For H₀: μ = 50 vs Hₐ: μ > 50 with n = 16, x̄ = 53.0, s = 6.4 and α = 0.05, compute t and df, compare with the given critical value t(0.05; 15) = 1.753, and bracket p between the row-15 columns 0.05 (1.753) and 0.025 (2.131).

Answer

import math n, xbar, s, mu0 = 16, 53.0, 6.4, 50 se = s / math.sqrt(n) t = (xbar - mu0) / se df = n - 1 critical = 1.753 decision = "reject H0" if t > critical else "do not reject H0" print(f"t={t:.3f} df={df} critical={critical} decision={decision}") row15 = [(0.10, 1.341), (0.05, 1.753), (0.025, 2.131), (0.01, 2.602)] for (a_hi, t_lo), (a_lo, t_hi) in zip(row15, row15[1:]): if t_lo < t < t_hi: print(f"p bracket: {a_lo} < p < {a_hi}")

IBS1 §9.4 (the t-test), CC BY 4.0; BCc ch 5; H-W9 (both routes; bracketing) — shape only; original dataset