Respuesta :
We don't have claim to strength of composite B is smaller than the strength of composite B by at least 2 [N].
From the given data we calculate μ₁s₁ and μ₂s₂ mean and standard deviation of samples A and B respectively.
μ₁ = 464.7, s₁ = 13.42 composite A
μ₂ = 479.49, s₂ = 13.38 composite B
n₁ : n₂ < 30 then we will use t-table.
Null hypothesis H₀:μ₂-μ₁ ≥ 2
Alternative hypothesis Hₐ : μ₂ - μ₁ < 2
assuming CI 95%
α = 5% = 0.05
and degree of freedom is n₁ + n₂ - 2 = 9 + 9 - 2 = 16
then for the one tail t(c) = 1.746
now we will compute t(s),
t(s) = (μ₂ - μ₁ - 2)/ sp×√1/n₁ + 1/n₂
sp²= (n₁ - 1) × s₁² + (n₂ - 1) × s₂² / n₁ + n₂ - 2
= 8 × (13.42)² + 8 × (13.38)²/ 16
= (8 × 180.1 + 8 × 179) / 16
sp² = 179.55
sp = 13.40
Therefore
t(s) = (479.49 - 464.7 - 2)/ 13.40 ×√1/9+1/9
= 12.79 / 13.40 × 0.4714
= 12.79 / 6.32
t(s) = 2.02
t(s) > t(c)
t(s) is rejection region.
Therefore we reject H₀
so we don't have evidence to claim to strength pf composite B is smaller than the strength of composite B at least 2 [N].
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