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mathstat oef - 2.5

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62 mathematicalstatistics/oefeningen.tex
@@ -387,6 +387,68 @@ \section{Estimation, basic concepts}
\section{Hypotheses testing}
+\begin{opgave}
+\end{opgave}
+\begin{oplossing}
+\end{oplossing}
+
+\begin{opgave}
+\end{opgave}
+\begin{oplossing}
+\end{oplossing}
+
+\begin{opgave}
+\end{opgave}
+\begin{oplossing}
+\end{oplossing}
+
+\begin{opgave}
+\end{opgave}
+\begin{oplossing}
+\end{oplossing}
+
+\begin{opgave}
+ Suppose a machine makes metal plates of a specific thickness.
+ In a particular company there are two such devices.
+ Engineers are worried whether both machines work equally consistent.
+ To test this, they have 10 plates from machine A and 12 platers from
+ machine B.
+ The thickness of each plate is measured and shown here.
+ $A: 22.3, 21.9, 21.8, 22.4, 22.3, 22.5, 21.6, 22.2, 21.8, 21.6$.
+ $B: 22.0, 21.7, 22.1, 21.9, 21.8, 22.0, 21.9, 22.1, 22.2, 21.9, 22.0, 22.1$.
+
+ Do both machines work equally consistent (that is: same variance)?
+ You may suppose that the thickness follows a normal distribution.
+ $\alpha = 0.05$.
+\end{opgave}
+\begin{oplossing}
+ For this we use the F-test of equality of variances, as worked out in the
+ previous exercise. First off we need the sample mean for each machine.
+ $\overline{A} = 22.04$, $\overline{B} = 21.975$. Sample variance
+ then becomes
+ \begin{align*}
+ \hat{s}_A^2 &= \frac{\sum (a_i - \overline{A})^2}{n_A-1} = 0.114\\
+ \hat{s}_B^2 &= \frac{\sum (b_i - \overline{B})^2}{n_B-1} = 0.0202
+ \end{align*}
+
+ The test statistic is
+ \[
+ T = \frac{\hat{s}_A^2}{\hat{s}_B^2} = \frac{0.114}{0.0202} = 5.64
+ \]
+ we use a table to look up the values
+ \begin{align*}
+ F_{9;11;0.975} &= 3.588\\
+ F_{9;11;0.025} &= 0.256
+ \end{align*}
+ It is clear that $5.64 > 3.588$, thus we are in the critical region
+ and the null hypothesis is rejected.
+\end{oplossing}
+
+\begin{opgave}
+\end{opgave}
+\begin{oplossing}
+\end{oplossing}
+
%\newpage
%\begin{thebibliography}{99}
% \bibitem{VOORBEELD} Voorbeeld, Ward Muylaert.
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