A study compared three display panels used by air traffic controllers. Each disp

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.

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.)

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.)

Which display panel minimizes the time required to stabilize an emergency condition? Does your answer depend on the emergency condition? Why?

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 MINITAB output of a two-way ANOVA of the data.

Explanation / Answer

A. To determine if interaction exists between display panel and emergency condition, we perform a partial F test of H0: bB2 bB3 bB4 bC2 bC3 bC4 0 (that is, there is no interaction) D4

Test: T1 Numerator: 2.9449 DF: 6 F Value: 0.5522 Denominator: 5.333333 DF: 9 Prob>F: 0.7581 Noting that the complete model for the partial F test is the entire two-factor model, specify the reduced model, interpret the output, and discuss why the output indicates that there is little or no interaction between display panel and emergency condition.

b Even though we have concluded that there is little or no interaction between display panel and emergency condition, we will leave the interaction terms in the two-factor model to compare the display panels and emergency conditions. Note that this is typically done when studying the effects of two qualitative factors on a response variable and is one reason for using the 1, 1, 0 dummy variables. Specifically, consider comparing the display panels by analyzing the following factor level means: It can be proven that using the 1, 1, 0 dummy variables implies that we can express mA., mB., and mC. in terms of the model parameters by setting each of these factor level means equal to (b0 bBDB bCDC) and ignoring the rest of the two-factor model. Specifically, mA. b0 bBDB bCDC b0 bB(1) bC(1) b0 bB bC mB. b0 bBDB bCDC b0 bB(1) bC(0) b0 bB mC. b0 bBDB bCDC b0 bB(0) bC(1) b0 bC To test H0: mA. mB. mC. (all display panels have the same effect on mean stabilization time), we perform a partial F test of H0: bB bC 0. The SAS output of this test is as follows: Test: T2 Numerator: 100.0541 DF: 2 F Value: 18.7601 Denominator: 5.333333 DF: 9 Prob>F: 0.0006 Noting that the complete model for the partial F test is the entire two-factor model, specify the reduced model, interpret the output, and discuss why the output indicates that at least two display panels have different effects on mean stabilization time.

c To more precisely compare the display panels, consider the following SAS output: T for HO: Pr > |T| Std Error of Parameter Estimate Parameter=0 Estimate MUB-MUA -3.5000000 -2.71 0.0240 1.29099445 MUC-MUA 4.0000000 2.95 0.0161 1.35400640 MUC-MUB 7.5000000 6.12 0.0002 1.22474487 mC. mC1 mC2 mC3 mC4 4

Using the least squares point estimates bB 3.6667 and bC 3.8333 in Figure G.1, SAS has calculated the point estimate of MUB MUA mB. mA. (b0 bB) (b0 bB bC) 2bB bC to be (2bB bC) (2(3.6667) 3.8333), or 3.5. Using the least squares point estimates, show how SAS has obtained the point estimates of MUC MUA mC. mA. (b0 bC) (b0 bB bC) 2bC bB and MUC MUB mC. mB. (b0 bC) (b0 bB) bC bB Which display panel seems to minimize mean stabilization time? d To compare the emergency conditions, consider the following factor level means By setting each factor level mean equal to (b0 b2D2 b3D3 b4D4), show that the four factor level means equal, respectively, (b0 b2 b3 b4), (b0 b2), (b0 b3), and (b0 b4). Furthermore, discuss why the following SAS output of a partial F test of H0: b2 b3 b4 0 implies that the different emergency conditions have different effects on mean stabilization time: Test: T3 Numerator: 371.1940 DF: 3 F value: 69.5989 Denominator: 5.333333 DF: 9 Prob>F: 0.0001 e Fully interpret the following SAS output, which gives (1) point estimates of and 95 percent confidence intervals for the mean stabilization times mB1, mB2, mB3, and mB4 and (2) point predictions of and 95 percent prediction intervals for the individual stabilization times yB1, yB2, yB3, and yB4: Panel and Predict Std Err Lower95% Upper95% Lower95% Upper95% Condition Value Predict Mean Mean Predict Predict B1 13.5000 1.633 9.8059 17.1941 7.1017 19.8983 B2 20.5000 1.633 16.8059 24.1941 14.1017 26.8983 B3 29.5000 1.633 25.8059 33.1941 23.1017 35.8983 B4 9.5000 1.633 5.8059 13.1941 3.1017 15.8983 m.3 mA3 mB3 mC3 3 m.4

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