Question

5 Using the data from problem 4, determine at the 0.05 level, which pairs of mean activity levels differ significantly from 1 another at the 0.05 level.

Answer 1

Soln

5)

	SS	df	MS	F
Between	21.652	4	5.413	24.40047
Within	7.7644	35	0.22184
Total	29.4164	39

I will be using Tukey HSD (Honestly Significant Difference):

HSD = q *Sqrt (MS _Within /n)          

q is calculated from q table using (No of Groups vs degrees of freedom of error)

where q is 4.04,

Hence, HSD = 0.67

µ₁₉₆ = 3.36

µ₂₀₀ = 0.226

µ₂₄₇ = 1.69

µ₄₃₄₇ = 2.03

µ₄₃₈₄ = 1.92

which is less than HSD, hence they are not different

Pairs:

196 and 200

Mod(µ₁₉₆ - µ₂₀₀)= mod(3.36 – 0.226) = 3.134 which is greater than HSD, hence they are different

196 and 247

Mod(µ₁₉₆ - µ₂₄₇)= mod(3.36 – 1.69) = 1.67 which is greater than HSD, hence they are different

196 and 4347

Mod(µ₁₉₆ - µ₄₃₄₇)= mod(3.36- 2.03) = 1.33 which is greater than HSD, hence they are different

196 and 4384

Mod(µ₁₉₆ - µ₄₃₈₄)= mod(3.36 – 1.92) = 1.44 which is greater than HSD, hence they are different

200 and 247

Mod(µ₂₀₀ - µ₂₄₇)= mod(0.226 – 1.69) = 1.46 which is greater than HSD, hence they are different

200 and 4347

Mod(µ₂₀₀ - µ₄₃₄₇)= mod(0.226 – 2.03) = 1.80 which is greater than HSD, hence they are different

200 and 4384

Mod(µ₂₀₀ - µ₄₃₈₄)= mod(0.226 – 1.92) = 1.69 which is greater than HSD, hence they are different

247 and 4347

Mod(µ₂₄₇ - µ₄₃₄₇)= mod(1.69 – 2.03) = 0.34 which is less than HSD, hence they are not different

247 and 4384

Mod(µ₂₄₇ - µ₄₃₈₄)= mod(1.69 – 1.92) = 0.23 which is less than HSD, hence they are not different

4347 and 4384

Mod(µ₄₃₄₇ - µ₄₃₈₄)= mod(2.03 – 1.92) = 0.11 which is less than HSD, hence they are not different