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Thread: Chi² reasonable?

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    Chi² reasonable?




    Dear Statistic-Professionals,

    I'm new to statistics and i have got a problem, that i cannot really solve on my own. I hope someone is out there who can explain Chi² test in more detail. Here is my problem

    The following frequency of answers is expected when asking people for there favourite letter (A,B, C or D) e.g.:
    A: 25%
    B: 25%
    C: 25 %
    D: 25%
    Then i make a real experiment with lets say 100 people and I get the following results for the frequencey values:
    A: 28%
    B: 22%
    C: 25 %
    D: 25%

    Am i right, when i want to do the Chi² test with such data to test wheter my experimental results are significant for my assumed distribution?

    But this is only the first part of the Problem. The next part is more tricky. Now my expected frequencenies are the following:
    A: 50%
    B: 30%
    C: 20%
    D: 0%
    I do the experiment with 100 people again and i get results as below:
    A: 49%
    B: 30%
    C: 20%
    D: 1%
    The question is now: I assume a frequency of zero for "D" now. When i am looking at the Chi² formular, i can see that this will be in the denominator of one of the terms. How do i handle this statistical outlier, because "D" occurs althoung i dont expect it?

    Thanks a lot for your help, maybe you can show me the calculation, esspecially for the last part.

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    Re: Chi² reasonable?


    Hi there,

    First general comment: Looking at your post, I can't help but think that you are trying to use an analogy to explain what your question is. It's generally a lot better to actually explain what your real research question is, and what your actual data looks like. Using an analogy tends to not work so well, because there may be differences between the analogy and the real question that are important, but that you don't realise to be important.

    In relation to your specific question: Interestingly, the data that you have obtained is sufficient to reject (with certainty) the null hypothesis that the population data have the distribution specified in the expected frequencies. You don't need a p value to test this: If the null hypothesis incorporates the specification that none of the population fall into category D, and you have observed even a single value in category D, then the null is obviously false. You don't need a statistical test to demonstrate this.

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