Question

A computer operations department decided to study the effect of the connection media used (either cable or fiber). The team designed a study in which a total of 30 subscribers were chosen. The subscribers were randomly assigned to one of the 3 messaging systems and measurements were taken on the updated time (in seconds). The data are shown below.

(With excel functions can you help in solving these questions)

a. State the null and alternate hypotheses that would be used to test whether there is a significant difference in the average update times for the three different messaging systems. Express symbolically if possible.

b. What type of study is this and what type of hypothesis test would be used to conduct the hypothesis test?

c. What is the p-value corresponding to this hypothesis test? (use cell reference to show answer and where it is found).

d. Conduct a hypothesis test to test whether there is a significant differencewhether there is a significant difference in the average update times for the three different messaging systems. Give a complete summary.

e. State the null and alternate hypotheses that would be used to test whether there is a significant difference in the mean update times of the messaging systems using the 2 different connection media. Express symbolically if possible.

f. What is the p-value corresponding to this hypothesis test? (use cell reference to show answer and where it is found).

g. Conduct a hypothesis test to test whether there is a significant effect due to messaging system used. Give a complete summary.

h. State the null and alternate hypotheses that would be used to test whether there is an interaction between the connection media and the messaging system. Express symbolically if possible.

i. What is the p-value corresponding to this hypothesis test? (use cell reference to show answer and where it is found).

j. Conduct a hypothesis test to determine whether there is an interaction between interaction between the connection media and the messaging system. State complete conclusions.

Technology	System1	System2	System3
Cable	4.56	4.17	3.53
Cable	4.90	4.28	3.77
Cable	4.18	4.00	4.10
Cable	3.56	3.96	2.87
Cable	4.34	3.60	3.18
Fiber	4.41	3.79	4.33
Fiber	4.08	4.11	4.00
Fiber	4.69	3.58	4.31
Fiber	5.18	4.53	3.96
Fiber	4.85	4.02	3.32

Answer 1

Solution)

In this case ,

We have 2 independent variables

Messaging system(System1,System2,System3)

and

Connection Media(Cable,Fiber)

and dependent variable is Updated time.

So, here we can use 2-way ANOVA . Because we have 2 factors(independent variables).

Let us define our hypothesis for the problem:

H01= there is not significant difference in the average update times for the three different messaging systems

H11= there is significant difference in the average update times for the three different messaging systems

H02= there is not significant difference in the average update times for the Two connection media.

H12= there is significant difference in the average update times for the Two connection media.

H03=there is not Significant effect of Interaction(Messaging systems*connection media) on updated time.

H13=there is Significant effect of Interaction(Messaging systems*connection media) on updated time.

Our Entered Data as

Technology	System1	System2	System3
Cable	4.56	4.17	3.53
Cable	4.90	4.28	3.77
Cable	4.18	4.00	4.10
Cable	3.56	3.96	2.87
Cable	4.34	3.60	3.18
Fiber	4.41	3.79	4.33
Fiber	4.08	4.11	4.00
Fiber	4.69	3.58	4.31
Fiber	5.18	4.53	3.96
Fiber	4.85	4.02	3.32

To perform 2 way ANOVA in excel:

Steps: Data tab --- data analysis Tool---anova two factor with replication---ok

           input range(select all data including labels)---row per sample(5)   because we have 5 row of cable and 5 rows of fiber

alpha =0.05

In output range (select any range where you want your anova result)

Click ok.

Result:

Anova: Two-Factor With Replication
SUMMARY	System1	System2	System3	Total
Cable
Count	5	5	5	15
Sum	21.54	20.01	17.45	59
Average	4.308	4.002	3.49	3.93333333
Variance	0.24772	0.06722	0.23315	0.27860952
Fiber
Count	5	5	5	15
Sum	23.21	20.03	19.92	63.16
Average	4.642	4.006	3.984	4.21066667
Variance	0.17617	0.12843	0.16693	0.2344781
Total
Count	10	10	10
Sum	44.75	40.04	37.37
Average	4.475	4.004	3.737
Variance	0.219383333	0.08696	0.24560111
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Sample	0.576853333	1	0.57685333	3.39451953	0.0778022	4.259677273
Columns	2.79258	2	1.39629	8.21653165	0.0019129	3.402826105
Interaction	0.312166667	2	0.15608333	0.91847943	0.4127035	3.402826105
Within	4.07848	24	0.16993667
Total	7.76008	29

From the above anova table:

Since ,P-value for Sample(connection media) = 0.0778>0.05

So ,we do not reject our Null hypothesis (H02)

Or

F < (F critical) , we do not reject our Null hypothesis(H02)

and conclude that there is not significant difference in the average update times for the Two connection media.

Also,P-value for Columns(messaging systems) =0.0019<0.05

So ,we reject our Null hypothesis(H01)

Or

F > (F critical) , we reject our Null hypothesis(H01)

and conclude that there is significant difference in the average update times for the three different messaging systems.

Also P-value for Interaction(Messaging system*connection media) =0.412>0.05

So we do not reject our Null hypothesis(H03)

Or

F < (F critical), we do not reject our null hypothesis(H03)

and conclude that there is not Significant effect of Interaction(Messaging systems*connection media) on updated time.