Question

Consider the next 1000 90% CIs for ? that a statistical consultant will obtain for various clients. Suppose the data sets on which the intervals are based are selected independently of one another. How many of these 1000 intervals do you expect to capture the corresponding value of ?


What is the probability that between 890 and 910 of these intervals contain the corresponding value of ? [Hint: Let Y = the number among the 1000 intervals that contain ?. What kind of random variable is Y?] (Round your answer to four decimal places.)

 

Answer 1

90% CI expects to capture u 90% of time

(a) This means 0.9 * 1000 = 900 intervals will capture u

(b) Here we treat CI as binomial random variable, having probability 0.9 for success

n = 1000 p = 0.9 For this case, applying normal approximation to binomial, we get: mean = n*p= 900 variance = n*p*(1-p) = 90 std dev = 9.4868

We want to Find : P(890 <= X <= 910) = P( 889.5 < X < 910.5) (integer continuity correction) We convert to standard normal form, Z ~ N(0,1) by z1 = (x1 - u )/s so z1 = (889.5 - 900 )/9.4868 = -1.11 so z2 = (910.5 - 900 )/9.4868 = 1.11 P( 889.5 < X < 910.5) = P(z1 < Z < z2) = P( Z < 1.11) - P(Z < -1.11) = 0.8665 - 0.1335 = 0.733 _ _ _ _ _ _ _ _ _ (From Z-table)