Question

(Aside from providing the results of the simulation or statistical tests, be sure to answer the questions – in bold.)

A tire manufacturer tested the braking performance of one of its tire models on a test track. The company tried the tires on 10 different cars, recording the stopping distance for each car on both wet and dry pavement. Is there a difference between dry and wet pavements for this type of tire in terms of braking performance? Include your null and alternative hypotheses; include a 95% confidence interval for the mean increase in stopping distance on wet pavement.

Vehicle	Dry	Wet
1	145	211
2	152	191
3	141	220
4	143	207
5	131	198
6	148	208
7	126	206
8	140	177
9	135	183
10	133	223

Answer 1

Ho :   µd=   0                  
Ha :   µd ╪   0                  
                          
Level of Significance ,    α =    0.05       claim:µd=0          
                          
sample size ,    n =    10                  
                          
mean of sample 1,    x̅1=   139.400                  
                          
mean of sample 2,    x̅2=   202.400                  
                          
mean of difference ,    D̅ =ΣDi / n =   -63.000                  
                          
std dev of difference , Sd =    √ [ (Di-Dbar)²/(n-1) =    17.5942                  
                          
std error , SE = Sd / √n =    17.5942   / √   10   =   5.5638      
                          
t-statistic = (D̅ - µd)/SE = (   -63   -   0   ) /    5.5638   =   -11.323
                          
Degree of freedom, DF=   n - 1 =    9                  
t-critical value , t* =    ±   2.262   [excel function: =t.inv.2t(α,df) ]               
                          
p-value =        0.000001   [excel function: =t.dist.2t(t-stat,df) ]               
Conclusion:     p-value <α , Reject null hypothesis                      

There is sufficient evidence that there is difference in both mean.

sample size ,    n =    10          
Degree of freedom, DF=   n - 1 =    9   and α =    0.05  
t-critical value =    t α/2,df =    2.2622   [excel function: =t.inv.2t(α/2,df) ]      
                  
std dev of difference , Sd =    √ [ (Di-Dbar)²/(n-1) =    17.5942          
                  
std error , SE = Sd / √n =    17.5942   / √   10   =   5.5638
margin of error, E = t*SE =    2.2622   *   5.5638   =   12.5861
                  
mean of difference ,    D̅ =   63.000          
confidence interval is                   
Interval Lower Limit= D̅ - E =   63.000   -   12.5861   =   50.414
Interval Upper Limit= D̅ + E =   63.000   +   12.5861   =   75.586
                  
so, confidence interval is (   50.4139   < µd <   75.5861   )  

Thanks in advance!

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