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19 lines
955 B
19 lines
955 B
2 years ago
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\begin{figure}
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\begin{align*}
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\sigma_{low} &= S_{n-1}^2 - z_{1-\alpha/2} \frac{\sqrt{2}S_{n-1}^2}{\sqrt{n}} \\
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\sigma_{high} &= S_{n-1}^2 + z_{1-\alpha/2} \frac{\sqrt{2}S_{n-1}^2}{\sqrt{n}} \\
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\mathcal{P}(\sigma_{low} \le \sigma \le \sigma_{high}) &\approx 1 - \alpha \\
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z_a \;\text{est tel que} &\int_{-\infty}^{z_a}{\frac{e^{-\frac{x^2}{2}}}{\sqrt{2\pi}}}dx=a \\
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\end{align*}
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\begin{align*}
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&I = [\mu^{\star} \pm \sigma_{high} \cdot z_{1 - \beta/2}] \\
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&\text{Avec un taux d'erreur de $\alpha$,} \\
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X \hookrightarrow \mathcal{N}(\mu, \sigma^2)
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&\Rightarrow \forall i, \mathcal{P}(\omega_i \in I) = 1 - \beta' \ge 1 - \beta \\
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&\Rightarrow S = \sum_{i=1}^{n}{\mathds{1}_{I}(\omega_i)} \hookrightarrow \mathcal{B}(1 - \beta', n) \\
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&\Rightarrow \mathbb{E}(S) = (1 - \beta') n \ge (1 - \beta) n
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\end{align*}
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\caption{Calcul du test de normalité par valeurs anormales}
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\label{normalitytest}
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\end{figure}
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