Question

In: Statistics and Probability

study compared three different display panels for use by air traffic controllers. Each display panel was...

  1. study compared three different display panels for use by air traffic controllers. Each display panel was test in a simulated emergency condition; 12 highly trained air traffic controllers took part in the study. Four controllers were randomly assigned to each display panel. The time (in seconds) needed to stabilize the emergency condition was recorded. The results of are given below. Let μ, μB, and μC represent the mean times to stabilize the emergency conditions when using display panels A, B, and C, respectively.
    1. State the null and alternative hypotheses
    1. µABC
    2. µA≠µB≠µC
    1. What is the test statistic?
    2. Using a .05 significance level, what is the decision rule?
    3. Show the test statistic and essential calculations.
    4. Interpret you results
    5. Should you conduct a Post Hoc Fisher or Tukey on this data? Why.
    6. Based on your answer for part f conduct a Post Hoc Tukey

Display Panel Study Data

A

B

C

21

24

40

27

21

36

24

18

35

26

19

32

Solutions

Expert Solution

The solution in details given below.

solutions:

a.

Null Hypothesis: All the means are equal

i.e. Ho: µA =µB =µC

VS

Alternative Hypothesis: Not all means are equal

i.e. H1: µA≠µB≠µC

b.

The test statistic is ANOVA

Method

Null hypothesis All means are equal
Alternative hypothesis Not all means are equal
Significance level α = 0.05

Equal variances were assumed for the analysis.

Factor Information

Factor Levels Values
Factor 3 A, B, C

Analysis of Variance

Source DF Adj SS Adj MS F-Value P-Value
Factor 2 500.17 250.083 30.11 0.000
Error 9 74.75 8.306
Total 11 574.92

C.

.Using a .05 significance level, the decision rule is that p-value =0.00

and alpha value=0.05 the alpha value is greater than p-value then Reject Ho

d.

Interprete: Then we can conclude that all the mean are not equal at  0.05 significance level.

e and f

Tukey Pairwise Comparisons

Grouping Information Using the Tukey Method and 95% Confidence

Factor N Mean Grouping
C 4 35.75 A
A 4 24.50 B
B 4 20.50 B

Means that do not share a letter are significantly different.

Tukey Simultaneous Tests for Differences of Means

Difference
of Levels
Difference
of Means
SE of
Difference
95% CI T-Value Adjusted
P-Value
B - A -4.00 2.04 (-9.69, 1.69) -1.96 0.177
C - A 11.25 2.04 (5.56, 16.94) 5.52 0.001
C - B 15.25 2.04 (9.56, 20.94) 7.48 0.000

Individual confidence level = 97.91%

Tukey Simultaneous 95% CIs

Fisher Pairwise Comparisons

Grouping Information Using the Fisher LSD Method and 95% Confidence

Factor N Mean Grouping
C 4 35.75 A
A 4 24.50 B
B 4 20.50 B

Means that do not share a letter are significantly different.

Fisher Individual Tests for Differences of Means

Difference
of Levels
Difference
of Means
SE of
Difference
95% CI T-Value Adjusted
P-Value
B - A -4.00 2.04 (-8.61, 0.61) -1.96 0.081
C - A 11.25 2.04 (6.64, 15.86) 5.52 0.000
C - B 15.25 2.04 (10.64, 19.86) 7.48 0.000

Simultaneous confidence level = 88.66%

Fisher Individual 95% CIs

Interval Plot of A, B, ...


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