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L19posted_rev3 - Confidence Intervals for µ1 − µ2 •...

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Unformatted text preview: Confidence Intervals for µ1 − µ2 • Compare responses in two groups • Can arise in two ways 1. 2. 1 2 Population 1 2 Sample Size n1 n2 Variable Mean St. dev. X Y µ1 µ2 σ1 σ2 Parameter of interest θ= Natural estimator Θ= Then ¯¯ E[X −Y ] = 3 Assume independent samples. Then ¯¯ Var [X −Y ] = Finally assume both populations are normal. Then ¯¯ X −Y ∼ Then standardized Z= Note. Make sure you understand why means, variances, and Z statements hold. These are properties of linear combinations or random variables and normals. 4 Therefore C.I. construction reduces to that for a single population in which Z= replaces the previous ¯ X −µ Z= √ σ/ n As before, start with P (−zα/2 ≤ Z ≤ zα/2) = 1 − α and use simple algebra to rewrite this as ¯¯ P (X − Y − zα/2 2 σ1 σ2 ¯¯ + 2 ≤ µ1 − µ2 ≤ X − Y + zα/2 n1 n2 2 σ1 σ2 + 2 = 1−α n1 n2 to end with the confidence interval ¯¯ X − Y ± zα/2 2 2 σ1 σ2 + n1 n2 5 What if population cannot be assumed normal? For large n1, n2 Central Limit Theorem. Replace σ1 with S1 and σ2 with S2 where S1, S2 are the sample standard deviations for X, Y samples respectively. to get approxi- mate 100(1 − α)% confidence interval 2 2 S1 S2 ¯¯ + X − Y ± zα/2 n1 n2 6 Example 4.2-5. Comparison of breaking strength of two fabrics. Find 95% C.I. for difference in population means. Fabric A Fabric B Sample Size 35 30 Sample Mean 25.2 28.5 Sample St. dev. 5.2 5.9 x − y = 25.2 − 28.5 = −3.3 ¯¯ 2 s2 1 + s2 M E = z0.025 n1 n2 1/2 5.22 5.92 = 1.96 + 35 30 = 1.96(1.3903) = 2.725 C.I. = −3.3 ± 2.725 = (−6.025, −0.575) Because C.I. is on the left side of the origin, we conclude that Fabric A has a smaller breaking strength than Fabric B. 7 Suppose want to be more certain of conclusion, say 99% confident. What is the C.I. now? x − y = 25.2 − 28.5 = −3.3 ¯¯ s2 s2 1+ 2 M E = z0.005 n1 n2 1/2 5.22 5.92 = 2.57 + 35 30 = 2.57(1.3903) = 3.5731 C.I. = −3.3 ± 3.573 = (−6.873, +0.273) Now, data don’t show significant difference. 8 ...
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This note was uploaded on 01/03/2012 for the course EE 1244 taught by Professor Drera during the Fall '10 term at Conestoga.

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L19posted_rev3 - Confidence Intervals for µ1 − µ2 •...

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