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ANOVA

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Chapter13

1. What does each of the following terms mean?

1. ANOVA

ANOVAis a statistical tool known as Analysis of Variance. The tool helpsin determining whether significant differences happen in anexperiment that contains two or more conditions

1. one-way design

thisis a design that is carried out when one and only one independentvariable is examined in an experiment

1. factor

thisis an independent variable in an experiment

1. level

thisrefers to each condition of an independent variable

1. treatment

justlike level, treatment also refers to each condition of theindependent variable

1. between subjects

thisis when an experiment tests for a factor with the se of samples thatare independent in all conditions.

1. within subjects.

Thisis when a factor is examined using samples that are related in alllevels

7.(a)When is it necessary to perform post hoc comparisons? Why?

Posthoc comparison is only performed when Fobt is significant and k isgrater than 2.In this case, the Fobt symbolizes that only two or more sample meansdiffer significantly. In this case, the post hoc test determineswhich treatment differ significantly

(b)When is it unnecessary to perform post hoc comparisons?

Posthoc comparison is a necessity when Fobt is not significant or simplywhen k=2

9.Whatdoes ƞ2indicate?

Itis a new correlation coefficient that describes the effect size, theproportion of variance in variables that are dependent which isaccounted for by changing the treatment of a factor in an ANOVA.

Toelaborate this it is good to give the formula of ƞ2

Thereforeƞ2 = SSbn/SStot where one can obtain SSbnand SStot from the tabke of ANOVA

18.Aresearcher investigated the number of viral infections peoplecontract as a function of the amount of stress they experiencedduring a 6-month period. She obtained the following data:

Amountof Stress

NegligibleStress Minimal Stress Moderate Stress Severe Stress

2 4 6 5

1 3 5 7

4 2 7 8

1 3 5 4

1. What are H0 and Ha?

Ho:the mean of viral infection people is same for all the differentamount of stress

Ha:the mean of viral infection people is not same for all the differentamount of stress

1. Compute Fobt and complete the ANOVA summary table.

Firstofall we calculate the totals in the table below

 Negligible stress Minimal stress Moderate stress Severe stress Totals 2 4 6 5 1 3 5 7 4 2 7 8 1 3 5 4 ∑X= 8 ∑X2=22 N1 = 4 Xbar =2 12 38 4 3 23 135 4 5.75 24 154 4 6 ∑Xtot = 67 ∑X2tot= 349 N=4 K=4

SStot=∑X2tot– (∑Xtot)2/N

=349-672/16

=349– 280.56

=68.44

SSbn= ∑[(∑Xi)2/ni]- (∑Xtot)2/N

=[64/4+144/4 +529/4 + 576/4] – 4489/16

=328.25 – 280.56

=47.69

Thereforethe sum of squares within groups is given as

SSwn= SStot– SSbn

=68.44 – 47.69 = 20.75

Allthese data will help us summarize the ANOVA table below

 Source Sum of squares Degrees of freedom Mean square F Between 47.69 k-1=3 MSbn =47.69/3 = 15.9 Fobt = MSbn/MSwn Within 20.75 N-k = 12 MSwn= 20.75/12 = 1.73 Total 68.44 N-1 = 15 Fobt = 15.9/1.73 = 9.19
1. With _ 5 .05, what is Fcrit?

SingF-distribtion tables we have Fcrit at alpha = 0.05, df (3, 12) as3.49

Consideringthe results of part c and b, we see that Fobt is greater than Fcrit,thus we reject the null hypothesis. This enables us to report that thmean number of viral infection people is not same for all thedifferent amount of stress at 0.05 significant level

1. Perform the appropriate post hoc comparisons.

Fromb and c, k = 4 Dfwn= 12 and alpha = 0.05

usingvalues of studentized range statistic table, qk = 4.2

HSD= qk()

=4.2 ()

=4.2* 0.6575 = 2.7615

Theabsolute differences between each pair of means are as follows

Negligibleand minimal = 2-3 = 1

Negligibleand moderate = 2 – 5.75 = 3.75

Minimaland moderate = 3 – 5.75 = 2.75

Minimaland severe = 3 – 6 = 3

Moderateand severe = 5.75 – 6 = 0.25

Negligibleand severe = 2 – 6 = 4

ItIs clear that the absolte differences 3, 3.75 and 4 are greater thanHSD. Therefore the means of pairs Minimal and severe, Negligible andmoderate, and Negligible and severe differ significantly

Frome above, we can conclude that there is no significant relationshipbetween the number of viral infection people and adjacent levels ofstress.

1. Compute the effect size and interpret it.

Effectsize = SSbn/SStot= 47.69/68.44 = 0.697

Thus,the proportion of variance in dependent scores acconted for by thelevel of the independent variable is 69.7%

1. Estimate the value of µ that is likely to be found in the severe stress condition.

Generally,the sample mean is the estimate of the population mean. Therefore,the estimate of the mean number of viral infected people under severestress = 6. From the table in b.

19.Hereare data from an experiment studying the effect of age on creativityscores:

Age4Age 6 Age 8 Age 10

39 97

5 11 12 7

7 14 96

4 10 84

3 10 95

1. Compute Fobt and create an ANOVA summary table.

Beforewe compute the Fobt and ANOVA table, we have to state the null andalternative hypothesis.

Ho:there is no effect of age on creative scores

H1:there is significant effect of age on creative scores

Totals

 Age 4 Age 6 Age 8 Age 10 Totals 3 9 9 7 5 11 12 7 7 14 9 4 4 10 8 4 3 10 9 5 ∑X= 22 ∑X2=108 N1 = 5 Xbar =4.4 54 598 5 10.8 47 451 5 9.4 29 175 5 5.8 ∑Xtot = 152 ∑X2tot= 1332 N=20 K=4

Usingthis table

SStot=∑X2tot– (∑Xtot)2/N

1332– 1522/20

=1332-1155.2

=176.8

SSbn= (222/5+542/5+472/5+292/5)– 1522/20

=1290 – 1155.2

=134.8

SSwn= 176.8 – 134.8

=42

ANOVAtable

 Source Sum of squares Degrees of freedom Mean square F Between 134.8 k-1=3 MSbn =134.8/3 = 44.933 Fobt = MSbn/MSwn Within 42 N-k = 16 MSwn= 42/16 = 2.625 Total 176.8 N-1 = 19 Fobt = 44.93/2.625 = 17.117
1. With α= .05, what do you conclude about Fobt?

Fromthe F-tables, Fcrit at 0.05 significance level for (3, 16) dfs = 3.24

Thereforefrom a above it is clear that Fobt is greater than Fcrit, thus wereject the null hypothesis. We therefore conclude that there is asignificant effect of age on creative scores

1. Perform the appropriate post hoc comparisons.

Turkey’sHSD test is an appropriate Post Hoc test here since all number ofsamples are equal

K=5,dfwn=16, alpha=0.05

Fromvales of Standardized Range Stat table, qk = 4.05

Xbarsare 4.4, 10.8. 9.4, and 5.8 respectively. MSwn=2.625 and n = 5

HSD= 4.05 (()

=4.05*0.7255

=2.935

Theabsolute differences between each pair are as follows

Age4/age6 = 6.4, age 6/age 8 = 1.4,

Age8/age 10 = 3.6, age 4/age 8 = 5

Age6/age 10 = 5, age4/age10 = 1.4

Wetherefore see that the absolute differences 3.6, 5 and 6.4 are biggerthan HSD. Therefor the means of the pairs of Age4/age 6, Age 8/age10, age 4/age 8 and Age 6/age 10 differ significantly. Thereforethere is no significant effect on the creative scores of age 6 andage 8 and age 4 and age 10.

ƞ2= SSbn/SStot

=134.8/176.8

=0.763

Thusthe proportion of variance in dependent scores accounted for by thelevel on the independent variable is 76.3%

1. Statistically, how important is the relationship in this study?

Fromd above, we can conclude that there is no significant relationshipbetween the effect of age on creativity scores and adjacent levels ofage.

1. Describe how you would graph these results.

Whengraphing, put the age factor on the X-axis and the mean creativescores on the Y-axis. Then plot the mean scores corresponding to thegiven ages and connect them with straight lines.

20.Ina study in which k=3,n=21,1= 45.3,Ẋ2= 16.9,and Ẋ3= 8.2,you compute the following sums of squares:

Source Sum of Squares df Mean Square F

Between 147.32 _ _ _

Within 862.99 _ _

Total 1010.31_

1. Complete the ANOVA summary table.

 Source Sum of squares Degrees of freedom Mean square F Between 147.32 k-1=2 MSbn =147.32/2 = 73.66 Fobt = MSbn/MSwn Within 862.99 N-k = 18 MSwn= 862.99/18 = 47.94 Total 1010.31 N-1 = 20 Fobt = 73.66/47.94 = 1.5364
1. (b) With α = .05, what do you conclude about Fobt?

Levelof significance = 0.05

UsingF-distribution tables, the critical value of F at α= .05 for (2, 18) degrees of freedom as Fcrit = 3.55

Wecan see that Fobt = 1.5364 is less than Fcrit = 3.55

Thisimplies that we fail to reject the null hypothesis

1. Report your results in the correct format.

Sincewe fail to reject the null hypothesis, the proportion means of allthe given three level are equal.

1. Perform the appropriate post hoc comparisons.

Itis hard to perform the post hoc comparison here since the teststatistic Fobt is not significant.

Sincethe test statistic is not significant, we conclude that all the meansare equal

1. What is the effect size in this study, and what does this tell you about the influence of the independent variable?

2. ƞ2 = SSbn/SStot

=147.32/1010.31 = 0.15

Theproportion of variance in dependent scores accounted for by the levelof independent variable is approximately 15%.

22.Aresearcher investigated the effect of volume of background noise onparticipants’ accuracy rates while performing a difficult task. Hetested three groups of randomly selected students and obtained thefollowing means and sums of squares:

LowVolumeModerate VolumeHighVolume

X61.5 65.5 48.25

n4 5 7

Source Sum of Squares df Mean Square F

Betweengroups652.16 _ __

Withingroups612.75 _ _

Total 1264.92_

1. Complete the ANOVA.

Degreesof freedom between = k – 1 = 3 -1 = 2

Degreesof freedom within = N – k = 16 – 3 = 13

Degreeof freedom within groups = N – 1

=16 – 1 = 15

MSbn=652.16/2 = 326.08

MSwn=612.75/13 = 47.13

Fobt= 326.08/47.13 = 6.92

ANOVAtable is summarized as below

 Source Sum of squares Degrees of freedom Mean square F Between 652.16 k-1=2 326.08 Fobt = MSbn/MSwn Within 612.75 N-k =13 47.13 Total 1264.92 N-1 = 15 6.92
1. At α = .05, what is Fcrit?

Fcritat α= .05 and DF (2, 13) = 3.8

1. Report the statistical results in the proper format.

Sincethe calculated F (6.92) is greater than the tabulated F (3.8) wereject the null hypothesis implying that the means of the conditionsare significantly different.

1. Perform the appropriate post hoc tests.

Sincethe number of samples in all levels are not equal, the appropriatepost hoc test is the Fisher’s protected test

ForX1 and X2

Tobt=(X1bar– X2bar)/

=(61.5 – 65.5)/ /

=- 0.8686

ForX2 and X 3

Tobt= (X2bar– X3bar)/

=(65.5 – 48.25)/

=4.29

ForX1 and X3

Tobt= (61.5 – 65.5)

=3.07

Usingthe t-table, tcrit at alpha = 0.05 and df = 13 is 2.16. from this weobserve that tobt (-0.899) is less that Tcrit (2.16) thus we concludethat the means from low volume and moderate volume do not differsignificantly

Forthe second part, tobt (4.29) is greater than Tcrit(2.16) thereforethe means from moderate volume and high volume are significantlydifferent

Forthe last lot, the means of low volume and high volume are alsosignificantly different since the calculated vale of t (3.07) islarger than the tabulated value of t (2.16)

1. Compute the effect size and interpret it.

Effectsize ƞ2= SSbn/SStot

=652.16/1264.92 = 0.52

Therefore,the effect size implies that we are approximately 52% of differencesin these volumes are accounted being caused by altering thetreatments of perceived difficulty

Chapter14

7.Fora 2X 2 ANOVA,describe the following in words:

(a)the statistical hypotheses for factor A,

Nullhypothesis, Ho: there is no significance differences existing betweenthe mean treatments of factor A in the population

StatisticallyHo: µA1= µA2A3

µrepresentsthe level means from factor A respectively

Alteernativehypothesis, Ha : not all µAareequal

(b)the statistical hypotheses for factor B,

Ho:there is no significant difference that exist between the meantreatments of factor B in the population

Ho:µB1= µB2B3

Ha:not all : µBareequal

(c)the statistical hypotheses for AX B.

Ho:there is no interaction that happens between the mean levels offactor A x B in the population

Ha:form an interaction effect

11.Beloware the cell means of three experiments. For each experiment, computethe main effect means and decide whether there appears to be aneffect of A, B, and/or

AX B.

Study1

A1 A2

 2 4 12 14

B1

B2

Study2

A1 A2

 10 5 5 10

B1

B2

Study3

A1 A2

 8 14 8 2

B2

B2

Fromstudy 1 we get the main effect of A1 and A2 being :

XA1bar= (2+12)/2 = 7

XA2bar= (4+14)/2 = 9

Andthe main effect of B1 and B2 are:

XB1bar= (2+4)/2 = 3

XB2bar= (12+14)/2 = 13

Aninteraction effect occurs when the effect of altering one factor isnot the same for each level of the other factor. In the first study,we have only effects of A and B, but not the interaction effect sincethe volume have similar effect for factor treatments B1 and B2

Inthe study 2

Themain effects of A1 and A2 are:

XA1bar= (10+5)/2 = 7.5

XA2bar= (5+10)/2 = 7.5

Forfactor B

XB1bar= (10+5)/2 = 7.5

XB2bar= (5+10)/2 = 7.5

Thisimplies that there is no effects on A and B since their means are notchanging, but there is an interaction effect since the increasingvolume does not have similar effect for factor levels B1 and B2

Forthe last study, the main effect A1 and A2 are:

XA1bar= (8+8)/2 = 8

XA2bar= (14+2)/2 = 9

ForB1 and B2 are:

XB1bar= (8+14)/2 = 11

XB2bar= (8+2)/2 = 5

Instudy 3, there is no effect of A. this is because the mean effect ofA1 and A2 are not changing for two levels. However, we have theinteraction effect since the increasing volme does not have similareffect for factor B1 and B2 and again we have the effect of B

16.Inan experiment, you measure the popularity of two brands of softdrinks (factor A), and for each brand you test males and females(factor B). The following table shows the main effect and cell meansfrom the study:

FactorA

LevelA1:Brand XLevel A2: Brand Y

LevelA1level A2

BrandX Brand Y

 14 23 25 1 2

LevelB1

Males

FactorB

LevelB2

Females

1. Describe the graph of the interaction when factor A is on the X axis.

Whenfactor A in on the X-axis, it implies that there is an interactioneffect that occur in the study and the effect of altering factor A isnot similar in all levels of the factor B. the reason behind this isthat increasing the popularity of the two brands does not have sameinfluence for males as it is for females

1. Does there appear to be an interaction effect? Why?

Yes.There is an interaction effect since the Two-way interaction effectexists when the association of one factor and dependent scoreschanges with the level of the other factor that exists. Looking atthe data, factor A does not depend on the treatment of factor B andthe effect of altering factor A is not similar in each level offactor B because increasing the fame of the two brands does not havesame effect for male as it does for females.

1. What are the main effect means and thus the main effect of changing brands?

Themain effect means are got through a procedure in which one collapsesacross the levels of the other factor. For instance, collapsingacross factor B fabricates the main effect means for factor B. fromthe study above, we calculate the overall mean of each level of A andexamine the means of each column

XA1bar= (14+25)/2 = 19.5

XA2bar= (23+12)/2 = 17.5

1. What are the main effect means and thus the main effect of changing gender?

Themain effect mean for B

XB1bar= (14+23)/2 = 18.5

XB2bar= (25+12)/2 = 18.5

1. Why will a significant interaction prohibit you from making conclusions based on the main effects?

Basingon the data and plotting a graph, the only thing that will be seen isthe interaction effects of brands and gender. Whereas in the case ofsignificance, the main effects are contradicted when the interactionis significant.

19.Astudy compared the performance of males and females tested by eithera male or a female experimenter. Here are the data:

FactorA Participants

LevelA1level A2

MalesFemales

 6 11 9 10 9 8 14 17 16 19 8 10 9 7 10 4 6 5 5 7

LevelB1

MalesExperimenter

FactorB Experimenter

LevelB2

FemalesExperimenter

1. Using α = .05, perform an ANOVA and complete the summary table.

Nullhypothesis: there is no mean effect and interaction effect

α1=α2=0, β1= β2 = 0, (α β)11=(α β)12=(α β)21=(α β)22=0

Ha:there are main and interaction effects being significant

Totals

FactorA Participants

LevelA1level A2

MalesFemales

 6 11 9 10 9 Xbar = 9 ∑X= 45 ∑X2=419 N=5 8 14 17 16 19 Xbar =14.8 ∑X= 74 ∑X2=1166 N=5 8 10 9 7 10 Xbar =8.8 ∑X= 44 ∑X2=394 N=5 4 6 5 5 7 Xbar = 5.4 ∑X= 27 ∑X2=151 N=5

LevelB1

MalesExperimenter

FactorB Experimenter

LevelB2

FemalesExperimenter

Forfactor B

Formale ∑X =119, Xbar fr male = 11.9

Forfemales, ∑X =71, Xbar for female = 7.1

Forfactor A

Formale, ∑X =89, Xbar = 8.9

Forfemale, ∑X= 101, Xbar =10.1

Fromthe above we can get the grand total whereby

totalsum = 190, total sum squared = 2130, N=20

sumof squares we have

=2130 = 2130– 1805 = 325

Sumof squares between the groups for factor A

=892/10+ 1012/10– 1902/20= 7.2

Forfactor B

=1192/10+ 712/10– 1902/20= 115.2

Totalsum of squares between groups

=452/5+742/5+442/5+272/5– 1902/20= 228.2

Sumof squares between groups for interaction

=228.2 – 7.2 – 113.2 = 105.8

Sumof squares within grops

=325 – 228.2

=96.8

DFsfor each factor = 1 and for within grops = 16

Meansquare between the groups for factor A is given by MSA=SSA/dfA

=7.2/1 = 7.2

Meansqare between groups for factor B

=115.2/1 = 115.2

Meansquare between the groups for interaction

MSAxB= SSAxB/dfAxB= 105.8/1 = 105.8

Withingroups, MSwn=SSwn/dfwn

=96.8/16 = 6.05

ThereforeFA= 7.2/6.05, FB= 115.2/6.05 = 19.04

FAxB=105.2/6.05= 17.49

Wenow summarize the ANOVA table as belpw

 Source Sum of sqares Df Mean squares F Factor A 7.2 1 7.2 1.19 Factor B 115.2 1 115.2 19.04 Interaction 105.8 1 105.8 17.49 Within groups 96.8 16 6.05 Total 325
1. Compute the main effect means and interaction means.

Thishs been done in a above

Forfactor B

Formale ∑X =119, Xbar fr male = 11.9

Forfemales, ∑X =71, Xbar for female = 7.1

Forfactor A

Formale, ∑X =89, Xbar = 8.9

Forfemale, ∑X= 101, Xbar =10.1

Forinteraction

XA1B1bar= 9, XA1B2bar = 8.8, XA2B1bar = 14.8, XA2B2bar = 5.4

1. Perform the appropriate post hoc comparisons.

Fromthe ANOVA, we realize that factor B is significant while Factor A isnot thus post hoc comparison is only possible for the interactioneffect which is given as

MSwn= 6.05, alpha = 0.05, k=3, dfwn = 16 qk = 3.65

=4.02

Thisimplies that the difference between females and males tested by maleis significant .

1. What do you conclude about the relationships this study demonstrates?

Thebest conlusion from this is that there is some association thatexists between the male, female and the performance when tested by amale and male experiments.

1. Compute the effect size where appropriate.

Effectsize for factor B = 115.2/325 = 0.35

Forinteraction AxB = 105.8/325 = 0.33

20.Youconduct an experiment involving two levels of self-confidence (A1islow and A2ishigh) and examine participants’ anxiety scores after they speak toone of four groups of differing sizes (B1throughB4representspeaking to a small, medium, large, or extremely large group,respectively). You compute the following sums of squares (n=4 andN=32):

Source Sum of Squares df MeanSquareF

Between

FactorA 8.42 __ _

FactorB76.79 _ _ _

Interaction 23.71 _ _ _

Within110.72 __

Total219.64 _

1. Complete the ANOVA summary table.

Usingthe formula and the given values, the ANOVA is as follows

Source Sum of Squares df MeanSquareF

Between

FactorA 8.42 __1____8.42__ 1.827_

FactorB76.79 ___3__ _25.6__ _5.553_

Interaction 23.71 __3___ _7.9___1.714_

Within110.72 __24___ 4.61_

Total219.64 __31___

1. With α= .05, what do you conclude about each Fobt?

Wefirst of all state the hypothesis,

Ho:there is no significance difference between the two factors

H1:there is significance difference between the two factors

Inorder to make a conclusion, we have to find the critical values.

Forfactor A Fcrit at dfA=1 and dfwn=24 Fcrit = 4.260

Forfactor B, dfB=3,dfwn=24,Fcrit=3.009

Forthe interaction of the two factors

DfAxB=3,dfwn=24,fcrit=3.009

Comparingthe Fcrit vale and Fobt, it is clear that for factor A, fobt (1.827)is smaller than Fcit(4.26). therefore we accept the null hypothesisand conclude that there is no difference in between theself-confidence levels and these are emanating from differentpopulation.

Forfactor B, Fobt(5.553) is greater than Fcrit(3.009), hence the nullhypothesis is false indicating that there is changing ofparticipant’s anxiety scores of four grops of different sizes

Forinteraction effect, F(1.714) is smaller than F(3.009) thus weconclude that there is no interaction between A and B

1. Compute the appropriate values of HSD.

Substitutingthe values from the above calculations, we have

=4.6914

1. For the levels of factor B, the means are X1 = 18.36, X2 = 20.02, X3 = 24.6, and X4 = 28.3. What should you conclude about the main effect of B?

Sincefactor B has the four levels B1, B2, B3, B4 their means are 12.38,20.2, 24.2, and 28.3 respectively.

Herethe main effects are in the increasing manner and also there existsdifference in between the factor levels in the population implyingthat population means have no effect.

1. How important is the size of the audience in determining a person’s anxiety score? How important is the person’s self-confidence?

Thesize of the audience in determining the anxiety scores of a person isincreased making the scores to also increase. Consequently, theresults will be more accurate. The audience size also determines theself-confidence level

21.Youmeasure the dependent variable of participants’ relaxation level asa function of whether they meditate before being tested, and whetherthey were shown a film containing a low, medium, or high amount offantasy. Perform all appropriate statistical analyses, and determinewhat you should conclude about this study.

Amountof Fantasy

Low Medium High

 5 6 2 2 5 7 5 6 9 5 9 8 10 10 10 10 10 9 10 10 2 5 4 3 2 5 6 5 7 6

Mediation

Nomediation

Solution

Nullhypothesis: there is no mean effect and interaction effect

α1=α2=0, β1= β2 = 0, (α β)11=(α β)12=(α β)21=(α β)22=0

Ha:there are main and interaction effects being significant

Amountof Fantasy

Low Medium High

 5 6 2 2 5 Xbar =4 ∑X= 20 ∑X2=94 N=5 7 5 6 9 5 Xbar =6.4 ∑X= 32 ∑X2=216 N=5 9 8 10 10 10 Xbar =9.4 ∑X= 47 ∑X2=445 N=5 10 10 9 10 10 Xbar =9.8 ∑X=49 ∑X2=481 N=5 2 5 4 3 2 Xbar =3.2 ∑X= 16 ∑X2=58 N=5 5 6 5 7 6 Xbar =5.8 ∑X= 29 ∑X2=171 N=5

Mediation

Forfactor A

Xlowbar= 6.9, Xmediumbar=4.8,Xhighbar=7.6

Forfactor B

Xmeditationbar=6.6,Xnomeditationbar=6.267

Grandtotals

∑Xtot=193

∑X2tot=1465

N=30

Totalsum of squares

=1465– {(193)2/30}

=1465 – 1241.63

=223.37

Sumof squares between the groups for factor A is: 692/10+482/10+762/10– 1932/30

=0.833

Totalsum of squares between groups

:202/5+322/5+472/5+492/5+162/5+292/5– 1932/30= 184.57

Sumof squares within groups = 223.37 – 184.57 = 38.8

Degreesof freedom

Foreach factor A = 3 -1 = 2

Foreach factor B = 2-1 = 1

Forbetween groups = 2

Withingroups = 30 -3-1-2 = 24

MeanSquares between the groups for factor A =

42.47/2= 21.23

Betweenthe groups for factor B = 0.833/1 = 0.833

Themean square between the groups for interaction = 141.27/2 = 70.63

Withingroups = 38.8/24 = 1.617

Fobtfor Factor A = 21.233/1.617 = 13.134

Fobtfor factor B = 0.833/1.617 = 0.5155

Fobtfor interaction = 70.633/1.617 = 43.671

of ANOVA table

 Source Sum of sqares Df Mean squares F Factor A 42.467 2 21.233 13.134 Factor B 0.833 1 0.833 0.516 Interaction 141.267 2 70.633 43.691 Within groups 38.8 24 1.617 Total 223.367

Nextwe get the critical values from the F-distribution table which is asfollows

Forfactor A, Fcrit at dfs 2, 24 and 0.05 significance level = 3.4

Forfactor B, Fcrit at dfs 1, 24 and 0.05 significance level = 4.26

Forinteraction effect, Fcrit at dfs 2, 24 and 0.05 significance level =3.4

Lookingat the F-ratios compared to the tabulated values, it is viable toconclude that factor B is significant and factor A and interactioneffect (AxB) are not significant.

Reference

Heiman,G. (1992).&nbspBasicstatistics for the behavioral sciences.Boston: Houghton Mifflin.