Can you use an n-1 chi-squared test to analyse the difference of one result in a data set? I am trying to establish if room colour influences how respondents interpret the 'mood' of that room. The sample size is relatively small. When applying a Pearson's chi-squared test across the data set there is no statistical difference. However, if I wanted to look at one specific 'mood' without regard for the data set to assess the increase could I use an n-1 chi-squared test? 
Example
     Happy     All other 'moods'    Total

Blue       ........ 11...............................83...............94
Black      ...... 38...............................70..............108
There is not or a very small difference between the other 'moods' dependent on the colour of the room hence I want to isolate this response.
Based on Campbell 2007 - http://www.iancampbell.co.uk/twobytwo/calculator.htm
 A: If I got it right you want to merge all of your columns but one so you can test one specific mood against all the others. This is possible, you end up with a table that has the same number of lines but only two columns. You can then perform a $\chi^2$ test on this table, and you would indeed have $(n-1)\times (2 - 1) = n - 1$ degrees of freedom where $n$ corresponds to the number of rooms.
However, you need to be carefull when doing that. One could be tempted to do all possible tests of "one mood vs all others" and only keep the test which has the lowest p-value to be shown as a result. This would be seen as cheating, since you should not select the relevant hypothesis to test based on the result of the test! 
If you don't want to be called a cheater you need to use a correction on the level of you test. Bonferroni correction is the easiest one to implement: if you want to test at a level $\alpha$ (like 10% or 5%), you will reject $\mathcal{H}_0$ if the p-value of your test is lower than $\frac{\alpha}{m}$ where $m$ is the total number of hypthesis you are testing. Here you would have $m$ equals the number of different moods so if you have 5 moods, in order to declare one of your test significative at 10%, you must get a p-value lower than $\frac{0.1}{5}= 0.02$.
