\begin{question}
A survey with 49 persons was conducted to analyze the
design of an advertising campaign. Each respondent was asked to
evaluate the overall impression of the advertisement on an
eleven-point scale from 0 (bad) to 10 (good). The evaluations are
summarized separately with respect to type of occupation of the
respondents in the following figure.
\setkeys{Gin}{width=0.8\textwidth}
\includegraphics{anova-002}
To analyze the influence of occupation on the evaluation of the
advertisement an analysis of variance was performed:
\begin{Schunk}
\begin{Soutput}
Res.Df RSS Df Sum of Sq F Pr(>F)
1 48 24.789
2 45 24.642 3 0.147 0.089 0.96565
\end{Soutput}
\end{Schunk}
Which of the following statements are correct?
\begin{answerlist}
\item A one-sided alternative was tested for the mean values.
\item It can be shown that the evaluation of the respondents depends on their occupation. (Significance level $5\%$)
\item The fraction of explained variance is larger than $59$\%.
\item The fraction of explained variance is smaller than $56$\%.
\item The test statistic is smaller than $1.2$.
\end{answerlist}
\end{question}
\begin{solution}
In order to be able to answer the questions the fraction of
explained variance has to be determined. The residual sum of squares
when using only a single overall mean value ($\mathit{RSS}_0$) as
well as the residual sum of squares when allowing different mean
values given occupation ($\mathit{RSS}_1$) are required. Both are
given in the \texttt{RSS}~column of the ANOVA~table. The
fraction of explained variance is given by $1 -
\mathit{RSS}_1/\mathit{RSS}_0 = 1 - 24.642/24.789 =
0.006$.
The statements above can now be evaluated as right or wrong.
\begin{answerlist}
\item False. An ANOVA always tests the null hypothesis, that all mean values are equal against the alternative hypothesis that they are different.
\item False. The $p$~value is $ 0.966 $ and hence \textit{not} significant. It can \textit{not} be shown that the evaluations differ with respect to the occupation of the respondents.
\item False. The fraction of explained variance is $0.006$ and hence \textit{not} larger than 0.59.
\item True. The fraction of explained variance is $0.006$ and hence smaller than 0.56.
\item True. The test statistic is $F = 0.089$ and hence smaller than $1.2$.
\end{answerlist}
\end{solution}
%% META-INFORMATION
%% \extype{mchoice}
%% \exsolution{00011}
%% \exname{Analysis of variance}