Chi-Squared Goodness of Fit

$E = \dfrac{200}{4} = 50$ per category, and $\sum E = 200 = \sum O$ ✓

$\chi^2 = \dfrac{64 + 25 + 121 + 64}{50} = \dfrac{274}{50} = 5.48$.

$\nu = 5 - 1 - 1 = 3$: one for the total, one for the estimated parameter.

$E = 300, 100$. $\chi^2 = \dfrac{225}{300} + \dfrac{225}{100} = 0.75 + 2.25 = 3.00$. With $\nu = 1$, $\chi^2_{0.05,1} = 3.841$, and $3.00 < 3.841$.

$3.1 + 1.2 = 4.3$, still under 5, so pool with the preceding class as well. Recount $\nu$ from the number of classes remaining after pooling — each merge reduces $\nu$ by one.

Nothing is estimated from the data, so $\nu = 6 - 1 = 5$. Had $p$ been estimated it would be 4.