96 s 2 3254 x julytodecember n y 128 y 9798 s

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Unformatted text preview: P(t ( n x + n y − 2) > tc ) = α 2 Chapter 9 Example: A stock market data base contains daily closing prices for the company Johnson & Johnson for the year 1999. For the first six months (January to June) observations are recorded for 124 trading days. For the period July to December observations are available for 128 trading days. An exercise of interest is to find a 95% confidence interval estimate for the difference between the population mean closing price in the two sample periods. Stock market prices adjust rapidly to the arrival of new information. Therefore, it is reasonable to consider that the two samples are independent. Summary statistics for the closing prices for the two sample periods are: January to June n x = 124 x = $89.96 s 2 = 32.54 x July to December n y = 128 y = $97.98 s 2 = 21.45 y The pooled variance estimate is: s = 26.91 2 10 Chapter 9 A confidence interval estimate is calculated as: (89.96 − 97.98 ) ± t c 26.91 26.91 + 124 128 A t‐distribution critical value t c for a 95% interval estimate is needed. The degrees of freedom is: (n x + n y − 2) = 124 +128 − 2 = 250 The Appendix Table for the t‐distribution does not have an entry for 250 degrees of freedom. However, for α 2 = 0.05 2 = 0.025 , P(t (250) > 1.96 ) ≅ P(Z > 1.96 ) = 0.025 where Z isthe standard normal random variable. With Microsoft Excel the Function TINV(0.05, 250) returns the answer: t c = 1.969 . 11 Chapter 9 Calculations give a 95% interval estimate for the difference in means for the closing prices in the two sample periods as: [ − 9.31 , − 6.73 ] A 99% interval estimate is wider. With α = 0.01 the Microsoft Excel Function TINV(0.01, 250) gives the critical value: t c = 2.596 . For a 99% interval estimate the lower and upper limits are: [ − 9.72 , − 6.32...
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