An experimental drug is known to cure 25% of the patients inflicted with a certa

Question

An experimental drug is known to cure 25% of the patients inflicted with a certain disease. A new drug is produced in hope to increase this proportion. Let us denote p the true proportion of patients cured with the new drug. The experiment consists in 20 independent patients. For the sake of the exercise, we also assume all patients to be identical. Recall B(x;n, p) denotes the cumulative distribution function of a Binomial random variable with parameters n, p. We give: B{7; 20, .25) = .898 and B(7;20, .3) = .771. 1. State the null and alternate hypothesis, and justify your choice. 2. We now test. H_0: p = 0.25 against H_a: p > 0.25. Let X be the number patients cured with the new drug. When H_0 is true, what is the distribution of X? 3. When H_0 is true, what is the expected number of patients cured with the new drug? 4. We consider the rejection region {x greaterthanorequalto 8}. Find alpha the probability of type I error. 5. Out of the 20 patients, 9 are cured. Does the data suggest, that p > 0.25? 6. What is the probability of type II error when p = 0.3?

Explanation / Answer

1. H0 : p = 0.25

Ha : p > 0.25

2. When H0 is true, it means that the proportion of cured people is 0.25

The distribution of X is a binomial random variable.

3. When H0 is true, p = 0.25

So, expected number of patients cured = np = 0.25*20 = 5

4. Rejection region is x>=8, proportion = 8/20 = 0.4

Std dev of p = sqrt(p*(1-p)/n) = sqrt(0.25*0.75/20) = 0.0968

So, z value = (0.4-0.25)/0.0968 = 1.549

So, corresponding probability = 0.9393

So, probability of type I error = 1-0.9393 = 0.0607

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