Exercise 2: An A/A Test Sanity Check — Possible Solution
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GIVEN
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Group 1: n1=25, mean1=200, s1=30
Group 2: n2=28, mean2=215, s2=35

STEP 1: THE STANDARD ERROR OF THE DIFFERENCE
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SE_diff = sqrt(s1^2/n1 + s2^2/n2)
        = sqrt(30^2/25 + 35^2/28)
        = sqrt(900/25 + 1225/28)
        = sqrt(36 + 43.75)
        = sqrt(79.75)
        ~= 8.930

STEP 2: THE T-STATISTIC
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t = (mean2 - mean1) / SE_diff = (215 - 200) / 8.930 ~= 1.680

STEP 3: THE CONSERVATIVE DEGREES OF FREEDOM
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df = min(n1, n2) - 1 = min(25, 28) - 1 = 24

STEP 4: DOES THE A/A TEST PASS ITS OWN SANITY CHECK?
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Given critical value at two-tailed alpha=0.05, df=24: 2.064

|t| = 1.680, which is less than 2.064.

Per this chapter's own decision rule, we FAIL TO REJECT H0 - there is
no statistically significant difference between the two identically-
treated groups. This A/A test PASSES its own sanity check: the
traffic-splitting mechanism shows no detectable artificial bias, so a
real A/B test can proceed with confidence in the underlying
infrastructure, exactly matching this chapter's own Step 4 outcome.

WHY THIS WORKS AS AN ANSWER
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Each quantity is computed using this chapter's own Step 4 Welch's-
t-test method, and the sanity-check interpretation is drawn directly
from comparing |t| against the given critical value, correctly
identifying a "fail to reject" result as the DESIRED outcome for an
A/A test specifically (unlike a real A/B test, where failing to reject
would instead mean no detected effect).