Question

The breaking strength of hockey stick shafts made of two different graphite-Kevlar composites yield the following results (in newtons):

Composite A: 487.3 444.5 467.7 456.3 449.7 459.2 478.9 461.5 477.2
Composite B: 488.5 501.2 475.3 467.2 462.5 499.7 470.0 469.5 481.5
485.2 509.3 479.3 478.3 491.5
Assuming normality, can you conclude that the mean breaking strength is smaller for hockey sticks made from composite B by at least 2 newtons? Carry out the appropriate test by hand.

Answer 1

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]

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]