Answer:
As the calculated F lies in the acceptance region therefore we conclude that there is not sufficient evidence to support the claim that the variability in concentration may differ for the two companies. Hence Ha is rejected and H0 is accepted.
Step-by-step explanation:
As we suspect the variability of concentration F - test is applied.
n1=10 s1=4.7
n2=16 s2=5.8
And α = 0.05.
The null and alternate hypothesis are
H0: σ₁²=σ₂² Ha: σ₁²≠σ₂²
The null hypothesis is the variability in concentration does not differ for the two companies.
against the claim
the variability in concentration may differ for the two companies
The critical region F∝(υ1,υ2) = F(0.025)9,15= 3.12
and 1/F∝(υ1,υ2) = 1/3.77= 0.26533
where υ1= n1-1= 10-1= 9 and υ2= n2-1= 16-1= 15
Test Statistic
F = s₁²/s₂²
F= 4.7²/5.8²=0.6566
Conclusion :
As the calculated F lies in the acceptance region therefore we conclude that there is not sufficient evidence to support the claim that the variability in concentration may differ for the two companies. Hence Ha is rejected and H0 is accepted.
