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RESITATION

RESITATION. HYPOTHESIS TESTS. June 13, 2012. Example : A study was conducted to see if a new therapeutic procedure is more effective than the standard treatment in improving the digital dexterity of certain handicapped persons .

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RESITATION

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  1. RESITATION HYPOTHESIS TESTS June 13, 2012

  2. Example:A studywasconductedtoseeif a newtherapeuticprocedure is moreeffectivethanthestandardtreatment in improvingthedigitaldexterity of certainhandicappedpersons. Twenty-fourpairs of twinswereused in thestudy, one of thetwinswasrandomlyassignedtoreceivethenewtreatment, whiletheotherreceivedthestandardtherapy. At theend of theexperimentalperiodeachindividualwasgiven a digitaldexterity test withscores as follows.

  3. Since • Thevariableconcerningdigitaldexterity test scores is continuous • Thesample size is greaterthan 10 • digitaldexterity test score is normallydistributed • Therearetwogroupsandtheyaredependent Pairedsample t-test

  4. H0: d = 0 New Standard Difference 49 54 -5 56 42 14 Ha: d> 0 70 63 7 83 77 6 83 83 0 68 51 17 84 82 2 63 54 9 67 62 5 79 71 8 88 82 6 48 50 -2 52 41 11 73 67 6 52 57 -5 73 70 3 78 72 6 64 62 2 71 64 7 42 44 -2 Since, rejectH0. 51 44 7 56 42 14 40 35 5 81 73 8 Total 129 Mean 65,46 60,08 5,38 SD 14,38 14,46 5,65 t(23,0.05)=1.7139 Weconcludethatthenewtreatmentis effective.

  5. SPSS Output

  6. Example:Totest whethertheweight-reducingdiet is effective, 9 personswereselected. Thesepersonsstayed on a dietfortwomonthsandtheirweightsweremeasuredbeforeandafterdiet. Thefollowingaretheweights in kg: • Since • Thevariableconcerningweight is continous. • Thesample size is lessthan 10 • Therearetwogroupsandtheyaredependent Wilcoxonsignedranks test

  7. T = 1.5 rejectH0 , p<0.05 T = 1.5 < T(n=9,a =0.05) = 6 rejectH0 , p<0.01 T = 1.5 < T(n=9,a =0.01) = 2 Weconclude 99% confidentthatdiet is effective.

  8. Example:35 patientswereevaluatedforarrhythmiawithtwodifferentmedicaldevices. Is thereanystatisticallysignificantdifferencebetweenthediagnose of twodevices?

  9. H0: P1=P2 Ha: P1 P2 Critical z value is ±1.96 Reject H0

  10. McNemar test approach: 2(1,0.05)=3.841<5.1 p<0.05; rejectH0.

  11. SPSS Output

  12. Example: NK cellactivitywasmeasuredforthreegroups of subjects: thosewho had low, medium, andhighscores on SocialReadjustmentRatingScale. Theoriginalobservations, samplesizes, meansandstandarddeviationsaregiven in table. Is themean NK cellactivitydifferent in threegroups?

  13. Thevariableconcerning NK cellactivitymeasures is continuous. • Therearethreegroups. • Variancesareequal in threegroups. • NK cellactivityvalues in eachgrouparenormallydistributed. • Thesample size of eachgroup is greaterthan 10 One-Way ANOVA

  14. 60 50 40 30 Mean 1 SD NKA 20 10 0 N = 13 12 12 Low Moderate High GROUP

  15. H0: 1=  2=3 Ha: Not allthei areequal. SSA=SST-SSW=7182.31-3394.01=3788.30

  16. MSW=SSW/34=3394.01/27=99.82 MSA=SSA/(3-1)=3788.30/2=1894.15 F=MSA/MSW=1894.15/99.82=18.98 We conclude that not all population means are equal.

  17. Hypothesis LSD Statistical Decision H0: 1= 2 22.32>8.12, reject H0. H0: 1= 3 19.87>8.12, reject H0. H0: 2= 3 2.46<8.28, accept H0. Since n1 n2, reject H0 if

  18. Example: Hamiltondepressionscoreswasmeasuredforthreegroups of subjectsandshown in table. Is theHamiltondepressionscoresdifferent in threegroups?

  19. Since • ThevariableconcerningHamiltondepressionscore is continuous • Thesample size is lessthan 10 • Therearethreegroupsandtheyareindependent KruskalWallis Test H0: Thepopulationdistributionsareallidentical. Ha: At leastone of thepopulationstendstoexhibitlargervaluesthan at leastone of theotherpopulations.

  20. KW=7.93 > χ2(2,a=0.05)= 5.99 Reject H0 Weconcludethat at leastone of thepopulationstendstoexhibitlargervaluesthan at leastone of theotherpopulations.

  21. MultipleComparisonsTable

  22. Example:A researcherwantstoknowwhetherthemothersage is affectingtheprobability of havingcongenitalabnormality of neonatalsor not. Thecollected data is given in thetable:

  23. H0: There is no relationbetweentheage of motherand presence of congenitalabnormality. Undertheassumptionthatnullhypothesis is true: (Expectedcount)

  24. Reject H0

  25. Omitthe >35 age group

  26. H0 isaccepted

  27. Inolderagegroups, risk of having a babywithcongenitalabnormality is higherthantheotheragegroups.

  28. Example: A specialdiet program wasgivento 20 clinicallyobesepeople. Subjects’ BMI weremeasuredbeforethedietandtheyhavebeenfollowedfortwomonths. BMI measuresbeforethedietandaftertheend of eachmonthfollowingthedietaregiven in thetable. Is thediet program effective?

  29. Thevariableconcerning BMI is continuous • Thesample size is greaterthan 10 • BMI valuesarenormallydistributed • Therearethreegroupsandtheyaredependent Repeatedmeasures ANOVA

  30. Sources of variation SS df MS F Sig. Times 16.79 2 8.39 95.92 0.000 Subjects 567.72 19 29.89 Error 3.33 38 0.09 Diet program is effective on obesesubjects’ BMI.

  31. Example: Tocomparetheeffects on theclotting time of plasma of fourdifferentmethods of treatment of theplasma. Samples of plasmafrom 8 subjectswereassignedtothefourtreatments.

  32. Since • Thevariableconcerningclotting time is continuous • Thesample size is lessthan 10 • Therearefourgroupsandtheyaredependent Friedman Test

  33. 2(3,0.05)= 7.815 < Fr=14.96, p<0.05 Reject H0. Weconcludethat at leastone of thetreatmentsweredifferentfromtheothertreatments.

  34. Example:Fourradiologistshaveevalueted 12 patientwhethertheyhavelesionor not; H0: There is no differencebetweenthedecision of radiologists • Since • Thevariableis categorical (0 and 1) • Therearefourgroupsandtheyaredependent Cochran Q Test

  35. QCalculated= 9.474 > 2(3,0.05) =7.815 p<0.05 Reject H0

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