Respuesta :
Answer:
We reject H₀
we don´t have evidence to claim the strength of composite B is smaller than the strength of composite B by at least 2 [N]
Step-by-step explanation:
From data, we calculate μ₁, s₁ and, μ₂, s₂ 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 should use t -student table
Test Hypothesis:
Null Hypothesis H₀ μ₂ - μ₁ >= 2
Alternative Hypothesis Hₐ μ₂ - μ₁ < 2
Assuming CI 95 % α = 5 % α = 0,05 and degree of freedom is
df = n₁ + n₂ - 2 df = 9 + 9 - 2 df = 16
Then for a one tail test t(c) = 1,746 and to compte t(s)
t(s) = ( μ₂ - μ₁ - 2 ) / sp * √ 1/n₁ + 1/n₂
sp² = ( n₁ - 1 )*s₁² + ( n₂ - 1 )*s₂² / n₁ + n₂ - 2
sp² = 8 * (13,42)² + 8 * (13,38)² / 16
sp² = (8 * 180,1 + 8 * 179) / 16
sp² = 179,55
sp = 13,40
then
t(s) = ( 479,49 - 464,7 - 2 ) / 13,40*√1/9 +1/9
t(s) = 12,79 / 13,40*0,4714
t(s) = 12,79/6,32
t(s) = 2,02
t(s) > t(c)
2,02 > 1,746
t(s) is in the rejection region, therefore we reject H₀ we don´t have evidence to claim the strength of composite B s smaller than the strength of composite B by at least 2 [N]
