How to do t-test with multiple probabilities?

Question

I have given some distribution among candy preferences. 30% of people like candy1, 30% of people like candy2, and 40% of people like candy3. The total amount of people is known, e.g. 100. Now the candy manufacturer says that all his candy flavors are being liked equally. This hypothesis shall be tested with a significance of 5%. I never did such a test with multiple probabilities.

My idea is to compare the mean value with the expected mean value (where 100/3 people would like each candy flavor equally). But I am not sure how to calculate the actual mean value.

Answer 1

Comment following @Ian's first Comment. This is not a formal answer to your Question, but thinking about what I have shown may get you started in a useful direction. [I am not talking about Ian's G test. And a t test is not what you want.]

If the information at the start of your Question is correct, then preferences of $n = 100$ people might be as in the vector x below, which I simulated in R.

set.seed(128)
x = sample(1:3, 100, rep=T, p = c(.3,.3,.4))
table(x)
x
 1  2  3 
33 28 39 

How would you use a chi-squared test to see if these counts are consistent with the manufacturer's null hypothesis that all three flavors are 'liked equally'?

Results from R, are as follows. How is the chi-squared statistic computed in chisq.test? (Can you find it using a calculator?) How is the P-value found? Is the null hypothesis rejected? Do you think $n = 100$ subjects is enough to investigate whether the null hypothesis is true?

chisq.test(c(33, 28, 39)) 

        Chi-squared test for given probabilities

data:  c(33, 28, 39)
X-squared = 1.82, df = 2, p-value = 0.4025

Note: If no other probabilities are provided for chisq.test, the given probabilities are assumed to be all equal: here $(1/3, 1/3, 1/3).$