Question

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
   14. µ_A =µ_B =µ_C
   15. µ_A≠µ_B≠µ_C

1. 2. What is the test statistic?
   3. Using a .05 significance level, what is the decision rule?
   4. Show the test statistic and essential calculations.
   5. Interpret you results
   6. Should you conduct a Post Hoc Fisher or Tukey on this data? Why.
   7. 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

Answer 1

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, ...