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Unformatted text preview: Statistics 511: Statistical Methods Dr. Levine Purdue University Spring 2011 Lecture 18: Inferences Based on Two Samples Devore: Section 9.19.3 Apr, 2011 Page 1 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 1 2 = 0 or 1 > 2 . It may also be appropriate to estimate 1 2 and compute its 100(1 )% confidence interval Assumptions 1. X 1 ,...,X m is a random sample from a population with mean 1 and variance 2 1 2. Y 1 ,...,Y n is a random sample from a population with mean 2 and variance 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 1 2 is X Y . To standardize this estimator, we need to find E ( X Y ) and V ( X Y ) . E ( X Y ) = 1 2 , so X Y is an unbiased estimator of 1 2 . The proof is elementary: E ( X Y ) = E ( X ) E ( Y ) = 1 2 Apr, 2011 Page 3 Statistics 511: Statistical Methods Dr. Levine Purdue University Spring 2011 The standard deviation of X Y is X Y = q 2 1 m + 2 2 n The proof is also elementary: V ( X Y ) = V ( X ) + V ( Y ) = 2 1 m + 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 ( 1 2 ) q 2 1 m + 2 2 n (1) has a standard normal distribution The null hypothesis 1 2 = 0 is a special case of the more general 1 2 = . Replacing 1 2 in (1) with gives us a test statistic. Apr, 2011 Page 5 Statistics 511: Statistical Methods Dr. Levine Purdue University Spring 2011 The following summary considers all possible types of alternatives: 1. H a : 1 2 > has the rejection region z z 2. H a : 1 2 < has the rejection region z  z 3. H a : 1 2 6 = has the rejection region z z / 2 or z  z / 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 coldrolled steel gives the sample average yield strength x = 29 . 8 ksi Another sample of n = 25 specimens of twosided galvanized steel gives us y = 34 . 7 ksi. The two variances are 1 = 4 . and 2 = 5 . Note that m 6 = n ...it is not important now but will be later......
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This note was uploaded on 02/20/2012 for the course STAT 511 taught by Professor Bud during the Fall '08 term at Purdue UniversityWest Lafayette.
 Fall '08
 BUD
 Statistics

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