Question

The deflection temperature under load for two different types of plastic pipe is being investigated. Two random samples of 12 pipe specimens are tested, and the deflection temperatures observed are shown as follows (in °F):

Type-1 206 188 205 187 199 192 185 213 192 194 178 205

Type-2 177 197 206 201 180 176 185 200 197 193    198 188

a. Use Excel/R to compute the sample mean and sample variance of each type.

b. Do the data support the claim that the deflection temperature under load for type 1 pipe exceeds that of type 2 at 5 % level of significance? (assume population standard deviation are equal)

Answer 1

a)

Type 1:

mean =    ΣX/n =    2344.000   /   12   =   195.3333
                      

                      
sample std dev =   √ [ Σ(X - X̄)²/(n-1)] =   √   (1180.6667/11)   =       10.3602

Type 2:

mean =    ΣX/n =    2298.000   /   12   =   191.5000
                      

                      
sample std dev =   √ [ Σ(X - X̄)²/(n-1)] =   √   (1115/11)   =       10.0680

2)

Ho :   µ1 - µ2 =   0                  
Ha :   µ1-µ2 >   0                  
                          
Level of Significance ,    α =    0.05                  
                          
Sample #1   ---->   Type 1   
mean of sample 1,    x̅1=   195.33                  
standard deviation of sample 1,   s1 =    10.36                  
size of sample 1,    n1=   12                  
                          
Sample #2   ---->   Type 2                  
mean of sample 2,    x̅2=   191.50                  
standard deviation of sample 2,   s2 =    10.07                  
size of sample 2,    n2=   12                  
                          
difference in sample means =    x̅1-x̅2 =    195.3333   -   191.5   =   3.83  
                          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    10.2151                  
std error , SE =    Sp*√(1/n1+1/n2) =    4.1703                  
                          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   3.8333   -   0   ) /    4.17   =   0.919
                          
Degree of freedom, DF=   n1+n2-2 =    22                  

p-value =        0.183981   [excel function: =T.DIST.RT(t stat,df) ]              
Conclusion:     p-value>α , Do not reject null hypothesis                      
                          
There is not enough evidencethat the deflection temperature under load for type 1 pipe exceeds that of type 2.

THANKS

revert back for doubt

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