I'm not sure If I understand what (b) is asking. Does it mean that the alternative hypothesis will be p<0.3, p<0.4, and p<0.5? If it is, then I have to find the probability of committing a type 2 error for each alternatives? If I am right then the setup that I have for p=0.3 in part (b) is correct.
Best Answer
You are on the right track, but you just need to go through and complete the calculation. Note that the sum $$\Pr[X > 3 \mid p = 0.3] = \sum_{x=4}^{10} \binom{10}{x} (0.3)^x (0.7)^{10-x}$$ contains a total of $10-4+1 = 7$ terms, whereas you could save work by exploiting the complementary probability: $$\Pr[X > 3 \mid p = 0.3] = 1 - \Pr[X \le 3 \mid p = 0.3] = 1 - \sum_{x=0}^3 \binom{10}{x} (0.3)^x (0.7)^{10-x},$$ and this latter sum has only $4$ terms, and they are easy to compute: $$\sum_{x=0}^3 \binom{10}{x} (0.3)^x (0.7)^{10-x} = (0.7)^{10} + 10(0.3)(0.7)^9 + 45(0.3)^2(0.7)^8 + 120(0.3)^3(0.7)^7.$$