Answer:
Since the calculated t = -3.9886 falls in the critical region t ∝/2 ≥±2.101 we reject the null hypothesis and conclude that variance of impact strength is different for the two suppliers
Step-by-step explanation:
The given data is
Sample1 mean = x1` = 289.3
Sample 1 size = n1= 10
Sample 1 std. dev= s1= 22.5
Sample2 mean = x2` = 321.5
Sample 2 size = n2= 16
Sample 2 std. dev= s2= 21
1)Formulate the null and alternate hypothesis as
H0: σ₁=σ₂ against the claim Ha: σ₁ ≠σ₂
2)The test statistic is
t= (x1`- x2`) / √ s1²/n1+ s2²/n2
where degrees of freedom = d.f=
υ = [s₁²/n1 + s₂²/n2]²/ (s₁²/n1 )²/ n1-1 + (s₂²/n2)²/n2-1
3) the significance level chosen is ∝= 0.05
4) Calculating the d.f
υ = [s₁²/n1 + s₂²/n2]²/ (s₁²/n1 )²/ n1-1 + (s₂²/n2)²/n2-1
= [21²/10 + 22.5²/16] / [(21/²10)² /9 + (22.5²/16)²/15]
= 18
5) the critical region for ∝= 0.05 for 18 d.f = ±2.101 for two sided test
6) t= (x1`- x2`) / √ s1²/n1+ s2²/n2
t= 289.3- 321.5/√21²/10 + 22.5²/16
t= -32.2/√44.1+31.64
t= -32.2/8.0729
t= -3.9886
7) Conclusion
Since the calculated t = -3.9886 falls in the critical region t ∝/2 ≥±2.101 we reject the null hypothesis and conclude that variance of impact strength is different for the two suppliers
