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

*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 [tex]\bar{x}_A -\bar{x}_B =4 ***even if** the two population means are equal. IN other words, "things are inconclusive!"

(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.**

(a)

$\displaystyle n_a = n_b =30$

$\displaystyle \sigma_{X_b - X_a}=\sqrt{\frac{5}{3}}$

$\displaystyle pr(z\geq \frac{4-0}\sqrt{\frac{5}{3}})$

$\displaystyle pr(z\geq 3.10)$

$\displaystyle =1-0.9990=0.0010$

(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.