Question

Referring to this data:

2.58     2.51     4.04     6.43     1.58     4.32     2.2       4.19    

4.79     6.2       1.52     1.38     3.87     4.54     5.12     5.15

5.5       5.92     4.56     2.46     6.9       1.47     2.11     2.32

6.75 5.84     8.8       7.4       4.72     3.62     2.46     8.75

Suppose we can only pass the quality test if there is evidence to suggest that the true mean ignition time is more than 0.04 seconds (or 4 hundredths of a second). please show work.

a. What is the null hypothesis for this test?

b. What is the alternative?

c. Conduct this hypothesis test at α=0.05 using the normal distribution. State all parts, including the critical value, test statistic, and conclusion in context of the problem.

d. Conduct this hypothesis test at α=0.05 using the t-distribution. State all parts, including the critical value, test statistic, and conclusion in context of the problem. Does your conclusion change?

Answer 1

Ho :   µ =   0.04                  
Ha :   µ >   0.04       (Right tail test)          
                          
Level of Significance ,    α =    0.05                  
population std dev ,    σ =    2.1040                  
Sample Size ,   n =    32                  
Sample Mean,    x̅ =   4.3750                  
                          
'   '   '                  
                          
Standard Error , SE = σ/√n =   2.1040   / √    32   =   0.3719      
Z-test statistic= (x̅ - µ )/SE = (   4.375   -   0.04   ) /    0.3719   =   11.66
                          
critical z value, z* =       1.6449   [Excel formula =NORMSINV(α/no. of tails) ]              
                                
Decision:   test stat > critical value, Reject null hypothesis

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d)

sample std dev ,    s =    2.1040                  
Sample Size ,   n =    32                  
Sample Mean,    x̅ =   4.3750                  
                          
degree of freedom=   DF=n-1=   31                  
                          
Standard Error , SE = s/√n =   2.1040   / √    32   =   0.3719      
t-test statistic= (x̅ - µ )/SE = (   4.375   -   0.04   ) /    0.3719   =   11.66
                          
critical t value, t* =        1.6955   [Excel formula =t.inv(α/no. of tails,df) ]              
                                
Decision: test stat > critical value, Reject null hypothesis       

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no change in conclusion

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THANKS

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