Question

2. The table below contains data on the enzyme activity of mannose-6-phosphate isomerase (MPI) for three different genotypes (SS, FS, FF) in the amphipod crustacean Platorchestia platensis. Because the effects of sex are not known, specimens were classified by sex. Conduct a two-way ANOVA on this dataset. Test for effect of genotype, the effect of sex, and whether there is an interaction effect between two factors. Show all calculations and state your conclusions. FS Females 3.633.57 3.93 2.894.14.26 3.22 3.345.02 2.744.253.91 3.87 Males 2.88 3.6 4.95 3.41 3.473.98 3.21 4.154.18 4.113.69 5.01 4.15 3.16 3.51 3.99 3.72 NOTE: Try to reduce rounding as much as possible.

Answer 1

This is a simple problem of testing of a hypothesis based on two way ANOVA .

To solve the problem let us first of all develop the hypothesis model .

Let the hypothesis be

H(οD) = The test results do not depend on the type of sex.
H(οT) = The test results do not depend on the type of genotype.

The example has two fsctors(factor sex, factor genotype) at a-2(Female and Male) and b-3(SS, FS and FF) levels. Thus, there are a b 3 2-6 different combination of sex and genotype with each combination. There arer-5 loads. is called the number ofreplicates). This sums up to n-a br 24 2335 loads in total. The amounts of Y(i ofimpactmade when usingtrials kk-1,2,3,4,5) with sex 1,2)at ganotypaii 1,2,3) are recorded in table below 5ex FS 3.63 2.89 3.22 2.74 3.16 2.88 3.41 3.21 4.11 3.99 3.57 4.11 3.34 4.25 3.51 3.61 3.47 4.15 3.69 3.72 3.93 4.26 5.02 3.91 3.87 4.95 3.98 4.18 5.01 4.15 Females Males Solution: 5ex FS Meanyi 3.63 2.89 3.22 2.74 3.16 3.128 2.88 3.41 3.21 4.11 3.99 3.52 3.57 4.11 3.34 4.25 3.51 3.756 3.61 3.47 4.15 3.69 3.72 3.728 3.93 3.71 4.26 3.75 5.02 3.86 3.91 3.63 3.87 3.51 3.69 4.95 3.81 3.98 3.62 4.18 3.85 5.014.27 4.153.95 3.90 Females Mean 4.198 3 Males Mean 4.4543 Me 3.32 of mean 3.74 4.33 3.80

We have calculated all the means like sex mean(Md), genotype means (Mt) and mean of every group combination.

Now what we only have to do is calculate the sum of) and degree of freedomf for genotype, sex and interaction between factor and levels. To do so firstly we ill calculate the SS(within)/gf(within) twe.way anevathe formula is different from thatused in one way ANOVA STEP 1: Formula for calculation of SS(within) is: Let us assume that Vjk is the elements in the groups YOU) is mean of combinations When we putthe values and do calculations with this formula we will get SSwithin) is SSwithin -(3.63-3.13) (2.89-313) (3.22-313)+(2.74-3.13)(3.16-3.13) 3.51-3.76) 3.87-4.20) 3.57-3.76)* 4.11 -3.76) 3.34-3.76) 4.25-3.76) 3.93-4.20) 4.26-4.20) 5.02-4.20)3.91-4.20) 2.050 (female) Similarly find it form males also We did it in excel and got the SS (within -male)-5.05 Total-5.05+2.05-7.10

Calculate the df(within) df(within) = (r-1)*a*b = 4*2*3=24 CalculateMS(within): MS(within) = SS(within)di(within-7. 10/24-031 41 STEP 2: Calculate SS(ser) and df(sex) and MS(sex) Y () is the mean ofsex (which is equal to 3.69 (female) and 3.90 (male) Yis the total mean sex andgenotype (which is equal to 3.79)

SS sex 5 33.69-3.79) 3.90-3.79-1 Calculate dflsex):| = 0.033 Calculate MS(se): MS(sex) SS(sex) d(sex) STEP 3: Simlilarlv Calculate the SS(genotype), df(genotype) and MS(genotype) = 0.033/1=0.033 Y (0 is the mean of sex Yis the total mean sex and genotype SS Genotype EI 5*2 3.32-3.79*+(3.74-3.79)+(4.33-3.79)1 10*0.515=5.15 Calculate df(genotype): d(genotype) b-1-3-1-2 CalculateMS(genotype): MS(genotype) SS(genotype) sf(genotype) = 5.15/2=2.575 STEP 4: Calculate SS(interaction), dfinteraction) andMS(interaction) YTİİ) is mean of combinations Yi) is the mean of sex YG) is the mean of genotype Y is the total mean sex and genotype

Calculate SS(interaction):

We have all the values.

ΣΣ(4-K-Y, +Y.)2 ss ssinler action - rx

Keepingvalues and summingthe differences we get SS (interaction)-0.2297 Calculate dfinteraction): dk interaction) )(b-1 (2-1)*3-1)-2 CalculateMS(interaction): MS interaction) SS(interaction) f(interaction) 0.2297/2 0.22972 = 0.1149 Now that we have all the date we will now move to the grand finale of the solutioni It is to conduct the F test which is related to copg the F statistic to the criticalF value. Its time to calculate the F-test: Calculate criticalF-value The formula for the variousF statistic are give below EXdf(sex), df(within)~MS(sex)MS(within) EKd(genotype), df(within)) ~MS(genotype) MS(within) Edf interaction), df(within)MS(interaction)MS(within) Thus keeping Values we gt F (Sex/within) 1.77 (xelated p value-0.1959 0.005 Hence not significant) F (Genotype/Within)-14.02 (xelated p value-0.00 0.005 Hence significant) F (Sex/Genotype interaction)4elated p value 0.53 0.005 Hence not significant)