Exercise 2: Standard Error at Three Sample Sizes — Possible Solution
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GIVEN
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sigma = 6

STEP 1: STANDARD ERROR AT EACH SAMPLE SIZE
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Per this chapter's own standard error formula, SE = sigma / sqrt(n):

  n = 9:   SE = 6 / sqrt(9)   = 6 / 3  = 2.0
  n = 36:  SE = 6 / sqrt(36)  = 6 / 6  = 1.0
  n = 144: SE = 6 / sqrt(144) = 6 / 12 = 0.5

STEP 2: THE PATTERN
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Going from n=9 to n=36 (a 4x increase in sample size) halves the
standard error, from 2.0 to 1.0. Going from n=36 to n=144 (again a 4x
increase) halves it again, from 1.0 to 0.5. In both cases, the sample
size had to QUADRUPLE to cut the standard error in HALF - a direct
consequence of the square root in this chapter's own SE = sigma/sqrt(n)
formula: halving SE requires the denominator sqrt(n) to double, which
requires n itself to increase by a factor of 2^2 = 4, not just 2.

WHY THIS WORKS AS AN ANSWER
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Each standard error is computed directly from this chapter's own
formula, and the pattern (quadrupling n to halve SE) is identified by
comparing consecutive results and tracing it back to the square root
in the formula itself, rather than just observing "SE gets smaller"
without explaining the specific rate at which it shrinks.