Sample Size for Comparing Two MeansThis is useful for planning studies with two independent groups. Assume 12 and 22 are known. Want equal sample size in each group. Level of significance = . Two-sided test. Power = 1 - . The appropriate sample size for each group is |μ|μΔΔzzσσnβα/12221212221wherewhere μ1 and μ2 are the two population meansσ12 and σ22 are the two population variances.Note that the sample size depends on Δ, the minimum detectable difference, the two variances, the level of significance (), and power (1 – β).Example: We want to test whether there is a significant difference between the mean blood-clotting times of persons using two different drugs. We wish to test for a significant difference, at the 0.05 level of significance, and with a 90% chance of detecting a true difference between population means as small as 0.5 minutes. Suppose the within-population variability, based on previous studies, is 0.52 min2 for each population. What sample size is needed for each drug?12 = 0.52 min2, 22 = 0.52 min2, z1-/2 = z.975 = 1.96, z1- = z.90 = 1.28, = .5 minutes 446743254976100415281961525222221212221or.........Δzzσσnβα/Sometimes, we expect a different sample size in each group. We may want one group to have k times the number of subjects in the other group where k 1. Let n2 = k * n1. Therefore, 121222121222122212122211nnk|μ|μΔΔzzσkσnΔzz/kσσnβα/βα/Example: For the same example, suppose we want 2 times as many subjects in sample 2 as in sample 1.12 = 0.52 min2, 22 = 0.52 min2, z1-/2 = z.975 = 1.96, z1- = z.90 = 1.28, = .5 minutes, k = 2 66332337532251881288252437852819612525222222212122211*n........./..Δzz/kσσnβα/orNote: If the variances of the two groups are the same, then for a given , , the smallest total sample size is achieved by the equal sample size rule. Note: For a one-tail test, substitute 1- for 1 - /2 in all above equations. This will result in smaller sample sizes. PowerSometimes we have a predetermined sample size and want to know how much power there is for detecting a specific alternative. Assume 12 and 22 are known. We want to conduct a two-tailed test with significance level . We want to know how much power there will be to test the hypothesisH0: 1 = 2H1: 1 2 for the specific alternative |1 - 2| = 1222211212221121wherePrPower/nnk/kσσΔnzzor/kσσΔnzΦα/α/For the example, suppose n1 = 60 and n2 = 60, and we want to estimate the power using a two-sidedtest with significance level = 0.05.12 = 0.52 min2, 22 = 0.52 min2, z1-/2 = z.975 = 1.96, = .5 minutes, k = 1, n1 = 60, n2 = 60 96718378179777396101980418729833396115252560961222121..Φ..Φ...Φ/..)(..Φ/kσσΔnzΦα/PowerThere is a 96.7% chance of detecting a significant different using a two-sided test with significancelevel = 0.05. Note: To calculate power for a one-sided test, substitute 1 - for 1 - /2 in each
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