In: Statistics and Probability
You must include all EXCEL printouts and the 6 step write-up for both problems. The questions are given below.
USING ANOVA: Assuming variances are equal.
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 |
Using Minitab software, (Stat -> ANOVA -> Test for equal variances), we get
b) Since p-value = 0.956 > 0.05, so we fail to reject H0 and we can conclude that there is no sufficient evidence of a difference in the variances in number of absences among the four plants.
c) Using Minitab software, (Stat -> ANOVA -> One way), we get
Since P-value = 0.004 < 0.05, so at 5% level of significance, we reject H0 and we can conclude that there is significant difference in the mean number of absences among the four plants.
e) Since there is a difference in the means, to figure out which ones are different, one need to use independent samples t test.