In: Statistics and Probability
A sunscreen company is attempting to improve upon their formula so that it lasts in water longer. They have 4 lead scientists who each came up with a different formulas. In order to see if there is a difference in the time the sunscreen lasts the CEO collects a random sample of each of the four sunscreens the data is shown below. Test the claim that at least one sunscreen has a different lifespan in water at a 0.10 level of significance.
Sunscreen A | Sunscreen B | Sunscreen C | Sunscreen D |
62 | 42 | 48 | 89 |
61 | 58 | 46 | 61 |
42 | 77 | 52 | 72 |
61 | 30 | 65 | 57 |
56 | 35 | 50 | 83 |
89 | 50 | 42 | 49 |
The hypotheses for this ANOVA test would be:
H0:μA=μB=μC=μDH0:μA=μB=μC=μD
HA:HA: At least one mean is different. (claim)
α=0.10α=0.10
Complete the ANOVA table below: (round answers to 3 decimal places)
SS | df | MS | F | p-value | |
Between | |||||
Within |
The decision of the test is to:
The final conclusion is:
Null and alternative hypothesis:
H0: μA = μB = μC = μD
HA: At least one mean is different. (claim)
ANOVA:
Sunscreen A | Sunscreen B | Sunscreen C | Sunscreen D | Total | |
Sum | 371 | 292 | 303 | 411 | 1377 |
Count | 6 | 6 | 6 | 6 | 24 |
Sum of square, Ʃ(xᵢ-x̅)² | 1166.83 | 1471.33 | 311.5 | 1211.5 |
Number of treatment, k = 4
Total sample Size, N = 24
df(between) = k-1 = 3
df(within) = k(n-1) = 20
df(total) = N-1 = 23
SS(between) = (Sum1)²/n1 + (Sum2)²/n2 + (Sum3)²/n3 + (Sum4)²/n4 - (Grand Sum)²/ N = 1600.46
SS(within) = SS1 + SS2 + SS3 + SS4 = 4161.17
SS(total) = SS(between) + SS(within) = 5761.63
MS(between) = SS(between)/df(between) = 533.486
MS(within) = SS(within)/df(within) = 208.058
F = MS(between)/MS(within) = 2.56412
p-value = F.DIST.RT(2.5641, 3, 20) = 0.0834
Source of Variation | SS | df | MS | F | P-value |
Between Groups | 1600.458 | 3 | 533.486 | 2.564 | 0.0834 |
Within Groups | 4161.167 | 20 | 208.058 | ||
Total | 5761.625 | 23 |
Decision:
The final conclusion is: