Two chemical companies can supply a raw material. The concentration of a particular element in this material is important. The mean concentration for both suppliers is the same, but you suspect that the variability in concentration may differ for the two companies. The standard deviation of concentration in a random sample of n1=10
batches produced by company 1 is
s1=4.7
grams per liter, and for company 2, a random sample of
n2=16
batches yields
s2=5.8
grams per liter. Is there sufficient evidence to conclude that the two population variances differ? Use α = 0.05.

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.

x= 149. 24+7=31. 180-31=149

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