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# Lecture19standard - Statistics 511 Statistical Methods Dr...

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Statistics 511: Statistical Methods Dr. Levine Purdue University Spring 2011 Lecture 18: Inferences Based on Two Samples Devore: Section 9.1-9.3 Apr, 2011 Page 1

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Statistics 511: Statistical Methods Dr. Levine Purdue University Spring 2011 z Tests and Confidence Intervals for a Difference Between Two Population Means An example of such hypothesis would be μ 1 - μ 2 = 0 or σ 1 > σ 2 . It may also be appropriate to estimate μ 1 - μ 2 and compute its 100(1 - α )% confidence interval Assumptions 1. X 1 , . . . , X m is a random sample from a population with mean μ 1 and variance σ 2 1 2. Y 1 , . . . , Y n is a random sample from a population with mean μ 2 and variance σ 2 2 3. The X and Y samples are independent of one another Apr, 2011 Page 2
Statistics 511: Statistical Methods Dr. Levine Purdue University Spring 2011 The natural estimator of μ 1 - μ 2 is ¯ X - ¯ Y . To standardize this estimator, we need to find E ( ¯ X - ¯ Y ) and V ( ¯ X - ¯ Y ) . E ( ¯ X - ¯ Y ) = μ 1 - μ 2 , so ¯ X - ¯ Y is an unbiased estimator of μ 1 - μ 2 . The proof is elementary: E ( ¯ X - ¯ Y ) = E ( ¯ X ) - E ( ¯ Y ) = μ 1 - μ 2 Apr, 2011 Page 3

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Statistics 511: Statistical Methods Dr. Levine Purdue University Spring 2011 The standard deviation of ¯ X - ¯ Y is σ ¯ X - ¯ Y = q σ 2 1 m + σ 2 2 n The proof is also elementary: V ( ¯ X - ¯ Y ) = V ( ¯ X ) + V ( ¯ Y ) = σ 2 1 m + σ 2 2 n The standard deviation is the root of the above expression Apr, 2011 Page 4
Statistics 511: Statistical Methods Dr. Levine Purdue University Spring 2011 The Case of Normal Populations with Known Variances As before, this assumption is a simplification. Under this assumption, Z = ¯ X - ¯ Y - ( μ 1 - μ 2 ) q σ 2 1 m + σ 2 2 n (1) has a standard normal distribution The null hypothesis μ 1 - μ 2 = 0 is a special case of the more general μ 1 - μ 2 = Δ 0 . Replacing μ 1 - μ 2 in (1) with Δ 0 gives us a test statistic. Apr, 2011 Page 5

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Statistics 511: Statistical Methods Dr. Levine Purdue University Spring 2011 The following summary considers all possible types of alternatives: 1. H a : μ 1 - μ 2 > Δ 0 has the rejection region z z α 2. H a : μ 1 - μ 2 < Δ 0 has the rejection region z ≤ - z α 3. H a : μ 1 - μ 2 6 = Δ 0 has the rejection region z z α/ 2 or z ≤ - z α/ 2 . Apr, 2011 Page 6
Statistics 511: Statistical Methods Dr. Levine Purdue University Spring 2011 Example I Analysis of a random sample of m = 20 specimens of cold-rolled steel gives the sample average yield strength ¯ x = 29 . 8 ksi Another sample of n = 25 specimens of two-sided galvanized steel gives us ¯ y = 34 . 7 ksi. The two variances are σ 1 = 4 . 0 and σ 2 = 5 . 0 Note that m 6 = n ...it is not important now but will be later... The normality suggestion is based on some exploratory data analysis The hypotheses are H 0 : μ 1 - μ 2 = 0 and H a : μ 1 - μ 2 6 = 0 Apr, 2011 Page 7

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Statistics 511: Statistical Methods Dr. Levine Purdue University Spring 2011 Example I The test statistic is z = ¯ x - ¯ y q σ 2 1 m + σ 2 2 n For a level of significance α = 0 . 01 , z α/ 2 = z . 005 = 2 . 58 and the rejection regions is z ≤ - 2 . 58 or z 2 . 58 .
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Lecture19standard - Statistics 511 Statistical Methods Dr...

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