Question

An experiment was conducted to investigate the machine performance of deep hole drilling when chip congestion exists. The length, in mm, of 50 drill chips had a sample mean of 71.2 mm. Assume the population standard deviation is 15.2 mm.  Conduct a test to determine whether the true mean drill chip length is less than 75 mm at α=.01.  

Null Hypothesis (u):

Alternative Hypothesis:

Standard Error:

Test Statistic (z):

p-value (lower tail):

p-value (upper tail):

p-value (two tail):

appropriate p value:

Conclusion in terms of alt. hypothesis:

Please include any excel formulas used. Thank you!

Answer 1

Solution:

Null Hypothesis (u): H0: µ = 75

Alternative Hypothesis: Ha: µ < 75

Standard Error: [σ/sqrt(n)] = 15.2/sqrt(50) = 2.1496

Test Statistic (z): Z = (Xbar - µ)/[σ/sqrt(n)]

Z = (71.2 - 75)/[15.2/sqrt(50)]

Z = -1.7678

p-value (lower tail): 0.0385

[Use excel command =NORMSDIST(-1.7678)]

p-value (upper tail): 0.9615

[Use excel command =1 - NORMSDIST(-1.7678)]

p-value (two tail): 0.0771

[Use excel command =2*NORMSDIST(-1.7678)]

appropriate p value: 0.0385

Conclusion in terms of alt. hypothesis: There is not sufficient evidence to conclude that the true mean drill chip length is less than 75 mm at α=.01, because p-value of 0.0385 is greater than alpha value 0.01.