Question

6 step hypothesis tests using ANOVA: Levene Test and ANOVA: single factor Info: Concerned about Friday...

6 step hypothesis tests using ANOVA: Levene Test and ANOVA: single factor

Info: Concerned about Friday absences, management examined the number of persons absent for each of the past three Fridays. Does this sample provide sufficient evidence to conclude there is a significant difference in the average number of absences? Use alpha = .05.

Plant 1

Plant 2

Plant 3

Plant 4

19

17

27

22

24

20

32

27

20

16

27

25

Homework Answers

Answer #1

For the given table

H1: not all means are equal

Using one way anova in excel

Anova: Single Factor
SUMMARY
Groups Count Sum Average Variance
plant 1 3 63 21 7
Plant 2 3 53 17.66667 4.333333
Plant 3 3 86 28.66667 8.333333
Plant 4 3 74 24.66667 6.333333
ANOVA
Source of Variation SS df MS F P-value F crit
Between Groups 202 3 67.33333 10.35897 0.003952 4.066181
Within Groups 52 8 6.5
Total 254 11

P-value:

P-value computed using Z table shown as

P-value= 0.0040

Conclusion:

Since the P-value calculatd is less than the level of significance 0.05 hence we reject the null hypothesis and conclude that there is sufficient evidence to conclude there is a significant difference in the average number of absences.

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