# ward/vub

### Subversion checkout URL

You can clone with
or
.

mathstat oef - 2.5

'
 @@ -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.