# Population is normally distributed and there is not

• Test Prep
• 5
• 97% (31) 30 out of 31 people found this document helpful

This preview shows page 2 - 5 out of 5 pages.

population is normally distributed, and there is not at least 30 participants.Step 2- Null Hypothesis:H0:µ1-µ2= 0Research Hypothesis:H1:µ1-Step 3-XX-M(X-M)210-0.330.118-2.335.43132.677.13s2X= ∑(X-M)2/ (N-1) = (0.11+5.43+7.13)/(3-1) = 12.67/2 = 6.34YY-M(Y-M)23-5.3328.41156.6744.497-1.331.77s2Y= ∑(Y-M)2/ (N-1) = (28.41+44.49+1.77)/(3-1) = 74.67/2 = 37.34dfX=N-1= 3-1 = 2Y=N-1= 3-1 = 2µ2≠ 0df
2
PS 307, Final Exam Part 2dftotal=dfX+dfY= 4s2pooled= (dfX/dftotal)s2Y+(dfY/dftotal)s2Y= (2/4)6.34 + (2/4)37.34 = 3.17+18.67 = 21.84s2Mx=s2pooled/NX= 21.84/3 = 7.28s2My=s2pooled/NY= 21.84/3 = 7.28s2difference=s2Mx+s2My= 14.56sdifference= √s2difference= √14.56 = 3.82Step 4- Critical Value for a two-tailed test withpvalue of 0.05 anddftotalof 4 = 2.365Step 5-t= (MX-MY)/sdifference= (10.33-8.33)/3.82 = 0.52Step 6-t(11) = 0.52,p< 0.05 ; fail to reject null hypothesis2.t(11) = 0.52,p< 0.05,d= =19.84Cohen’sd= (MX-MY)/spooled= (10.33-8.33)/21.84 = -19.84Question 3 - 60 PointsYou wish to compare how a low-fat diet, low-fat diet and a brisk four-milewalk each day, and a low-fat diet and a four-mile jog each day affectcholesterol levels. You randomly assign six subjects to each group. Thescores in the table below show the reduction in cholesterol count aftertwo months.1.Conduct all 6 steps of the appropriate statistic. Be sure to labelall six steps.2.Calculate and interpret the effect size for this test.3.Calculate and interpret any post-hoc test if necessary.Low-Fat DietDiet and WalkDiet and Run1014198121515181412161691318617201.Step 1- Thepopulationsare (1) individuals on a low-fat diet only, (2) individuals who are on a low-fatdiet and walk four miles every day, and (3) individuals who are on a low-fat diet and run four miles every day.Thecomparison distributionwill be anFdistribution. The hypothesis test will be a one-way between-groupsANOVA. This study meets one of the three assumptions and may meet the others: (1) The data is randomly3
PS 307, Final Exam Part 2selected, (2) we don’t know whether the population is normally distributed, and (3) We will checkhomoscedasticity.Step 2-H0:µ1=µ2=µ2Research Hypothesis:H1: At least oneµis different from anotherµ.Step 3-dfbetween=Ngroups– 1 = 3-1 = 2df1=n1– 1 = 6-1 = 5df2=n2– 1 = 6-1 = 5df3=n3– 1 = 6-1 = 5dfwithin=df1+df2+df3= 5+5+5 = 15Step 4

Upload your study docs or become a

Course Hero member to access this document

End of preview. Want to read all 5 pages?

Upload your study docs or become a

Course Hero member to access this document

Term
Spring
Professor
Blaydes,KathyJ.
Tags
Null hypothesis, Statistical hypothesis testing, Statistical significance