Problem 22. There are two hypotheses about the probability of heads for a given

Question

Problem 22. There are two hypotheses about the probability of heads for a given coin: = 0.5 (hypothesis Ho) and = 0.6 (hypothesis H1). Let X be the number of heads obtained in n tosses, where n is large enough so that normal approximations are appropriate. We test Ho against Hi by rejecting Ho if X is greater than some suitably chosen threshold kn (a) What should be the value of kn so that the probability of false rejection is less than or equal to 0.05? false acceptance can be made less than or equal to 0.05? acceptance if we were to use a LRT with the same probability of false rejection? (b) What is the smallest value of n for which both probabilities of false rejection arn (c) For the value of n found in part (b), what would be the probability of false

Explanation / Answer

Question 22

(a) Here we can approximately normalise the binomial distribution.

Here Expected number of Heads = 0.5n

Standard deviation of number of heads = sqrt [0.5 * 0.5 * n ] = 0.5 n

so as the false rejection is less than or equal to 0.05 so Z value for such probability is 1.645 (one tailed)

so kn = 0.5n +- 1.645 * 0.5  n = 0.5n + 0.8225  n.

(b) Here now as we know that if X is greater than kn , then we will be able to reject the null hypothesis.

So now if true null hypothesis is 0.6

then standard error of sample proportion se0 = sqrt (0.6 * 0.4 * n) = (0.24n)

Now false acceptance probability = 0.05

Pr(X >  0.5n + 0.8225  n ; 0.6 ; (0.24n) ) = 0.05

THe Z - value is -1.645

-1.645 = (0.5n + 0.8225n - 0.6n)/ ((0.24n)

-1.645 = (0.8225n - 0.1n)/ ((0.24n)

1.6284n = 0.1n

n = 265.1628 or say 266

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.