: 6000. Using this "bootstrap" estimate, we compute the SE of the sum of the tickets in the sample to be 6000 x 15, and the SE of their average to be
6000x15/225 = 6000/15 = 2000/5 = 400
A 95% confidence interval is computed to be 3700 +/ 2x400 = 3700 +/ 800 .
So statement (i) is correct.
Statement (ii) looks plausible. If to say that 3700 +/ 800 is a 95% confidence interval for the box average meant that 95% of the tickets in the box are marked with numbers within 800 units of 3700, then (ii) would be correct. But that isn't what it means, and the statement is false. To see this, recall that about 68% of the tickets in the box have numbers within one SD of the box average. Since the box average is about 6000, about 100 - 68 = 32% of the numbers in the box have numbers which are either smaller that 3700 - 6000 = -2300 (which is impossible since a college can't have a negative enrollment) or bigger than 3700 + 6000 = 9700. But over 25% of the enrollments can't be bigger than 9700 if 95% of them are between 2900 and 4500.