1 / 30

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

cecile
Download Presentation

Chapter 5

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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

More Related