Question

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 0.01 level of significance.

Sunscreen A	Sunscreen B	Sunscreen C	Sunscreen D
64	33	57	69
80	58	38	73
61	44	47	57
55	62	47	71
44	37	61	47
69	35	46	54

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=0.01α=0.01

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:

• reject H0H0
• do not reject H0H0

The final conclusion is:

• There is enough evidence to reject the claim that at least one sunscreen lasts a different amount of time.
• There is enough evidence to support the claim that at least one sunscreen lasts a different amount of time.
• There is not enough evidence to reject the claim that at least one sunscreen lasts a different amount of time.
• There is not enough evidence to support the claim that at least one sunscreen lasts a different amount of time.

Answer 1

Null and alternative hypothesis:

H₀: μ_A = μ_B = μ_C = μ_D

H_A: At least one mean is different. (claim)

ANOVA:

	Sunscreen A	Sunscreen B	Sunscreen C	Sunscreen D	Total
Sum	373	269	296	371	1309
Count	6	6	6	6	24
Sum of square, Ʃ(xᵢ-x̅)²	750.833	766.833	345.333	564.833

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 = 1396.13

SS(within) = SS1 + SS2 + SS3 + SS4 = 2427.83

SS(total) = SS(between) + SS(within) = 3823.96

MS(between) = SS(between)/df(between) = 465.375

MS(within) = SS(within)/df(within) = 121.392

F = MS(between)/MS(within) = 3.83367

p-value = F.DIST.RT(3.8337, 3, 20) = 0.0256

ANOVA
Source of Variation	SS	df	MS	F	P-value
Between Groups	1396.125	3	465.375	3.834	0.0256
Within Groups	2427.833	20	121.392
Total	3823.958	23

The decision of the test is to:

• As P-value = 0.0256 > 0.01, do not reject H0.

The final conclusion is:

• There is not enough evidence to support the claim that at least one sunscreen lasts a different amount of time.