: 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 <p>

6000x15/225 = 6000/15 = 2000/5 = 400  <p>

A 95% confidence interval is computed to be
 
 	
 3700 +/  2x400 = 3700 +/  800 .  <p>

So statement (i) is correct. <p>

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. <p>