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 uA – uB: [  ,  ] uA – uC: [  ,   ] uB – uC: [  ,  ]

(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.)

 u1 – u2: [  ,  ] u1 – u3: [  ,  ] u1 – u4: [  ,   ] u2 – u3: [  ,  ] u2 – u4: [  ,  ] u3 – u4: [  ,  ]

(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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