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 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 34 13 B 16 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 24.500000 C 25.625000 3 32.166670 4 13.500000 Analysis of Variance Source DF Sum of Squares Mean Square F Ratio Model 11 1,458.3333 132.576 32.4675 Error 12 49.0000 4.083 Prob > F C. Total 23 1,507.3333 <.0001* Effect Tests Source Nparm DF Sum of Squares F Ratio Prob > F Panel 2 2 211.5833 25.9082 <.0001* Condition 3 3 1,230.6667 100.4626 <.0001* Panel* Condition 6 6 16.0833 0.6565 0.6857 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 1.010363 3.09 0.0235* 0.4296 5.82040 A C −4.12500 1.010363 −4.08 0.0040* −6.8204 −1.42960 B C −7.25000 1.010363 −7.18 < .0001* −9.9454 −4.55460 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 −7.3333 1.166667 −6.29 0.0002* −10.7969 −3.8697 1 3 −15.0000 1.166667 −12.86 < .0001* −18.4636 −11.5364 1 4 3.6667 1.166667 3.14 0.0370* 0.2031 7.1303 2 3 −7.6667 1.166667 −6.57 0.0001* −11.1303 −4.2031 2 4 11.0000 1.166667 9.43 < .0001* 7.5364 14.4636 3 4 18.6667 1.166667 16.00 < .0001* 15.2031 22.1303 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.)

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 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 34 13 B 16 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 24.500000 C 25.625000 3 32.166670 4 13.500000 Analysis of Variance Source DF Sum of Squares Mean Square F Ratio Model 11 1,458.3333 132.576 32.4675 Error 12 49.0000 4.083 Prob > F C. Total 23 1,507.3333 <.0001* Effect Tests Source Nparm DF Sum of Squares F Ratio Prob > F Panel 2 2 211.5833 25.9082 <.0001* Condition 3 3 1,230.6667 100.4626 <.0001* Panel* Condition 6 6 16.0833 0.6565 0.6857 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 1.010363 3.09 0.0235* 0.4296 5.82040 A C −4.12500 1.010363 −4.08 0.0040* −6.8204 −1.42960 B C −7.25000 1.010363 −7.18 < .0001* −9.9454 −4.55460 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 −7.3333 1.166667 −6.29 0.0002* −10.7969 −3.8697 1 3 −15.0000 1.166667 −12.86 < .0001* −18.4636 −11.5364 1 4 3.6667 1.166667 3.14 0.0370* 0.2031 7.1303 2 3 −7.6667 1.166667 −6.57 0.0001* −11.1303 −4.2031 2 4 11.0000 1.166667 9.43 < .0001* 7.5364 14.4636 3 4 18.6667 1.166667 16.00 < .0001* 15.2031 22.1303 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.)

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Question

	Emergency Condition
Display Panel	1	2	3	4
A	17	25	31	14
	14	24	34	13
B	16	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	24.500000
C	25.625000	3	32.166670
		4	13.500000

Analysis of Variance
Source	DF	Sum of Squares	Mean Square	F Ratio
Model	11	1,458.3333	132.576	32.4675
Error	12	49.0000	4.083	Prob > F
C. Total	23	1,507.3333		<.0001*

Effect Tests
Source	Nparm	DF	Sum of Squares	F Ratio	Prob > F
Panel	2	2	211.5833	25.9082	<.0001*
Condition	3	3	1,230.6667	100.4626	<.0001*
Panel* Condition	6	6	16.0833	0.6565	0.6857

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	1.010363	3.09	0.0235*	0.4296	5.82040
A	C	−4.12500	1.010363	−4.08	0.0040*	−6.8204	−1.42960
B	C	−7.25000	1.010363	−7.18	< .0001*	−9.9454	−4.55460

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	−7.3333	1.166667	−6.29	0.0002*	−10.7969	−3.8697
1	3	−15.0000	1.166667	−12.86	< .0001*	−18.4636	−11.5364
1	4	3.6667	1.166667	3.14	0.0370*	0.2031	7.1303
2	3	−7.6667	1.166667	−6.57	0.0001*	−11.1303	−4.2031
2	4	11.0000	1.166667	9.43	< .0001*	7.5364	14.4636
3	4	18.6667	1.166667	16.00	< .0001*	15.2031	22.1303

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.)

Definition Definition Measure of central tendency that is the average of a given data set. The mean value is evaluated as the quotient of the sum of all observations by the sample size. The mean, in contrast to a median, is affected by extreme values. Very large or very small values can distract the mean from the center of the data. Arithmetic mean: The most common type of mean is the arithmetic mean. It is evaluated using the formula: μ = 1 N ∑ i = 1 N x i Other types of means are the geometric mean, logarithmic mean, and harmonic mean. Geometric mean: The nth root of the product of n observations from a data set is defined as the geometric mean of the set: G = x 1 x 2 ... x n n Logarithmic mean: The difference of the natural logarithms of the two numbers, divided by the difference between the numbers is the logarithmic mean of the two numbers. The logarithmic mean is used particularly in heat transfer and mass transfer. ln x 2 − ln x 1 x 2 − x 1 Harmonic mean: The inverse of the arithmetic mean of the inverses of all the numbers in a data set is the harmonic mean of the data. 1 1 x 1 + 1 x 2 + ...

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