= 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> uses
$$
X^2=\sum_{j=1}^3\frac{(O_j-100)^2}{100}
=\frac{(-8)^2+(-11)^2+19^2}{100}
=\boxed{5.46}.
$$
No 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 $5.46<5.99$. \b[At the 5% significance level there is insufficient evidence to reject equal popularity.]
The asymptotic <P-value> is $\Pr(\chi_2^2\ge5.46)=e^{-5.46/2}\approx0.0652$. 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.
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