Answer
(a) $X^2=11.987$
(b) $d.f.=4$
(c) $X_α^2=X_{0.05}^2=9.488$
(d) $X^2\gt X_α^2$: null hypothesis is rejected.
There is enough evidence to conclude that the random variable $X$ does not follow a binomial distribution with $n=4$ and $p=0.3$
Work Step by Step
(a) $X^2=Σ\frac{(O_i-E_i)^2}{E_1}=\frac{(260-240.1)^2}{240.1}+\frac{(400-411.6)^2}{411.6}+\frac{(280-264.6)^2}{264.6}+\frac{(50-75.6)^2}{75.6}+\frac{(10-8.1)^2}{8.1}=11.987$
(b) $k=5$. So, $d.f.=4$
(c) $X_α^2=X_{0.05}^2=9.488$
(According to Table VII, for d.f. = 4 and area to the right of critical value = 0.05)
(d) Since $X^2\gt X_α^2$, we do not reject the null hypothesis.
