SOLUTION: Question 1. Why is the F-distribution centered around a value of 1?

Question 1. Why is the F-distribution centered around a value of 1?

Because the F distribution is about ratio, we divide the variances of the 2 groups to see if they are equal or different. When they are equal, the ratio is close to 1, but when they are different, the ratio is far away to 1.

Question 2. why the application of the variance ratio, Fs, as described above, is always a one-tailed test.

That is because when the F statistic is unusually large, that means that there is a difference between the averages of the groups. But when the F statistic is small, that suggests that the averages of the groups are very close.

Question 3. ANOVA table for data Example 10.2

First, we need to calculate the mean for each group and the grand mean.

Mean1 = (34 + 36 + 34 + 35 + 34) / 5 = 34.6

Mean2 = (37 + 36 + 35 + 37 + 37) / 5 = 36.4

Mean3 = (34 + 37 + 35 + 37 + 36) / 5 = 35.8

Mean4 = (36 + 34 + 37 + 34 + 35) / 5 = 35.2

Grand mean = (34.6 + 36.4 + 35.8 + 35.2) / 4 = 35.5

SS total = 9 + 20 = 29

Df between = # groups – 1 = 4 – 1 = 3

Df within = N – # groups = 20 – 4 = 16

Df total = N – 1 = 20 – 1 = 19

MS between = SS between / df between = 9 / 3 = 3

MS within = SS within / df within = 20 / 16 = 1.25

F stat = MS between / MS within = 3 / 1.25 = 2.4

Variation

Between

2.4

Within

1.25

Total

Question 4. ANOVA table for the squirrel data. Is there a significant difference among treatments?

First, we need to calculate the mean for each group and the grand mean.

Mean marbs = (0.137009+0.115348+0.094108+0.095733+0.089406+0.098226+0.076334+0.115456+0.09342+0.099996) / 10 = 0.101504

Mean marbslog = (0.121946123+0.141123584+0.11057181+0.104269178+0.113961107+0.095912508+0.139546153+0.120853428+0.138269344+0.056030807) / 10 = 0.114248404

Mean truffs = (1.629663+1.723097+1.716988+1.72573+1.70367+1.674631+1.718007+1.68641+1.695043+1.691949) / 10 = 1.696519

Mean truffslog = (1.76049042+1.763138076+1.764124177+1.762978585+1.749805277+1.76770056+1.764817055+1.737063292+1.718714469+1.749057504) / 10 = 1.753788941

Grand mean = (0.101504+0.114248404+1.696519+1.753788941) / 4 = 0.916515

SS total = 26.17309 + 0.018227 = 26.19132

Df between = # groups – 1 = 4 – 1 = 3

Df within = N – # groups = 40 – 4 = 36

Df total = N – 1 = 40 – 1 = 39

MS between = SS between / df between = 26.17309 / 3 = 8.724363

MS within = SS within / df within = 0.018227 / 36 = 0.000506

F stat = MS between / MS within = 8.724363 / 0.000506 = 17231.88

Variation

Between

8.724363

17231.88

Within

0.018227

0.000506

Total

We can say that there is a significant difference among treatments because the F value is unusually large.

Question 5. ANOVA table for newt data. Is there a significant difference in the mean number of D. longispina among the months?

First, we need to calculate the mean for each group and the grand mean.

Mean june = (71+92+86+101+104+120+79+101+110+72+113+106+95+93+104+90+104+86+74+98+120) / 21 = 96.14286

Mean august = (157+137+91+99+121+104+133+128+156+117+132+103+151+111+129+105+141+102+98+145+115+106+93+101+154) / 25 = 121.16

Mean truffs = (32+33+3+27+27+3+15+40+40+35+29+14) / 12 = 24.8333

Grand mean = (96.14286+121.16+24.8333) / 3 = 92.17241

SS total = 75752.68 + 16917.6 = 92670.28

Df between = # groups – 1 = 3 – 1 = 2

Df within = N – # groups = 58 – 3 = 55

Df total = N – 1 = 58 – 1 = 57

MS between = SS between / df between = 75.753 / 2 = 37876.34

MS within = SS within / df within = 16917.6 / 55 = 307.5927

F stat = MS between / MS within = 37876.34 / 307.5927 = 123.138

Variation

Between

75.753

37876.34

123.138

Within

16917.6

307.5927

Total

92.670

We can say that there is a significant difference among treatments because the F value is unusually large.

Question 6.

By looking the p value for each treatment at each analysis, we can say that all of the samples for these 2 analyses meet the assumption of normality. All p values are greater than 0.05

Question 7.

From Example10.2 dataset.

Fmax = 1.7 / 0.8 = 2.125

F critical from F table is 6.31

As the Fmax is lower than F critical we cannot reject Ho and we conclude that the assumption of homogeneity of variance is met.

From squirrel dataset.

Fmax = 0.000841 / 0.000243 = 3.459

F critical from F table is 6.31

As the Fmax is lower than F critical we cannot reject Ho and we conclude that the assumption of homogeneity of variance is met.

From newt data set

F max = 450.9733 / 170.5152 = 2.644

F critical from F table is 4.16

As the Fmax is lower than F critical we cannot reject Ho and we conclude that the assumption of homogeneity of variance is met.

Question 8.

Making ANOVA using data 10.11

First, we need to calculate the mean for each group and the grand mean.

Mean1 = (1+2+3.5+3.5+8+10+10+17) / 8 = 6.875

Mean2 = (6+10+13.5+13.5+20+20+23.5+26) / 8 = 16.5625

Mean3 = (13.5+16+18+20+23.5+26+28) / 7 = 20.714

Mean4 = (6+6+13.5+2+26+29+30+31) / 8 = 20.4375

Grand mean = (6.875+16.5625+20.714+20.4375) / 4 = 16.14732

SS total = 982.4317 + 1484.24

Df between = # groups – 1 = 4 – 1 = 3

Df within = N – # groups = 31 – 4 = 27

Df total = N – 1 = 31 – 1 = 30

MS between = SS between / df between = 982.4317 / 3 = 327.4772

MS within = SS within / df within = 1484.24 / 27 = 54.972

F stat = MS between / MS within = 327.4772 / 54.972 = 5.957

Variation

Between

982

327.4772

5.957176252

Within

1484.24

54.97189

Total

2.467

Question 9.

Doing ANOVA using sculpin data

First, we need to calculate the mean for each group and the grand mean.

Mean1 = 1

Mean2 = 1.1667

Mean3 = 0.0769

Grand mean = (1+1.1667+0.0769) / 3 = 0.75

SS total = 8.91 + 2.59

Df between = # groups – 1 = 3 – 1 = 2

Df within = N – # groups = 40 – 3 = 37

Df total = N – 1 = 40 – 1 = 39

MS between = SS between / df between = 8.910 / 2 = 4.455

MS within = SS within / df within = 2.59 / 37 = 0.069993

F stat = MS between / MS within = 4.455 / 0.06999 = 63.651

Variation

Between

8.910

4.455128

63.65099

Within

2.59

0.069993

Total

Choose one of the data sets for which you obtained a significant result, and perform the Tukey comparisons, comparing qs to the critical values in Table B.5 in your textbook (the table for α =0.05 starts on page 723 in the 5th edition) for the appropriate number of groups (remember that your text uses k instead of a), and using the degrees of freedom for MSwithin (i.e., dfwithin) as v.
Remember to start with the largest difference between means, work your way through the next largest difference, and stop comparing when you get a non-significant result.

I will use “squrrel” dataset which we obtained significant results.

Q between marbs and marbslog = (0.101504 – 0.114248) / 0.0007113 = 1.791761

Q between marbs and truffs = (0.101504 – 1.696519) / 0.0007113 = 224.2395

Q between marbs and truffslog = (0.101504 – 1.753789) / 0.0007113 = 232.2909

Q between marbslog and truffs = (0.114248 – 1.696519) / 0.0007113 = 222.4477

Q between marbslog and truffslog = (0.114248 – 1.753789) / 0.0007113 = 230.4992

Q between truffs and truffslog= (1.696519 – 1.753789) / 0.0007113 = 8.051461

Ordering from largest to smallest

marbs – truffslog

232.2909

marbslog – truffslog

230.4992

marbs – truffs

224.2395

marbslog – truffs

222.4477

truffs – truffslog

8.051461

marbs – marbslog

1.791761

Using table to get critical value, using alpha = 0.05, k = 4, and df error = 36

Critical value = 3.79

As you can see the Tukey Q values that are greater than critical value are the first 5, and the last one is lower than critical value, so all comparisons are significant, except for marbs – marbslog.

Bar plot with standard errors and Tukey Letters.

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