In: Statistics and Probability
A study compared three display panels used by air traffic controllers. Each display panel was tested for four different simulated emergency conditions. Twenty-four highly trained air traffic controllers were used in the study. Two controllers were randomly assigned to each display panel-emergency condition combination. The time (in seconds) required to stabilize the emergency condition was recorded. The following table gives the resulting data and the MINITAB output of a two-way ANOVA of the data.
Emergency Condition | ||||
Display Panel | 1 | 2 | 3 | 4 |
A | 20 | 26 | 33 | 11 |
20 | 25 | 35 | 11 | |
B | 15 | 20 | 29 | 12 |
10 | 19 | 30 | 9 | |
C | 22 | 30 | 35 | 10 |
23 | 29 | 36 | 16 | |
Two-way ANOVA: Time versus Panel, Condition | |||||
Source | DF | SS | MS | F | P |
Panel | 2 | 280.583 | 140.292 | 43.73 | .0000 |
Condition | 3 | 1,427.46 | 475.819 | 148.31 | .0000 |
Interaction | 6 | 20.42 | 3.403 | 1.06 | .4361 |
Error | 12 | 38.50 | 3.208 | ||
Total | 23 | 1,766.96 | |||
Tabulated statistics: Panel, Condition | |||||
Rows: | Panel | Columns: | Condition | ||
1 | 2 | 3 | 4 | All | |
A | 17.00 | 25.50 | 34.00 | 11.50 | 22.00 |
B | 12.50 | 19.50 | 29.50 | 9.50 | 17.75 |
C | 22.50 | 29.50 | 35.50 | 17.00 | 26.13 |
All | 17.67 | 24.83 | 33.00 | 12.67 | 21.96 |
Figure 12.12
(a) Interpret the interaction plot in the above table. Then test for interaction with α = .05.
Panel B requires (Click to select) more time less time to stabilize the emergency condition. | |
F(int)= 1.06, p-value= .436;; (Click to select) cannot can reject H_{0}, no interaction exists. |
(b) Test the significance of display panel effects with α = .05.
F = 43.73, p-value = .0000; (Click to select) do not reject reject H_{0}
(c) Test the significance of emergency condition effects with α = .05.
F = 148.31, p-value = .0000; (Click to select) do not reject reject H_{0}
(d) Make pairwise comparisons of display panels A, B , and C by using Tukey simultaneous 95 percent confidence intervals. (Round your answers to 2 decimal places. Negative amounts should be indicated by a minus sign.)
Tukey q_{.05} = , MSE = 3.208 | |
u_{A} – u_{B}: | [ , ] |
u_{A} – u_{C}: | [ , ] |
u_{B} – u_{C}: | [ , ] |
(e) Make pairwise comparisons of emergency conditions 1, 2, 3, and 4 by using Tukey simultaneous 95 percent confidence intervals. (Round your answers to 2 decimal places. Negative amounts should be indicated by a minus sign.)
u_{1} – u_{2}: | [ , ] | |
u_{1} – u_{3}: | [ , ] | |
u_{1} – u_{4}: | [ , ] | |
u_{2} – u_{3}: | [ , ] | |
u_{2} – u_{4}: | [ , ] | |
u_{3} – u_{4}: | [ , ] | |
(f) Which display panel minimizes the time required to stabilize an emergency condition? Does your answer depend on the emergency condition? Why?
(Click to select) Panel
A Panel C Panel
B minimizes the time required to stabilize an emergency
condition. (Click to select) No Yes , there is (Click to select) no some interaction. |
(g) Calculate a 95 percent (individual) confidence interval for the mean time required to stabilize emergency condition 4 using display panel B. (Round your answers to 2 decimal places.)
Confidence interval [ , ]
a) panel B require (less time) to stablize the emergency
condition
cannot reject Ho, no interaction eists
b) reject Ho (because p value <α)
c) Reject Ho (because p value < α)
d)
q(0.05)=3.7728
confidence interval | ||||
population mean difference | critical value | lower limit | upper limit | |
µ1-µ2 | 4.63 | 2.39 | 2.24 | 7.01 |
µ1-µ3 | -2.500 | 2.39 | -4.89 | -0.11 |
µ2-µ3 | -7.13 | 2.39 | -9.51 | -4.74 |
e)
confidence interval | ||||
population mean difference | critical value | lower limit | upper limit | |
µ1-µ2 | -6.50 | 3.06998 | -9.57 | -3.43 |
µ1-µ3 | -14.67 | 3.07 | -17.74 | -11.60 |
µ1-µ4 | 6.83 | 3.07 | 3.76 | 9.90 |
µ2-µ3 | -8.17 | 3.07 | -11.24 | -5.10 |
µ2-µ4 | 13.33 | 3.07 | 10.26 | 16.40 |
µ3-µ4 | 21.50 | 3.07 | 18.43 | 24.57 |
f)
Panel B, minimizes the time required…
No, there is no interaction
g)
Xbar= 10.500
MSE= 3.2080
n= 2
df error= 12
t(α/2,df error) = 2.17881283
Confidence interval = Xbar ± t(α/2,dfe) *√(MSE/n)=
( 7.74 ,
13.26 )