d'apparition du caractère 
  dans un échantillon de taille $n$.

 $$\text{Si } \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad J=\Big[ \ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots , \ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots \Big] $$
\end{itemize}
}
\newpage

\textbf{Question 1 :}  L'échantillon de Norbert est-il représentatif ?

\begin{center}
 \begin{minipage}{9cm}
  \includegraphics[scale=0.8]{Act1_ProgEchantillons.eps}
 \end{minipage}
 \begin{minipage}{9cm}
  \includegraphics[scale=0.8]{Act1_Echantillons.eps}
 \end{minipage}
\end{center}

\textbf{\underline{Observations} : }

\vspace{6cm}

\Cadre[Critère de décision]{
  Pour une étude statistique sur certains caractères connus de la population 
  (âge, sexe, taille, etc), on consi\-dère qu'un échantillon est \textbf{représentatif}, si 
  \vspace{4cm}
  }

\newpage

\Rqs{
\item Il s'agit d'un intervalle dans lequel se situent $95\%$ des échantillons établis dans ces conditions.\\
Cet intervalle dépend évidemment de la taille $n$ de l'échantillon et de la probabilité $p$ d'apparition 
  du caractère dans la population.
  \item Dans la pratique, on connait $n$ et on peut faire en sorte que notre échantillon soit établi au hasard.\\
  Par contre, pour $p$, à moins d'avoir regarder la population entière (et dans ce cas, ce chapitre n'a plus d'intérêt)
  il s'agit d'une hypothèse.\\
  Cet intervalle de fluctuation servira donc surtout à vérifier la crédibilité d'une hypothèse sur $p$.
\item Ainsi, si la fréquence observée n'appartient pas à l'intervalle de fluctuation au seuil de $95\%$, on
considérera que l'échantillon n'est pas représentatif, ie non établi au hasard, 
ou encore que l'hypothèse de départ sur $p$ est \textbf{mauvaise} et on la rejettera.\\
On a un risque de se tromper dans $5\%$ des cas, puisque $5\%$ des fréquences établies dans ces conditions 
ne sont pas dans cet intervalle. On a donc refusé $5\%$ d'échantillons \og en trop \fg. \\
Par contre, si $f$ est dans l'intervalle de fluctuation au seuil de $95\%$, on considérera l'échantillon comme
représentatif, ie établi au hasard, ou encore que notre hypothèse sur $p$ est \textbf{crédible}, 
sans connaître le risque d'erreur, ie le nombre d'échantillons acceptés \og en trop \fg.\\
Dans tous les cas, on est sûr de rien ! Donc on évitera le vocabulaire \og vrai \fg ou \og faux \fg.
\item Tout est une question d'équilibre : \\
Si l'on veut diminuer l'erreur de rejet, par exemple en prenant un intervalle de fluctuation au seuil de $100\%$, 
on ne rejettera donc aucun échantillon \og en trop \fg, par contre, on les acceptera tous, donc évidemment, beaucoup trop.
Toute hypothèse sur $p$ semblera crédible, ce qui n'a aucun intérêt.\\
Ainsi, dans la pratique, on utilise surtout les seuils de $95\%$ et de $99\%$. 
}

\vspace{1cm}

\Cadre[En seconde]{
 Pour un caractère donné, on note $p$ sa probabilité d'apparition et $f$ sa fréquence d'apparition dans un échantillon.\\
 La fréquence $f$ se situe dans au moins $95\%$ des cas dans l'intervalle 
 $$\Big[ \ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots , \ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots \Big]$$
 \begin{itemize}
  \item[\textbullet] \textbf{Avantages} : \\\\
  \item[\textbullet] \textbf{Inconvénients} :  \\\\
  \textit{Sans parler du fait que cet intervalle
 n'a jamais été justifié par votre enseignant ...}
 \end{itemize}
}

\textbf{\underline{Applications} : }

\newpage

\Cadre[En Première]{
  On appelle $X$ la variable aléatoire qui compte le nombre de personnes d'un échantillon de taille $n$,
  possèdant un certain caratère, de probabilité d'apparition $p$. On a donc $X\hookrightarrow B(n,p)$\\
  On note $f=\dfrac{X}{n}$ la fréquence d'apparition de caractère.\\
  La fréquence $f$ se situe dans au moins $95\%$ des cas dans l'intervalle 
  $\Big[ \ldots\ldots\ldots , \ldots\ldots\ldots \Big]$ ,
  où :\\
  \begin{itemize}
   \item $a$ désigne le plus petit entier $k$ à partir duquel $P(X\leq k)$ dépasse $\ldots\ldots\ldots$.
   \item $b$ désigne le plus petit entier $k$ à partir duquel $P(X\leq k)$ dépasse $\ldots\ldots\ldots$.
  \end{itemize}
  \begin{itemize}
  \item[\textbullet] \textbf{Avantages} :\\\\
  \textit{Sans parler du fait que cet intervalle
  est justifié par vos connaissances sur la loi binomiale ...}
  \item[\textbullet] \textbf{Inconvénients} : \\\\
   \end{itemize}
}

%\vspace{1cm}

\textbf{\underline{Applications} : } $X \hookrightarrow $\\

%\begin{center}
\begin{tabular}{*{2}{|>{\centering}m{1.5cm}}|}\hline
 $k$ & $P(X\leq k)$\tabularnewline\hline
 $\ldots$ & \ldots \tabularnewline\hline
 54 & 0.00953 \tabularnewline\hline
 55 & 0.01375 \tabularnewline\hline
 56 & 0.01944 \tabularnewline\hline
 57 & 0.02699 \tabularnewline\hline
 58 & 0.03678 \tabularnewline\hline
 59 & 0.04924  \tabularnewline\hline
 $\ldots$ & \ldots \tabularnewline\hline
\end{tabular} 
\qquad 
\begin{tabular}{*{2}{|>{\centering}m{1.5cm}}|}\hline
 $k$ & $P(X\leq k)$\tabularnewline\hline
 $\ldots$ & \ldots \tabularnewline\hline
 83 & 0.94587 \tabularnewline\hline
 84 & 0.94587 \tabularnewline\hline
 85 & 0.95825 \tabularnewline\hline
 86 & 0.96821 \tabularnewline\hline
 87 & 0.97609 \tabularnewline\hline
 88 & 0.98225 \tabularnewline\hline
 $\ldots$ & \ldots \tabularnewline\hline
\end{tabular} 
\quad 
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\end{minipage}


%\end{center}






\vspace{2.5cm}

\textbf{Question 2 :} L'étude de Norbert montre que dans cet échantillon, $29\%$ des éléphants sont daltoniens.\\
  Estimer la proportion d'éléphants daltoniens dans la population totale.

\Cadre[En Seconde]{
Pour un caractère donné, on note $f$ sa fréquence d'apparition dans un échantillon de taille $n$.\\
 Sa probabilité $p$ d'apparition dans la population totale se situe dans au moins $95\%$ des cas dans 
 l'intervalle $$\Big[\ldots\ldots\ldots\ldots\ldots\ldots  , \ldots\ldots\ldots\ldots\ldots\ldots\Big]$$ 
 \begin{itemize}
  \item[\textbullet] \textbf{Avantages} : \\\\
  \item[\textbullet] \textbf{Inconvénients} : \\\\\\
  \textit{Sans parler du fait que cet intervalle
 n'a jamais été justifié par votre enseignant ...}
 \end{itemize}
}


\textbf{\underline{Applications} : } \\



\newpage

\Expl{
  On admet que dans la population d’enfants de 11 à 14 ans d’un département français le pourcentage
d’enfants ayant déjà eu une crise d’asthme dans leur vie est de $13\%$.\\
Un médecin d’une ville de ce département est surpris du nombre important d’enfants le consultant
ayant des crises d’asthme et en informe les services sanitaires. Ceux–ci décident d’entreprendre une
étude et d’évaluer la proportion d’enfants de 11 à 14 ans ayant déjà eu des crises d’asthme.\\
Ils sélectionnent de manière aléatoire 100 jeunes de 11 à 14 ans de la ville.\\
La règle de décision prise est la suivante : si la proportion observée est supérieure à la borne
supérieure de l’intervalle de fluctuation asymptotique au seuil de $95\%$ alors une investigation plus
complète sera mise en place afin de rechercher les facteurs de risque pouvant expliquer cette
proportion élevée.
\begin{enumerate}
 \item Déterminer l’intervalle de fluctuation asymptotique au seuil de $95\%$ de la proportion de jeunes de
11 à 14 ans ayant eu une crise d’asthme dans un échantillon de taille 100. 
\item  L’étude réalisée auprès des 100 personnes a dénombré 19 jeunes ayant déjà eu des crises d’asthme.\\
Que pouvez-vous conclure ?
\item  Le médecin n’est pas convaincu par cette conclusion et déclare que le nombre de personnes
interrogées était insuffisant pour mettre en évidence qu’il y avait plus de jeunes ayant eu des crises
d’asthme que dans le reste du département.\\
Combien faudrait-il prendre de sujets pour qu’une proportion observée de $19\%$ soit en dehors de
l’intervalle de fluctuation asymptotique ?
\end{enumerate}
}

\vspace{2cm}

\Cadre[Résumé]{
\begin{tabular}{|>{\centering}m{2cm}|>{\centering}m{8cm}|>{\centering}m{5cm}|}\hline
 & Intervalle de \textbf{fluctuation}\\ Au seuil 0.95\\ ($p$ connue)
 & Intervalle de \textbf{confiance} \\ Au seuil 0.95 \\ ($p$ inconnue) \tabularnewline\hline
 \textbf{SECONDE} & $n\geq 25$ , $0.2 \leq p \leq 0.8$ \\  $I=\left[ p-\dfrac{1}{\sqrt{n}} , p +\dfrac{1}{\sqrt{n}}\right]$ \\ & Sensibilisation\tabularnewline\hline
 \textbf{PREMIERE} & Avec la loi binomiale \\ $I=\left[\dfrac{a}{n} ; \dfrac{b}n \right]$\\ & \cellcolor{gray} \tabularnewline\hline
 \textbf{TERMINALE} &$n\geq 30$ , $np \leq 5$ et $n(1-p)  \leq 5$ \\ Asymptotique \\ $I_n=\left[p-1.96 \dfrac{\sqrt{p(1-p)}}{\sqrt{n}} ; p+1.96 \dfrac{\sqrt{p(1-p)}}{\sqrt{n}}\right]$\\
 & $J = \left[ f-\dfrac{1}{\sqrt{n}} , f +\dfrac{1}{\sqrt{n}}\right]$
 \tabularnewline\hline
\end{tabular} 
}

\Expl{
  Une entreprise désire contrôler la qualité de sa production. \\
  Un échantillon de 370 produits a
été testé. Il ressort de cette étude que 26 produits sur 370 ne sont pas conformes au cahier des
charges. \\
On considère que la production de l’entreprise est suffisamment importante pour
qu’on puisse assimiler le tirage de l’échantillon à un tirage avec remise.
\begin{enumerate}
 \item A partir de cette étude, estimer la probabilité qu’un produit ne soit pas conforme au cahier
des charges à l’aide d’un intervalle de confiance au niveau asymptotique $95\%$.
\item Combien faudrait-il prélever de produit pour avoir un intervalle de confiance au niveau
$95\%$ de longueur inférieure à 0.05 ?
\end{enumerate}
}

\begin{center}
 \includegraphics[scale=1]{Ppe_Echantillonnage.eps}
\end{center}

\end{document}