# 2 large populations σ 2 1 σ 2 2 unknown two

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2. Large Populations, σ 2 1 = σ 2 2 (unknown) Two Independent Samples ¤ § ¥ ƒ Confidence Interval for μ 1 - μ 2 , Independent Samples y 1 - ¯ y 2 ) ± t α/ 2 s p r 1 n 1 + 1 n 2 where s p = s ( n 1 - 1) s 2 1 + ( n 2 - 1) s 2 2 n 1 + n 2 - 2 and df = n 1 + n 2 - 2 . 4

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Sample 1 Sample 2 10.2 10.6 9.8 9.7 10.5 10.7 9.6 9.5 10.3 10.2 10.1 9.6 10.8 10.0 10.2 9.8 9.8 10.6 10.1 9.9 Table 1: Potency Reading for Two Samples ¤ § ¥ ƒ s 2 p , a weighted average The quantity s p in the confidence interval is an estimate of the standard deviation σ for the two populations and is formed by combining (pooling) in- formation from the two samples. In fact, s 2 p is a weighted average of the sample variances s 2 1 and s 2 2 . For the special case where the sample sizes are the same ( n 1 = n 2 ), the formula for s 2 p reduces to s 2 p = ( s 2 1 + s 2 2 ) / 2, the mean of the two sample variances. The degrees of freedom for the confidence inter- val are a combination of the degrees of freedom for the two samples; that is, df = ( n 1 - 1) + ( n 2 - 1) = n 1 + n 2 - 2. Recall that we are assuming that the two populations from which we draw the samples have normal distributions with a common variance σ 2 . If the confidence interval presented were valid only when these assumptions were met exactly, the estimation procedure would be of limited use. Fortunately, the confidence coefficient remains relatively stable if both distributions are mound-shaped and the sample sizes are approximately equal. More discussion about these assumptions is presented at the end of this section. / £ ¡ ¢ EXAMPLE 6.5 Company officials were concerned about the length of time a particular drug product retained its potency. A random sample, sample 1, of n 1 = 10 bottles of the product was drawn from the production line and analyzed for potency. A second sample, sample 2, of n 2 = 10 bottles was obtained and stored in a regulated environment for a period of one year. The readings obtained from each sample are given in Table 1. Suppose we let μ 1 denote the mean potency for all bottles that might be sampled coming off the production line and μ 2 denote the mean potency for all bottles that may be retained for a period of one year. Estimate μ 1 - μ 2 by using a 95% confidence interval. 5
Sample 1 Sample 2 j y 1 j = 103 . 7 j y 2 j = 98 . 3 j y 2 1 j = 1076 . 31 j y 2 2 j = 966 . 81 Then ¯ y 1 = 103 . 7 10 = 10 . 37 ¯ y 2 = 98 . 3 10 = 9 . 83 s 2 1 = 1 9 1076 . 31 - (103 . 7) 2 10 = . 105 s 2 2 = 1 9 966 . 81 - (98 . 3) 2 10 = . 058 . The estimate of the common standard deviation σ is s p = s ( n 1 - 1) s 2 1 + ( n 2 - 1) s 2 2 n 1 + n 2 - 2 = r 9( . 105) + 9( . 058) 18 = . 285 . The t -value based on df = n 1 + n 2 - 2 = 18 and α = . 025 is 2.101. A 95% confidence interval for the difference in mean potencies is (10 . 37 - 9 . 83) ± 2 . 101( . 285) p 1 / 10 + 1 / 10 or . 54 ± . 268 . We estimate that the difference in mean potencies for the bottles from the production line and those stored for 1 year, μ 1 - μ 2 , lies in the interval .272 to .808.

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