Accept the null hypothesis that the two treatments

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accept the null hypothesis that the two treatments have equal variances. 5.(a) We can use the formula n = ( z α + z β ) 2 σ 2 ( μ 0 - μ A ) 2 with α = . 05 so z α = 1 . 645, β = 1 - . 90 = . 10 so that z β = 1 . 282, σ = 18. μ 0 = 10 μ g SO 2 /m 3 is the tested value and μ A = 30 is the value we want to detect (if true). We get n = 6 . 93 so that n = 7 sites need to be sampled. (b) If the test is to be done at level α = 0 . 01 and if we keep sample size n = 7, then the rejection region will be smaller and the power will decrease. In order to keep the same power, it is necessary to increase the sample size. Another way to see that is to use the formula for n above. Decreasing α means increasing z α so that n also increases. (c) 20 μ g SO 2 /m 3 will have a 90% chance of being de- tected (if true). We can use the formula above three times, plugging 7.5, 15, or 20 for μ A , and see which value gets us n = 4 * 7 = 28. Or we can see that mul- tiplying n by 4 means dividing μ 0 - μ A by 2. This was 10-30=20 in (a), so when we divide it by 2 it is 10, which mean that the detectable μ A is now μ 0 +10, i.e. 20. Summary of grades: Frequency 40 50 60 70 80 90 100 0 10 20 30
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