Suppose that we want to compare the cleansing action of three detergents on the

Question

Suppose that we want to compare the cleansing action of three detergents on the basis of the following whiteness readings made on 15 swatches of white cloth, which were first soiled with India ink and then washed in an agitator - type machine with the respective detergents: Detergent A: 77 81 71 76 80 Detergent B: 72 58 74 66 70 Detergent C: 76 85 82 77 80 (a). The means of these three samples are 77, 68, and 80, determine whether the differences among them are significant for alpha = 0.01. (b). Use Tukey's test at alpha = 0.02 to determine the nature of the differences among treatment means.

Explanation / Answer

(a) H0: All detergents are same

H1: There is a difference between the detergents

The R-code is:

data <- c(77,81,71,76,80,72,58,74,66,70,76,85,82,77,80)
detergent <- c(rep("Detergent 1",5),rep("Detergent 2",5),rep("Detergent 3",5))
anova = aov(data~detergent)
summary(anova)

The result is:

Df Sum Sq Mean Sq
detergent 2 390 195
Residuals 12 276 23
F value Pr(>F)   
detergent 8.478 0.00507 **
Residuals   
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Signif. codes:
0 ‘***’ 0.001 ‘**’ 0.01 ‘*’
0.05 ‘.’ 0.1 ‘ ’ 1

Since p-value is less than 0.05, the null hypothesis is rejected and it is concluded that there is a difference between the detergents.

(b)

The R-code is:

TukeyHSD(anova)

The output is:

Tukey multiple comparisons of means
95% family-wise confidence level

Fit: aov(formula = data ~ detergent)

$detergent
diff
Detergent 2-Detergent 1 -9
Detergent 3-Detergent 1 3
Detergent 3-Detergent 2 12
lwr
Detergent 2-Detergent 1 -17.092031
Detergent 3-Detergent 1 -5.092031
Detergent 3-Detergent 2 3.907969
upr
Detergent 2-Detergent 1 -0.9079689
Detergent 3-Detergent 1 11.0920311
Detergent 3-Detergent 2 20.0920311
p adj
Detergent 2-Detergent 1 0.0293959
Detergent 3-Detergent 1 0.5972708
Detergent 3-Detergent 2 0.0050063

Since p-value of Detergent 3-Detergent 2 is less than 0.05, there is a significant difference between Detergent 2 and detergent 3.

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