Question 754212: ) In order to test a new production method, 15 employees were selected randomly to try the new production method. The mean production rate for the sample was 80 parts her hour with a sample standard deviation 10 parts per hour. The mean production rate under the old method is known to be 70 parts per hour. Is the new method rate significantly different, or is this just the luck of the employees?
a) What is the hypothesis scenario?
b) What is the test statistic?
c) Run a 1-sample hypothesis test at 95% level of significance
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! In order to test a new production method, 15 employees were selected randomly to try the new production method. The mean production rate for the sample was 80 parts her hour with a sample standard deviation 10 parts per hour.
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The mean production rate under the old method is known to be 70 parts per hour.
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Is the new method rate significantly different, or is this just the luck of the employees?
a) What is the hypothesis scenario?
Ho: u = 70
Ha: u # 70
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b) What is the test statistic?
z(80) = (80-70)/[10/sqrt(15)] = 3.8730
p-value = 2*P(z > 3.8730) = 2*normalcdf(3.8730,100) = 0.000108
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Conclusion: Since the p-value is less than 5% reject Ho.
The results do not support the claim the mean is 70.
c) Run a 1-sample hypothesis test at 95% level of significance
Done above.
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Cheers,
Stan H.
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