Question

Suppose that a new blood screening procedure is believed to reduce the incidence of hepatitis. Blood screened using this procedure is given to n = 200 randomly selected blood recipients. Only 6 of the 200 patients contract hepatitis. Recall that the hepatitis rate was given as 0.07. First, what is the typical distance between the sample proportion and 0.07 when n = 200? Give your answer to 5 decimal places.

Does the result of 6 out of 200 contractions indicate the new screening procedure is better, or could we just chalk this up to random chance? Calculate the probability of finding 6 or fewer out of 200 patients having hepatitis if the rate of contracting it is still 0.07. Calculate your answer to 4 decimal places.

Answer 1

(a) p' = 6/200 = 0.03, p = 0.07

SE = √(p(1 - p)/n) = √(0.07 * 0.93/200) = 0.01804

z = (p' - p)/SE = (0.03 - 0.07)/0.01804 = -2.217295

(b) P(x ≤ 6) = P(p < 0.03) = P(z < -2.217295) = 0.0133

Since the probability < 0.05, it indicates that he new screening procedure is better