(1 pt) A new treatment is claimed to have a success rate better than 0.2. Let pl
ID: 3205171 • Letter: #
Question
(1 pt) A new treatment is claimed to have a success rate better than 0.2. Let pl be the success rate of the treatment. Here, the null hypothesis is p 0.24 and the alternate hypothesis is p 0.2. A trial is conducted with 6 patients. If the treatment is effective for at least m people, then the null hypothesis is rejected the treatment is superior. Otherwise, the null hypothesis is accepted the treatment is no better than previous ones. lfm then what is the probability of a Type 1 error? If m 54 then what is the probability of a Type 1 error? suppose the treatment is actually effective with probability p 0.35 6 then using p 0.35, what is the probability of a Type 2 error? If m 5l then using p 0.35, what is the probability of a Type 2 error? If mExplanation / Answer
Solution
P(Type I error) = P(rejecting H0 when H0 is true)
P(Type II error) = P(accepting H0 when HA is true)
Let X =number of patients on whom the treatment was effective. Then, X B(n, p) i.e., X has a binomial distribution with parameters n (= sample size) and p (= success rate).
H0p = 0.2 vs HAp > 0.2
Decision criterion: Reject H0, if X m.
Part (i)
Given, n = 6, p = 0.2 and m = 6, P(Type I error) = P(X 6) = P(X = 6)
= 6C6(.2)6(.8)0 = 0.26 = 0.000064 ANSWER
Part (ii)
Given, n = 6, p = 0.2 and m = 5, P(Type I error) = P(X 5) = P(X = 5) + P(X = 6)
= 6C5(.2)5(.8)1 + 6C6(.2)6(.8)0 = (6x0.00032x0.8) + 0.26 = 0.001536 + 0.000064
= 0.0016 ANSWER
Part (iii)
Given, n = 6, p = 0.35 and m = 6, P(Type II error) = P(X < 6) = 1 - P(X = 6)
= 1 - 6C6(.35)6(.65)0 = 1 - 0.356 = 1 - 0.001838 = 0.998162 ANSWER
Part (iv)
Given, n = 6, p = 0.35 and m = 5, P(Type II error) = P(X < 5) = 1 – {P(X = 5) + P(X = 6)}
= 1 – {6C5(.35)5(.65)1 + 6C6(.35)6(.35)0 = 1 - (0.020483 + 0.001838 = 0.977679 ANSWER
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