paraphrase

Problem Definition: To determine whether there is any significant difference between average whiteness readings by the three detergents Tide, Gain, All.

Hypothesis:

H0: µTide=µGain=µAll

H1: There are at least 2 detergents which have different average whiteness.

Method: As there are more than two groups whose average is to compare, so we will apply one-way ANOVA with post-hoc Tukey HSD test if ANOVA is significant.

Decision Rule: If the F-test Statistic is greater than critical value we reject the null hypothesis, otherwise do not.

Numerator degree of freedom =k-1=3-1=2, where k is number of groups.

Denominator degree of freedom =n-k= 24-3=21

At 0.05 level of significance F0.05(2, 21)= 3.4668

Inverse Cumulative Distribution Function

F distribution with 2 DF in numerator and 21 DF in denominator

P( X ≤ x ) x

0.95 3.46680

So, we reject H0 if calculated F>3.4668.

If one-way ANOVA is signficant, we compare groups using Tukey test and found significant differences if p-value is less than 0.05.

Test of Assumptions:

First of all we have to test the null hypothesis

H0: σTide2= σGain2= σAll2

….versus the alternative hypothesis

H1: at least two variances are not equal.

The results of Bartlett test are as follows from where we have not enough evidence to reject the null hypothesis as p-value=0.595>0.05.

Test for Equal Variances: Tide, Gain, All

Method

Null hypothesis All variances are equal

Alternative hypothesis At least one variance is different

Significance level α = 0.05

Bartlett’s method is used. This method is accurate for normal data only.

95% Bonferroni Confidence Intervals for Standard Deviations

Sample N StDev CI

Tide 8 4.55914 (2.77066, 11.1659)

Gain 8 6.71353 (4.07992, 16.4423)

All 8 5.22186 (3.17341, 12.7890)

Individual confidence level = 98.3333%

Tests

Test

Method Statistic P-Value

Bartlett 1.04 0.595

From the following box plot of we observe that thy are symmetric and hence seems to be normally distributed. .

The results of one-way ANOVA are as follows from where we observe that we have not enough evidence to reject the null hypothesis as F(2, 21)=2.880, F>3.4668 and p=0.083>0.05. Hence, we conclude that the average whiteness reading of three detergents are not significantly different.

One-way ANOVA: Tide, Gain, All

Method

Null hypothesis All means are equal

Alternative hypothesis At least one mean is different

Significance level α = 0.05

Equal variances were assumed for the analysis.

Factor Information

Factor Levels Values

Factor 3 Tide, Gain, All

Analysis of Variance

Source DF Adj SS Adj MS F-Value P-Value

Factor 2 174.1 87.04 2.80 0.083

Error 21 651.9 31.04

Total 23 826.0

Model Summary

S R-sq R-sq(adj) R-sq(pred)

5.57150 21.08% 13.56% 0.00%

Means

Conclusion:

We conclude that conglomerate claims that its detergent “whitens and brightens better than all the rest.” Is not valid and all the detergents have the same whitens and brightens.

AllGainTide

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Boxplot of Tide, Gain, ...