 Solution

Under the null hypothesis of equal population probabilities, the three counts have a multinomial distribution with probabilities $1/3$ and expected counts $100$ each. The Pearson chi-squared goodness-of-fit test usesNo probabilities have been estimated from the data, and the three counts sum to a fixed total. Hence the null reference chi-squared distribution has $3-1=2$ statistical degrees of freedom. The supplied $5\%$ upper critical value is $5.99$, and $5.46<5.99$. At the 5% significance level there is insufficient evidence to reject equal popularity.

The asymptotic P-value is $\mathrm{Pr}({\chi }_2^2\ge 5.46)=e^{-5.46/2}\approx 0.0652$. The expected counts are all large, so the approximation is appropriate if the sampled choices are independent and representative of this branch. The conclusion depends on the chosen significance level; for instance, a $10\%$ test would reject. Nonrejection does not establish the null hypothesis, and a survey at one branch does not by itself justify a population claim about every branch.