# ch 15 - Chapter 15 Solutions 15.1(a If = 12 the...

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Chapter 15 Solutions 15.1. (a) If µ = 12, the distribution is approximately Normal with mean µ = 12 g / dl and standard de- viation σ/ 50 . = 0 . 2263 g / dl. (b) A result like x = 11 . 3g / dl lies out toward the low tail of the curve, while 11 . 8g / dl is fairly close to the middle. If µ = 12 g / dl, observing a value similar to 11 . / dl would not be too surprising, but 11 . / dl is less likely, and it therefore provides some evidence that µ< 12 g / dl. 12.23 12.45 12.68 11.77 11.55 11.32 12 11.3 11.8 15.2. (a) If µ = 115, the distribution is approximately Normal with mean µ = 115 and standard deviation 25 = 6. (b) The actual result lies out toward the high tail of the curve, while 118.6 is fairly close to the middle. If µ = 115, observing a value similar to 118.6 would not be too surprising, but 125.8 is less likely, and it therefore provides some evidence that µ> 115. 121 127 133 109 103 97 115 125.8 118.6 15.3. H 0 : µ = 12 g / dl ; H a : 12 g / dl. 15.4. H 0 : µ = 115 ; H a : 115. 15.5. H 0 : µ = 51 mpg ; H a : 51 mpg. 15.6. H 0 : µ = 20 min ; H a : µ 6= 20 min. 15.7. Hypotheses should be stated in terms of µ , not x . Note: Students who think about this problem a bit more might also point out that 1000 g (2.2 lb) is a dangerously low birth weight; babies smaller than this are classifed as extremely low birth weight (ELBW). 15.8. For x = 0 . 3, z = 0 . 3 0 1 / 10 . = 0 . 95. 15.9. For x = 11 . / dl, z = 11 . 3 12 1 . 6 / 50 . =− 3 . 09. For x = 11 . / dl, z = 11 . 8 12 1 . 6 / 50 . 0 . 88. 15.10. For x = 118 . 6, z = 118 . 6 115 30 / 25 = 0 . 60. For x = 125 . 8, z = 125 . 8 115 30 / 25 = 1 . 80. 175

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176 Chapter 15 Tests of SigniFcance: The Basics 15.11. The applet’s output for n = 18, σ = 60, and x = 17 is shown on the right. (This Normal curve has mean 0 and standard deviation σ/ 18 . = 14 . 14.) 15.12. In Exercise 15.8, the test statistic z was found to be 0.95, so for the alternative µ> 0, the P -value is P ( Z 0 . 95 ) = 0 . 1711. 15.13. For x = 11 . 3g / dl, z =− 3 . 09 and P = P ( Z < 3 . 09 ) = 0 . 0010. For x = 11 . 8g / dl, z 0 . 88 and P = P ( Z < 0 . 88 ) = 0 . 1894. The ±rst P -value tells us that values of x similar to 11.3 g / dl would rarely occur when H 0 is true, while the second P -value is larger, indicating that results similar to this one give little reason to doubt H 0 . 15.14. For x = 118 . 6, the P -value is P ( x > 118 . 6 ) = P ( Z > 0 . 6 ) = 0 . 2743. For x = 125 . 8, P = P ( x > 125 . 8 ) . = P ( Z > 1 . 80 ) = 0 . 0359. A small P -value (such as 0.0359) tells us that values of x similar to 125.8 would rarely occur when H 0 is true, while P = 0 . 2743 indicates that results similar to 118.6 give little reason to doubt H 0 . 15.15. For x = 11 . / dl, we found z 0 . 88 and P = 0 . 1894; this is not signi±cant at α = 0 . 05 or α = 0 . 01. 15.16. For x = 125 . 8, we found z = 1 . 80 and P = 0 . 0359; this is signi±cant at α = 0 . 05, but not at α = 0 . 01. 15.17. (a) The results observed in this study would rarely have occurred by chance if vitamin C were ineffective. (b) P < 0 . 01 means that results similar to those observed would occur less than 1% of the time if vitamin C supplements had no effect. 15.18. We test H 0 : µ = 5mgversus H a : µ< 5mg. The test statistic is z = 4 . 62 5 0 . 92 / 45 . 2 . 77, and the P -value is P = P ( Z < 2 . 77 ) = 0 . 0028. This is very strong evidence that 5mg . 15.19. State: Do student SATM scores improve on their second attempt?
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ch 15 - Chapter 15 Solutions 15.1(a If = 12 the...

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