Exercise 2: Welch's Two-Sample t-test — Possible Solution
====================================================================

GIVEN
------------------------------
Group 1: n1=15, x-bar1=48, s1=6
Group 2: n2=18, x-bar2=53, s2=7

STEP 1: THE STANDARD ERROR OF THE DIFFERENCE
------------------------------
SE_diff = sqrt(s1^2/n1 + s2^2/n2)
        = sqrt(6^2/15 + 7^2/18)
        = sqrt(36/15 + 49/18)
        = sqrt(2.4 + 2.722)
        = sqrt(5.122)
        ~= 2.263

STEP 2: THE T-STATISTIC
------------------------------
t = (x-bar2 - x-bar1) / SE_diff = (53 - 48) / 2.263 = 5 / 2.263 ~= 2.209

(Taking the difference as group2 minus group1, since group 2 has the
higher mean - the sign convention doesn't change the magnitude
comparison against the critical value.)

STEP 3: THE CONSERVATIVE DEGREES OF FREEDOM
------------------------------
df = min(n1, n2) - 1 = min(15, 18) - 1 = 15 - 1 = 14

STEP 4: THE DECISION
------------------------------
Given critical value at two-tailed alpha=0.05, df=14: 2.145

|t| = 2.209, which is greater than 2.145.

Per this chapter's own decision rule, we REJECT H0 - the two groups'
means differ by more than would be expected from sampling error
alone, at this significance level.

WHY THIS WORKS AS AN ANSWER
------------------------------
Each quantity is computed directly using this chapter's own Welch's
t-test formula and its conservative min(n1,n2)-1 degrees-of-freedom
simplification, and the decision follows directly from comparing the
resulting |t| against the given critical value, consistent with the
chapter's own decision-rule convention from the one-sample test.