In: Statistics and Probability
An amusement park studied methods for decreasing the waiting time (minutes) for rides by loading and unloading riders more efficiently. Two alternative loading/unloading methods have been proposed. To account for potential differences due to the type of ride and the possible interaction between the method of loading and unloading and the type of ride, a factorial experiment was designed. Use the following data to test for any significant effect due to the loading and unloading method, the type of ride, and interaction. Use = .05. Factor A is method of loading and unloading; Factor B is the type of ride.
The p-value for Factor A is Selectless than .01between .01 and .025between .025 and
.05between .05 and .10greater than .10Item 21 The p-value for Factor B is Selectless than .01between .01 and .025between .025 and
.05between .05 and .10greater than .10Item 23 The p-value for the interaction of factors A and B is Selectless than .01between .01 and .025between .025 and
.05between .05 and .10greater than .10Item 25 What is your recommendation to the amusement park? |
The Summary Statistics are as below
A | 2 |
B | 3 |
n | 2 |
N | 12 |
Total | Roller | Demon | Flume | Total Row |
Method1 | 88 | 114 | 90 | 292 |
Method2 | 102 | 98 | 100 | 300 |
Total Column | 190 | 212 | 190 | 592 |
(1) (Overall Total)2 / N = (592)2/12 = 29205.23
(2) Sum of Squares of Individual Observations = 432 + 562 + .....+ 482 = 29454
(3) SUM (Row Total)2 / (b*n); b * n = 3 * 2 = 6
(292)2/6 + (300)2/6 = 29210.67
(4) SUM (Column Total)2 / (a * n); a * n = 2 * 2 = 4
(190)2/4 + (212)2/4 + (190)2/4 = 29286
(5) SUM(Cell)2/n; n = 2
(88)2/2 + (114)2/2 + (90)2/2 + (102)2/2 + (98)2/2 + (100)2/2 = 29424
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SS A = (3) - (1) = 29210.67 - 29205.23 = 5.33
SS B = (4) - (1) = 29286 - 29205.23 = 80.67
SS AB = (5) + (1) - (3) - (4) = 29424 + 29205.23 - 29210.67 - 29286 = 132.67
SS Error = (2) - (5) = 29454 - 29424 = 30
SS Total = 5.33 + 80.67 + 132.67 + 30 = 248.67
df A = a - 1 = 3 - 1 = 2
df B = b - 1 = 2 - 1 = 1
df AB = df A * df B = 2 * 1 = 2
df Error = N - (a * b) = 12 - (3 * 2) = 6
MS A = SS A/df A = 5.33
MS B = SS B/df B = 40.34
MS AB = SS AB/df AB = 66.34
MS Error = SS Error / df Error = 5
F A = MS A/ MS error = 1.07
F B = MS B/ MS error = 8.07
F AB = MS AB/ MS error = 13.27
The Complete ANOVA Table with the p values are as below.
SS | df | MS | F | P value | |
A | 5.33 | 1 | 5.33 | 1.07 | 0.3408 |
B | 80.67 | 2 | 40.34 | 8.07 | 0.0199 |
AB | 132.67 | 2 | 66.34 | 13.27 | 0.0063 |
Error | 30 | 6 | 5 | ||
Total | 248.67 | 11 | |||
(1) The p value for Factor A is > 0.10
The Factor A is not significant (Since p value is > )
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(2) The p value for Factor B is between 0.01 and 0.05
The Factor A is significant (Since p value is < )
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(3) The p value for Factor A and B is < 0.01
The Factor AB is significant (Since p value is < )
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The Recommendation: Since method is not significant use either method 1 or method 2 for loading and unloading.
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