Statistical Sciences 1024A/B Chapter 18: Chapter+18
24 Sep 2015
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18. 1: the standard error of the mean is s/ n = 63. 9/ 1000 = 2. 0207 minutes. 18. 2: the standard error of the mean is s/ n =15/ 10 = 4. 743 beats per minute. 18. 3: (a) t* = 2. 132. (b) t* = 2. 479. = 18. 66 4. 8096 = 13. 8504 to 23. 4696. x = 18. 66, s = 10. 2768, and. Plan: we will estimate with a 90% confidence interval. Solve: we are told to view the observations as an srs. A stemplot shows some left-skewness; however, for such a small sample, the data are not unreasonably skewed. With t* = 1. 860 (df = 8), the 90% confidence interval for is 59. 5889 1. 860. Conclude: we are 90% confident that the mean percent of nitrogen in ancient air is between 55. 71% and 63. 47%. x = 59. 5889% and s = 6. 2553% nitrogen, and. 9 t* = 2. 093 (with right-tail probability 0. 025). 18. 8: (a) df = 20 1 = 19. (b) t = 1. 84 is bracketed by.
