σ0=240N=8s=300We assume that the breaking strengths has normal distribution.a)α=0.05H0:σ2=σ02=2402,H1:σ2=σ02=2402We will use the following random variable:χ2=σ02(N−1)s2 where S2 is a random valueof corrected variance.This random variable has χ2-distribution with k=N−1=8−1=7 degree of freedom.Observed value:χ2≈10.9375.Critical values:χcriticalleft2=χcritical2(1−α/2;k)≈1.6899.χcriticalright2=χcritical2(α/2;k)≈16.013.Critical region: (−∞;1.6899)∪(16.013;∞)Our observed value does not fall into the critical region.So we accept H0.Variability did not change after a changein the process of manufacture.b)α=0.01Critical values:χcriticalleft2≈0.9893χcriticalright2≈20.278Critical region: (−∞;0.9893)∪(20.278;∞)Our critical value does not fall into the critical region.So we accept H0.Variability did not change after a changein the process of manufacture.
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