Answer:
So with the p value obtained and using the significance level given 
we have 
so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the true average strength for the 1078 grade exceeds that for the 1064 grade by more than 10 kg/mm2.
Step-by-step explanation:
The statistic is given by this formula:
Where t follows a t distribution with 
degrees of freedom
The system of hypothesis on this case are:
Null hypothesis: 
Alternative hypothesis: 
Or equivalently:
Null hypothesis: 
Alternative hypothesis: 
Our notation on this case :
represent the sample size for group AISI 1078
represent the sample size for group AISI 1064
represent the sample mean for the group AISI 1078
represent the sample mean for the group AISI 1064
represent the sample standard deviation for group 1 AISI 1078
represent the sample standard deviation for group AISI 1064
And now we can calculate the statistic:
Now we can calculate the degrees of freedom given by:
And now we can calculate the p value using the altenative hypothesis:
So with the p value obtained and using the significance level given we have so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the true average strength for the 1078 grade exceeds that for the 1064 grade by more than 10 kg/mm2.
