SOLUTION: In repairing a helicopter component, a recent sample of 18 repairs showed that the mean time between failure (MTBF) is 520 hours, with a SD of 17 hours. An unsolicited modification

Question 1008686: In repairing a helicopter component, a recent sample of 18 repairs showed that the mean time between failure (MTBF) is 520 hours, with a SD of 17 hours. An unsolicited modification proposal is received which is supposed to increase the MTBF, thereby saving the government money. Tests are run on 10 components using this proposal resulting in the following time to failure data:
{518, 548, 561, 523, 536, 499, 538, 557, 528, 563}.
Determine if there is something to this proposed mod and if it has any possible future.
What significance level will you use and why?
Test the Theory at your SL and comment on the results, telling your manager what you think of this unsolicited proposal.
Test the claim by using a Confidence Interval. Interpret the results with respect to the possible political ramifications.

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
In repairing a helicopter component, a recent sample of 18 repairs showed that the mean time between failure (MTBF) is 520 hours, with a SD of 17 hours. An unsolicited modification proposal is received which is supposed to increase the MTBF, thereby saving the government money. Tests are run on 10 components using this proposal resulting in the following time to failure data:
{518, 548, 561, 523, 536, 499, 538, 557, 528, 563}.
Determine if there is something to this proposed mod and if it has any possible future.
What significance level will you use and why?
Test the Theory at your SL and comment on the results, telling your manager what you think of this unsolicited proposal.
Test the claim by using a Confidence Interval. Interpret the results with respect to the possible political ramifications.
------
x-bar = 537.1
s = 20.7
------
Ho:: u = 520
Ha:: u > 520
----
t(537.1) = (537.1-520)/(17/sqrt(10)) = 3.1809
p-value = P(t > 3.1809 when df = 9) = tcdf(3.1809,100,9) = 0.0056
-----------
Conclusion:: Since the p-value is less than 5%, reject Ho.
The new procedure does increase the MTBF.
----------------
95% CI::
ME = 2.2622[17/sqrt(10)] = 12.16
----
537.1-12.16 < u < 537.1+12.16
----
524.94 < u < 549.26
Conclusion:; Since 520 is not in the confidence interval,
reject Ho.
---------------
Cheers,
Stan H.

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