Question

A tire manufacturer has been producing tires with an average life expectancy of 26,000 miles. Now the company is advertising that its new tires' life expectancy has increased. In order to test the legitimacy of the advertising campaign, an independent testing agency tested a sample of 6 of their tires and has provided the following data. The p-value is equal to

Life Expectancy (In Thousands of Miles)

28

27

25

28

29

25

Select one:

a. 0.102

b. 0.072

c. 0.203

d. 1.46

Answer 1

Ho :   µ =   26000                  
Ha :   µ >   26000       (Right tail test)          
                          
Level of Significance ,    α =    0.05                  
sample std dev ,    s = √(Σ(X- x̅ )²/(n-1) ) =   1673.3201                  
Sample Size ,   n =    6                  
Sample Mean,    x̅ = ΣX/n =    27000.0000                  
                          
degree of freedom=   DF=n-1=   5                  
                          
Standard Error , SE = s/√n =   1673.3201   / √    6   =   683.1301      
t-test statistic= (x̅ - µ )/SE = (   27000.000   -   26000   ) /    683.1301   =   1.46
                          
critical t value, t* =        2.0150   [Excel formula =t.inv(α/no. of tails,df) ]              
                          
p-Value   =   0.102 [Excel formula =t.dist(t-stat,df) ]             

ANSWER option (A)

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