## 3 Results
### Audibility  
There were 30 respondents to the audibility form which consisted of telling which of two alternatives of the same song was steganography for 10 different MIDI songs. The 10 songs were as listed in Section 2.8 and the results of this experiment are illustrated in the bar charts in Figure 3.  
We tested the hypothesis that the suggested method left audible traces a test of whether the probability \(\pi=P\) (a respondent cannot tell apart the stego MIDI file from the carrier MIDI file) exceeds 0.5 against the null hypothesis that the probability \(\pi=0.5\) (corresponding to the respondent choosing one of the alternatives entirely at random). Letting each guess be coded as 0 if it is wrong and 1 if it is right, a binomial test is an obvious possibility since the sum \(S\) of correct guesses over all pairs of MIDI songs and all respondents is a sum of 0-1-variables which, assuming independence between songs and respondents and that all guesses are correct with equal probability \(\pi\), is binomially distributed with parameters \(N\) and \(\pi\). 30 respondents were signing up for the experiment and 10 pairs of songs making \(n=300\). In total, 153 correct guesses made the \(p\)-value of the binomial test  
\[P(S>153\mid H_{0})\ =\ \sum_{k=154}^{N}\binom{N}{k}\pi^{k}(1-\pi)^{N-k}\Big{|} _{ \begin{subarray}{c}N=300\\ \pi=0.5\end{subarray} }\ =\ 0.3431\]  
which is well and truly above than any reasonable level of significance, i.e., there are no indications of standing a better chance of guessing which of the songs is steganography after listening to both MIDI carrier and stego MIDI file. The more common test for this kind of question is a \(\chi^{2}\)-test. Letting each song constitute a class, the number of correct guesses for each class was calculated. Under the \begin{table}
\begin{tabular}{c c c c} \hline \hline
**Method** & **MAE** & **PSNR** & **F\({}_{\mathbf{r}}\)** \\ \hline Velody 2, 4 bits & 6.27 (0.36) & 24.64 (0.43) & 0.00\% (0.00\%) \\ Velody 2, 5 bits & 12.52 (0.81) & 18.71 (0.50) & 0.00\% (0.00\%) \\ Velody 2, 6 bits & 25.37 (1.88) & 12.62 (0.55) & 0.00\% (0.00\%) \\ [PERSON], [PERSON] and [PERSON] [8] & unknown & 24.99  (0.60) & small \\ [PERSON] and [PERSON] [7] & unknown & unknown & 40.47\% 1 \\ [PERSON], [PERSON] and [PERSON] [3] & unknown & unknown & 0.00\% (0.00\%) \\ \hline \hline \end{tabular}
\end{table} 
Table 2: Table of Mean Absolute Error (MAE), Peak Signal-to-Noise Ratio (PSNR) and file-size change-rates (\(F_{r}\)) for the methods considered. The file-size change-rates are as defined by [PERSON] and [PERSON] [7] and briefly explained in the text. Values within parentheses are standard errors. Figure 3: Barplots of the distribution of guesses divided into the 10 songs in the audibility experiment. A total of 30 respondents guessed each and every of the 10 pairs of songs about which one was steganography. The bars show the number of incorrect guesses in turquoise and the number correct guesses in orange for each pair of songs. assumption of independence between guesses and that each song was guessed to be steganography correctly with probability \(\pi\) the total number \(X\) of correct guesses for one respondent could be subjected to a \(\chi^{2}\)-test of whether \(X\) is binomially distributed with parameters \(n=10\) and \(\pi=0.5\) or not. After merging classes so that the expected number of observations in each class exceeded 2, there were 5 classes and the test statistic turned out 5.8252 rendering a high \(p\)-value of 0.8171. So, what does it mean that the null hypothesis \(X\in Bin(10,0.5)\) can not be rejected? Certainly, it does not prove that \(X\in Bin(10,0.5)\) and that \(\pi=0.5\) which corresponds to that respondents can not tell the stego MIDI file apart from the carrier file, only that no deviance in the distribution of \(X\) from \(Bin(10,0.5)\) can be found. How large would that deviance have had to be for the hypothesis test to prove it?