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.

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]

answer: c

because if you look at the chart you can see a pattern and compare it to the question he simplify out your

but i think the formula is f(x)=6x

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step-by-step explanation: