Question

An airline wants to select a computer software package for its reservation system. Four software packages (1, 2, 3, and 4) are commercially available. The airline will choose the package that bumps as few passengers, on the average, as possible during a month. An experiment is set up in which each package is used to make reservations for 5 randomly selected weeks. (A total of 20 weeks was included in the experiment.) The number of passengers bumped each week is given below. Please use the appropriate test to see if there is sufficient evidence to conclude that the average numbers of customers bumped by the 4 packages are not all the same. (14 points, without test procedure and excel output, you will get zero point). Level of significance is assumed to be 0.01.

Package 1:​ 12, 14, 9, 11, 16

Package 2: ​ 2, 4, 7, 3, 1

Package 3:​ 10, 9, 6, 10, 12

Package 4:​ 7, 6, 6, 15, 12

Answer 1

The sample statistics from the above data is as below:

	Package 1	Package 2	Package 3	Package 4
Total	62	17	47	47
n	5	5	5	5
Mean	12.400	3.400	9.4	9.4
Sum Of Squares	29.2	21.2	19.2	79.2

The Hypothesis:

H0: There is no difference in the average number of customers bumped by the 4 packages.

Ha: There is a difference in the average number of customers bumped by the 4 packages.

The Test Statistic:

.The ANOVA table is as below.

The critical value is calculated for = 0.01 for df1 = 3 and df2 = 16; Use the Excel Formula FINV(0.01,3,11)

Critical value = 5.2922

The p value is calculated for F = 7.66 for df1 = 3 and df2 = 16; Using the Excel Formula FDIST(7.66,3,16)

p value = 0.0021

Source	SS	DF	Mean Square	F	Fcv	p
Between	213.75	3	71.25	7.66	5.2922	0.0021
Within/Error	148.80	16	9.30
Total	362.55	19

F_observed = 7.66

The Decision Rule: If F_observed is > F_critical, Then reject H0.

Also if p-value is < , Then reject H0.

The Decision: Since Fobserved (7.66) is > F_critical (5.2922), We reject H0.

Also since p-value (0.0021) is < (0.05), We reject H0.

The Conclusion: There is sufficient evidence at the 99% level of significance to conclude that there is a difference in the average number of customers bumped by the 4 packages.

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Calculations For the ANOVA Table:

Overall Mean = (62 + 17 + 47 + 47) / 20 = 8.65

SS treatment = SUM n* ( - overall mean)²

= 5 * (12.4 - 8.65)² + 5 * (3.4 - 8.65)² + 5 * (9.4 - 8.65)² + 5 * (9.4 - 8.65)² = 213.75

df1 = k - 1 = 4 - 1 = 3

MSTR = SS treatment / df1 = 213.75 / 3 = 71.25

SSerror = SUM (Sum of Squares) = 29.2 + 21.2 + 19.2 + 79.2 = 148.8

df2 = N - k = 20 - 4 = 16

Therefore MS error = SSerror / df2 = 148.8 / 16 = 9.3

F = MSTR/MSE = 71.25 / 9.3 = 7.66

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