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Chapter 5

Chapter 5. Exercise 1. Fail to reject Ho. Exercise 2. Fail to reject Ho. Exercise 3. CI contains μ, so fail to reject is justified. . Exercise 4. pnorm (-1.265) [1] 0.1029357

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Chapter 5

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  1. Chapter 5

  2. Exercise 1 Fail to reject Ho

  3. Exercise 2 Fail to reject Ho

  4. Exercise 3 CI contains μ, so fail to reject is justified.

  5. Exercise 4 pnorm(-1.265) [1] 0.1029357 For a one tailed test, P<0.103, meaning that the probability of obtaining a sample mean of 78 or lower when μ>80 is 0.103.

  6. Exercise 5 This is a two tailed test, so P now relects p<-1.265 + 1-p>1.265 = 2(p<-1.265 ) pnorm(-1.265) [1] 0.1029357 For a two tailed test, P<0.206, meaning that the probability of obtaining a sample mean of 78 or lower or 82 or higher when μ>80 is 0.206.

  7. Exercise 6 Reject Ho

  8. Exercise 7 Reject

  9. Exercise 8 CI does not contain contains μ (130), so decision to reject is justified.

  10. Exercise 9 Yes, because the sample mean of 23 is already lower than the value stated by the null hypothesis of μ<25. To reject this hypothesis, we must have a sample mean that is higher than 25.

  11. Exercise 10 Reject Ho

  12. Exercise 11 Reject

  13. Exercise 12 A CI can be a good alternative to a two tailed test. If the CI range does not contain μ, you can reject Ho of μ=546.

  14. Exercise 13 R function: power.t.test(25,4/5,type="one.sample",alternative="one.sided",sig.level=0.01) returns: power = 0.9254881 Use case# 2 on P. 188 pnorm(1.67) [1] 0.9525403

  15. Exercise 14 R function power.t.test(36,3/8,type="one.sample",alternative="one.sided",sig.level=0.025) returns power = 0.5901872 Apply case #1 P.188 pnorm(-0.29) [1] 0.3859081

  16. Exercise 15 R function: power.t.test(49,3/10,type="one.sample",alternative="two.sided",sig.level=0.05) returns:power = 0.5390021 Third Case: pnorm(-0.14) [1] 0.44433 pnorm(-4.06) [1] 2.453636e-05

  17. Exercise 16 The sample size is two small, so failure to reject can be the result of insufficient power.

  18. Exercise 17 • power.t.test(10,2/5,type="one.sample",alternative="one.sided",sig.level=0.05) power = 0.3174914 pnorm(-0.38) [1] 0.3519727

  19. Exercise 18

  20. Exercise 19 Increase α, but this will also increase type I error, which is highly undesirable.

  21. Exercise 20 qt(0.975,24) [1] 2.063899 Only in the last case T>C Fail to reject Fail to reject Reject

  22. Exercise 21 Power depends on the variance; larger variance lower power.

  23. Exercise 22 Fail to reject Fail to reject Reject qt(0.95,15) [1] 1.75305 Only in the last case T>C

  24. Exercise 23 The sample means are consistent with Ho: μ>42, so we fail to reject without testing.

  25. Exercise 24 qt(0.975,9) [1] 2.262157 Fail to reject

  26. Exercise 25 qt(0.025,99) [1] -1.984217 T<C therefore reject.

  27. Exercise 26 C is set by n-2g-1 df=11 qt(0.025,11) [1] -2.200985 Fail to reject in all cases

  28. Exercise 27 qt(0.95,9) [1] 1.833113 Fail to reject in all cases C is set by n-2g-1 df=9

  29. Exercise 28 > qt(0.975,5) [1] 2.570582 Reject C is set by n-2g-1 df=5

  30. Exercise 29 qt(0.995,14) [1] 2.976843 Fail to reject C is set by n-2g-1 df=14

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