A study is made of amino acids in the hemolymph of millipedes. For a sample of f

Question

A study is made of amino acids in the hemolymph of millipedes. For a sample of four males and four females of each of the three species, the following concentrations of the amino acid alanine (in mg/100ml) are determined: Test the hypothesis that there is no difference in mean hemolymph alanine concentration among the three species. Test the hypothesis that there is no difference between males and females in mean hemolymph alannine concentration. Test the hypothesis that there is no interaction between sex and species in the mean concentration of alanine in hemolymph. If the hypothesis of part a, above, is rejected, then find out which species has different alanine concentration?

Explanation / Answer

SSwithin = (21.5-21.2)^2 + (19.6-21.2)^2 + ..... + (13-14.225)^2 = 31.4175

dfwithin = (r-1)*a*b = (4-1)* 2 * 3 = 18

MSwithin = 31.4175/18 = 1.745

SSgender = 4 * 3 * [(18.383- 16.1875)^2 + ( 13.992 - 16.1875)^2] = 115.6853

dfgender = 2-1 = 1

MSgender = 115.6853/1 = 115.6853

SSspecies = 4 * 2 * [(18.1375 - 16.1875)^2 + (14.425 - 16.1875)^2 + (16 - 16.1875)^2 ] = 55.5525

dfspecies = 3-1 = 2

MSspecies = 55.5525/2 = 27.776

SSintercation = 4 * [(21.2 - 18.383 - 18.1375 + 16.1875)^2 + (16.175 - 18.383 - 14.425 + 16.1875)^2 + (17.775 - 18.383 - 16 + 16.1875)^2 + (15.075 - 13.992 - 18.1375 + 16.1875)^2 + (12.675 - 13.992 - 14.425 + 16.1875)^2 + (14.225 - 13.992 - 16 + 16.1875)^2 + (21.2 - 18.1375 + 18.383 - 16.1875)^2 + (15.075 - 18.1375 + 13.992 - 16.1875)^2 + (16.175 - 14.425 + 18.383 - 16.1875)^2 + (12.675 - 14.425 + 13.992 - 16.1875)^2 + (17.775 - 16 + 18.383 - 16.1875)^2 + (14.225 - 16 + 13.992 - 16.1875)^2 ] = 480.8431

Dfinteration = (3-1)*(2-1) = 2

MSintercation = 480.8431/2 = 240.4215

(a) F test for species = MSspecies/MSwithin = 27.776/1.745 = 15.91

Critical value of F for significance level 0.05 and df = 2,18 is 3.55

As observed F is greater than critical value, we reject the null hypothesis and conclude that species is significant factor.

(b) F test for gender = MSgender/MSwithin = 115.6853/1.745 = 66.2953

Critical value of F for significance level 0.05 and df = 1,18 is 4.41

As observed F us greater than critical value, we reject the null hypothesis and conclude that gender is significant factor.

(c) F test for interaction = MSinteraction/MSwithin = 240.4215/1.745 = 137.77

Critical value of F for significance level 0.05 and df = 2,18 is 3.55

As observed F us greater than critical value, we reject the null hypothesis and conclude that interaction between gender and species is is significant factor.

(d) Critical value of F at significance level of 0.05 and df=1,28 is 4.32

LSD = sqrt(2*MSspecied* F/r) = sqrt(2*1.745*4.32/4) = 1.941

Difference between Species 1 and Species 2 = 18.1375 - 14.425 = 3.7125

Difference between Species 1 and Species 3 = 18.1375 - 16 = 2.1375

Difference between Species 2 and Species 3 = 16 - 14.425 = 1.575

As all the differences of Species 1 with other species is gretaer than LSD, species 1 have different alanine concentration.

Species 1 Species 2 Species 3 Mean Male 21.5,19.6,20.9,22.8 (21.2) 14.5,17.4,15,17.8 (16.175) 16,19.3,17.5,18.3 (17.775) 18.383 Female 14.8,15.6,13.5,16.4 (15.075) 12.1,11.4,12.7,14.5 (12.675) 14.4,14.7,14.8,13 (14.225) 13.992 Mean 18.1375 14.425 16 16.1875

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