Question

A new paint process has been installed in a manufacturing facility. With this process, the facility has to wait only 58 minutes for drying before the next coat can be applied. That is, with the new process, the average dry time was 58 minutes and the standard deviation was 15.9 minutes for a sample of 30 automobiles. Compare this to the process that was replaced. This process (old process) was studied for 50 automobiles and the average dry time was 64 minutes with a standard deviation of 16.8 minutes. Use an alpha of .05 for your hypothesis test. Call your present process “sample one” and the replaced process (old process) “sample two.”   You want the mean of sample 1 minus the mean of sample 2 to be negative.

Answer the following :

1- Give H0:      and   H1:

2- Level of Significance

30 Confidence coefficient is ?  

4- Sample One the Current Process       

Sample Size   

Sample mean

Sample Standard Deviation   

5- Sample Two the previous Process     

Sample Size   

Sample mean

Sample Standard Deviation

6- Current Sample Degrees of Freedom

7- Old process Sample Degrees of Freedom     

8- Total Degrees of Freedom     

9- Pooled Variance        

10- The difference in sample means  

11- Test Statistic  

12- Critical Value                                                         

13- p-value       

14- Conclusion:

. . .

What do you mean by "tail"?

Answer 1

1

Hypothesis:

2.

Level of Significance = 0.05

3.

Confidence coefficient is 0.95

4.

Sample One the Current Process       

Sample Size = 30

Sample mean = 58

Sample Standard Deviation = 15.9

5.

Sample Two the previous Process     

Sample Size = 50

Sample mean = 64

Sample Standard Deviation = 16.8

6.

Current Sample Degrees of Freedom = n1 -1 = 30 - 1 = 29

7.

Old process Sample Degrees of Freedom = n2-1 = 50 - 1 = 49  

8.

Total Degrees of Freedom = n1+n2-2 = 30 + 50 -2 = 78

9.

Pooled Variance :

10.

The difference in sample means = 58 - 64 = -6

11.

Test Statistic ,

12.

Critical Value = -1.665   

13.

p-value = 0.0594

It is obtained using excel with function =TDIST(1.577,78,1)

14.

Conclusion: Since p-value  is greater than 0.05, we accept null hypothesis at 5% level of significance and conclude that mean waiting time for new process is less than old process.