Question

Two different box-filling machines are used to fill

cereal boxes on an assembly line. The critical measurement
influenced by these machines is the weight of the
product in the boxes. Engineers are quite certain that
the variance of the weight of product is σ2=1 ounce.
Experiments are conducted using both machines with
sample sizes of 36 each. The sample averages for machines
A and B are x¯¯¯A=4.5 ounces and x¯¯¯B=4.7
ounces. Engineers are surprised that the two sample
averages for the filling machines are so different.
(a) Use the Central Limit Theorem to determine
P(X^B−X^A≥0.2)
under the condition that μA=μB.
(b) Do the aforementioned experiments seem to, in any
way, strongly support a conjecture that the population
means for the two machines are different?
Explain using your answer in (a).

Answer 1

a) 0.1984

b) No. Is not too small to support the conjecture with confidence

Explanation:

1) Previous concepts and data given

The sample sizes are
`n_A=n_B=36`

We can find the distribution for
`\hat X_A -\hat X_B`, and since is a sampling distribution and
`\hat X_A` and
`\hat X_B` follows normal distributions then
`\hat X_A -\hat X_B` follows a normal distribution too.

The mean and the deviation for
`\hat X_A -\hat X_B` is given by:

`\mu_(\hat X_A)-\mu_(\hat X_B)=\mu_A -\mu_B` and since
`\mu_A =\mu_B` then
`\mu_A -\mu_B=0`

`\sigma_(\hat X_A -\hat X_B)=\sqrt{(\sigma^2_A)/(n_A)+(\sigma^2_B)/(n_B)}`

Since both A and B have the same deviation and variance then:

`\sigma_(\hat X_A -\hat X_B)=\sqrt{(\sigma^2)/(n_A)+(\sigma^2)/(n_B)}=\sqrt{(1)/(36)+(1)/(36)}=\sqrt{(1)/(18)}=0.236`

2) Part a

We want to find:

`P(\hat X_A -\hat X_B \geqslant 0.2)`

The random variable

`Z=(\hat X_A -\hat X_B -\mu_(\hat X_A)-\mu_(\hat X_B))/(\sigma_(\hat X_A -\hat X_B))` follows a normal distribution with mean 0 and deviation 1 we can use this:

`P(\hat X_A -\hat X_B \geqslant 0.2)=P((\hat X_A -\hat X_B -\mu_(\hat X_A)-\mu_(\hat X_B))/(\sigma_(\hat X_A -\hat X_B))\geqslant (0.2-\mu_(\hat X_A -\hat X_B))/(\sigma_(\hat X_A -\hat X_B)))`

`P(\hat X_A -\hat X_B \geqslant 0.2)=P(Z\geqslant (0.2-0)/(0.236))=1-P(Z<0.8475)=1-0.8016=0.1984`

3) Part b

From part a we found that we have a change of 19.84% that the experiment would give a difference between the sample means which is greater or equal than 0.2

Analyzing the probability obtained we can say that is not too small to support the conjecture with confidence that the population means for the two machines are different