Chi-square test problem (and some more maybe) A colleague of mine has asked me for help on this problem, but I am confused with it as well, the main question he asked me was that the chi-square result he was getting was very high.
He conducted a survey, and among other questions, one of them is a question which has 8 sub-questions about certain processes in company of examinees and the examinees have to give their grade of the quality of those processes by choosing values from 1 (bad) to 5 (great) (classical Likert scale).
There is one other value (sixth) that can be selected which is causing me trouble aside from the main problem. That sixth option is titled "N/A" and means that the examinees don't have that process in their company. Enough for intro (hopefully).
So the data table he made in Excel (for one subquestion) looks like this (there were 195 examinees in total):

I wonder why did he assume that theoretical frequencies (f$\_t$) are fixed to 32,5 (shown as rounded to 35 in data table) for all values? My opinions is that he is wrong, and that the theoretical frequencies should be these (and if not these, for sure not the same for every mark):

The chi-square value is even greater now. I am not even sure why does he calculate it in this fashion in Excel.
My opinion that for this data set chi-square value should be calculated with CHITEST function of the following syntax (cell addresses are as in Image 2.):
=CHITEST(C9:C14;D9:D14)

That way i get a chi-square value of 1,26177E-44, which I think is ok.
Please help me, all I thought I know or knew on topic seems to be vanished now.
Sorry if any terms are not understandable, English is not my primary language.
If any clarification is required, I'll do my best.
Thank you in advance for your time and knowledge and sorry that images are not automatically displayed, I need to gain more reputation in order to enable them.
 A: The first table is testing the assumption that all possible answers are equally likely, the 33 comes from $\frac{195}{6} = 32.5$, which then is being rounded to 33 for display.  The testing looks correct after that (though as @whuber points out, this is probably a meaninless test to do).
In the second table you compute the "expected" values ($f_t$) in a different way, possibly from a normal distribution, but the numbers sum to 100 instead of 195.  Your higher $\chi^2$ value is due to the fact that the columns add up to different numbers (195 vs. 100) and does not in any way fit in with the idea of a $\chi^2$ test.  The results are fairly meaningless.
The value you report from the chitest looks more like it would be the p-value than the $\chi^2$ value, are you sure that you are not getting the 2 confused?  That value is essentially 0 which would not be the test statistic for the differences in pattern seen there.
Figure out the question first, then look to find the test that answers that question. 
