In: Statistics and Probability
The computer operations department has a business objective of reducing the amount of time to fully update each subscriber’s set of messages in a special secure email system. A team designed a factorial experiment involving two factors, i.e., Interface and Media, with the goal of understanding the importance of these factors in explaining the variability in Update Time. The factor Interface has three levels, i.e., System I, System II, and System III, while the factor Media has two levels, i.e., Cable and Fiber. Consequently, there are 3x2=6 different factor-level combinations. Subscribers were randomly assigned to one of the 6 factor-level combinations and Update Time was recorded for each observation.
1. Analyze the experimental data. Report the ANOVA Table and comment on the importance of the two factors and their interaction in explaining the variability in Update Time.
2. If there are model terms with p-values > 0.05, drop them from the model and re-analyze the data under your reduced model. Refer to this as your final model.
3. Report and interpret the coefficient of determination, ? , for your final model.
4. What level(s) of the important factor(s) appears to produce the shortest Update Time?
Subscriber ID | Update Time | Interface | Media |
1 | 4.56 | System I | Cable |
2 | 4.9 | System I | Cable |
3 | 4.18 | System I | Cable |
4 | 3.56 | System I | Cable |
5 | 4.34 | System I | Cable |
6 | 4.17 | System II | Cable |
7 | 4.28 | System II | Cable |
8 | 4 | System II | Cable |
9 | 3.96 | System II | Cable |
10 | 3.6 | System II | Cable |
11 | 3.53 | System III | Cable |
12 | 3.77 | System III | Cable |
13 | 4.1 | System III | Cable |
14 | 2.87 | System III | Cable |
15 | 3.18 | System III | Cable |
16 | 4.41 | System I | Fiber |
17 | 4.08 | System I | Fiber |
18 | 4.69 | System I | Fiber |
19 | 5.18 | System I | Fiber |
20 | 4.85 | System I | Fiber |
21 | 3.79 | System II | Fiber |
22 | 4.11 | System II | Fiber |
23 | 3.58 | System II | Fiber |
24 | 4.53 | System II | Fiber |
25 | 4.02 | System II | Fiber |
26 | 4.33 | System III | Fiber |
27 | 4 | System III | Fiber |
28 | 4.31 | System III | Fiber |
29 | 3.96 | System III | Fiber |
30 | 3.32 | System III | Fiber |
Solution:
Given
Subscriber ID | Update Time | Interface | Media |
1 | 4.56 | System I | Cable |
2 | 4.9 | System I | Cable |
3 | 4.18 | System I | Cable |
4 | 3.56 | System I | Cable |
5 | 4.34 | System I | Cable |
6 | 4.17 | System II | Cable |
7 | 4.28 | System II | Cable |
8 | 4 | System II | Cable |
9 | 3.96 | System II | Cable |
10 | 3.6 | System II | Cable |
11 | 3.53 | System III | Cable |
12 | 3.77 | System III | Cable |
13 | 4.1 | System III | Cable |
14 | 2.87 | System III | Cable |
15 | 3.18 | System III | Cable |
16 | 4.41 | System I | Fiber |
17 | 4.08 | System I | Fiber |
18 | 4.69 | System I | Fiber |
19 | 5.18 | System I | Fiber |
20 | 4.85 | System I | Fiber |
21 | 3.79 | System II | Fiber |
22 | 4.11 | System II | Fiber |
23 | 3.58 | System II | Fiber |
24 | 4.53 | System II | Fiber |
25 | 4.02 | System II | Fiber |
26 | 4.33 | System III | Fiber |
27 | 4 | System III | Fiber |
28 | 4.31 | System III | Fiber |
29 | 3.96 | System III | Fiber |
30 | 3.32 | System III | Fiber |
1) Null hypothesis Ho : The mean update time for the three different systems are same / The mean update time for the two different media are same. The mean update time for the interaction between three different systems and two different medias are same.
Alternative hypothesis Ha : The mean update time for at least one systems is different / The mean update time for at least one media is different . The mean update time for the interaction between at least one systems and media is different.
Level of significance = 0.05
R software:
The first two way model shows that the Media and Interaction between Media and Interface are not significant (p-value > 0.05), we remove both the terms from explanatory variable and do ANOVA analysis with Interface variable alone.
2) The second two way model shows that the variable Interface is statistically significant at 0.05 level of significance.
3) The R square for the reduced final model is
SSE/SST = 4.967 / (2.793 + 4.967)
= 0.6701
Nearly, 67.01% of variation in Update_Time is explained by the variable Interface.
4) Tukey Pairwise comparison is performed which system pair in Interface produce the shortest Update_Time.
From pairwise comparison Test,
we conclude that the pair System 1 - System 3 produces the shortest Update_Time(Since the p value is < 0.05).
Please give upvote.
Thank you.