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Unformatted text preview: INFERENCES ABOUT μ 1 μ 2 • In many situations one wishes to compare the means of two different populations. • Let μ 1 and μ 2 denote means two populations. • Most often, the comparison of interest is the difference μ 1 μ 2 . • We will make the comparison based on a random sample from each population. Suppose samples of sizes n 1 and n 2 are drawn. 1 • Let ¯ Y 1 and ¯ Y 2 denote sample mean random variables and S 1 and S 2 be the sample standard deviation random variables computed from each sample. • Also assume that the samples are drawn independently from each population, and that the populations have the same variance σ 2 . • For large sample sizes, the random variables ¯ Y 1 and ¯ Y 2 are approximately normal . • Thus the random variable ¯ Y 1 ¯ Y 2 is also approximately normal. 2 • The mean of the random variable ¯ Y 1 ¯ Y 2 is μ 1 μ 2 and standard deviation is σ ¯ Y 1 ¯ Y 2 = s σ 2 1 n 1 + σ 2 2 n 2 where σ 2 1 and σ 2 2 are the variances of each of each population. • However, under the assumption that σ 2 1 = σ 2 2 ≡ σ 2 , the standard deviation becomes σ ¯ Y 1 ¯ Y 2 = s σ 2 n 1 + σ 2 n 2 = σ s 1 n 1 + 1 n 2 3 • An estimator of σ ¯ Y 1 ¯ Y 2 is S ¯ Y 1 ¯ Y 2 = S p s 1 n 1 + 1 n 2 • S 2 p is the pooled estimator of the common variance σ 2 of the two populations. • This is obtained under the assumption that σ 2 1 = σ 2 2 ≡ σ 2 , the common variance . • Later we will talk about what to do if we can’t reasonably make this assumption in a given situation. 4 • Consider the random variable defined as: T n 1 + n 2 2 = ( ¯ Y 1 ¯ Y 2 ) ( μ 1 μ 2 ) S ¯ Y 1 ¯ Y 2 • When sampling is from Normal distributions, the distribution of the above statistic is distributed exactly as Student’s t with df = n 1 + n 2 2 . • For large samples we may use the CLT and say that T n 1 + n 2 2 will be approximately distributed as Student’s t with df = n 1 + n 2 2 • For small samples, however, we must verify if the samples 5 were actually drawn from Normal distributions, using tools such as boxplots and normal probability plots • We shall do that before calculating confidence intervals for μ 1 μ 2 , and for testing hypotheses using the t distribution. • The pooled estimator of the common variance σ 2 , S 2 p , is obtained using the weighted average S 2 p = ( n 1 1) S 2 1 + ( n 2 1) S 2 2 n 1 + n 2 2 6 • Here S 2 1 and S 2 2 are sample variance random variables . S 2 1 = 1 n 1 1 n 1 X j =1 ( Y 1 j ¯ Y 1 ) 2 S 2 2 = 1 n 2 1 n 1 X j =1 ( Y 2 j ¯ Y 2 ) 2 • Thus a 100(1 α )% Confidence Interval for μ 1 μ 2 is ¯ y 1 ¯ y 2 ± t α/ 2 ,df · s p r 1 n 1 + 1 n 2 • Here df = n 1 + n 2 2 and s p = s ( n 1 1) s 2 1 + ( n 2 1) s 2 2 n 1 + n 2 2 7 Example 6.1 Company officials, concernerned about potency of a product retained after storage, drew a random sample of n 1 = 10 bottles from the production line and tested for potency....
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This note was uploaded on 01/23/2010 for the course STAT 213 taught by Professor Hao during the Spring '10 term at Internet2.
 Spring '10
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