Chi Square Test – How to Compare Proportions in Hypothesis Testing

chi-squared-testhypothesis testingproportion;

I've read that the chi square test is useful to see if a sample is significantly different from a set of expected values.

For example, here is a table of results of a survey regarding people's favourite colours (n=15+13+10+17=55 total respondents):

red, blue, green, yellow

15, 13, 10, 17

A chi square test can tell me if this sample is significantly different from the null hypothesis of equal probability of people liking each colour.

Question: Can the test be run on the proportions of total respondents who like a certain colour? Like below:

red, blue, green, yellow

0.273, 0.236, 0.182, 0.309

Where, of course, $0.273 + 0.236 + 0.182 + 0.309=1$.

If the chi square test is not suitable in this case, what test would be?

Edit: I tried @Roman Luštrik answer below, and got the following output, why am I not getting a p-value and why does R say "Chi-squared approximation may be incorrect"?

chisq.test(c(0, 0, 0, 8, 6, 2, 0, 0), p = c(0.406197174, 0.088746395, 
             0.025193306, 0.42041479, 0.03192905, 0.018328576, 
             0.009190708, 0))
    
        Chi-squared test for given probabilities
    
    data:  c(0, 0, 0, 8, 6, 2, 0, 0) 
    X-squared = NaN, df = 7, p-value = NA
    
    Warning message:
    In chisq.test(c(0, 0, 0, 8, 6, 2, 0, 0), p = c(0.406197174, 
           0.088746395,  :
      Chi-squared approximation may be incorrect

Best Answer

Correct me if I'm wrong, but I think this can be done in R using this command

chisq.test(c(15, 13, 10, 17))
    
        Chi-squared test for given probabilities
    
    data:  c(15, 13, 10, 17) 
    X-squared = 1.9455, df = 3, p-value = 0.5838

This assumes proportions of 1/4 each. You can modify expected values via argument p. For example, you think people may prefer (for whatever reason) one color over the other(s).

chisq.test(c(15, 13, 10, 17), p = c(0.5, 0.3, 0.1, 0.1))
    
        Chi-squared test for given probabilities
    
    data:  c(15, 13, 10, 17) 
    X-squared = 34.1515, df = 3, p-value = 1.841e-07
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