Let Q = ln(3 + |Q1| + 2|Q2|). Then T = 5 sin2(100Q) satisfies:— (A) 0 ≤ T < 1. — (B) 1 ≤ T < 2. — (C) 2 ≤ T < 3. — (D) 3 ≤ T < 4. — 4 ≤ T ≤ 5

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16. Let Q1 be the lower and Q2 the upper end point of the 96 percent confidence
interval for the sample proportion ˆp if the success count in the sample is x = 512 and
n = 698 is the sample size. Let Q = ln(3 + |Q1| + 2|Q2|). Then T = 5 sin2(100Q)
satisfies:— (A) 0 ≤ T < 1. — (B) 1 ≤ T < 2. — (C) 2 ≤ T < 3. — (D) 3 ≤ T < 4. — 4 ≤ T ≤ 5

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