Question

1) Suppose that four independent companies collect data on their employees, regarding the number of sick days used per year: Company 1 Company 2 Company 3 Company 4 4 5 6 9 7 11 7 12 1 6 7 11 5 9 10 6 10 5 9 8 8 8 8 14 7 4 12 6 9 2 4 8 4 10 2 9 At the 0.05 significance level, test for the average equality of means. Assume the populations are all normally distributed, with equal variances.

Answer 1

For the given data using Anova single factor in Excel we get output as

Anova: Single Factor
SUMMARY
Groups	Count	Sum	Average	Variance
company 1	9	55	6.111111	8.111111
company 2	9	60	6.666667	9
company 3	9	65	7.222222	9.194444
company 4	9	83	9.222222	7.194444
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	49.63888889	3	16.5463	1.975677	0.137425	2.90112
Within Groups	268	32	8.375
Total	317.6388889	35

So from the above output

Noll hypothers (Ho): lill, legally Alternative hypothesis (AG): Not all the equal. Test statistici means are = 1.975677 16.5463 8.375 F. MS 6/6 Ms within F = 1.98 we eget P-value: with F = 1.98, dof=(3,32) P-value = 0.1374 Decision: P-value>los i-e., 0.1374>0.05 So fail to reject to Conclusion: At x = 0.05 lors there is enough evidence to claim that the call the means are equal.