Answer
(a) $X^2=2.2$
(b) $d.f.=4$
(c) $X_α^2=X_{0.05}^2=9.488$
(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) $X^2=Σ\frac{(O_i-E_i)^2}{E_1}=\frac{(38-40)^2}{40}+\frac{(45-40)^2}{40}+\frac{(41-40)^2}{40}+\frac{(33-40)^2}{40}+\frac{(43-40)^2}{40}=2.2$
(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 $X^2\lt X_α^2$, we do not reject the null hypothesis.
