Answer
(a) $X^2=2.72$
(b) $d.f.=3$
(c) $X_α^2=X_{0.05}^2=7.815$
(d) $X^2\lt X_α^2$: null hypothesis is not rejected.
There is not enough evidence to conclude that the distribution of proportions is different from what is stated.
Work Step by Step
(a) $X^2=Σ\frac{(O_i-E_i)^2}{E_1}=\frac{(30-25)^2}{25}+\frac{(20-25)^2}{25}+\frac{(28-25)^2}{25}+\frac{(22-25)^2}{25}=2.72$
(b) $k=4$. So, $d.f.=4-1=3$
(c) $X_α^2=X_{0.05}^2=7.815$
(According to Table VII, for d.f. = 3 and area to the right of critical value = 0.05)
(d) Since $X^2\lt X_α^2$, we do not reject the null hypothesis.
