# July to december 1999 128 y n observations econ 325

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July to December, 1999, 128 = y n observations. Econ 325 – Chapter 10 10 Denote the population means of the daily closing prices in the two sample periods by X μ and Y μ . Test the null hypothesis: Y X 0 : H μ = μ population means are equal against the one-sided alternative hypothesis: Y X 1 : H μ < μ higher mean in the second sample period That is, test: 0 : H Y X 0 = μ - μ against 0 : H Y X 1 < μ - μ The sample means and variances for the closing prices in the two sample periods are denoted by x , y and 2 x s , 2 y s . By using the statistical results given in the lecture notes for Chapter 8.2 a test statistic is calculated as: y 2 x 2 n s n s y x t + - = where 2 s is the pooled sample variance constructed by combining all the observations in the two sample periods: ) 2 n n ( s ) 1 n ( s ) 1 n ( s y x 2 y y 2 x x 2 - + - + - = The test statistic can be compared with the t-distribution with ) 2 n n ( y x - + degrees of freedom.

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Econ 325 – Chapter 10 11 For the comparison of the closing prices for the two sample periods the calculated test statistic is: 12.27 128 26.91 124 26.91 97.98 89.96 - = + - = t The degrees of freedom is (124 + 128 – 2) = 250 . For this one-tailed test, the p-value is calculated as: ) t ( P ) 250 ( 12.27 - < By symmetry, this is the same as: ) t ( P ) 250 ( 12.27 > With Microsoft Excel the p-value is calculated with the function: TDIST(12.27, 250, 1) This gives the answer: p-value = 1.05E–27 = 0.00005 10 1.05 27 < The calculated p-value is lower than any reasonable significance level to give strong evidence to reject the null hypothesis. The information in the data suggests that, for 1999, the mean daily closing price for the company Johnson & Johnson was higher in the second half of the year compared to the first half of the year.

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