SOLVED Normal distribution of the difference between two averages (2nd problem)

Oct 2009
Two alloys A and B are being used to manufacture a certain steel product. An experiment needs to be designed to compare the two in terms of maximum load capacity in tons. This is a maximum that can be tolerated without breaking. It is known that the two standard deviations in load capacity are equal at 5 tons each. An experiment is conducted in which 30 specimens of each alloy (A and B) are tested and the results gave

\(\displaystyle \bar{x}_A =49.5\); \(\displaystyle \bar{x}_B =45.5\)
\(\displaystyle \bar{x}_A - \bar{x}_B =4\)

The manufacturers of alloy A are convinced that this evidence shows conclusively that \(\displaystyle \mu_A > \mu_B\) and strongly supports their alloy. Manufacturers of alloy B claim that the experiment could easily have given \(\displaystyle \bar{x}_A -\bar{x}_B =4 even if the two population means are equal. IN other words, "things are inconclusive!"\)\(\displaystyle

(a)Make an argument that manufacturers of alloy B are wrong. Do it by computing
\(\displaystyle Pr(\bar{X}_A - \bar{X}_B > 4|\mu_A = \mu_B)\)

(b) Do you think these data strongly support alloy A?

Simple to solve, just like the previous problem I posted.

\(\displaystyle n_a = n_b =30\)
\(\displaystyle \sigma_{X_b - X_a}=\sqrt{\frac{5}{3}}\)

\(\displaystyle pr(z\geq \frac{4-0}\sqrt{\frac{5}{3}})\)
\(\displaystyle pr(z\geq 3.10)\)
\(\displaystyle =1-0.9990=0.0010\)

(b)Yes, the experiment supports alloy A's case. Given the the population means of the two alloys are equal, there is a 0.1% probability that the difference between the sample means of alloy A and alloy B would be more than 4 tons.\)
Last edited:
Oct 2009
I think I've solved this problem. I'll post the solution after we go over the problem in class, which will probably be before the end of the week. If my solution is correct, I'll post it here so that others may learn from it. In the mean time, I'd love to see anyone else's approach to solving the problem. Thanks