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. 
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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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