
Dattaprabhakar G. answered 08/08/14
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Christina:
One quick and easy way to solve this problem is via simultaneous confidence intervals for multinomial proportions. I am assuming that the sample is random with replacement so that the underlying distribution is multinomial.
First, count the number k of "cells". Here k = 4, namely "BK", "W", "M" , and "Other (O)" .
Find the sample proportion for "O". It is 1 - (0.22 + 0.19 + 0.15) = 0.64
Your problem does NOT explain how the margin of error for the overall sample is obtained. More precisely what significance level is used. It should be the prescribed level of significance divided by k. Let us assume that that is done correctly. Otherwise, the answers below WILL NOT APPLY. OK?
Calculate confidence intervals for each population proportion by the formula pj plus or minus the margin of error for the overall sample, given to be 0.044.
BK : 0,22 minus /plus 0.044, that is, 0.176, 0.264
W : 0.19 minus/plus 0.044 that is 0.146, 0.234
M : 0.15 minus/plus 0.044 that is 0.106, 0.194
O : 0.64 minus/plus 0.044 that is 0.596, 0.684
If there is an overlap between one pair (one is enough), then the result is NOT statistically significant. Here there is.
In context, this means that the data does not provide sufficient evidence to claim that in at least one pair of the population proportions of "favorite foods", these proportions are significantly different. Which, in turn, means that if you consider preference to other foods (such as "In-and-Out, Arby's", etc) it can NOT be concluded at the given level of significance that BK (or M or W for that matter) is more preferred.
Of course, you can do what is called a conditional test, looking only at BK, W and M, but then the observed proportions in that experimental design must add to 1. Here they do not.
Christina, there are many more advanced methods of answering your question. Take it from me, I know. My areas of specialization are discrete data analysis and information theory. But these methods are beyond the scope of an intro course.
Dattaprabhakar ("Dr. G.")

Dattaprabhakar G.
08/08/14