# Thread: Normal distribution of the difference between two averages

1. ## Normal distribution of the difference between two averages

To different box-filling machines are used to fill cereal boxes on the assembly line. The critical measurement influenced by these machines is the wieght of the product in the macines. Engineers are quite certain that the variance of the weight of product is $\displaystyle \sigma^2=1$ ounce. Experiments are conducted using both machines with sample sizes of 36 each. The sample averages for machine A and B are $\displaystyle \bar{x}_A =4.5$ ounces and $\displaystyle \bar{x}_A =4.5$ ounces. Engineers for the filling machines were so different.

(a) Use the central limit theorem to determine

$\displaystyle P(\bar{X}_B - \bar{X}_A \geq 0.2)$

Under the condition $\displaystyle \mu_A = \mu_B$

Ok, so I have:

$\displaystyle n_a = n_b =36$
$\displaystyle \sigma_{X_b - X_a}=\sqrt{\frac{1}{18}}$

$\displaystyle pr(z\geq \frac{0.2-0}\sqrt{\frac{1}{18}})$
$\displaystyle pr(z\geq 0.85)$
$\displaystyle =1-0.8023=0.1977$

Do the aforementioned experiments seem to, in any way, strongly support a conjecture that the two population means for the two machines are different? Explain using your answer in (a).

(b) No, the aforementioned experiments do not support a conjecture that the two population means for the two machines are different. Given the that the two population means for the the two machines are the same, there is a 19.77% probability that the difference between their two sample means will be more than 0.2 ounces.

2. 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