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.

Frogs	Toads	Salamanders
7	7	3
5	6	3
6	11	2
5	8	5
7	2	4
5	10	2
5	6	5
5	2	2
5	8
6
8

The researchers made a qq plot of the data. Does the normality assumption hold?

  
Yes, the normality assumption holds
No, the normality assumption does not hold

What is the correct null hypothesis? Let μ₁ be the mean for toads, μ₂ for frogs and μ₃ for salamanders.

H₀: μ₁ = μ₂ + μ₃
H₀: μ₁ > μ₂ > μ₃
H₀: μ₁ = μ₂ = μ₃
H₀: μ₁ = μ₂ = μ₃ = 10
What is the correct alternative hypothesis? Let μ₁ be the mean for toads, μ₂ for frogs and μ₃ for salamanders.

H_a: μ_i ≠ μ_j for some i ≠ j
H_a: μ_i ≠ μ_j where i = j
H_a: μ₁ − μ₂ − μ₃ = 0
H_a: μ₁ ≠ μ₂ ≠ μ3

What is the correct distribution for the test statistic?

Poisson distribution
χ² distribution
t distribution
F distribution

What is the average rank for the toad group?

What is the average rank for the frog group?

What is the average rank for the salamander group?

Calculate the test statistic

Which interval in the table contains the p value for the test?

p-value ≤ 0.005
0.005 < p-value ≤ 0.01
0.01 < p-value ≤ 0.025
0.025 < p-value ≤ 0.05
0.05 < p-value ≤ 0.1
p-value > 0.1

Answer 1

Excel output foe ANOVA is attached below:

########### Answers:

* Null hypothesis is:

H₀: μ₁ = μ₂ = μ₃

_{* Alternative hypothesis:}

H_a: μ₁ ≠ μ₂ ≠ μ3

# Distribution to be used:

F distribution

# P-value is:

0.005 < p-value ≤ 0.01