STAT. FOR BEHAVIORAL SCIENCES WEBASSIGN
STAT. FOR BEHAVIORAL SCIENCES WEBASSIGN
3rd Edition
ISBN: 9781544317823
Author: PRIVITERA
Publisher: Sage Publications
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Chapter 14, Problem 24CAP
To determine

Complete the F table for the given study.

Expert Solution & Answer
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Answer to Problem 24CAP

The completed F table is,

Source of VariationSSdfMSFobt
Exam1902956.33
Student Class603201.33
Exam × Student Class5406906.00
Error1,62010815 
Total2,410119  

Explanation of Solution

Calculation:

From the information given that, the study of grades for each participants are considered in which the students of four categories from local college were freshmen, sophomores, juniors, and seniors. That is, q=4, who were in the same statistics class were given one of three tests (recall, recognition, or a mix of both). That is, p=3. Also, the sum of squares for student class is 60, the sum of squares for exam × student class is 540, the total sum of squares is 2,410, the degrees of freedom for exam are 2, and the mean sum of squares for exam is 95.

The formulas for computing the F table are,

Source of VariationSSdfMSFobt
Factor ASSAp1SSAdfAMSAMSE
Factor BSSBq1SSBdfBMSBMSE
A×BSSA×B(p1)(q1)SSA×BdfA×BMSA×BMSE
Error (within groups)SSEpq(n1)SSEdfE
TotalSSTnpq1

Substituting dfExam=2,MSExam=95 in the sum of squared for main effect exam formula

SSExam=95×2=190

The sum of squared for main effect exam is 190.

Substituting q=4 in the degrees of freedom for main effect student class formula

dfS=(41)=3

The degrees of freedom for main effect student class are 3.

Substituting dfS=3,SSS=60 in the mean square for main effect student class formula

MSS=603=20

The mean square for main effect student class is 20.

Substituting p=3,q=4 in the degrees of freedom for exam × student class formula

dfExam×S=(31)(41)=2×3=6

The degrees of freedom for interaction exam × student class are 6.

Substituting dfExam×S=6,SSExam×S=540 in the mean square for exam × student class formula

MSExam×S=5406=90

The mean square for exam × student class is 90.

Substituting p=3,q=4,n=10 in the degrees of freedom for error formula

dfError=3×4×(101)=3×4×9=108

The degrees of freedom for error are 108.

Substituting SSExam=190,SSS=60,SSExam×S=540,SST=2,410 in the sum of squared for error formula

SSE=SST(SSExam+SSS+SSExam×S)=2,410(190+60+540)=2,410790=1,620

The sum of squared for error is 1,620.

Substituting dfE=108,SSE=1,620 in the mean square for error formula

MSE=1,620108=15

The mean square for error is 15.

Substituting dfExam=2,dfS=3,dfExam×S=6,dfE=108 in the total degrees of freedom formula

dfT=dfExam+dfS+dfExam×S+dfE=2+3+6+108=119

The total degrees of freedom are 119.

Substituting MSExam=95,MSE=15 in the F obtained value for exam formula

FExam=9515=6.33

The F obtained value for exam is 6.33.

Substituting MSS=20,MSS=15 in the F obtained value for student class formula

FS=2015=1.33

The F obtained value for student class is 1.33.

Substituting MSExam×S=90,MSE=15 in the F obtained value for interaction formula

FExam×S=9015=6.00

The F obtained value for interaction is 6.00.

The completed F table is,

Source of VariationSSdfMSFobt
Exam1902956.33
Student Class603201.33
Exam × Student Class5406906.00
Error1,62010815 
Total2,410119  

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