Question

Samples of three different species of amphibian (frog, toad and salamander) were collected and dissected and the number of parasites found within each was counted and recorded. The researchers are interested in determining whether the distribution of parasites across species is different. The number of parasites within each sample was counted and recorded. Toads Frogs Salamanders 4 3 9 2 9 9 3 6 4 3 5 9 2 8 8 5 4 14 3 5 8 The researchers made a qq plot of the data. Does the normality assumption hold? 14 12 10 * Sample Quantiles 8 *** 6 *** 4 *** * * * * 1 1.5 2 2 * -2 -1.5 -1 -0.5 0 0.5 Theoretical Quantiles Yes, the normality assumption holds O No, the normality assumption does not hold [1 pt(s)] Submit Answer Tries 0/2

Answer 1

Yes its following normal assumption, but very weakly, the QQplot must have a 45 degree line connecting the points.
Correct hypothesis is H₀ : $\mu$ ₁ =  $\mu$₂ = $\mu$ ₃
and correct laternative hypothesis is H_a : $\mu$ _i$\neq$$\mu$_j , for some i$\neq$j

Correct answer is the distribution for F-test statistic which is the F-distribtion
Notation:
A: Toads
B: Forgs
C: Salamanders

Solution: B C 4 3 9 2 9 9 3 6 4 3 5 9 2 8 8 5 4 14 3 5 αΟ ΣΑ = 22 ΣΒ = 40 Σc= 61 Α2 B2 16 9 81 4 81 81 9 36 16 9 25 81 4 64 64 25 16 196 9 25 64 ΣΑ? = 76 ΣΒ' = 256 ΣC = 583


Data table Group B C Total N 11 = 7 = 7 = 7 1 = 21 ΣΑ, Τ, = Σχ, = 22 Ι, = Σx) = 40 T, = Σχή = 61 Σ = 76 Σ3 = 256 Σ = 583 Mean x; Χι = 3.1429 X) = 5.7143 X3 = 8.7143 S, = 1.069 S, = 2.1381 = 2.9277 Σx = 123 Σ' = 915 Overall x = 5.8571 Std Dev Si S3 Let k = the number of different samples = 3 η = 1, +n, tns = 7 + 7 + 7 = 21 123 Overall x = 5.8571 21 Σx = T + T, +T, = 22 + 40 + 61 = 123 – (1) (Σ)? 1232 21 = 720.4286 - (2) n 222 402 612 + + = 829.2857 – (3) n: 7 7 7. Σ? - Σ+Σ+ Σ = 76 + 256 + 583 = 915 – (4) ANOVA: Step-1: sum of squares between samples


(2x)2 SSB = (Σ (27) = (3)-(2) = 829.2857 - 720.4286 = 108.8571 Or SSB = En; · (ā; - ) = 7 x (3.1429 - 5.8571)2 + 7 x (5.7143 - 5.8571)2 + 7 x (8.7143 - 5.8571)2 = 108.8571 Step-2: sum of squares within samples SSW = {x2-13 ni = (4)-(3) = 915 - 829.2857 = 85.7143 Step-3 : Total sum of squares SST = SSB + SSW = 108.8571 +85.7143 = 194.5714 Step-4 : variance between samples SSB MSB k-1 = 108.8571 2 = 54.4286


Step-5 : variance within samples SSW MSW = n-k = 85.7143 21 - 3 85.7143 18 = 4.7619 Step-6 : test statistic F for one way ANOVA test MSB F= MSW 54.4286 4.7619 = 11.43 the degree of freedom between samples k-1 = 2 Now, degree of freedom within samples n-k= 21 - 3 = 18


ANOVA table Source of Variation Sums of Squares Degrees of freedom Mean Squares DF F SS p-value MS Between samples SSB = 108.8571 k-1 = 2 MSB = 54.4286 11.43 0.0006 Within samples SSW = 85.7143 n-k = 18 MSW = 4.7619 Total SST = 194.5714 n-1 = 20 Ho : There is no significant differentiating between samples H1 : There is significant differentiating between samples F(2, 18) at 0.05 level of significance = 3.5546 As calculated F = 11.43 > 3.5546 So, Ho is rejected, Hence there is significant differentiating between samples

therefore p<0.005 is the range where p=0.0006 lies. option 1 is the correct answer.