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	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         [  ,  ]

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(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   )