Question

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.

Do not use any software. I need to see how you came to the answer. Please be very detailed.

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. 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, R2, for your final model.

4. What level(s) of the important factor(s) appears to produce the shortest Update Time?

Answer 1

Solve using Two-way ANOVA method

Solve using Two-way ANOVA method

Observation	A	B	C
1	4.56,4.9,4.18,3.56,4.34	4.17,4.28,4,3.96,3.6	3.53,3.77,4.1,2.87,3.18
2	4.41,4.08,4.69,5.18,4.85	3.79,4.11,3.58,4.53,4.02	4.33,4,4.31,3.96,3.32

Here, 1 represents Cable and 2 represents Fibre

A,B and C represent system I,II and III respectively.

Row and column sums

	A	B	C	Row total (xa)
1	21.54	20.01	17.45	59
2	23.21	20.03	19.92	63.16
Col total (xb)	44.75	40.04	37.37	122.16

∑x²=4.56²+4.9²+4.18²+...+4.31²+3.96²+3.32²=505.1956→(A)

$\sum \frac{X^2_b}{ra}$ =(44.75²+40.04²+37.37²)/(5*2)

=(2002.5625+1603.2016+1396.5169)/10

=(5002.281)/10

=500.2281→(B)

$\sum \frac{X^2_b}{rb}$ =(59²+63.16²)/(5*3)

=(3481+3989.1856)/15

=7470.1856/15

=498.0124→(C)

$\sum \frac{X^2_{ab}}{r}$ =(21.54²+20.01²+17.45²+23.21²+20.03²+19.92²)/5

= (463.9716+400.4001+304.5025+538.7041+401.2009+396.8064)/5

= (2505.5856)/5

= 501.1171 →(C)

$\frac{({\sum X})^2}{rab}$=(122.16)²/(5*2*3)

=14923.0656/30

=497.4355→(D)

Sum of squares total
SST=∑x²-(∑x)²/n=(A)-(D)

=505.1956-497.4355

=7.7601

Sum of squares between rows
SSA=∑x_a²/rb-(∑x)²/n=(C)-(D)

=498.0124-497.4355

Sum of squares between columns
SSB=∑xb²/ra-(∑x)²n=(B)-(D)

=500.2281-497.4355

=2.7926

Sum of squares between rows and columns
SSAB=∑∑x²ab/r-(∑x)²n-SSA-SSB=(B)-(D)-SSA-SSB

=501.1171-497.4355-0.5769-2.7926

=0.3122

Sum of squares Error (residual)
SSE=SST-SSA-SSB-SSAB

=7.7601-0.5769-2.7926-0.3122

=4.0785

ANOVA table

Source of Variation	Sums of Squares SS	Degrees of freedom DF	Mean Squares MS	F	p-value
A	SSA=0.5769	a-1=1	MSR=0.57691=0.5769	0.57690.1699=3.3945	0.0778
B	SSB=2.7926	b-1=2	MSC=2.79262=1.3963	1.39630.1699=8.2165	0.0019
AB	SSAB=0.3122	(a-1)(b-1)=2	MSAB=0.31222=0.1561	0.15610.1699=0.9185	0.4127
Error (residual)	SSE=4.0785	rab-ab=24	MSE=4.078524=0.1699
Total	SST=7.7601	rab-1=29

Conclusion:
1. F for between columns
F(1,2) at 0.05 level of significance

=4.2597

As calculated FR=3.3945<4.2597

So, H0 is accepted, Hence there is no significant differentiating between rows or there is no significant difference between Media 'Cable' and 'Fibre'

2. F for between columns
F(2,2) at 0.05 level of significance

=3.4028

As calculated FC=8.2165>3.4028

So, H0 is rejected, Hence there is significant differentiating between columns or there is significant difference between three systems.

3. Interaction term has p-value = 0.4127 > 0.05 or interaction is not significant.

Please post other parts as another question and rate my answer.

Thanks