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Worked solution

Compare two independent sample means

Two firms A and B manufacture components with mean breaking strengths 6 kN and 5.5 kN and standard deviations 0.4 kN and 0.2 kN respectively. Random samples of 100 components from A and 50 from B are tested. Find the probability that the mean strength from A is between 0.45 kN and 0.55 kN more than that from B.

CJC · Lecture · Example 9

Apply CLT separately, then subtract

Let A,BA,B denote individual component strengths. For independent observations within and between the two samples, Aˉ≈N(6,0.42/100)\bar A\approx N(6,0.4^2/100) and Bˉ≈N(5.5,0.22/50)\bar B\approx N(5.5,0.2^2/50) by CLT. Thus D=Aˉ−Bˉ≈N(0.5,0.42100+0.2250)=N(0.5,0.0024).\begin{aligned}&D\\ & =\bar A\\ &{}-\bar B\approx N\left(0.5,\frac{0.4^2}{100}+\frac{0.2^2}{50}\right)\\ & =N(0.5,0.0024).\end{aligned}

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Use both interval endpoints

P(0.45<D<0.55)≈Φ(0.050.0024)−Φ(−0.050.0024)=0.6925658341≈0.693.\begin{aligned}&P(0.45<D<0.55)\approx\Phi\left(\frac{0.05}{\sqrt{0.0024}}\right)\\ &{}-\Phi\left(\frac{-0.05}{\sqrt{0.0024}}\right)\\ & =0.6925658341\approx0.693.\end{aligned} Variances add in an independent difference; subtracting the means does not subtract the variances.

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Common mistakes

The source implicitly uses independence between samples as well as within each sample. State it before adding variances.

Read the hint