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 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	19	25	34	12
	19	26	37	12
B	16	20	28	14
	11	19	28	7
C	22	30	33	10
	25	30	39	16

Two-way ANOVA: Time versus Panel, Condition
Source	DF	SS	MS	F	P
Panel	2	361.750	180.875	29.73	.0000
Condition	3	1,381.50	460.500	75.70	.0000
Interaction	6	28.25	4.708	.77	.6052
Error	12	73.00	6.083
Total	23	1,844.50

Tabulated statistics: Panel, Condition
Rows:	Panel	Columns:	Condition
	1	2	3	4	All
A	16.00	25.50	35.50	13.00	22.50
B	13.50	19.50	28.00	8.50	17.38
C	23.50	30.00	36.00	18.00	26.88
All	17.67	25.00	33.17	13.17	22.25

  

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)   less time   more time  to stabilize the emergency condition.

F(int)= .77, p-value= .605;;   (Click to select)   can   cannot  reject H0, no interaction exists.

(b) Test the significance of display panel effects with α = .05.

F = 29.73, p-value = .0000;   (Click to select)   do not reject   reject  H₀

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

F = 75.70, 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 = 6.083
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 C   Panel B   Panel A  minimizes the time required to stabilize an emergency condition.   (Click to select)   Yes   No  , 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         [  ,  ]

there is no plot. its based on the information in the table

Answer 1

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-statistic value(α,k,N-k) = 3.7728

			confidence interval
population mean difference	critical value	lower limit	upper limit
µ1-µ2	5.12	3.29	1.83	8.41
µ1-µ3	-4.380	3.29	-7.67	-1.09
µ2-µ3	-9.50	3.29	-12.79	-6.21

e)

			confidence interval
population mean difference	critical value	lower limit	upper limit
µ1-µ2	-7.33	3.15599	-10.49	-4.17
µ1-µ3	-15.50	3.16	-18.66	-12.34
µ1-µ4	4.50	3.16	1.34	7.66
µ2-µ3	-8.17	3.16	-11.33	-5.01
µ2-µ4	11.83	3.16	8.67	14.99
µ3-µ4	20.00	3.16	16.84	23.16

f)

Panel B, minimizes the time required…  
No, there is no interaction  

g)

Xbar=   10.500          
MSE=   4.1985          
n=   2          
df error=   12          
t(α/2,df error) =    2.17881283          
Confidence interval = Xbar ± t(α/2,dfe) *√(MSE/n)=              
(   7.34   ,   13.66   )