ANOVA
YourName
Chapter13

What does each of the following terms mean?

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

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

factor
thisis an independent variable in an experiment

level
thisrefers to each condition of an independent variable

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

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

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 = SS_{bn}/SS_{tot} where one can obtain SS_{bn}and SS_{tot} 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 6month 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

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

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
∑X^{2}=22
N1 = 4
Xbar =2
12
38
4
3
23
135
4
5.75
24
154
4
6
∑X_{tot} = 67
∑X^{2}_{tot}= 349
N=4
K=4
SS_{tot}=∑X^{2}_{tot}– (∑X_{tot})^{2}/N
=34967^{2}/16
=349– 280.56
=68.44
SS_{bn}= ∑[(∑X_{i})^{2}/n_{i}] (∑X_{tot})^{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
SS_{wn}= SS_{tot}– SS_{bn}
=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
k1=3
MS_{bn} =47.69/3 = 15.9
Fobt = MS_{bn}/MS_{wn}
Within
20.75
Nk = 12
MS_{wn}= 20.75/12 = 1.73
Total
68.44
N1 = 15
Fobt = 15.9/1.73 = 9.19

With _ 5 .05, what is Fcrit?
SingFdistribtion tables we have Fcrit at alpha = 0.05, df (3, 12) as3.49

Report your statistical results.
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

Perform the appropriate post hoc comparisons.
Fromb and c, k = 4 Df_{wn}= 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 = 23 = 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

What do you conclude about this study?
Frome above, we can conclude that there is no significant relationshipbetween the number of viral infection people and adjacent levels ofstress.

Compute the effect size and interpret it.
Effectsize = SS_{bn}/SS_{tot}= 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%

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

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
∑X^{2}=108
N1 = 5
Xbar =4.4
54
598
5
10.8
47
451
5
9.4
29
175
5
5.8
∑X_{tot} = 152
∑X^{2}_{tot}= 1332
N=20
K=4
Usingthis table
SS_{tot}=∑X^{2}_{tot}– (∑X_{tot})^{2}/N
1332– 152^{2}/20
=13321155.2
=176.8
SS_{bn}= (22^{2}/5+54^{2}/5+47^{2}/5+29^{2}/5)– 152^{2}/20
=1290 – 1155.2
=134.8
SS_{wn}= 176.8 – 134.8
=42
ANOVAtable

Source
Sum of squares
Degrees of freedom
Mean square
F
Between
134.8
k1=3
MS_{bn} =134.8/3 = 44.933
Fobt = MS_{bn}/MS_{wn}
Within
42
Nk = 16
MS_{wn}= 42/16 = 2.625
Total
176.8
N1 = 19
Fobt = 44.93/2.625 = 17.117

With α= .05, what do you conclude about Fobt?
Fromthe Ftables, 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

Perform the appropriate post hoc comparisons.
Turkey’sHSD test is an appropriate Post Hoc test here since all number ofsamples are equal
K=5,df_{wn}=16, alpha=0.05
Fromvales of Standardized Range Stat table, qk = 4.05
Xbarsare 4.4, 10.8. 9.4, and 5.8 respectively. MS_{wn}=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.

What should you conclude about this relationship?
ƞ2= SS_{bn}/SS_{tot}
=134.8/176.8
=0.763
Thusthe proportion of variance in dependent scores accounted for by thelevel on the independent variable is 76.3%

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.

Describe how you would graph these results.
Whengraphing, put the age factor on the Xaxis and the mean creativescores on the Yaxis. Then plot the mean scores corresponding to thegiven ages and connect them with straight lines.
20.Ina study in which k=3,n=21,Ẋ1= 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_

Complete the ANOVA summary table.

Source
Sum of squares
Degrees of freedom
Mean square
F
Between
147.32
k1=2
MS_{bn} =147.32/2 = 73.66
Fobt = MS_{bn}/MS_{wn}
Within
862.99
Nk = 18
MS_{wn}= 862.99/18 = 47.94
Total
1010.31
N1 = 20
Fobt = 73.66/47.94 = 1.5364

(b) With α = .05, what do you conclude about Fobt?
Levelof significance = 0.05
UsingFdistribution 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

Report your results in the correct format.
Sincewe fail to reject the null hypothesis, the proportion means of allthe given three level are equal.

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

What do you conclude about this relationship?
Sincethe test statistic is not significant, we conclude that all the meansare equal

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

ƞ2 = SS_{bn}/SS_{tot}
=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_

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
MS_{bn}=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
k1=2
326.08
Fobt = MS_{bn}/MS_{wn}
Within
612.75
Nk =13
47.13
Total
1264.92
N1 = 15
6.92

At α = .05, what is Fcrit?
Fcritat α= .05 and DF (2, 13) = 3.8

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.

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
T_{obt}=(X_{1}bar– X_{2}bar)/
=(61.5 – 65.5)/ /
= 0.8686
ForX2 and X 3
Tobt= (X_{2}bar– X_{3}bar)/
=(65.5 – 48.25)/
=4.29
ForX1 and X3
Tobt= (61.5 – 65.5)
=3.07

What do you conclude about this study?
Usingthe ttable, 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)

Compute the effect size and interpret it.
Effectsize ƞ2= SS_{bn}/SS_{tot}
=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: µ_{A1}= µ_{A2}=µ_{A3}
µ_{}representsthe level means from factor A respectively
Alteernativehypothesis, Ha : not all µ_{A}areequal
(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:µ_{B1}= µ_{B2}=µ_{B3}
Ha:not all : µ_{B}areequal
(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 :
X_{A1}bar= (2+12)/2 = 7
X_{A2}bar= (4+14)/2 = 9
Andthe main effect of B1 and B2 are:
X_{B1}bar= (2+4)/2 = 3
X_{B2}bar= (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:
X_{A1}bar= (10+5)/2 = 7.5
X_{A2}bar= (5+10)/2 = 7.5
Forfactor B
X_{B1}bar= (10+5)/2 = 7.5
X_{B2}bar= (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:
X_{A1}bar= (8+8)/2 = 8
X_{A2}bar= (14+2)/2 = 9
ForB1 and B2 are:
X_{B1}bar= (8+14)/2 = 11
X_{B2}bar= (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

Describe the graph of the interaction when factor A is on the X axis.
Whenfactor A in on the Xaxis, 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

Does there appear to be an interaction effect? Why?
Yes.There is an interaction effect since the Twoway 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.

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
X_{A1}bar= (14+25)/2 = 19.5
X_{A2}bar= (23+12)/2 = 17.5

What are the main effect means and thus the main effect of changing gender?
Themain effect mean for B
X_{B1}bar= (14+23)/2 = 18.5
X_{B2}bar= (25+12)/2 = 18.5

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

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 ∑X^{2}=419 N=5 
8 14 17 16 19 Xbar =14.8 ∑X= 74 ∑X^{2}=1166 N=5 
8 10 9 7 10 Xbar =8.8 ∑X= 44 ∑X^{2}=394 N=5 
4 6 5 5 7 Xbar = 5.4 ∑X= 27 ∑X^{2}=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
=89^{2}/10+ 101^{2}/10– 190^{2}/20= 7.2
Forfactor B
=119^{2}/10+ 71^{2}/10– 190^{2}/20= 115.2
Totalsum of squares between groups
=45^{2}/5+74^{2}/5+44^{2}/5+27^{2}/5– 190^{2}/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 MS_{A}=SS_{A}/df_{A}
=7.2/1 = 7.2
Meansqare between groups for factor B
=115.2/1 = 115.2
Meansquare between the groups for interaction
MS_{AxB}= SS_{AxB}/df_{AxB}= 105.8/1 = 105.8
Withingroups, MS_{wn}=SS_{wn}/df_{wn}
=96.8/16 = 6.05
ThereforeF_{A}= 7.2/6.05, F_{B}= 115.2/6.05 = 19.04
F_{AxB}=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 

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

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 .

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.

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 selfconfidence (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 _

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___

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 df_{A}=1 and df_{wn}=24 Fcrit = 4.260
Forfactor B, df_{B}=3,df_{wn}=24,F_{crit}=3.009
Forthe interaction of the two factors
Df_{AxB}=3,df_{wn}=24,f_{crit}=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 theselfconfidence 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

Compute the appropriate values of HSD.
Substitutingthe values from the above calculations, we have
=4.6914

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.

How important is the size of the audience in determining a person’s anxiety score? How important is the person’s selfconfidence?
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 theselfconfidence 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 ∑X^{2}=94 N=5 
7 5 6 9 5 Xbar =6.4 ∑X= 32 ∑X^{2}=216 N=5 
9 8 10 10 10 Xbar =9.4 ∑X= 47 ∑X^{2}=445 N=5 
10 10 9 10 10 Xbar =9.8 ∑X=49 ∑X^{2}=481 N=5 
2 5 4 3 2 Xbar =3.2 ∑X= 16 ∑X^{2}=58 N=5 
5 6 5 7 6 Xbar =5.8 ∑X= 29 ∑X^{2}=171 N=5 
Mediation
Forfactor A
X_{low}bar= 6.9, X_{medium}bar=4.8,X_{high}bar=7.6
Forfactor B
X_{meditation}bar=6.6,X_{nomeditation}bar=6.267
Grandtotals
∑X_{tot}=193
∑X^{2}_{tot}=1465
N=30
Totalsum of squares
=1465– {(193)^{2}/30}
=1465 – 1241.63
=223.37
Sumof squares between the groups for factor A is: 69^{2}/10+48^{2}/10+76^{2}/10– 193^{2}/30
=0.833
Totalsum of squares between groups
:20^{2}/5+32^{2}/5+47^{2}/5+49^{2}/5+16^{2}/5+29^{2}/5– 193^{2}/30= 184.57
Sumof squares within groups = 223.37 – 184.57 = 38.8
Degreesof freedom
Foreach factor A = 3 1 = 2
Foreach factor B = 21 = 1
Forbetween groups = 2
Withingroups = 30 312 = 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 Fdistribution 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 Fratios 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). Basicstatistics for the behavioral sciences.Boston: Houghton Mifflin.