Question

In: Statistics and Probability

A study compared three display panels used by air traffic controllers. Each display panel was tested...

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 H0, 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  H0

(c) Test the significance of emergency condition effects with α = .05.

F = 148.31, p-value = .0000;   (Click to select)   do not reject   reject  H0

(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
uAuB: [  ,  ]
uAuC: [  ,   ]
uBuC: [  ,  ]

(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.)

u1u2: [  ,  ]
u1u3: [  ,  ]
u1u4: [  ,   ]
u2u3: [  ,  ]
u2u4: [  ,  ]
u3u4: [  ,  ]

(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         [  ,  ]

Solutions

Expert Solution

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   )


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