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I'm running chi-squared goodness-of-fit tests on answers to a series of questions. I originally had 5 options (Strongly Agree, Agree, Neither Agree nor Disagree, Disagree, Strongly Disagree) but I decided to consolidate the Strongly Agrees with the Agrees and the Strongly Disagrees with the Disagrees.

Here's sample of my original data

N = 56
                   Observed  Expected
Strongly Agree      25       11.2
Agree               22       11.2
Neither             7        11.2
Disagree            3        11.2
Strongly Disagree   0        11.2

And here's a sample of when I consolidated. I changed the expected frequencies to reflect that I had originally given 5 options.

N = 56
                   Observed  Expected
Agree               47       22.4
Neither             7        11.2
Disagree            3        22.4

My question is: was I correct to double the expected frequencies for the groups that I consolidated? And also do I change my df from 4 to 2 because I technically only have 3 groups instead of 5? Or do I keep it the same?

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  • $\begingroup$ If you aggregate to 3 groups, then you now have 2 degrees of freedom. The d.f. argument is the same as before: if you know the total frequency and the combined frequency of two groups, then the third frequency is redundant. But in this example, there is no need to aggregate, however. If the null hypothesis of equal frequencies is of interest, then a chi-square test with 5 groups is fine. The expected frequencies are all $> 5$ and $>> 1$ and the result is overwhelmingly conclusive: the null is rejected. It's a straw person hypothesis, however. $\endgroup$ – Nick Cox Apr 10 '15 at 0:25
  • $\begingroup$ On what basis would you give equal proportions to each of the original groups? Why should the proportion of Strongly Agree equal the proportion of Neither? Since I can't think of a situation where that makes sense as a null hypothesis, perhaps you can clarify -- how does that arise? What are you trying to find out? $\endgroup$ – Glen_b Apr 10 '15 at 0:25
  • $\begingroup$ I gave equal proportions to the original groups using the chi-squared goodness-of-fit formula since I want to see if the answers are randomly distributed and thus there would be more or less the same number of people who chose Strongly Agree, Agree, Neither, etc. Basically I'm trying to find out if the trends are significant. Like is the number of people who chose Agree or Strongly Agree significantly higher than the other ones. Should I have chosen a different type of statistical test? $\endgroup$ – linao Apr 10 '15 at 0:58
  • $\begingroup$ I can see sense in trying to answer a question like "do people tend to agree rather than disagree" (where you combine the two agree options and the two disagree options) or "do at least a majority of people agree rather than neutral or worse". But unless you imagine something bizarre, like 'people are unable to see what answer they're giving' I don't see a mechanism by which you'd expect that p(Neutral) would equal p(Strongly Agree). ...(ctd) $\endgroup$ – Glen_b Apr 10 '15 at 1:29
  • $\begingroup$ (ctd)... If you had a null that people on average don't lean one way or another, I could see p(A+SA)=P(D+SD) for a hypothesis, but I see no basis on which to think that P(A)=P(SA) or P(SA)=p(N) unless you really think something like people can't see their answers (what could cause people to answer at random? If you're not sure what you prefer, why would p(SA) be as likely as p(N)? If I wasn't sure what I preferred, I sure wouldn't pick "strongly" anything. ) $\endgroup$ – Glen_b Apr 10 '15 at 1:29

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