Question

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

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.