Question: 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

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 panelemergency condition combination. The time (in seconds) required to stabilize the emergency condition was recorded. The following table gives the resulting data and the JMP output of a two-way ANOVA of the data.

Emergency Condition
Display Panel 1 2 3 4
A 17 25 31 14
14 24 35 13
B 15 22 28 9
12 19 31 10
C 21 29 32 15
24 28 37 19

Least Squares Means Estimates
Panel Estimate Condition Estimate
A 21.500000 1 17.166670
B 18.375000 2 24.666670
C 25.625000 3 32.166670
4 13.333300

Analysis of Variance
Source DF Sum of Squares Mean Square F Ratio
Model 11 1,480.3333 134.576 30.4700
Error 12 53.0000 4.417 Prob > F
C. Total 23 1,533.3333 <.0001*

Effect Tests
Source Nparm DF Sum of Squares F Ratio Prob > F
Panel 2 2 211.5833 23.9528 <.0001*
Condition 3 3 1,253.0000 94.5660 <.0001*
Panel* Condition 6 6 15.7500 0.5943 0.7298

Tukey HSD All Pairwise Comparisons

Quantile = 2.66776, Adjusted DF = 12.0, Adjustment = Tukey

Panel -Panel Difference Std Error t Ratio Prob>|t| Lower 95% Upper 95%
A B 3.12500 1.050793 2.97 0.0290* 0.3217 5.92826
A C 4.12500 1.050793 3.93 0.0053* 6.9283 1.32174
B C 7.25000 1.050793 6.90 < .0001* 10.0533 4.44674

Tukey HSD All Pairwise Comparisons

Quantile = 2.9688, Adjusted DF = 12.0, Adjustment = Tukey

Condition -Condition Difference Std Error t Ratio Prob>|t| Lower 95% Upper 95%
1 2 7.5000 1.213352 6.18 0.0002* 11.1022 3.8978
1 3 15.0000 1.213352 12.36 < .0001* 18.6022 11.3978
1 4 3.8333 1.213352 3.16 0.0359* 0.2311 7.4355
2 3 7.5000 1.213352 6.18 0.0002* 11.1022 3.8978
2 4 11.3333 1.213352 9.34 < .0001* 7.7311 14.9355
3 4 18.8333 1.213352 15.52 < .0001* 15.2311 22.4355

(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 = (?,?)

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