Question

In a clinical trial, 50 patients suffering from an illness will be randomly assigned to one of two groups so that 25 receive an experimental treatment and 25 receive the best available treatment. The random variable X is the response of a patient to the experimental medicine, and the random variable B is the response of a patient to the best currently available treatment. The random variables X and B are normally distributed with = = 500  X  B under both the null and alternative distributions. The null hypothesis to be tested is that E(X) − E(B) = 0 against the alternative that E(X) − E(B)  0 at the 0.01 level of significance. What is the probability of a Type II error for the test of the null hypothesis when E(X ) − E(B) = 500 ?

Answer 1

Null hypothesis H0: E(X ) - E(B) = 0

Alternative hypothesis Ha: E(X ) - E(B) > 0

Probability of a Type II error = P(Fail to reject H0 | H0 is False)

As, E(X ) − E(B) = 500 , null hypothesis is False.

Probability of a Type II error = P(Fail to reject H0 | E(X ) − E(B) = 500)

Now, standard error of E(X ) − E(B) = 500 is

SE = $\sqrt{(\sigma_X^2 / n_X) +(\sigma_B^2 / n_B) } = \sqrt{(500^2 / 25) +(500^2 / 25) } = 141.4214$

At 0.01 level of significance, z = 2.326

Hypothesized value of E(X ) − E(B) is 0

Critical value to reject H0 = 0 + 2.326 * 141.4214 = 328.9462

Thus, we fail to reject H0 if observed E(X ) − E(B) < 328.9462.

Now,

Probability of a Type II error = P(Fail to reject H0 | E(X ) − E(B) = 500)

= P(observed E(X ) − E(B) < 328.9462 | E(X ) − E(B) = 500)

= P(Z < (328.9462 - 500) / 141.4214)

= P(Z < -1.21)

= 0.1132