Q9 Tensile strength tests were carried out on two different

Q9. Tensile strength tests were carried out on two different grades of wire rod, resulting in the following data. (a) Does the data show that the true average strength for the B grade exceeds that for the A grade by more than 10kg/mm^2? Test the appropriate hypotheses at the level of significance alpha = 0.01. (b) Find the 95% confidence interval for Mu 1 - Mu 2.

Solution

Formulating the null and alternative hypotheses,              
              
Ho:   u1 - u2   >=   -10  
Ha:   u1 - u2   <   -10  
At level of significance =    0.01          
As we can see, this is a    left   tailed test.      
Calculating the means of each group,              
              
X1 =    107.6          
X2 =    123.6          
              
Calculating the standard deviations of each group,              
              
s1 =    1.3          
s2 =    2          
              
Thus, the standard error of their difference is, by using sD = sqrt(s1^2/n1 + s2^2/n2):              
              
n1 = sample size of group 1 =    129          
n2 = sample size of group 2 =    129          
Thus, df = n1 + n2 - 2 =    256          
Also, sD =    0.210020302          
              
Thus, the t statistic will be              
              
t = [X1 - X2 - uD]/sD =    -28.5686667          
              
where uD = hypothesized difference =    -10          
              
Now, the critical value for t is              
              
tcrit =    -   2.341002248      
              
Thus, comparing t and tcrit, we decide to   WE REJECT THE NULL HYPOTHESIS.          
              
Also, using p values,              
              
p =    6.85671E-82          
              
As p < 0.01,    WE REJECT THE NULL HYPOTHESIS.  

There is significant evidence at 0.01 level that the average strength of B grade exceeds that of A grade by more than 10 kg/mm^2.   [conclusion]
              
              
              
******************************************              
              
For the   0.95   confidence level, then      
              
alpha/2 = (1 - confidence level)/2 =    0.025          
t(alpha/2) =    1.96927389          
              
lower bound = [X1 - X2] - t(alpha/2) * sD =    -16.4135875          
upper bound = [X1 - X2] + t(alpha/2) * sD =    -15.5864125          
              
Thus, the confidence interval is              
              
(   -16.4135875   ,   -15.5864125   ) [answer]