Question

In: Statistics and Probability

You are asked to analyze the below speed data collected on a highway on different days...

You are asked to analyze the below speed data collected on a highway on different days of the week for similarity. Use ANOVA (analysis of variance) to support your conclusion by performing the following tasks:

Sample 1 (mph)

Sample 2 (mph)

Sample 3 (mph)

50

45

55

55

50

60

50

55

50

50

60

45

60

40

40

State your null hypothesis (H0) and alternative hypothesis (H1)

State your assumed level of significance (αα)

Calculate mean for sample 1

Calculate mean for sample 2

Calculate grand mean for sample 1 & 2

Calculate SSB (between sample sum of square errors) and the associated degree of freedom

Calculate SSW (within sample sum of square errors) and the associated degree of freedom

Calculate SST (total sample sum of square errors) and the associated degree of freedom (Hint: this validates the accuracy of your calculated SSB and SSW values)

Calculate the F value

State the critical F value (table value based on your assumed level of significance)

Make your conclusion whether or not the speed reduction strategy was effective and state why?

Solutions

Expert Solution

using excel>data>data analysis>ANOVA single factor

we have

Anova: Single Factor
SUMMARY
Groups Count Sum Average Variance
Sample 1 (mph) 5 265 53 20
Sample 2 (mph) 5 250 50 62.5
Sample 3 (mph) 5 250 50 62.5
ANOVA
Source of Variation SS df MS F P-value F crit
Between Groups 30 2 15 0.310345 0.738906 3.885294
Within Groups 580 12 48.33333
Total 610 14

null hypothesis (H0) :

the alternative hypothesis (H1) :at least two means are different .

assumed level of significance (α) = 0.05

mean for sample 1 =53

mean for sample 2 =50

grand mean for sample 1 & 2 = (50+53)/2 = 51.5

SSB (between sample sum of square errors) =30 and the associated degree of freedom =2

SSW (within sample sum of square errors) = 580 and the associated degree of freedom =12

SST (total sample sum of square errors) = 610 and the associated degree of freedom =14

the F value =0.3104

the critical F value = 3.885

Do not reject H0 the speed reduction strategy was not effective because their average are same


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