SOLUTION: 10.64 A cognitive retraining clinic assists outpatient victims of head injury, anoxia, or other conditions that result in cognitive impairment. Each incoming patient is evaluated t

Question 131650: 10.64 A cognitive retraining clinic assists outpatient victims of head injury, anoxia, or other conditions that result in cognitive impairment. Each incoming patient is evaluated to establish an appropriate treatment program and estimated length of stay. To see if the evaluation teams are consistent, 12 randomly chosen patients are separately evaluated by two expert teams (A and B) as shown. At
the .10 level of significance, are the evaluator teams consistent in their stimates? State your hypotheses and show all steps clearly.

Estimated Length of Stay in Weeks
Patient
Team 1 2 3 4 5 6 7 8 9 10 11 12
A 24 24 52 30 40 30 18 30 18 40 24 12
B 24 20 52 36 36 36 24 36 16 52 24 16

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
A cognitive retraining clinic assists outpatient victims of head injury, anoxia, or other conditions that result in cognitive impairment. Each incoming patient is evaluated to establish an appropriate treatment program and estimated length of stay.
------------------
To see if the evaluation teams are consistent, 12 randomly chosen patients are separately evaluated by two expert teams (A and B) as shown.
------------------
At the .10 level of significance, are the evaluator teams consistent in their estimates? State your hypotheses and show all steps clearly.
------------------
Estimated Length of Stay in Weeks
Patient
Team 1. 2. 3. 4. 5. 6. 7. 8. 9 10. 11 12
A.. 24 24 52 30 40 30 18 30 18 40. 24 12
B.. 24 20 52 36 36 36 24 36 16 52. 24 16
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D....0..4..0.-6..4.-6.-6.-6..2.-12..0.-4
d-bar = -2.5
The data consists of 12 dependent pairs.
------------
Ho: mu-d = 0
Ha: mu-d is not zero
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Critical value for df=11, alpha=10%, 2-tail test: t= +/- 1.796
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Test Statistic: t = d-bar /[s/(sqrt(12))]
t(-2.5) = -2.5/[4.9082/sqrt(12)] = -1.76444...
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p-value = 2*P(-10 < t < -1.76444) = 2*0.0526 = 0.105369...
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Conclusion: p-value is slightly greater than alpha=10%.
You could Fail to reject Ho but more testing should be
done to get more evidence.
===================
Cheers,
Stan H.

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