By Christopher F. Baum

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**Extra resources for A review of Stata 8.1 and its time series**

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Stem-and-leaf display for Problem 6-15 cycles: unit = 100 1 1 5 10 22 33 (15) 22 11 5 2 1|2 represents 1200 0T|3 0F| 0S|7777 0o|88899 1*|000000011111 1T|22222223333 1F|444445555555555 1S|66667777777 1o|888899 2*|011 2T|22 b) No, only 5 out of 70 coupons survived beyond 2000 cycles. 6-19. 0 Stem-and-leaf display for Problem 6-25. 0 Note: Minitab has dropped the value to the right of the decimal to make this display. 15 Section 6-5 6-43. 931 c) 5 PMC 4 3 2 6-47. 692 c) 52 51 temperatur 50 49 48 47 46 45 44 43 The data appear to be slightly skewed.

05. 866 ≤ µ Since the lower limit of the CI is just slightly below 100, we are confident that the mean speed is not less than 100 m/s. 9-29 a) 1) The parameter of interest is the true average battery life, µ. 05. 05 − 0 . 05. Section 9-3 9-31 a. 1) The parameter of interest is the true mean female body temperature, µ. 4821 n=25 t0 = 98 . 264 − 98 . 6 = − 3 . 48 0 . 05. 24, and n = 25, we get β ≅ 0 and power of 1−0 ≅ 1. 9), * n = 20 . 5 and n=11. 025, 24 n n 0 . 4821 0 . 4821 98 .

1759 Supplemental Exercises 7-49. 14. 93%. 14. 36%. 14. 78%. ) The 99% CI on the mean calcium concentration would be longer. b). No, that is not the correct interpretation of a confidence interval. 82 is either 0 or 1. c). Yes, this is the correct interpretation of a confidence interval. The upper and lower limits of the confidence limits are random variables. 005 n n 3162 3162 . . 96 solve for n. 78 3 Therefore, n=267. ) The data appear to be normally distributed based on examination of the normal probability plot below.