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1. Problem 1: To compare the effectiveness of three different types of phosphorescent coatings of airplane instrument dials,

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Answer #1
Type 1 Type 2 Type 3 Total
Sum 456.8 473.4 547.6 1477.8
Count 8 8 8 24
Mean, Sum/n 57.1 59.175 68.45
Sum of square, Ʃ(xᵢ-x̅)² 144.48 81.155 134.38

Number of treatment, k = 3

Total sample Size, N = 24

df(between) = k-1 = 2

df(within) = N-k = 21

df(total) = N-1 = 23

SS(between) = (Sum1)²/n1 + (Sum2)²/n2 + (Sum3)²/n3 - (Grand Sum)²/ N = 584.41

SS(within) = SS1 + SS2 + SS3 = 360.015

SS(total) = SS(between) + SS(within) = 944.425

MS(between) = SS(between)/df(between) = 292.205

MS(within) = SS(within)/df(within) = 17.1436

a) Null and Alternative Hypothesis:

Ho: µ1 = µ2 = µ3

H1: At least one mean is different.

F = MS(between)/MS(within) = 17.0446

p-value = F.DIST.RT(17.0446, 2, 21) = 0.0000

As p-value = 0.0000 < 0.01, we reject the null hypothesis.

ANOVA
Source of Variation SS df MS F P-value
Between Groups 584.4100 2 292.2050 17.0446 0.0000
Within Groups 360.0150 21 17.1436
Total 944.4250 23

b) Assumptions:

  • The groups are independent of each other.
  • The data in each cell should be approximately normally distributed.
  • The variance in each cell should be similar.

c) Tukey's method:

Q statistic at α = 0.01, k = 3, N-k = 21, Q = 4.62

Samples in each treatment, n =8

Comparison Absolute Diff. = |xi - xj| Critical Range, CV = Q* √(MS(Within)/n) Results
µ1-µ2 2.075 6.7631 Means are not different
µ1-µ3 11.35 6.7631 Means are different
µ2-µ3 9.275 6.7631 Means are different
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