Question

The computer operations department had a business objective of reducing the amount of time to fully update each subscriber's set of messages in a special secured email system. An experiment was conducted in which 23 subscribers were selected and three different messaging systems were used. Eight subscribers were assigned to each system, and the update times were measured as follows:

System A	System B	System C
38.8	41.8	36.9
42.1	36.4	36.1
45.2	39.1	39.2
34.8	28.7	35.6
48.3	36.4	41.9
37.8	36.1	31.7
41.1	35.8	35.2
43.6	-	38.1

Given Sample Means: x̅a = 41.46, x̅b = 36.33, x̅c = 35.84, Grand Mean = 38.29

Given Sample Standard Deviations: sa = 4.32, sb = 4.00, sc = 3.02, S = 4.34

A) At the 0.05 level of significance, is there evidence of a difference in the variance of the update times between Systems B and C? (Show your work: hypotheses, test statistic, critical value, and decision).

B) Fill out the following summary table for One-Way ANOVA:

Source of Variation	SS	df	MS	F
Among Groups
Within Groups			14.54	-
Total		22	-	-

C) Using the Tukey-Kramer method, determine which pair of the designs have the difference in mean distances at the 0.05 level of significance by filling out the following table (the upper-tail critical value from the studentized range distribution with 3 and 20 degrees of freedom as Qα = 3.578)

Pair (i,j)	|Xbari -Xbarj |	Comparison (> or <)	Critical Range	Difference (Yes or No)
(A,B)
(A,C)
(B,C)

Answer 1

A)

The hypotheses are :

Test is two tailed.

Here we have

Test statistics will be

Degree of freedom of numerator is df1=7-1=6 and degree of freedom of denominator is df2=8-1=7.

So critical value for 0.05 level of significance using excel function "=FINV(0.025,6,7)" is 5.119.

Since F < 5.119 so we fail to reject the null hypothesis.

That is we cannot conclude that there is evidence of a difference in the variance of the update times between Systems B and C.

(b)

The SS among groups is:

There are 3 groups so degree of freedom among groups is df=n-12

Source of Variation	SS	df	MS	F
Among Groups	124.326	2	62.163	4.275
Within Groups	14.54*20=290.8	23-3=20	14.54	-
Total	415.126	22	-	-

C)

Here we have 3 groups. The degree of freedom is

df= 20

So Tukey's HSD will be

Following is the completed table:

groups (i-j)	xbari	xbarj	ni	nj	HSD	xbari-xbarj	Lower limit	Upper limit	Significant(Yes/No)
mu1-mu2	41.46	36.33	8	7	4.993	5.13	0.137	10.123	Yes
mu1-mu3	41.46	35.84	8	8	4.8237	5.62	0.7963	10.4437	yes
mu2-mu3	36.33	35.81	7	8	4.993	0.52	-4.473	5.513	No